I n s tr u c to r ’s
M a n u a l
NOTICE
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-------------------------------
Digital Image Processing
Third Edition
Instructor's Manual
Version 3.0
Rafael C. Gonzalez
Richard E. Woods
Prentice Hall
Upper Saddle River, NJ 07458
www.imageprocessingplace.com
Copyright ©1992-2008 R. C. Gonzalez and R. E. Woods
NOTICE
This manual is intended for your personal use only.
Copying, printing, posting, or any form of printed or electronic
distribution of any part of this manual constitutes a violation of
copyright law.
A s a security measure, this manual was encrypted during download
with the serial number of your book, and with your personal information.
A ny printed or electronic copies of this file will bear that encryption,
which will tie the copy to you.
Please help us defeat piracy of intellectual property, one of the principal
reasons for the increase in the cost of books.
-------------------------------
Chapter 1
Introduction
T h e p u r p o s e o f th is c h
r ia l fr o m D ig ita l Im a g
a ls o d is c u s s u s e o f th e
th e w e b s ite o ffe r s , a
c o m p u te r p r o je c ts th
D e ta ile d s o lu tio n s to
in g c h a p te r s o f th is m
1.1
a p te r is to p re s e n t s
e P ro c e ssin g a t th e s
b o o k w e b s ite . A lth
m o n g o th e r th in g s
a t c a n b e a s s ig n e d
a ll p r o b le m s in th e
a n u a l.
u g g e s
e n io r
o u g h
, c o m
in c o
b o o k
te d g u id e lin e s fo r te a c h
a n d fi rs t-y e a r g ra d u a te
th e b o o k is to ta lly s e lf- c
p le m e n ta r y r e v ie w m a
n ju n c tio n w ith c la s s r o
a ls o a r e in c lu d e d in th
in g m a
le v e ls .
o n ta in
te r ia l a
o m w o
e re m a
te W e
e d ,
n d
rk .
in -
Teaching Features of the Book
U n d e r g r a d u a te p r o g r a m s th a t o ffe r d ig ita l im a g e p r o c e s s in g ty p ic a lly lim it c o v e r a g e to o n e s e m e s te r. G r a d u a te p r o g r a m s v a r y , a n d c a n in c lu d e o n e o r tw o
s e m e s te r s o f th e m a te r ia l. In th e fo llo w in g d is c u s s io n w e g iv e g e n e r a l g u id e lin e s
fo r a o n e - s e m e s te r s e n io r c o u r s e , a o n e - s e m e s te r g r a d u a te c o u r s e , a n d a fu lly e a r c o u r s e o f s tu d y c o v e r in g tw o s e m e s te r s . W e a s s u m e a 1 5 - w e e k p r o g r a m p e r
s e m e s te r w ith th r e e le c tu r e s p e r w e e k . In o r d e r to p r o v id e fl e x ib ility fo r e x a m s
a n d r e v ie w s e s s io n s , th e g u id e lin e s d is c u s s e d in th e fo llo w in g s e c tio n s a r e b a s e d
o n fo r ty , 5 0 - m in u te le c tu r e s p e r s e m e s te r. T h e b a c k g r o u n d a s s u m e d o n th e p a r t
o f th e s tu d e n t is s e n io r - le v e l p r e p a r a tio n in m a th e m a tic a l a n a ly s is , m a tr ix th e o r y , p r o b a b ility , a n d c o m p u te r p r o g r a m m in g . T h e T u to r ia ls s e c tio n in th e b o o k
w e b s ite c o n ta in s r e v ie w m a te r ia ls o n m a tr ix th e o r y a n d p r o b a b ility , a n d h a s a
b r ie f in tr o d u c tio n to lin e a r s y s te m s . P o w e r P o in t c la s s r o o m p r e s e n ta tio n m a te r ia l o n th e r e v ie w to p ic s is a v a ila b le in th e F a c u lty s e c tio n o f th e w e b s ite .
T h e s u g g e s te d te a c h in g g u id e lin e s a r e p r e s e n te d in te r m s o f g e n e r a l o b je c tiv e s , a n d n o t a s tim e s c h e d u le s . T h e r e is s o m u c h v a r ie ty in th e w a y im a g e p r o c e s s in g m a te r ia l is ta u g h t th a t it m a k e s little s e n s e to a tte m p t a b r e a k d o w n o f th e
m a te r ia l b y c la s s p e r io d . In p a r tic u la r, th e o r g a n iz a tio n o f th e p r e s e n t e d itio n o f
1
C H A P T E R 1 . IN T R O D U C T IO N
2
th e b o o k is s u c h th a t it m a k e s it m u c h e a s ie r th a n b e fo r e to a d o p t s ig n ifi c a n tly
d iffe r e n t te a c h in g s tr a te g ie s , d e p e n d in g o n c o u r s e o b je c tiv e s a n d s tu d e n t b a c k g r o u n d . F o r e x a m p le , it is p o s s ib le w ith th e n e w o r g a n iz a tio n to o ffe r a c o u r s e
th a t e m p h a s iz e s s p a tia l te c h n iq u e s a n d c o v e r s little o r n o tr a n s fo r m m a te r ia l.
T h is is n o t s o m e th in g w e re c o m m e n d , b u t it is a n o p tio n th a t o fte n is a ttr a c tiv e
in p r o g r a m s th a t p la c e little e m p h a s is o n th e s ig n a l p r o c e s s in g a s p e c ts o f th e
fi e ld a n d p r e fe r to fo c u s m o r e o n th e im p le m e n ta tio n o f s p a tia l te c h n iq u e s .
1.2 One Semester Senior Course
A b a s ic s tr a te g y in te a c h in g a s e n io r c o u r s e is to fo c u s o n a s p e c ts o f im a g e p r o c e s s in g in w h ic h b o th th e in p u ts a n d o u tp u ts o f th o s e p r o c e s s e s a re im a g e s .
In th e s c o p e o f a s e n io r c o u r s e , th is u s u a lly m e a n s th e m a te r ia l c o n ta in e d in
C h a p te r s 1 th r o u g h 6 . D e p e n d in g o n in s tr u c to r p r e fe r e n c e s , w a v e le ts (C h a p te r 7 ) u s u a lly a r e b e y o n d th e s c o p e o f c o v e r a g e in a ty p ic a l s e n io r c u r r ic u lu m .
H o w e v e r, w e r e c o m m e n d c o v e r in g a t le a s t s o m e m a te r ia l o n im a g e c o m p r e s s io n (C h a p te r 8 ) a s o u tlin e d b e lo w .
W e h a v e fo u n d in m o re th a n th re e d e c a d e s o f te a c h in g th is m a te r ia l to s e n io r s in e le c tr ic a l e n g in e e r in g , c o m p u te r s c ie n c e , a n d o th e r te c h n ic a l d is c ip lin e s ,
th a t o n e o f th e k e y s to s u c c e s s is to s p e n d a t le a s t o n e le c tu r e o n m o tiv a tio n
a n d th e e q u iv a le n t o f o n e le c tu r e o n r e v ie w o f b a c k g r o u n d m a te r ia l, a s th e n e e d
a r is e s . T h e m o tiv a tio n a l m a te r ia l is p r o v id e d in th e n u m e r o u s a p p lic a tio n a r e a s
d is 1 .2 O n e S e m e s te r S e n io r C o u r s e c u s s e d in C h a p t e r 1 . T h is c h a p t e r w a s p r e p a r e d w ith th is o b je c tiv e in m in d . S o m e o f th is m a te r ia l c a n b e c o v e r e d in c la s s
in th e fi r s t p e r io d a n d th e re s t a s s ig n e d a s in d e p e n d e n t re a d in g . B a c k g r o u n d re v ie w s h o u ld c o v e r p r o b a b ility th e o r y (o f o n e r a n d o m v a r ia b le ) b e fo r e h is to g r a m
p r o c e s s in g ( S e c t io n 3 .3 ) . A b r ie f r e v ie w o f v e c t o r s a n d m a tr ic e s m a y b e r e q u ir e d
la te r, d e p e n d in g o n th e m a te r ia l c o v e r e d . T h e r e v ie w m a te r ia l in th e b o o k w e b
s ite w a s d e s ig n e d fo r ju s t th is p u r p o s e .
C h a p te r 2 s h o u ld b e c o v e r e d in its e n tir e ty . S o m e o f th e m a te r ia l (S e c tio n s
2 .1 t h r o u g h 2 .3 .3 ) c a n b e a s s ig n e d a s in d e p e n d e n t r e a d in g , b u t m o r e d e t a ile d
e x p la n a tio n (c o m b in e d w ith s o m e a d d itio n a l in d e p e n d e n t r e a d in g ) o f S e c tio n s
2 .3 .4 a n d 2 .4 t h r o u g h 2 .6 is t im e w e ll s p e n t . T h e m a t e r ia l in S e c t io n 2 .6 c o v e r s
c o n c e p ts th a t a re u s e d th r o u g h o u t th e b o o k a n d p r o v id e s a n u m b e r o f im a g e
p r o c e s s in g a p p lic a tio n s th a t a r e u s e fu l a s m o tiv a tio n a l b a c k g r o u n d fo r th e r e s t
o f th e b o o k
C h a p te r 3 c o v e r s s p a tia l in te n s ity tr a n s fo r m a tio n s a n d s p a tia l c o r r e la tio n a n d
c o n v o lu tio n a s th e fo u n d a tio n o f s p a tia l fi lte r in g . T h e c h a p te r a ls o c o v e r s a
n u m b e r o f d iffe r e n t u s e s o f s p a tia l tr a n s fo r m a tio n s a n d s p a tia l fi lte r in g fo r im a g e e n h a n c e m e n t. T h e s e te c h n iq u e s a r e illu s tr a te d in th e c o n te x t e n h a n c e m e n t
1 .2 . O N E S E M E S T E R S E N I O R C O U R S E
(a s
th e
n io
3 .4
o n
3
m o tiv a tio n a l a id s ), b u t it is p o in te d o u t s e v e r a l tim e s in th e c h a p te r th a t
m e th o d s d e v e lo p e d h a v e a m u c h b r o a d e r r a n g e o f a p p lic a tio n . F o r a s e r c o u r s e , w e r e c o m m e n d c o v e r in g S e c t io n s 3 .1 t h r o u g h 3 .3 .1 , a n d S e c t io n s
t h r o u g h 3 .6 . S e c t io n 3 .7 c a n b e a s s ig n e d a s in d e p e n d e n t r e a d in g , d e p e n d in g
tim e .
T h e k e y o b je c tiv e s o f C h a p te r 4 a r e (1 ) to s ta r t fr o m b a s ic p r in c ip le s o f s ig n a l
s a m p lin g a n d fr o m th e s e d e r iv e th e d is c r e te F o u r ie r tr a n s fo r m ; a n d (2 ) to illu s tr a te th e u s e o f fi lte r in g in th e fr e q u e n c y d o m a in . A s in C h a p te r 3 , w e u s e m o s tly
e x a m p le s fr o m im a g e e n h a n c e m e n t, b u t m a k e it c le a r th a t th e F o u r ie r tr a n s fo r m h a s a m u c h b r o a d e r s c o p e o f a p p lic a tio n . T h e e a r ly p a r t o f th e c h a p te r
th r o u g h S e c t io n 4 .2 .2 c a n b e a s s ig n e d a s in d e p e n d e n t r e a d in g . W e r e c o m m e n d
c a r e fu l c o v e r a g e o f S e c t io n s 4 .2 .3 t h r o u g h 4 .3 .4 . S e c t io n 4 .3 .5 c a n b e a s s ig n e d a s
in d e p e n d e n t r e a d in g . S e c tio n 4 .4 s h o u ld b e c o v e r e d in d e t a il. T h e e a r ly p a r t
o f S e c tio n 4 .5 d e a ls w ith e x te n d in g to 2 - D th e m a te r ia l d e r iv e d in th e e a r lie r
s e c t io n s o f th is c h a p te r. T h u s , S e c tio n s 4 .5 .1 th r o u g h 4 .5 .3 c a n b e a s s ig n e d a s
in d e p e n d e n t r e a d in g a n d th e n d e v o te p a r t o f th e p e r io d fo llo w in g th e a s s ig n m e n t to s u m m a r iz in g th a t m a te r ia l. W e r e c o m m e n d c la s s c o v e r a g e o f th e r e s t
o f t h e s e c t io n . In S e c t io n 4 .6 , w e r e c o m m e n d t h a t S e c t io n s 4 .6 .1 - 4 .6 .6 b e c o v e r e d in c la s s . S e c t io n 4 .6 .7 c a n b e a s s ig n e d a s in d e p e n d e n t r e a d in g . S e c tio n s
4 .7 .1 - 4 .7 .3 s h o u ld b e c o v e r e d a n d S e c t io n 4 .7 .4 c a n b e a s s ig n e d a s in d e p e n d e n t
r e a d in g . In S e c t io n s 4 .8 t h r o u g h 4 .9 w e r e c o m m e n d c o v e r in g o n e fi lt e r ( lik e t h e
id e a l lo w p a s s a n d h ig h p a s s fi lte r s ) a n d a s s ig n in g th e r e s t o f th o s e tw o s e c tio n s
a s in d e p e n d e n t r e a d in g . In a s e n io r c o u r s e , w e r e c o m m e n d c o v e r in g S e c t io n 4 .9
t h r o u g h S e c t io n 4 .9 .3 o n ly . In S e c t io n 4 .1 0 , w e a ls o r e c o m m e n d c o v e r in g o n e
fi lt e r a n d a s s ig n in g t h e r e s t a s in d e p e n d e n t r e a d in g . In S e c t io n 4 .1 1 , w e r e c o m m e n d c o v e r in g S e c t io n s 4 .1 1 .1 a n d 4 .1 1 .2 a n d m e n t io n in g t h e e x is t e n c e o f F F T
a lg o r ith m s . T h e lo g 2 c o m p u ta tio n a l a d v a n ta g e o f th e F F T d is c u s s e d in th e e a r ly
p a r t o f S e c t io n 4 .1 1 .3 s h o u ld b e m e n t io n e d , b u t in a s e n io r c o u r s e t h e r e t y p ic a lly
is n o tim e to c o v e r d e v e lo p m e n t o f th e F F T in d e ta il.
C h a p te r 5 c a n b e c o v e r e d a s a c o n tin u a tio n o f C h a p te r 4 . S e c tio n 5 .1 m a k e s
th is a n e a s y a p p r o a c h . T h e n , it is p o s s ib le to g iv e th e s tu d e n t a “ fl a v o r ” o f w h a t
r e s to r a tio n is (a n d s till k e e p th e d is c u s s io n b r ie f ) b y c o v e r in g o n ly G a u s s ia n a n d
im p u ls e n o is e in S e c t io n 5 .2 .1 , a n d t w o o f t h e s p a t ia l fi lt e r s in S e c t io n 5 .3 . T h is
la tte r s e c tio n is a fr e q u e n t s o u r c e o f c o n fu s io n to th e s tu d e n t w h o , b a s e d o n d is c u s s io n s e a r lie r in th e c h a p te r, is e x p e c tin g to s e e a m o r e o b je c tiv e a p p r o a c h . It
is w o r th w h ile to e m p h a s iz e a t th is p o in t th a t s p a tia l e n h a n c e m e n t a n d r e s to r a tio n a r e th e s a m e th in g w h e n it c o m e s to n o is e r e d u c tio n b y s p a tia l fi lte r in g .
A g o o d w a y to k e e p it b r ie f a n d c o n c lu d e c o v e r a g e o f r e s to r a tio n is to ju m p a t
th is p o in t to in v e r s e fi lte r in g (w h ic h fo llo w s d ir e c tly fr o m th e m o d e l in S e c tio n
5 .1 ) a n d s h o w t h e p r o b le m s w it h t h is a p p r o a c h . T h e n , w it h a b r ie f e x p la n a t io n
C H A P T E R 1 . IN T R O D U C T IO N
4
re g a rd in g th e fa c t th a t m u c h o f re s to r a tio n c e n te r s a
h e r e n t in in v e r s e fi lte r in g , it is p o s s ib le to in tr o d u c e th
W ie n e r fi lte r in E q . ( 5 .8 - 3 ) a n d d is c u s s E x a m p le s 5 .1 2
w e re c o m m e n d a b r ie f d is c u s s io n o n im a g e re c o n s tr u
5 .1 1 .1 - 5 .1 1 - 2 a n d m e n t io n in g t h a t t h e r e s t o f S e c t io
g e n e r a te d p r o je c tio n s in w h ic h b lu r is m in im iz e d .
ro u n
e “ in
a n d
c tio n
n 5 .1
d th e in s ta b
te r a c tiv e ” fo
5 .1 3 . A t a m
b y c o v e r in g
1 d e a ls w ith
ilitie s in r m o f th e
in im u m ,
S e c tio n s
w a y s to
C o v e r a g e o f C h a p te r 6 a ls o c a n b e b r ie f a t th e s e n io r le v e l b y fo c u s in g o n
e n o u g h m a te r ia l to g iv e th e s tu d e n t a fo u n d a tio n o n th e p h y s ic s o f c o lo r (S e c t io n 6 .1 ) , t w o b a s ic c o lo r m o d e ls ( R G B a n d C M Y / C M Y K ) , a n d t h e n c o n c lu d in g
w it h a b r ie f c o v e r a g e o f p s e u d o c o lo r p r o c e s s in g ( S e c t io n 6 .3 ) . W e t y p ic a lly c o n c lu d e a s e n io r c o u r s e b y c o v e r in g s o m e o f th e b a s ic a s p e c ts o f im a g e c o m p r e s s io n (C h a p te r 8 ). In te r e s t in th is to p ic h a s in c r e a s e d s ig n ifi c a n tly a s a r e s u lt o f
th e h e a v y u s e o f im a g e s a n d g r a p h ic s o v e r th e In te r n e t, a n d s tu d e n ts u s u a lly a r e
e a s ily m o tiv a te d b y th e to p ic . T h e a m o u n t o f m a te r ia l c o v e r e d d e p e n d s o n th e
tim e le ft in th e s e m e s te r.
1.3 One Semester Graduate Course (No Background
in DIP)
T h e m
n e ith e
o f th e
a n d fe
a d d th
a in d iffe re n c e b e tw e e n a s e n io r a n d a fi r s t- y e a
r g r o u p h a s fo r m a l b a c k g r o u n d in im a g e p r o c e
m a te r ia l c o v e r e d , in th e s e n s e th a t w e s im p ly g o
e l m u c h fre e r in a s s ig n in g in d e p e n d e n t re a d in
e fo llo w in g m a te r ia l to th e m a te r ia l s u g g e s te d
r g ra d
s s in g
fa s te
g . In
in th e
u
is
r
a
a t
m
in
g r
p re
e c o u rs
o s tly in
a g ra d u
a d u a te
v io u s s
e in w h ic
th e s c o p
a te c o u rs
c o u rs e w
e c tio n .
h
e
e
e
S e c t io n s 3 .3 .2 - 3 .3 .4 a r e a d d e d a s is S e c t io n 3 .3 .8 o n fu z z y im a g e p r o c e s s in g .
W e c o v e r C h a p te r 4 in its e n tire ty (w ith a p p r o p r ia te s e c tio n s a s s ig n e d a s in d e p e n d e n t r e a d y in g , d e p e n d in g o n th e le v e l o f th e c la s s ). T o C h a p te r 5 w e a d d S e c t io n s 5 .6 - 5 .8 a n d c o v e r S e c t io n 5 .1 1 in d e t a il. In C h a p t e r 6 w e a d d t h e H S I m o d e l
( S e c t io n 6 .3 .2 ) , S e c t io n 6 .4 , a n d S e c t io n 6 .6 . A n ic e in t r o d u c t io n t o w a v e le t s
(C h a p te r 7 ) c a n b e a c h ie v e d b y a c o m b in a tio n o f c la s s r o o m d is c u s s io n s a n d in d e p e n d e n t r e a d in g . T h e m in im u m n u m b e r o f s e c tio n s in th a t c h a p te r a r e 7 .1 ,
7 .2 , 7 .3 , a n d 7 .5 , w it h a p p r o p r ia t e ( b u t b r ie f ) m e n t io n o f t h e e x is t e n c e o f fa s t
w a v e le t t r a n s fo r m s . S e c t io n s 8 .1 a n d 8 .2 t h r o u g h S e c t io n 8 .2 .8 p r o v id e a n ic e
in tr o d u c tio n to im a g e c o m p re s s io n .
im
fro
p u
d e
If
a g
m
ts
fi n
a d d itio n a l tim e is a
e p r o c e s s in g (C h a p
m e th o d s w h o s e in p
a re im a g e s , b u t th e
e d in S e c t io n 1 .1 . W
v a ila b
te r 9 ).
u ts a n
o u tp u
e re c o
le , a n a tu r a
T h e m a te r
d o u tp u ts a
ts a re a ttr ib
m m e n d c o v
l to p ic to
ia l in th is
re im a g e s
u te s a b o u
e ra g e o f S
c o v e r n e x t is m o r p h o lo g ic a l
c h a p te r b e g in s a tr a n s itio n
to m e th o d s in w h ic h th e in t th o s e im a g e s , in th e s e n s e
e c t io n s 9 .1 t h r o u g h 9 .4 , a n d
1 .4 . O N E S E M E S T E R G R A D U A T E C O U R S E ( W I T H S T U D E N T H A V I N G B A C K G R O U N D I N D I P ) 5
s o m e o f t h e a lg o r it h m s in S e c t io n 9 .5 .
1.4 One Semester Graduate Course (with Student Having Background in DIP)
S o m e p r o g r a m s h a v e a n u n d e r g r a d u a te c o u r s e in im a g e p r o c e s s in g a s a p re re q u is ite to a g r a d u a te c o u r s e o n th e s u b je c t, in w h ic h c a s e th e c o u r s e c a n b e b ia s e d
to w a r d th e la tte r c h a p te r s . In th is c a s e , a g o o d d e a l o f C h a p te r s 2 a n d 3 is r e v ie w ,
w it h t h e e x c e p t io n o f S e c t io n 3 .8 , w h ic h d e a ls w it h fu z z y im a g e p r o c e s s in g . D e p e n d in g o n w h a t is c o v e re d in th e u n d e r g r a d u a te c o u r s e , m a n y o f th e s e c tio n s
in C h a p te r 4 w ill b e r e v ie w a s w e ll. F o r C h a p te r 5 w e r e c o m m e n d th e s a m e le v e l
o f c o v e r a g e a s o u tlin e d in th e p r e v io u s s e c tio n .
In C h a p te r 6 w e a d d fu ll- c o lo r im a g e p r o c e s s in g ( S e c t io n s 6 .4 t h r o u g h 6 .7 ) .
C h a p te r s 7 a n d 8 a r e c o v e r e d a s o u tlin e d in th e p r e v io u s s e c tio n . A s n o te d in
th e p re v io u s s e c tio n , C h a p te r 9 b e g in s a tr a n s itio n fr o m m e th o d s w h o s e in p u ts
a n d o u tp u ts a re im a g e s to m e th o d s in w h ic h th e in p u ts a re im a g e s , b u t th e o u tp u ts a re a ttr ib u te s a b o u t th o s e im a g e s . A s a m in im u m , w e re c o m m e n d c o v e r a g e
o f b in a r y m o r p h o lo g y : S e c t io n s 9 .1 t h r o u g h 9 .4 , a n d s o m e o f th e a lg o r it h m s in
S e c t io n 9 .5 . M e n t io n s h o u ld b e m a d e a b o u t p o s s ib le e x t e n s io n s t o g r a y - s c a le
im a g e s , b u t c o v e r a g e o f th is m a te r ia l m a y n o t b e p o s s ib le , d e p e n d in g o n th e
s c h e d u le . In C h a p t e r 1 0 , w e r e c o m m e n d S e c t io n s 1 0 .1 t h r o u g h 1 0 .4 . In C h a p t e r
1 1 w e t y p ic a lly c o v e r S e c t io n s 1 1 .1 t h r o u g h 1 1 .4 .
1.5 Two Semester Graduate Course (No Background
in DIP)
In a tw o - s e m e s te r c o u r s e it is p o s s ib le to c o v e r m a te r ia l in a ll tw e lv e c h a p te r s
o f th e b o o k . T h e k e y in o r g a n iz in g th e s y lla b u s is th e b a c k g r o u n d th e s tu d e n ts
b r in g to th e c la s s . F o r e x a m p le , in a n e le c tr ic a l a n d c o m p u te r e n g in e e r in g c u r r ic u lu m g r a d u a te s tu d e n ts h a v e s tr o n g b a c k g r o u n d in fr e q u e n c y d o m a in p r o c e s s in g , s o C h a p te r 4 c a n b e c o v e r e d m u c h q u ic k e r th a n w o u ld b e th e c a s e in
w h ic h th e s tu d e n ts a re fr o m , s a y , a c o m p u te r s c ie n c e p r o g r a m . T h e im p o r ta n t
a s p e c t o f a fu ll y e a r c o u r s e is e x p o s u r e to th e m a te r ia l in a ll c h a p te r s , e v e n w h e n
s o m e to p ic s in e a c h c h a p te r a re n o t c o v e re d .
1.6 Projects
O n e o f th e m o s t in te re s tin g a s p e c ts o f a c o u r s e in d ig ita l im a g e p r o c e s s in g is th e
p ic to r ia l n a tu r e o f th e s u b je c t. It h a s b e e n o u r e x p e r ie n c e th a t s tu d e n ts tr u ly
e n jo y a n d b e n e fi t fr o m ju d ic io u s u s e o f c o m p u te r p r o je c ts to c o m p le m e n t th e
C H A P T E R 1 . IN T R O D U C T IO N
6
m a te
w o rk
b r ie f
th e w
r ia
a n
a s
a y
l c o v e r
d h o m
p o s s ib
p r o je c
e d in
e w o rk
le . In
t re p o
c la
a s
o r
r ts
s s . B e c a u
s ig n m e n t
d e r to fa c
a re p re p a
s e c
s , w
ilita
re d
o m p
e tr y
te g r
. A u
u te r p r o je
to k e e p th
a d in g , w e
s e fu l re p o
c ts
e fo
tr y
r t fo
a re
r m
to
r m
in
a l
a c
a t
a d d itio n to
p r o je c t r e p o
h ie v e u n ifo r
is a s fo llo w s
c o u rs e
r tin g a s
m ity in
:
P a g e 1 : C o v e r p a g e .
•
P r o je c t title
•
P r o je c t n u m b e r
•
C o u rs e n u m b e r
•
S t u d e n t ’s n a m e
•
D a te d u e
•
D a te h a n d e d in
•
A b s tra c t (n o t to e x c e e d 1 / 2 p a g e )
P a g e 2 : O n e to tw o p a g e s (m a x ) o f te c h n ic a l d is c u s s io n .
P a g e 3 (o r 4 ): D is c u s s io n o f r e s u lts . O n e to tw o p a g e s (m a x ).
R e s u lts : Im a g e r e s u lts (p r in te d ty p ic a lly o n a la s e r o r in k je t p r in te r ). A ll im a g e s
m u s t c o n ta in a n u m b e r a n d title r e fe r r e d to in th e d is c u s s io n o f r e s u lts .
A p p e n d ix : P r o g r a m lis tin g s , fo c u s e d o n a n y o r ig in a l c o d e p r e p a r e d b y th e s tu d e n t. F o r b re v ity , fu n c tio n s a n d r o u tin e s p r o v id e d to th e s tu d e n t a re re fe r re d to
b y n a m e , b u t th e c o d e is n o t in c lu d e d .
L a y o u t: T h
U .S . o r A 4
fo r m a b o o
S e m e s te r S
lis t
im a
w ill
w e b
P r o je
o f s u
g e s ,
fi n d
s ite
.
c t
g g
a n
a d
e e n tire
in E u r o
k le t, o r
e n io r C
re
p e
b o
o u
re s o u rc e s
e s te d p ro
d M A T L A
d itio n a l s
p o r t m u s t b e o n a s ta n d a r d s h e e t s iz e ( e .g ., le t te r s iz e in t h e
), s ta p le d w ith th r e e o r m o r e s ta p le s o n th e le ft m a r g in to
u n d u s in g c le a r p la s t ic s t a n d a r d b in d in g p r o d u c t s .1 .2 O n e
rs e
a v a ila b le
je c ts fr o m
B fu n c tio n
o ftw a re s u
in
w
s .
g g
th e b o o
h ic h th e
In s tr u c
e s tio n s
k w e
in s t
to rs
in th
b
r u
w
e
s ite
c to
h o
S u p
in
r c
d o
p o
c lu
a n
n o
r t/
d e a s a m p
s e le c t, b o o
t w is h to u
S o ftw a re s e
le p r o je
k a n d o
s e M A T
c tio n o
c t, a
th e r
L A B
f th e
1 .7 . T H E B O O K W E B S I T E
7
1.7 The Book Web Site
T h e c o m p a n io n w e b s ite
w w w .p r e n h a ll.c o m / g o n z a le z w o o d s
(o r its m ir r o r s ite )
w w w .im a g e p r o c e s s in g p la c e .c o m
is a v a lu a b le te a c h in g a id , in th e s e n s e th a t it in c lu d e s m a te r ia l th a t p r e v io u s ly
w a s c o v e r e d in c la s s . In p a r tic u la r, th e r e v ie w m a te r ia l o n p r o b a b ility , m a tr ic e s , v e c to r s , a n d lin e a r s y s te m s , w a s p r e p a r e d u s in g th e s a m e n o ta tio n a s in
th e b o o k , a n d is fo c u s e d o n a r e a s th a t a r e d ir e c tly r e le v a n t to d is c u s s io n s in th e
te x t. T h is a llo w s th e in s tr u c to r to a s s ig n th e m a te r ia l a s in d e p e n d e n t r e a d in g ,
a n d s p e n d n o m o r e th a n o n e to ta l le c tu r e p e r io d r e v ie w in g th o s e s u b je c ts . A n o th e r m a jo r fe a tu r e is th e s e t o f s o lu tio n s to p r o b le m s m a r k e d w ith a s ta r in th e
b o o k . T h e s e s o lu tio n s a r e q u ite d e ta ile d , a n d w e r e p r e p a r e d w ith th e id e a o f
u s in g th e m a s te a c h in g s u p p o r t. T h e o n - lin e a v a ila b ility o f p r o je c ts a n d d ig ita l
im a g e s fre e s th e in s tr u c to r fr o m h a v in g to p re p a re e x p e r im e n ts , d a ta , a n d h a n d o u ts fo r s tu d e n ts . T h e fa c t th a t m o s t o f th e im a g e s in th e b o o k a r e a v a ila b le fo r
d o w n lo a d in g fu r th e r e n h a n c e s th e v a lu e o f th e w e b s ite a s a te a c h in g r e s o u r c e .
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Chapter 2
Problem Solutions
Problem 2.1
T h e d ia m e te r, x , o f th e re tin a l im a g e c o r re s p o n d in g to th e d o t is o b ta in e d fr o m
s im ila r t r ia n g le s , a s s h o w n in F ig . P 2 .1 . T h a t is ,
(d / 2 )
=
0 .2
(x / 2 )
0 .0 1 7
w h ic h g iv e s x = 0 .0 8 5 d . F r o m t h e d is c u s s io n in S e c t io n 2 .1 .1 , a n d t a k in g s o m e
lib e r tie s o f in te r p r e ta tio n , w e c a n th in k o f th e fo v e a a s a s q u a r e s e n s o r a r r a y
h a v in g o n t h e o r d e r o f 3 3 7 ,0 0 0 e le m e n t s , w h ic h t r a n s la t e s in t o a n a r r a y o f s iz e
5 8 0 × 5 8 0 e le m e n ts . A s s u m in g e q u a l s p a c in g b e tw e e n e le m e n ts , th is g iv e s 5 8 0
e le m e n ts a n d 5 7 9 s p a c e s o n a lin e 1 .5 m m lo n g . T h e s iz e o f e a c h e le m e n t a n d
e a c h s p a c e is t h e n s = [(1 .5 m m )/ 1 , 1 5 9 ] = 1 .3 × 1 0 − 6 m . If t h e s iz e ( o n t h e fo v e a )
o f th e im a g e d d o t is le s s th a n th e s iz e o f a s in g le r e s o lu tio n e le m e n t, w e a s s u m e
th a t th e d o t w ill b e in v is ib le to th e e y e . In o th e r w o r d s , th e e y e w ill n o t d e te c t a
d o t if it s d ia m e t e r, d , is s u c h t h a t 0 .0 8 5 (d ) < 1 .3 × 1 0 − 6 m , o r d < 1 5 .3 × 1 0 − 6 m .
Ed g e v iew o f d o t
Imag e o f th e d o t
o n th e fo v ea
d /2
d
0 .2 m
0 .0 1 7 m
F ig u r e P 2 .1
9
x
x/2
C H A P T E R 2 . P R O B L E M S O L U T IO N S
1 0
Problem 2.2
B r ig h tn e s s a d a p ta tio n .
Problem 2.3
T h e s o lu tio n is
λ
c / v
=
2 .9 9 8 × 1 0 8 (m / s )/ 6 0 (1 / s )
=
=
4 .9 9 7 × 1 0
6
m = 4 9 9 7 K m .
Problem 2.4
(a ) F ro m th e
s o u rc e o f th e
s a m e s iz e o r s
s h a p e a n d n o
m in a tio n s o u
b e a b le to d e t
im a g e th e s p e
d is c u s s io n o n th e e le c tr o m a g n e tic s p e c tr u m in S e
illu m in a tio n r e q u ir e d to s e e a n o b je c t m u s t h a v e w
m a lle r th a n th e o b je c t. B e c a u s e in te r e s t lie s o n ly o n
t o n o th e r s p e c tr a l c h a r a c te r is tic s o f th e s p e c im e n s
r c e in th e fa r u ltr a v io le t (w a v e le n g th o f .0 0 1 m ic r o n
e c t a ll o b je c ts . A fa r - u ltr a v io le t c a m e r a s e n s o r w o u ld
c im e n s .
c t io n 2 .2 , t h e
a v e le n g th th e
th e b o u n d a r y
, a s in g le illu s o r le s s ) w ill
b e n e e d e d to
(b ) N o a n s w e r is re q u ire d b e c a u s e th e a n s w e r to (a ) is a ffi r m a tiv e .
Problem 2.5
F ro m t
S o th e
lin e , s o
d iv id e
h e
ta r
th
b y
g e o m e tr y
g e t s iz e is
e r e s o lu tio
2 , g iv in g a
o f F ig .
1 0 0 m
n o f 1
n a n s w
2 .3 , (7
m o n
lin e is
e r o f 5
m m )/ (3 5 m m ) = (z )/ (5 0 0 m m ), o r z = 1 0 0 m m .
th e s id e . W e h a v e a to ta l o f 1 0 2 4 e le m e n ts p e r
1 0 2 4 / 1 0 0 = 1 0 e le m e n ts / m m . F o r lin e p a ir s w e
lp / m m .
Problem 2.6
O n e p o s s ib le s o lu tio n is to e q u ip a m o n o c h r o m e c a m e r a w ith a m e c h a n ic a l d e v ic e th a t s e q u e n tia lly p la c e s a r e d , a g r e e n a n d a b lu e p a s s fi lte r in fr o n t o f th e
le n s . T h e s tr o n g e s t c a m e r a r e s p o n s e d e te r m in e s th e c o lo r. If a ll th r e e r e s p o n s e s
a r e a p p r o x im a te ly e q u a l, th e o b je c t is w h ite . A fa s te r s y s te m w o u ld u tiliz e th r e e
d iffe r e n t c a m e r a s , e a c h e q u ip p e d w ith a n in d iv id u a l fi lte r. T h e a n a ly s is th e n
w o u ld b e b a s e d o n p o llin g th e r e s p o n s e o f e a c h c a m e r a . T h is s y s te m w o u ld b e
a little m o r e e x p e n s iv e , b u t it w o u ld b e fa s te r a n d m o r e r e lia b le . N o te th a t b o th
s o lu tio n s a s s u m e th a t th e fi e ld o f v ie w o f th e c a m e r a (s ) is s u c h th a t it is c o m -
1 1
In ten sity
2 5 5
0
( x 0, y 0)
(a)
In ten sity
DG
2 5 5
Eq u ally
sp aced
su b d iv isio n s
0
(b )
F ig u r e P 2 .7
p le te ly
th e v e h
q u ire d
s o lv in g
fi lle d
ic le w
to is o
th is p
b y a u n
h e re o n
la te th e
r o b le m
ifo r m c o lo r [i.e ., t h e c a m e r a ( s ) is ( a r e ) fo c u s e d o n a p a r t o f
ly its c o lo r is s e e n . O th e r w is e fu r th e r a n a ly s is w o u ld b e r e r e g io n o f u n ifo r m c o lo r, w h ic h is a ll th a t is o f in te r e s t in
].
Problem 2.7
T h e im a g e in q u e s tio n is g iv e n b y
f (x ,y ) = i (x ,y )r (x ,y )
= 2 5 5 e
= 2 5 5 e
A c ro
tiz e d
4 G =
b e d e
w o rd
)2 + (y − y 0 )2 ]
− [(x − x
− [(x − x
0
0
2
2
) + (y − y 0 ) ]
s s s e c tio n o f th e im a g e is s h o w n in F ig .
u s in g m b its , th e n w e h a v e th e s itu a tio
(2 5 5 + 1 )/ 2 m . S in c e a n a b r u p t c h a n g e o
te c ta b le b y th e e y e , it fo llo w s th a t 4 G =
s , 3 2 , o r fe w e r, in te n s ity le v e ls w ill p r o d u c
P 2
n
f 8
8
× 1 .0
.7 (a ). If t
s h o w n in
in te n s ity
= 2 5 6 / 2 m
e v is ib le fa ls
h e in te n s ity is q u a
F ig . P 2 .7 ( b ) , w h e
le v e ls is a s s u m e d
, o r m = 5 . In o th
e c o n to u r in g .
n re
to
e r
C H A P T E R 2 . P R O B L E M S O L U T IO N S
1 2
In ten sity
2 5 5
1 9 1
1 2 7
6 3
0
2 5 5
1 9 1
1 2 7
6 3
Imag e
q u an tized
in to fo u r
lev els
F ig u r e P 2 .8
Problem 2.8
T h e u
in th e
0 a n d
o n . T
c o u rs
s e o f tw
ra n g e 0
6 3 b e c
h e im a g
e , th e re
o b its (m
to 2 5 5 . O
o d e d a s 6
e r e s u ltin
a re o th e r
= 2 ) o f in te n s ity re s o
n e w a y to s u b d iv id e
3 , a ll le v e ls b e tw e e n
g fr o m th is ty p e o f s
w a y s to s u b d iv id e th
lu tio n p r
th is r a n g
6 4 a n d 1
u b d iv is io
e ra n g e [
o d u c e s fo
e is to le t
2 7 b e c o d
n is s h o w
0 ,2 5 5 ] in t
u r in
a ll le
e d a
n in
o fo u
te n s ity le
v e ls b e tw
s 1 2 7 , a n
F ig . P 2 .8
r b a n d s .
v e ls
e e n
d s o
. O f
Problem 2.9
( a ) T h e to ta l a m o u n t o f d a ta (in c lu d in g th e s ta r t a n d s to p b it) in a n 8 - b it, 1 0 2 4 ×
1 0 2 4 im a g e , is (1 0 2 4 )2 × [8 + 2 ] b its . T h e to ta l tim e re q u ire d to tr a n s m it th is im a g e
o v e r a 5 6 K b a u d lin k is (1 0 2 4 )2 × [8 + 2 ]/ 5 6 0 0 0 = 1 8 7 .2 5 s e c o r a b o u t 3 .1 m in .
( b ) A t 3 0 0 0 K t h is t im e g o e s d o w n t o a b o u t 3 .5 s e c .
Problem 2.10
T h e w id th - to 1 1 2 5 lin e s (o r,
g iv e n th a t th e
th e r e s o lu tio n
T h e s y s te m “ p
h e ig h t r a t
w h a t is th
r e s o lu tio n
in th e h o r
a in ts ” a fu
io is
e s a
in t
iz o n
ll 1 1
1 6 / 9 a n
m e th in
h e h o r iz
ta l d ire c
2 5 × 2 0 0
d th
g , 1 1
o n ta
tio n
0 , 8 -
e r e s o lu tio n in
2 5 p ix e ls in th e
l d ire c tio n is in
is (1 1 2 5 ) × (1 6 / 9
b it im a g e e v e r y
th e
v e r
th e
) =
1 / 3
v e
tic
1 6
2 0
0 s
r tic a l
a l d ire
/ 9 p ro
0 0 p ix
e c fo r
d ire c tio n is
c tio n ). It is
p o r tio n , s o
e ls p e r lin e .
e a c h o f th e
1 3
F ig u r e P 2 .1 1
re d , g
to ta l
1 .1 6 6
s h o w
re e n , a n d b
d ig ita l d a ta
× 1 0 1 3 b its
w h y im a g e
lu e
g e n
, o r
d a t
c o m p o
e ra te d
1 .4 5 8 ×
a c o m p
n e n t im a g
in th is tim
1 0 12 b y te
re s s io n (C
e s . T h e re a r
e in te r v a l is
s ( i.e ., a b o u
h a p te r 8 ) is
e 7
(1
t 1
s o
2 0 0
1 2 5 )
.5 t e
im p
s e c in tw o h o u r s , s o th e
(2 0 0 0 )(8 )(3 0 )(3 )(7 2 0 0 ) =
ra b y te s ). T h e s e fi g u re s
o r ta n t.
Problem 2.11
L e
b e
th
th
t p a
c a u s
e s e t
e s e t
n d
e q
N 8
N 4
q b e
is n
(p );
(p ) \
a s s h o
o t in th
(c ) S 1 a
N 4 (q )
w n in F ig . P 2 .1 1 . T h e n , ( a ) S 1 a n d S 2 a r e n o t 4 - c o n n e c t e d
e s e t N 4 (p ); (b ) S 1 a n d S 2 a re 8 - c o n n e c te d b e c a u s e q is in
n d S 2 a re m - c o n n e c te d b e c a u s e (i) q is in N D (p ), a n d (ii)
is e m p ty .
Problem 2.12
T h e s o lu tio n o f th is
to g o fr o m a d ia g o n
P 2 .1 2 illu s t r a t e s . T h
e r y tim e a d ia g o n a l
p ro
a l s
e a
s e g
b le m c
e g m e n
lg o r ith
m e n ts
o n s is ts
t to a c o
m th e n
is e n c o
o f d e fi n in g a ll p o s s
r re s p o n d in g 4 - c o n
s im p ly lo o k s fo r th
u n te re d in th e b o u
ib le
n e c
e a p
n d a
n e ig h b o r h o o d s h a p e s
te d s e g m e n ts a s F ig .
p r o p r ia te m a tc h e v r y .
Problem 2.13
T h e s o lu t io n t o t h is p r o b le m is th e s a m e a s fo r P r o b le m 2 .1 2 b e c a u s e c o n v e r t in g
fr o m a n m - c o n n e c te d p a th to a 4 - c o n n e c te d p a th s im p ly in v o lv e s d e te c tin g d ia g o n a l s e g m e n ts a n d c o n v e r tin g th e m to th e a p p r o p r ia te 4 - c o n n e c te d s e g m e n t.
Problem 2.14
T h e d iffe r e n c e b e tw e e n th e p ix e ls in th e b a c k g r o u n d th a t a r e h o le s a n d p ix e ls
th a t a r e n o t h o le s is th a n n o p a th s e x is t b e tw e e n h o le p ix e ls a n d th e b o u n d a r y
o f th e im a g e . S o , th e d e fi n itio n c o u ld b e r e s ta te d a s fo llo w s : T h e s u b s e t o f p ix e ls
C H A P T E R 2 . P R O B L E M S O L U T IO N S
1 4
Þ
or
Þ
or
Þ
or
Þ
or
F ig u r e P 2 .1 2
o f (R U )c th a t a r e c o n n e c te d to th e b o r d e r o f th e im a g e is c a lle d th e b a c k g r o u n d .
A ll o th e r p ix e ls o f (R U )c a r e c a lle d h o le p ix e ls .
Problem 2.15
(a )
s ib
a ls
to
le n
a re
W h e n
le to g
o h a v e
g e t to
g th o f
u n iq u
(b
its
b e
s h
d a
) O
le
tw
o w
s h
V = f 0 ,1 g
e t fro m p
v a lu e s fr o
q . T h e s h
th e s h o r te
e in th is c
n e p o s s ib
n g th is 6 .
e e n p a n
n in F ig .
e d ) is 6 . T
, 4
to
m
o r
s t
a s
-p
q
V
te
m
e .
a th
b y
. F
s t 8
- p
d o e s
tra v e
ig u re
-p a th
a th (s
n o t
lin g
P 2 .1
is s
h o w
ility fo r th e s h o r te s t
It is e a s ily v e r ifi e d
d q . O n e p o s s ib ilit
P 2 .1 5 ( d ) ; it s le n g t h
h is p a th is n o t u n iq
e x is t b
a lo n g
5 (a ) s h
h o w n
n d a s h
4 -p a
th a t
y fo r
is 4 .
u e .
e tw e e n p a n d q b e c a u s e
p o in ts th a t a re b o th 4 - a
o w s th is c o n d itio n ; it is
in F ig . P 2 .1 5 ( b ) ; it s le n g
e d ) is 5 . B o th o f th e s e s h
th w h e
a n o th e
th e s h
T h e le
n V =
r 4 -p
o r te s
n g th
f 1
a th
t 8
o f
,2 g is
o f th
-p a th
a s h o
s h o w n
e s a m e
(it is n
r te s t m
it is im p o s d ja c e n t a n d
n o t p o s s ib le
th is 4 . T h e
o r te s t p a th s
in F ig . P
le n g th
o t u n iq
-p a th (s
2 .1 5
e x is
u e )
h o w
(c );
ts
is
n
1 5
F ig u r e P.2 .1 5
Problem 2.16
(a ) A s h o r te s t 4 - p a th b e tw e e n a p o in t p w ith c o o rd in a te s (x ,y ) a n d a p o in t q
w it h c o o r d in a t e s (s , t ) is s h o w n in F ig . P 2 .1 6 , w h e r e t h e a s s u m p t io n is t h a t a ll
p o in ts a lo n g th e p a th a r e fr o m V . T h e le n g th o f th e s e g m e n ts o f th e p a th a r e
jx − s j a n d y − t , r e s p e c t iv e ly . T h e t o t a l p a t h le n g t h is jx − s j + y − t , w h ic h w e
r e c o g n iz e a s t h e d e fi n it io n o f t h e D 4 d is t a n c e , a s g iv e n in E q . ( 2 .5 - 2 ) . ( R e c a ll t h a t
t h is d is t a n c e is in d e p e n d e n t o f a n y p a t h s t h a t m a y e x is t b e t w e e n th e p o in t s .)
T h e D 4 d is ta n c e o b v io u s ly is e q u a l to th e le n g th o f th e s h o r te s t 4 - p a th w h e n th e
le n g t h o f t h e p a t h is jx − s j + y − t . T h is o c c u r s w h e n e v e r w e c a n g e t f r o m p
to q b y fo llo w in g a p a th w h o s e e le m e n ts (1 ) a r e fr o m V , a n d (2 ) a r e a r r a n g e d in
s u c h a w a y th a t w e c a n tr a v e r s e th e p a th fr o m p to q b y m a k in g tu r n s in a t m o s t
t w o d ir e c t io n s ( e .g ., r ig h t a n d u p ) .
( b ) T h e p a th m a y o r m a y n o t b e u n iq u e , d e p e n d in g o n V a n d th e v a lu e s o f th e
p o in ts a lo n g th e w a y .
Problem 2.17
(a ) T h e D
8
d is t a n c e b e t w e e n p a n d q [s e e E q . ( 2 .5 - 3 ) a n d F ig . P 2 .1 6 ] is
D
R e c a
m e n
D 4 d
in th
th e p
ll
ts
is
e
a
th a t th
th e s a
ta n c e ,
p re v io
th le n g
e D 8 d
m e a s
is in d e
u s p ro
th is m
8
p ,q
is ta n c e (u n
h o r iz o n ta l
p e n d e n t o f
b le m , th e s
a x jx − s j ,
=
m a x
lik e th e E u
a n d v e r tic
w h e th e r o
h o r te s t 8 y − t . T
jx − s j , y − t
.
c lid e a n d is ta n c e ) c o u n
a l s e g m e n ts , a n d , a s in
r n o t a p a th e x is ts b e tw
p a th is e q u a l to th e D 8
h is o c c u r s w h e n w e c a n
ts d ia g o n a l s e g th e c a s e o f th e
e e n p a n d q . A s
d is ta n c e w h e n
g e t fro m p to q
C H A P T E R 2 . P R O B L E M S O L U T IO N S
1 6
F ig u r e P 2 .1 6
b y fo llo
a w a y t
o n e d ir
th e h o r
w in g a
h a t w e
e c tio n
iz o n ta
p a th w h o s e e le
c a n tra v e rs e th
a n d , w h e n e v e r
l o r v e r tic a l (b u t
m e
e p
d ia
n o
n ts
a th
g o n
t b o
(1 ) a re f
fro m p
a l tra v e
th ) d ire
ro m
to q
l is n
c tio n
V , a n d (2 ) a re a r r a n g e d in s u c h
b y tr a v e lin g d ia g o n a lly in o n ly
o t p o s s ib le , b y m a k in g tu r n s in
.
( b ) T h e p a th m a y o r m a y n o t b e u n iq u e , d e p e n d in g o n V a n d th e v a lu e s o f th e
p o in ts a lo n g th e w a y .
Problem 2.18
W it h r e fe r e n c e t o E q . ( 2 .6 - 1 ) , le t H d e n o t e t h e s u m o p e r a t o r, le t S 1 a n d S 2 d e n o te tw o d iffe r e n t s m a ll s u b im a g e a r e a s o f th e s a m e s iz e , a n d le t S 1 + S 2 d e n o te
th e c o r r e s p o n d in g p ix e l- b y - p ix e l s u m o f th e e le m e n ts in S 1 a n d S 2 , a s e x p la in e d
in S e c t io n 2 .6 .1 . N o t e t h a t t h e s iz e o f t h e n e ig h b o r h o o d ( i.e ., n u m b e r o f p ix e ls )
is n o t c h a n g e d b y th is p ix e l- b y - p ix e l s u m . T h e o p e r a to r H c o m p u te s th e s u m
o f p ix e l v a lu e s in a g iv e n n e ig h b o r h o o d . T h e n , H (a S 1 + b S 2 ) m e a n s : (1 ) m u ltip ly th e p ix e ls in e a c h o f th e s u b im a g e a r e a s b y th e c o n s ta n ts s h o w n , (2 ) a d d
th e p ix e l- b y - p ix e l v a lu e s fr o m a S 1 a n d b S 2 (w h ic h p r o d u c e s a s in g le s u b im a g e
a r e a ), a n d (3 ) c o m p u te th e s u m o f th e v a lu e s o f a ll th e p ix e ls in th a t s in g le s u b im a g e a r e a . L e t a p 1 a n d b p 2 d e n o te tw o a r b itr a r y (b u t c o r r e s p o n d in g ) p ix e ls fr o m
a S 1 + b S 2 . T h e n w e c a n w r ite
H (a S
1
+ b S
2
) =
a p
p
1
2 S
1
p
1
2 S
1
a n d p
=
a p
2 S
= a H (S
1
1
1
+ b p
1
2
+
1
p
1
2 S
b p
p
= a
p
2
2
2 S
2
2
+ b
p
p
) + b H (S
2 S
2
2
)
2
2
2
1 7
w h ic h , a c c o r d in g t o E q . ( 2 .6 - 1 ) , in d ic a t e s t h a t H is a lin e a r o p e r a t o r.
Problem 2.19
T h e m e d ia n , ζ , o f a s e t
b e lo w ζ a n d th e o th e r h
t h a t E q . ( 2 .6 - 1 ) is v io la
f 4 ,5 ,6 g , a n d a = b = 1 .
H (S 1 + S 2 ) = m e d ia n f 5 , 3
s u m o f S 1 a n d S 2 . N e x t,
m e d ia n f 4 ,5 ,6 g = 5 . T h e
t h a t E q . ( 2 .6 - 1 ) is v io la t e
o f
a lf
te d
In
,9 g
w e
n ,
d a
n u m b e r s is s u c h th a t h a lf th e v a lu e s in
a r e a b o v e it. A s im p le e x a m p le w ill s u ffi
b y th e m e d ia n o p e r a to r. L e t S 1 = f 1 ,−
th is c a s e H is th e m e d ia n o p e r a to r. W e
= 5 , w h e re it is u n d e r s to o d th a t S 1 + S 2
c o m p u te H (S 1 ) = m e d ia n f 1 , − 2 , 3 g = 1 a
b e c a u s e H ( a S 1 + b S 2 ) 6= a H ( S 1 ) + b H ( S 2
n d th e m e d ia n is a n o n lin e a r o p e r a to r.
th e s e t a re
c e to s h o w
2 ,3 g , S 2 =
th e n h a v e
is th e a r r a y
n d H (S 2 ) =
), it fo llo w s
Problem 2.20
F r o m E q . ( 2 .6 - 5 ) , a t a n y p o in t (x , y ),
g =
K
1
i
th e f i a r
o m e a n ,
o f E q . (2
ro v e th e
e th e
s o E
.6 - 6 )
v a lid
.
i = 1
K
i = 1
E f η ig .
s a m e im a g e , s o E f f i g = f . A ls o , it is g iv e n th a t th e n o is e
f η i g = 0 . T h u s , it fo llo w s th a t E f g g = f , w h ic h p r o v e s th e
it y o f E q . ( 2 .6 - 7 ) c o n s id e r t h e p r e c e d in g e q u a t io n a g a in :
g =
It is k n o w n fr o m
r e la te d r a n d o m v
[1 9 9 1 ]). B e c a u s e
u n c o r r e la te d , th e
K
i = 1
K
1
E f f ig +
η i.
K
i = 1
K
1
E f g g =
K
1
f i +
K
i = 1
T h e n
B u t a ll
h a s z e r
v a lid ity
T o p
=
g
K
K
1
K
1
g
K
=
i
K
1
f i +
K
i = 1
i = 1
1
K
K
η i.
i = 1
r a n d o m - v a r ia b le th e o r y th a t th e v a r ia n c e o f th e s u m o f u n c o r a r ia b le s is th e s u m o f th e v a r ia n c e s o f th o s e v a r ia b le s (P a p o u lis
it is g iv e n th a t th e e le m e n ts o f f a r e c o n s ta n t a n d th e η i a r e
n
σ
2
g
= σ
2
f
+
1
K
2
[σ
2
η
1
+ σ
2
η
2
+ ··· + σ
2
η
K
].
T h e fi r s t te r m o n th e r ig h t s id e is 0 b e c a u s e th e e le m e n ts o f f a r e c o n s ta n ts . T h e
v a r i o u s σ η2 i a r e s i m p l y s a m p l e s o f t h e n o i s e , w h i c h i s h a s v a r i a n c e σ η2 . T h u s ,
C H A P T E R 2 . P R O B L E M S O L U T IO N S
1 8
σ
2
η
i
= σ
2
η
a n d w e h a v e
σ
2
g
=
K
K
2
σ
2
η
=
1
σ
K
2
η
w h ic h p r o v e s t h e v a lid it y o f E q . ( 2 .6 - 7 ) .
Problem 2.21
( a ) P ix e ls a r e in te g e r v a lu e s , a n d 8 b its a llo w r e p r e s e n ta tio n o f 2 5 6 c o n tig u o u s in te g e r v a lu e s . In o u r w o r k , th e r a n g e o f in te n s ity v a lu e s fo r 8 - b it im a g e s is [0 , 2 5 5 ].
T h e s u b tr a c tio n o f v a lu e s in th is r a n g e c o v e r th e r a n g e [− 2 5 5 , 2 5 5 ]. T h is r a n g e o f
v a lu e s c a n n o t b e c o v e r e d b y 8 b its , b u t it is g iv e n in th e p r o b le m s ta te m e n t th a t
th e r e s u lt o f s u b tr a c tio n h a s to b e r e p r e s e n te d in 8 b its a ls o , a n d , c o n s is te n t w ith
th e r a n g e o f v a lu e s u s e d fo r 8 - b it im a g e s th r o u g h o u t th e b o o k , w e a s s u m e th a t
v a lu e s o f th e 8 - b it d iffe r e n c e im a g e s a r e in th e r a n g e [0 , 2 5 5 ]. W h a t th is m e a n s is
th a t a n y s u b tr a c tio n o f 2 p ix e ls th a t y ie ld s a n e g a tiv e q u a n tity w ill b e c lip p e d a t
0 .
T h e p r o c e s s o f re p e a te d s u b tr a c tio n s o f a n im a g e b (x ,y ) fr o m a n im a g e a (x ,y )
c a n b e e x p re s s e d a s
K
d
K
(x ,y ) = a (x ,y ) −
b (x ,y )
k = 1
= a (x ,y ) − K × b (x ,y )
w h e r e d K (x , y ) is th e d iffe r e n c e im a g e r e s u ltin g a fte r K s u b tr a c tio n s . B e c a u s e
im a g e s u b tr a c tio n is a n a r r a y o p e r a tio n (s e e S e c tio n 2 .6 .1 ), w e c a n fo c u s a tte n tio n o n th e s u b tr a c tio n o f a n y c o r r e s p o n d in g p a ir o f p ix e ls in th e im a g e s . W e
h a v e a lr e a d y s ta te d th a t n e g a tiv e r e s u lts a r e c lip p e d a t 0 . O n c e a 0 r e s u lt is o b ta in e d , it w ill r e m a in s o b e c a u s e s u b tr a c tio n o f a n y n o n n e g a tiv e v a lu e fr o m 0 is
a n e g a tiv e q u a n tity w h ic h , a g a in , is c lip p e d a t 0 . S im ila r ly , a n y lo c a tio n x 0 , y 0
fo r w h ic h b x 0 , y 0 = 0 , w ill p r o d u c e th e r e s u lt d K x 0 , y 0 = a x 0 , y 0 . T h a t
is , r e p e a te d ly s u b tr a c tin g 0 fr o m a n y v a lu e r e s u lts in th a t v a lu e . T h e lo c a tio n s
in b (x , y ) th a t a r e n o t 0 w ill e v e n tu a lly d e c r e a s e th e c o r r e s p o n d in g v a lu e s in
d K (x ,y ) u n til th e y a re 0 . T h e m a x im u m n u m b e r o f s u b tr a c tio n s in w h ic h th is
ta k e s p la c e in th e c o n te x t o f th e p r e s e n t p r o b le m is 2 5 5 , w h ic h c o r r e s p o n d s to
th e c o n d itio n a t a lo c a tio n in w h ic h a (x , y ) is 2 5 5 a n d b (x , y ) is 1 . T h u s , w e c o n c lu d e fr o m th e p r e c e d in g d is c u s s io n th a t r e p e a te d ly s u b tr a c tin g a n im a g e fr o m
a n o th e r w ill r e s u lt in a d iffe r e n c e im a g e w h o s e c o m p o n e n ts a r e 0 in th e lo c a tio n s in b (x , y ) th a t a r e n o t z e r o a n d e q u a l to th e o r ig in a l v a lu e s o f a (x , y ) a t
th e lo c a tio n s in b (x , y ) th a t a r e 0 . T h is r e s u lt w ill b e a c h ie v e d in , a t m o s t, 2 5 5
s u b tr a c tio n s .
1 9
(b
o r
a
re
) T h e
d in a te
x 0 ,y 0
s u lt in
o rd e r d o e s m a tte r. F o r e x a m
s , x 0 ,y 0 , a x 0 ,y 0 = 1 2 8 a n
w ill r e s u lt in d K x 0 , y 0 = 1 2
a v a lu e o f 0 in th a t s a m e lo c a
p le ,
d b
8 in
tio n
s u p p o s e th a t a t a p a ir o f a r b itr a r y c o x 0 ,y 0 = 0 . S u b tr a c tin g b x 0 ,y 0 fr o m
th e lim it. R e v e r s in g th e o p e r a tio n w ill
.
Problem 2.22
L e t g (x , y ) d e n o te th e g o ld e n im a g e , a n d le t f (x , y ) d e n o te a n y in p u t im a g e a c q u ire d d u r in g r o u tin e o p e r a tio n o f th e s y s te m . C h a n g e d e te c tio n v ia s u b tr a c tio n is b a s e d o n c o m p u tin g th e s im p le d iffe r e n c e d (x , y ) = g (x , y ) − f (x , y ). T h e
r e s u ltin g im a g e , d (x , y ), c a n b e u s e d in tw o fu n d a m e n ta l w a y s fo r c h a n g e d e te c tio n . O n e w a y is u s e p ix e l- b y - p ix e l a n a ly s is . In th is c a s e w e s a y th a t f (x , y ) is
“ c lo s e e n o u g h ” to th e g o ld e n im a g e if a ll th e p ix e ls in d (x , y ) fa ll w ith in a s p e c ifi e d th r e s h o ld b a n d [T m in , T m a x ] w h e r e T m in is n e g a tiv e a n d T m a x is p o s itiv e .
U s u a lly , th e s a m e v a lu e o f th r e s h o ld is u s e d fo r b o th n e g a tiv e a n d p o s itiv e d iffe r e n c e s , s o th a t w e h a v e a b a n d [− T , T ] in w h ic h a ll p ix e ls o f d (x , y ) m u s t fa ll in
o r d e r fo r f (x , y ) to b e d e c la r e d a c c e p ta b le . T h e s e c o n d m a jo r a p p r o a c h is s im p ly to s u m a ll th e p ix e ls in d (x , y ) a n d c o m p a r e th e s u m a g a in s t a th r e s h o ld Q .
N o te th a t th e a b s o lu te v a lu e n e e d s to b e u s e d to a v o id e r r o r s c a n c e lin g o u t. T h is
is a m u c h c r u d e r te s t, s o w e w ill c o n c e n tr a te o n th e fi r s t a p p r o a c h .
T h e re a re th re e fu n d a m e n ta l fa c to r s th a t n e e d tig h t c o n tr o l fo r d iffe re n c e b a s e d in s p e c tio n to w o r k : (1 ) p r o p e r r e g is tr a tio n , (2 ) c o n tr o lle d illu m in a tio n ,
a n d (3 ) n o is e le v e ls th a t a r e lo w e n o u g h s o th a t d iffe r e n c e v a lu e s a r e n o t a ffe c te d
a p p r e c ia b ly b y v a r ia tio n s d u e to n o is e . T h e fi r s t c o n d itio n b a s ic a lly a d d r e s s e s
th e r e q u ir e m e n t th a t c o m p a r is o n s b e m a d e b e tw e e n c o r r e s p o n d in g p ix e ls . T w o
im a g e s c a n b e id e n tic a l, b u t if th e y a r e d is p la c e d w ith r e s p e c t to e a c h o th e r,
c o m p a r in g th e d iffe re n c e s b e tw e e n th e m m a k e s n o s e n s e . O fte n , s p e c ia l m a r k in g s a r e m a n u fa c tu r e d in to th e p r o d u c t fo r m e c h a n ic a l o r im a g e - b a s e d a lig n m e n t
C o n tr o lle d illu m in a tio n (n o te th a t “ illu m in a tio n ” is n o t lim ite d to v is ib le lig h t)
o b v io u s ly is im p o r ta n t b e c a u s e c h a n g e s in illu m in a tio n c a n a ffe c t d r a m a tic a lly
th e v a lu e s in a d iffe r e n c e im a g e . O n e a p p r o a c h u s e d o fte n in c o n ju n c tio n w ith
illu m in a tio n c o n tr o l is in te n s ity s c a lin g b a s e d o n a c tu a l c o n d itio n s . F o r e x a m p le , th e p r o d u c ts c o u ld h a v e o n e o r m o r e s m a ll p a tc h e s o f a tig h tly c o n tr o lle d
c o lo r, a n d th e in te n s ity (a n d p e r h a p s e v e n c o lo r ) o f e a c h p ix e ls in th e e n tir e im a g e w o u ld b e m o d ifi e d b a s e d o n th e a c tu a l v e r s u s e x p e c te d in te n s ity a n d / o r
c o lo r o f th e p a tc h e s in th e im a g e b e in g p r o c e s s e d .
F in a lly , th e n o is e c o n te n t o f a d iffe r e n c e im a g e n e e d s to b e lo w e n o u g h s o
th a t it d o e s n o t m a te r ia lly a ffe c t c o m p a r is o n s b e tw e e n th e g o ld e n a n d in p u t im a g e s . G o o d s ig n a l s tr e n g th g o e s a lo n g w a y to w a r d r e d u c in g th e e ffe c ts o f n o is e .
C H A P T E R 2 . P R O B L E M S O L U T IO N S
2 0
F ig u r e P 2 .2 3
A n o th e r (s o m e tim e s c o m p le m e n ta r y ) a p p r o a c h is to im p le m e n t im a g e p r o c e s s in g t e c h n iq u e s ( e .g ., im a g e a v e r a g in g ) t o r e d u c e n o is e .
O b v io u s ly th e r e a r e a n u m b e r if v a r ia tio n s o f th e b a s ic th e m e ju s t d e s c r ib e d .
F o r e x a m p le , a d d itio n a l in te llig e n c e in th e fo r m o f te s ts th a t a r e m o r e s o p h is tic a te d th a n p ix e l- b y - p ix e l th r e s h o ld c o m p a r is o n s c a n b e im p le m e n te d . A te c h n iq u e u s e d o fte n in th is r e g a r d is to s u b d iv id e th e g o ld e n im a g e in to d iffe r e n t
r e g io n s a n d p e r fo r m d iffe r e n t (u s u a lly m o r e th a n o n e ) te s ts in e a c h o f th e r e g io n s , b a s e d o n e x p e c te d re g io n c o n te n t.
Problem 2.23
( a ) T h e a n s w e r is s h o w n in F ig . P 2 .2 3 .
( b ) W ith r e fe r e n c e to th e s e ts in th e p r o b le m s ta te m e n t, th e a n s w e r s a r e , fr o m
le ft to r ig h t,
(A \ B \ C ) − (B \ C ) ;
(A \ B \ C ) [ (A \ C ) [ (A \ B ) ;
c
B \ (A [ C )
[ f (A \ C ) − [(A \ C ) \ (B \ C )]g .
Problem 2.24
U s in g tr ia n g u la r r e g io n s m e a n s th r e e tie p o in ts , s o w e c a n s o lv e th e fo llo w in g s e t
o f lin e a r e q u a tio n s fo r s ix c o e ffi c ie n ts :
y
x
0
0
=
c 1 x + c 2 y + c
3
=
c 4 x + c 5 y + c
6
2 1
to im p le m e n t s p a tia l tr a n s fo r m a tio n s . In te n s ity in te r p o la tio n is im p le m e n te d
u s in g a n y o f t h e m e t h o d s in S e c t io n 2 .4 .4 .
Problem 2.25
T h e F o u r ie r tr a n s fo r m a tio n k e r n e l is s e p a r a b le b e c a u s e
r (x ,y ,u ,v )
=
− j 2 π
e
=
e
=
(
)
u x / M + v y / N
− j 2 π (u x / M )
e
− j 2 π
(
v y / N
)
r 1 (x ,u )r 2 (y ,v ).
It is s y m m e tr ic b e c a u s e
e
− j 2 π
(
u x / M + v y / N
)
=
e
=
− j 2 π (u x / M )
e
− j 2 π
(
v y / N
)
r 1 (x ,u )r 1 (y ,v ).
Problem 2.26
F r o m E q . ( 2 .6 - 2 7 ) a n d t h e d e fi n it io n o f s e p a r a b le k e r n e ls ,
M − 1 N − 1
T (u ,v )
f (x ,y )r (x ,y ,u ,v )
=
x = 0
y = 0
M − 1
N − 1
r 1 (x ,u )
=
x = 0
f (x ,y )r 2 (y ,v )
y = 0
M − 1
=
T (x ,v )r 1 (x ,u )
x = 0
w h e re
N − 1
T (x ,v ) =
f (x ,y )r 2 (y ,v ).
y = 0
F o r a fi x e d v a lu e o f x , th is e q u a tio n is
o n e r o w o f f (x , y ). B y le ttin g x v a r y fr o m
T (x ,v ). T h e n , b y s u b s titu tin g th is a r r a y
tio n w e h a v e th e 1 - D tr a n s fo r m a lo n g t
w h e n a k e r n e l is s e p a r a b le , w e c a n c o m
o f th e im a g e . T h e n w e c o m p u te th e 1 - D
te r m e d ia te r e s u lt to o b ta in th e fi n a l 2 - D
r e s u lt b y c o m p u tin g th e 1 - D tr a n s fo r m a
th e 1 - D tr a n s fo r m a lo n g th e r o w s o f th e
r e c o g n iz e d a s th e 1 - D tr a n s fo r m a lo n g
0 to M − 1 w e c o m p u te th e e n tire a r r a y
in to th e la s t lin e o f th e p r e v io u s e q u a h e c o lu m n s o f T (x , v ). In o th e r w o r d s ,
p u te th e 1 - D tr a n s fo r m a lo n g th e r o w s
tr a n s fo r m a lo n g th e c o lu m n s o f th is in tr a n s fo r m , T (u ,v ). W e o b ta in th e s a m e
lo n g th e c o lu m n s o f f (x , y ) fo llo w e d b y
in te r m e d ia te r e s u lt.
C H A P T E R 2 . P R O B L E M S O L U T IO N S
2 2
T h is r e s u lt p la y s a n im p o r ta n t r o le in C h a p te r 4 w h e n w e d is c u s s th e 2 - D
F o u r ie r t r a n s fo r m . F r o m E q . ( 2 .6 - 3 3 ) , t h e 2 - D F o u r ie r t r a n s fo r m is g iv e n b y
M − 1 N − 1
x = 0
(
− j 2 π
f (x ,y )e
T (u ,v ) =
).
u x / M + v y / N
y = 0
It is e a s ily v e r ifi e d t h a t t h e F o u r ie r t r a n s fo r m k e r n e l is s e p a r a b le ( P r o b le m 2 .2 5 ) ,
s o w e c a n w r ite th is e q u a tio n a s
M − 1 N − 1
T (u ,v )
=
f (x ,y )e
x = 0
(
u x / M + v y / N
)
y = 0
N − 1
M − 1
=
− j 2 π
e
− j 2 π (u x / M )
x = 0
− j 2 π
f (x ,y )e
(
v y / N
)
y = 0
M − 1
T (x ,v )e
=
− j 2 π (u x / M )
x = 0
w h e re
N − 1
f (x ,y )e
T (x ,v ) =
− j 2 π
(
v y / N
)
y = 0
is th e 1 - D F o u r ie r tr a n s fo r m a lo n g th e r o w s o f f (x , y ), a s w e le t x = 0 , 1 , . . . , M − 1 .
Problem 2.27
T h e g e o m e t r y o f t h e c h ip s is s h o w n in F ig . P 2 .2 7 ( a ) . F r o m F ig . P 2 .2 7 ( b ) a n d t h e
g e o m e t r y in F ig . 2 .3 , w e k n o w t h a t
Δ x =
λ × 8 0
λ − z
w h e re Δ x is th e s id e d im e n s io n o f th e im a g e (a s s u m e d s q u a re b e c a u s e th e
v ie w in g s c r e e n is s q u a r e ) im p in g in g o n th e im a g e p la n e , a n d th e 8 0 m m r e fe r s
to th e s iz e o f th e v ie w in g s c r e e n , a s d e s c r ib e d in th e p r o b le m s ta te m e n t. T h e
m o s t in e x p e n s iv e s o lu tio n w ill r e s u lt fr o m u s in g a c a m e r a o f r e s o lu tio n 5 1 2 × 5 1 2 .
B a s e d o n t h e in fo r m a t io n in F ig . P 2 .2 7 ( a ) , a C C D c h ip w it h t h is r e s o lu t io n w ill
b e o f s iz e (1 6 μ ) × (5 1 2 ) = 8 m m o n e a c h s id e . S u b s titu tin g Δ x = 8 m m in th e
a b o v e e q u a tio n g iv e s z = 9 λ a s th e r e la tio n s h ip b e tw e e n th e d is ta n c e z a n d th e
fo c a l le n g th o f th e le n s , w h e r e a m in u s s ig n w a s ig n o r e d b e c a u s e it is ju s t a c o o r d in a te in v e r s io n . If a 2 5 m m le n s is u s e d , th e fr o n t o f th e le n s w ill h a v e to b e
lo c a te d a t a p p r o x im a te ly 2 2 5 m m fr o m th e v ie w in g s c r e e n s o th a t th e s iz e o f th e
2 3
F ig u r e P 2 .2 7
im a
m m
b u t
p o s
g e o f th e s c r e e n p r o je c te d o n to th e C C D im a g e p la n e d o e s n o t e x c e e d th e 8
s iz e o f th e C C D c h ip fo r th e 5 1 2 × 5 1 2 c a m e r a . T h is v a lu e o f z is r e a s o n a b le ,
a n y o th e r g iv e n le n s s iz e s w o u ld b e a ls o ; th e c a m e r a w o u ld ju s t h a v e to b e
itio n e d fu r th e r a w a y .
A s s u m in g a 2 5 m m le n s , th e n e x t is s u e is to d e te r m in e if th e s m a lle s t d e fe c t
w ill b e im a g e d o n , a t le a s t, a 2 × 2 p ix e l a r e a , a s r e q u ir e d b y th e s p e c ifi c a tio n . It
is g iv e n th a t th e d e fe c ts a r e c ir c u la r, w ith th e s m a lle s t d e fe c t h a v in g a d ia m e te r
o f 0 .8 m m . S o , a ll t h a t n e e d s t o b e d o n e is t o d e t e r m in e if t h e im a g e o f a c ir c le o f
d ia m e t e r 0 .8 m m o r g r e a t e r w ill, a t le a s t , b e o f s iz e 2 × 2 p ix e ls o n t h e C C D im a g in g p la n e . T h is c a n b e d e t e r m in e d b y u s in g t h e s a m e m o d e l a s in F ig . P 2 .2 7 ( b )
w ith th e 8 0 m m r e p la c e d b y 0 .8 m m . U s in g λ = 2 5 m m a n d z = 2 2 5 m m in t h e
a b o v e e q u a tio n y ie ld s Δ x = 1 0 0 μ . In o th e r w o r d s , a c ir c u la r d e fe c t o f d ia m e te r
0 .8 m m w ill b e im a g e d a s a c ir c le w it h a d ia m e t e r o f 1 0 0 μ o n t h e C C D c h ip o f a
5 1 2 × 5 1 2 c a m e r a e q u ip p e d w ith a 2 5 m m le n s a n d w h ic h v ie w s th e d e fe c t a t a
d is ta n c e o f 2 2 5 m m .
If, in o rd e r fo r a C C D re c e p to r to b e a c tiv a te d , its a re a h a s to b e e x c ite d in
its e n tir e ty , th e n , it c a n b e s e e n fr o m F ig . P 2 .2 7 (a ) th a t to g u a r a n te e th a t a 2 ×
2 a r r a y o f s u c h r e c e p to r s w ill b e a c tiv a te d , a c ir c u la r a r e a o f d ia m e te r n o le s s
th a n (6 )(8 ) = 4 8 μ h a s to b e im a g e d o n to th e C C D c h ip . T h e s m a lle s t d e fe c t is
im a g e d a s a c ir c le w ith d ia m e te r o f 1 0 0 μ , w h ic h is w e ll a b o v e th e 4 8 μ m in im u m
re q u ire m e n t.
2 4
C H A P T E R 2 . P R O B L E M S O L U T IO N S
T h e r e fo r e , w e c o n c lu d e th a t a C C D c a m e r a o f r e s o lu tio n 5 1 2 × 5 1 2 p ix e ls ,
u s in g a 2 5 m m le n s a n d im a g in g th e v ie w in g s c r e e n a t a d is ta n c e o f 2 2 5 m m , is
s u ffi c ie n t to s o lv e th e p r o b le m p o s e d b y th e p la n t m a n a g e r.
Chapter 3
Problem Solutions
Problem 3.1
L e t f d e n o te th e o r ig in a l im a g e . F ir s t s u b tr a c t th e m in im u m v a lu e o f f d e n o te d
f m in f r o m f t o y i e ld a f u n c t i o n w h o s e m i n i m u m v a lu e i s 0 :
g
=
1
f −
f
m in
N e x t d iv id e g 1 b y its m a x im u m v a lu e to y ie ld a fu n c tio n in th e r a n g e [0 , 1 ] a n d
m u ltip ly th e r e s u lt b y L − 1 to y ie ld a fu n c tio n w ith v a lu e s in th e r a n g e [0 , L − 1 ]
=
g
=
K e e p in m in d th a t f
m in
L − 1
g 1
m a x g 1
L − 1
m a x f − f m
f −
f
m in
in
is a s c a la r a n d f is a n im a g e .
Problem 3.2
2
( a ) G e n e r a l fo r m : s = T (r ) = A e − K r . F o r th e c o n d itio n s h o w n in th e p r o b le m
2
fi g u r e , A e − K L 0 = A / 2 . S o lv in g fo r K y ie ld s
− K L
2
0
= ln (0 .5 )
K = 0 .6 9 3 / L
T h e n ,
−
s = T (r ) = A e
(b ) G e n e ra l fo r m : s = T (r ) = B (1 − e
− K r
2 5
2
0 .6 9 3
L 2
0
2
.
0
r
2
.
). F o r th e c o n d itio n s h o w n in th e p r o b -
C H A P T E R 3 . P R O B L E M S O L U T IO N S
2 6
F ig u r e P 3 .3
le m fi g u r e , B (1 − e
2
− K L
0
) = B / 2 . T h e s o lu tio n fo r K is th e s a m e a s in (a ), s o
−
s = T (r ) = B (1 − e
(c ) G e n e ra l fo r m : s = T (r ) = (D − C )(1 − e
− K r
2
0 .6 9 3
L 2
0
2
r
)
) + C .
Problem 3.3
(a ) s = T (r ) =
1
1 + (m / r )
E
.
( b ) S e e F ig . P 3 .3 .
(c ) W e w a n t s to b e 0 fo r r < m , a n d s to b e 1 fo r
s = 1 / 2 . B u t, b e c a u s e th e v a lu e s o f r a r e in te g e r s ,
⎧
w h e n r ≤
⎨ 0 .0
0 .5
w h e n r =
s = T (r ) =
⎩
1 .0
w h e n r ≥
v a lu e s o f r > m . W h e n r = m ,
th e b e h a v io r w e w a n t is
m − 1
m
m + 1 .
T h e q u e s tio n in th e p r o b le m s ta te m e n t is to fi n d th e s m a lle s t v a lu e o f E th a t
w ill m a k e th e th r e s h o ld b e h a v e a s in th e e q u a tio n a b o v e . W h e n r = m , w e s e e
fr o m ( a ) t h a t s = 0 .5 , r e g a r d le s s o f t h e v a lu e o f E . If C is t h e s m a lle s t p o s it iv e
2 7
n u m b
th e n a
s m a lle
fo r E ,
e r re p re
n y v a lu
s t v a lu e
u s in g th
s e n ta b le
e o f s le s
o f E fo r
e g iv e n v
in th
s th a
w h ic
a lu e
e c o m
n C / 2
h th is
m = 1
p u te r, a n d k e e p in g in m in d th a t s is p o s itiv e ,
w ill b e c a lle d 0 b y th e c o m p u te r. T o fi n d th e
h a p p e n s , s im p ly s o lv e th e fo llo w in g e q u a tio n
2 8 :
1
1 + [m / (m − 1 )]
E
< C / 2 .
B e c a u s e th e fu n c tio n is s y m m e tr ic a b o u t m , th e r e s u ltin g v a lu e o f E w ill y ie ld
s = 1 fo r r ≥ m + 1 .
Problem 3.4
T h e tr a n s fo r m a tio n s r e q u ir e d to p r o d u c e th e in d iv id u a l b it p la n e s a r e n o th in g
m o r e th a n m a p p in g s o f th e tr u th ta b le fo r e ig h t b in a r y v a r ia b le s . In th is tr u th
ta b le , th e v a lu e s o f th e 8 th b it a r e 0 fo r b y te v a lu e s 0 to 1 2 7 , a n d 1 fo r b y te v a lu e s
1 2 8 to 2 5 5 , th u s g iv in g th e tr a n s fo r m a tio n m e n tio n e d in th e p r o b le m s ta te m e n t.
N o te th a t th e g iv e n tr a n s fo r m e d v a lu e s o f e ith e r 0 o r 2 5 5 s im p ly in d ic a te a b in a r y
im a g e fo r th e 8 th b it p la n e . A n y o th e r tw o v a lu e s w o u ld h a v e b e e n e q u a lly v a lid ,
th o u g h le s s c o n v e n tio n a l.
C o n tin u in g w ith th e tr u th ta b le c o n c e p t, th e tr a n s fo r m a tio n r e q u ir e d to p r o d u c e a n im a g e o f th e 7 th b it p la n e o u tp u ts a 0 fo r b y te v a lu e s in th e r a n g e [0 ,
6 3 ], a 1 fo r b y te v a lu e s in th e r a n g e [6 4 , 1 2 7 ], a 0 fo r b y te v a lu e s in th e r a n g e
[1 2 8 , 1 9 1 ], a n d a 1 fo r b y te v a lu e s in th e r a n g e [1 9 2 , 2 5 5 ]. S im ila r ly , th e tr a n s fo r m a tio n fo r th e 6 th b it p la n e a lte r n a te s b e tw e e n e ig h t r a n g e s o f b y te v a lu e s ,
th e tr a n s fo r m a tio n fo r th e 5 th b it p la n e a lte r n a te s b e tw e e n 1 6 r a n g e s , a n d s o
o n . F in a lly , th e o u tp u t o f th e tr a n s fo r m a tio n fo r th e lo w e s t- o r d e r b it p la n e a lte r n a te s b e tw e e n 0 a n d 2 5 5 d e p e n d in g o n w h e th e r th e b y te v a lu e s a r e e v e n o r
o d d . T h u s , th is tr a n s fo r m a tio n a lte r n a te s b e tw e e n 1 2 8 b y te v a lu e r a n g e s , w h ic h
e x p la in s w h y a n im a g e o f th a t b it p la n e is u s u a lly th e “ b u s ie s t” lo o k in g o f a ll th e
b it p la n e im a g e s .
Problem 3.5
(a
th
th
th
in
) T h e n u m b e r o f p ix e ls
u s c a u s in g th e n u m b e r
e n u m b e r o f p ix e ls w o u
e re m a in in g h is to g r a m
in te n s ity le v e l v a lu e s w
h a
o f
ld
p e
ill
v in g d iffe re n t in te n s ity
c o m p o n e n ts in th e h is
n o t c h a n g e , th is w o u ld
a k s to in c re a s e in g e n e
re d u c e c o n tra s t.
le v e l v a lu
to g ra m to
c a u s e th
r a l. T y p ic
e s w o u
d e c re a
e h e ig h
a lly , le s
ld d e
s e . B
t o f s
s v a r
c re
e c a
o m
ia b
a s e ,
u s e
e o f
ility
( b ) T h e m o s t v is ib le e ffe c t w o u ld b e s ig n ifi c a n t d a r k e n in g o f th e im a g e . F o r e x a m p le , d r o p p in g th e h ig h e s t b it w o u ld lim it th e b r ig h te s t le v e l in a n 8 - b it im -
C H A P T E R 3 . P R O B L E M S O L U T IO N S
2 8
a g e to b
o f s o m e
to g ra m
lo c a te d
e 1 2 7 . B e c a u s e th e n u m b e r o f p ix e ls w o u ld r e m a in c o n s ta n t, th e h e ig h t
o f th e h is to g r a m p e a k s w o u ld in c r e a s e . T h e g e n e r a l s h a p e o f th e h is w o u ld n o w b e ta lle r a n d n a r r o w e r, w ith n o h is to g r a m c o m p o n e n ts b e in g
p a s t 1 2 7 .
Problem 3.6
A ll th a t h is to g r a m e q u a liz a tio n d o e s is r e m a p h is to g r a m c o m p o n e n ts o n th e in te n s ity s c a le . T o o b ta in a u n ifo r m (fl a t) h is to g r a m w o u ld r e q u ir e in g e n e r a l th a t
p ix e l in te n s itie s a c tu a lly b e r e d is tr ib u te d s o th a t th e r e a r e L g r o u p s o f n / L p ix e ls
w ith th e s a m e in te n s ity , w h e r e L is th e n u m b e r o f a llo w e d d is c r e te in te n s ity le v e ls a n d n = M N is th e to ta l n u m b e r o f p ix e ls in th e in p u t im a g e . T h e h is to g r a m
e q u a liz a tio n m e th o d h a s n o p r o v is io n s fo r th is ty p e o f (a r tifi c ia l) in te n s ity r e d is tr ib u tio n p r o c e s s .
Problem 3.7
L e t n = M N b e th e to ta l n u m b e r o f p ix e ls a n d le t n rj b e th e n u m b e r o f p ix e ls
in th e in p u t im a g e w ith in te n s ity v a lu e r j . T h e n , th e h is to g r a m e q u a liz a tio n
tr a n s fo r m a tio n is
k
s
k
= T (r k ) =
n
r
j
k
1
/ n =
n
n
j = 0
r
j
.
j = 0
B e c a u s e e v e r y p ix e l (a n d n o o th e r s ) w ith v a lu e r k is m a p p e d to v a lu e s k , it fo llo w s th a t n sk = n rk . A s e c o n d p a s s o f h is to g r a m e q u a liz a tio n w o u ld p r o d u c e
v a lu e s v k a c c o r d in g to th e tr a n s fo r m a tio n
v
B u t, n
s
j
= n
r
j
k
= T (s
k
k
1
) =
n
n
s
j
.
j = 0
, s o
v
k
= T (s
k
) =
1
n
k
n
r
j
= s
k
j = 0
w h ic h s h o w s th a t a s e c o n d p a s s o f h is to g r a m e q u a liz a tio n w o u ld y ie ld th e s a m e
r e s u lt a s th e fi r s t p a s s . W e h a v e a s s u m e d n e g lig ib le r o u n d - o ff e r r o r s .
2 9
Problem 3.8
T h e g e n e r a l h is to g r a m e q u a liz a tio n tr a n s fo r m a tio n fu n c tio n is
r
s = T (r ) =
p
(w ) d w .
r
0
T h e re a re tw o im p o r ta n t p o in ts a b o u t w h ic h th e s tu d e n t m u s t s h o w a w a re n e s s
in a n s w e r in g th is p r o b le m . F ir s t, th is e q u a tio n a s s u m e s o n ly p o s itiv e v a lu e s fo r
r . H o w e v e r, th e G a u s s ia n d e n s ity e x te n d s in g e n e r a l fr o m − 1 to 1 . R e c o g n itio n
o f th is fa c t is im p o r ta n t. O n c e re c o g n iz e d , th e s tu d e n t c a n a p p r o a c h th is d iffi c u lty in s e v e r a l w a y s . O n e g o o d a n s w e r is to m a k e s o m e a s s u m p tio n , s u c h a s th e
s ta n d a r d d e v ia tio n b e in g s m a ll e n o u g h s o th a t th e a r e a o f th e c u r v e u n d e r p r (r )
fo r n e g a tiv e v a lu e s o f r is n e g lig ib le . A n o th e r is to s c a le u p th e v a lu e s u n til th e
a r e a u n d e r th e n e g a tiv e p a r t o f th e c u r v e is n e g lig ib le . T h e s e c o n d m a jo r p o in t
is to r e c o g n iz e is th a t th e tr a n s fo r m a tio n fu n c tio n its e lf,
r
s = T (r ) =
p
1
2 π σ
e
−
(w − m )2
2 σ 2
d w
0
h
G
u
h
h
A
p
d
a s n o c lo s e d - fo r m s o lu tio n . T h is is th e c u m u la tiv e d is tr ib u tio n fu n c tio n o f th e
a u s s ia n d e n s ity , w h ic h is e ith e r in te g r a te d n u m e r ic a lly , o r its v a lu e s a r e lo o k e d
p in a ta b le . A th ir d , le s s im p o r ta n t p o in t, th a t th e s tu d e n t s h o u ld a d d r e s s is th e
ig h - e n d v a lu e s o f r . A g a in , th e G a u s s ia n P D F e x te n d s to + 1 . O n e p o s s ib ility
e re is to m a k e th e s a m e a s s u m p tio n a s a b o v e re g a rd in g th e s ta n d a rd d e v ia tio n .
n o th e r is to d iv id e b y a la r g e e n o u g h v a lu e s o th a t th e a r e a u n d e r th e p o s itiv e
a r t o f th e P D F p a s t th a t p o in t is n e g lig ib le (th is s c a lin g r e d u c e s th e s ta n d a r d
e v ia tio n ).
A n o th e r a p p r o a c h th e s tu d e n t c a n ta k e is to w o r k w ith h is to g r a m s , in w h ic h
c a s e th e tr a n s fo r m a tio n fu n c tio n w o u ld b e in th e fo r m o f a s u m m a tio n . T h e is s u e o f n e g a tiv e a n d h ig h p o s itiv e v a lu e s m u s t s till b e a d d r e s s e d , a n d th e p o s s ib le
a n s w e r s s u g g e s te d a b o v e r e g a r d in g th e s e is s u e s s till a p p ly . T h e s tu d e n t n e e d s to
in d ic a te th a t th e h is to g r a m is o b ta in e d b y s a m p lin g th e c o n tin u o u s fu n c tio n ,
s o s o m e m e n tio n s h o u ld b e m a d e r e g a r d in g th e n u m b e r o f s a m p le s (b its ) u s e d .
T h e m o s t lik e ly a n s w e r is 8 b its , in w h ic h c a s e th e s tu d e n t n e e d s to a d d r e s s th e
s c a lin g o f th e fu n c tio n s o th a t th e r a n g e is [0 , 2 5 5 ].
Problem 3.9
W e a r e in te r e s te d in ju s t o n e e x a m p le in o r d e r to s a tis fy th e s ta te m e n t o f th e
p r o b le m . C o n s id e r t h e p r o b a b ilit y d e n s it y fu n c t io n in F ig . P 3 .9 ( a ) . A p lo t o f
C H A P T E R 3 . P R O B L E M S O L U T IO N S
3 0
2(L - 1)
L - 1
0
L /4
L /2
3L /4
L - 1
0
L /4
L /2
3L /4
L - 1
L - 1
(L - 1)/2
F ig u r e P 3 .9 .
t h e t r a n s fo r m a t io n T (r ) in E q . ( 3 .3 - 4 ) u s in g t h is p a r t ic u la r d e n s it y fu n c t io n is
s h o w n in F ig . P 3 .9 ( b ) . B e c a u s e p r (r ) is a p r o b a b ilit y d e n s it y fu n c t io n w e k n o w
fr o m th e d is c u s s io n in S e c tio n 3 .3 .1 th a t th e tr a n s fo r m a tio n T (r ) s a tis fi e s c o n d it io n s ( a ) a n d ( b ) s t a t e d in t h a t s e c t io n . H o w e v e r, w e s e e fr o m F ig . P 3 .9 ( b ) t h a t
th e in v e r s e tr a n s fo r m a tio n fr o m s b a c k to r is n o t s in g le v a lu e d , a s th e r e a r e a n
in fi n ite n u m b e r o f p o s s ib le m a p p in g s fr o m s = (L − 1 ) / 2 b a c k to r . It is im p o r ta n t to n o te th a t th e re a s o n th e in v e r s e tr a n s fo r m a tio n fu n c tio n tu r n e d o u t n o t
to b e s in g le v a lu e d is th e g a p in p r (r ) in th e in te r v a l [ L / 4 , 3 L / 4 ] .
Problem 3.10
( a ) W e n e e d to s h o w th a t th e tr a n s fo r m a tio n fu n c tio n in E q . (3 .3 - 8 ) is m o n o to n ic
in t h e r a n g e [0 , L − 1 ], a n d t h a t it s v a lu e s a r e in t h e r a n g e [0 , L − 1 ]. F r o m E q . ( 3 .3 8 )
k
s
k
= T (r k ) = (L − 1 )
p
r
(r j )
j = 0
B e c a u s e a ll th e p r (r j ) a r e p o s itiv e , it fo llo w s th a t T (r k ) is m o n o to n ic in th e r a n g e
k 2 [0 , L − 1 ]. B e c a u s e th e s u m o f a ll th e v a lu e s o f p r (r j ) is 1 , it fo llo w s th a t 0 ≤
s k ≤ L − 1 .
3 1
( b ) If n o n e o f th e in te n s ity le v e ls r k , k = 1 , 2 , . . . , L − 1 , a r e 0 , th e n T (r k ) w ill b e
s tr ic tly m o n o to n ic . T h is im p lie s a o n e - to - o n e m a p p in g b o th w a y s , m e a n in g th a t
b o th fo r w a r d a n d in v e r s e tr a n s fo r m a tio n s w ill b e s in g le - v a lu e d .
Problem 3.11
F ir s t, w e o b ta in th e h is to g r a m e q u a liz a tio n tr a n s fo r m a tio n :
r
r
s = T (r ) =
p
r
(w ) d w =
0
2
(− 2 w + 2 ) d w = − r
+ 2 r .
0
N e x t w e fi n d
z
v = G (z ) =
z
p
(w ) d w =
z
2 w d w = z
0
2
.
0
F in a lly ,
− 1
z = G
p
(v ) = ±
v .
B u t o n ly p o s itiv e in te n s ity le v e ls a r e a llo w e d , s o z =
s , w h ic h in tu r n is − r 2 + 2 r , a n d w e h a v e
z =
− r
2
p
v . T h e n , w e r e p la c e v w ith
+ 2 r .
Problem 3.12
T h e v a lu e o f th e h is to g r a m c o m p o n e n t c o r r e s p o n d in g to th e k th in te n s ity le v e l
in a n e ig h b o r h o o d is
n k
p r (r k ) =
n
fo r k = 1 , 2 , . . . , K − 1 , w h e r e n k is th e n u m b e r o f p ix e ls h a v in g in te n s ity le v e l r k ,
n is th e to ta l n u m b e r o f p ix e ls in th e n e ig h b o r h o o d , a n d K is th e to ta l n u m b e r
o f p o s s ib le in te n s ity le v e ls . S u p p o s e th a t th e n e ig h b o r h o o d is m o v e d o n e p ix e l
to th e r ig h t (w e a r e a s s u m in g r e c ta n g u la r n e ig h b o r h o o d s ). T h is d e le te s th e le ftm o s t c o lu m n a n d in tr o d u c e s a n e w c o lu m n o n th e r ig h t. T h e u p d a te d h is to g r a m
th e n b e c o m e s
1
p r0 ( r k ) =
[n k − n L k + n R k ]
n
fo r k = 0 , 1 , . . . , K − 1 , w h e r e n L k is th e n u m b e r o f o c c u r r e n c e s o f le v e l r k o n th e
le ft c o lu m n a n d n R k is th e s im ila r q u a n tity o n th e r ig h t c o lu m n . T h e p r e c e d in g
e q u a tio n c a n b e w r itte n a ls o a s
p
r
0
(r k ) = p
r
(r k ) +
1
n
[n
R
k
− n
L
k
]
C H A P T E R 3 . P R O B L E M S O L U T IO N S
3 2
fo r k = 0 , 1 , . . . , K − 1 . T h e s a m e c o n c e p t a p p lie s to o th e r m o d e s o f n e ig h b o r h o o d
m o tio n :
1
[b k − a k ]
p r0 ( r k ) = p r ( r k ) +
n
fo r k = 0 , 1 , . . . , K − 1 , w h e r e a k is th e n u m b e r o f p ix e ls w ith v a lu e r k in th e n e ig h b o r h o o d a r e a d e le te d b y th e m o v e , a n d b k is th e c o r r e s p o n d in g n u m b e r in tr o d u c e d b y th e m o v e .
Problem 3.13
T h e p u r p o s e o f th is s im p le p r o b le m is to m a k e th e s tu d e n t th in k o f th e m e a n in g
o f h is to g r a m s a n d a r r iv e a t th e c o n c lu s io n th a t h is to g r a m s c a r r y n o in fo r m a tio n
a b o u t s p a tia l p r o p e r tie s o f im a g e s . T h u s , th e o n ly tim e th a t th e h is to g r a m o f th e
im a g e s fo r m e d b y th e o p e r a tio n s s h o w n in th e p r o b le m s ta te m e n t c a n b e d e te r m in e d in te r m s o f th e o r ig in a l h is to g r a m s is w h e n o n e (b o th ) o f th e im a g e s
is (a re ) c o n s ta n t. In (d ) w e h a v e th e a d d itio n a l re q u ire m e n t th a t n o n e o f th e
p ix e ls o f g (x , y ) c a n b e 0 . A s s u m e fo r c o n v e n ie n c e th a t th e h is to g r a m s a r e n o t
n o r m a liz e d , s o th a t, fo r e x a m p le , h f (r k ) is th e n u m b e r o f p ix e ls in f (x , y ) h a v in g
in te n s ity le v e l r k . A s s u m e a ls o th a t a ll th e p ix e ls in g (x , y ) h a v e c o n s ta n t v a lu e c .
T h e p ix e ls o f b o th im a g e s a r e a s s u m e d to b e p o s itiv e . F in a lly , le t u k d e n o te th e
in te n s ity le v e ls o f th e p ix e ls o f th e im a g e s fo r m e d b y a n y o f th e a r ith m e tic o p e r a tio n s g iv e n in th e p r o b le m s ta te m e n t. U n d e r th e p r e c e d in g s e t o f c o n d itio n s ,
th e h is to g r a m s a r e d e te r m in e d a s fo llo w s :
(a ) W e o b ta in
h sum (u k ) = h
n e n ts o f h sum
in te n s ity a x is
th e
f ( r k
a re
a re
h is to
) fo r
th e s
s h ifte
g ra
a ll
a m
d r
m h su m (u k ) o f th e s u m b y le ttin g u k = r k + c , a n d a ls o
k . In o th e r w o r d s , th e v a lu e s (h e ig h t) o f th e c o m p o e a s th e c o m p o n e n ts o f h f , b u t th e ir lo c a tio n s o n th e
ig h t b y a n a m o u n t c .
( b ) S i m i la r ly , t h e h i s t o g r a m h d iff( u k ) o f t h e d i f f e r e n c e h a s t h e s a m e c o m p o n e n t s
a s h f b u t th e ir lo c a tio n s a r e m o v e d le ft b y a n a m o u n t c a s a r e s u lt o f th e s u b tr a c tio n o p e r a tio n .
( c ) F o llo w in g th e s a m e r e a s o n
to g r a m h p ro d (u k ) o f th e p r o d
u k = c × r k . N o te th a t w h ile
h is to g r a m s in (a ) a n d (b ) w a s
h p ro d (u k ) w ill b e s p r e a d o u t b
(d
th
c o
p r
) F in
o s e o
m p o
e c e d
a lly , a s s
f h f , b u
n e n ts o
in g s o lu
u m in g th a t
t th e ir lo c a t
f h d iv ( u k ) w
tio n s a re a p
in g , th e v a lu e s (h
u c t a re th e s a m e
th e s p a c in g b e tw
n o t a ffe c te d , th e
y a n a m o u n t c .
c =6
io n
ill b
p lic
0 , th e c o m p
s w ill b e a t u k
e c o m p re s s e d
a b le if im a g e
e ig h ts ) o
a s h f , b
e e n c o m
s p a c in g
o n e n ts
= r k / c
b y a n
f (x , y )
f th e
u t th
p o n e
b e tw
c o m p
e ir lo
n ts o
e e n c
o f h d iv ( u k )
. T h u s , th e
a m o u n t e q
is c o n s ta n t
o n e n ts o f h is c a tio n s a re a t
f th e r e s u ltin g
o m p o n e n ts o f
a re th
s p a c in
u a l to
a ls o . I
e s a m e
g b e tw e
1 / c . T
n th is c a
a s
e n
h e
s e
3 3
F ig u r e P 3 .1 4
th e fo u r h is to g r a m s ju s t d is c u s s e d w o u ld e a c h h a v e o n ly o n e c o m p o n e n t. T h e ir
lo c a tio n w o u ld b e a ffe c te d a s d e s c r ib e d (a ) th r o u g h (d ).
Problem 3.14
(a ) T h e n
la r g e r in
p o in ts w
r ig h t, s o
(b ) T
a s s u
fro m
a 3 ×
o f v a
m e n
s u m
c o n s
th e r
u m
th e
ill g
th e
b e r o f
im a g
iv e r is
h is to g
b o u n
e o n
e to a
ra m s
d a r y
th e r
la r g
o f th
p o
ig h
e r
e t
in ts b e tw e e n th e b la c k a n d w h ite r e g io n s is m u c h
t. W h e n th e im a g e s a r e b lu r r e d , th e b o u n d a r y
n u m b e r o f d iffe r e n t v a lu e s fo r th e im a g e o n th e
w o b lu r r e d im a g e s w ill b e d iffe r e n t.
o h a n d le th e b o r d e r e ffe c ts , w e s u r r o u n d th e im a g e w ith a b o r d e r o f 0 s . W e
m e th a t im a g e is o f s iz e N × N (th e fa c t th a t th e im a g e is s q u a re is e v id e n t
th e r ig h t im a g e in th e p r o b le m s ta te m e n t). B lu r r in g is im p le m e n te d b y
3 m a s k w h o s e c o e ffi c ie n t s a r e 1 / 9 . F ig u r e P 3 .1 4 s h o w s th e d iffe r e n t t y p e s
lu e s th a t th e b lu r r e d le ft im a g e w ill h a v e (s e e im a g e in th e p r o b le m s ta te t ) . T h e v a lu e s a r e s u m m a r iz e d in T a b le P 3 .1 4 - 1 . It is e a s ily v e r ifi e d t h a t t h e
o f th e n u m b e r s o n th e le ft c o lu m n o f th e ta b le is N 2 . A h is to g r a m is e a s ily
tr u c te d fr o m th e e n tr ie s in th is ta b le . A s im ila r (te d io u s ) p r o c e d u r e y ie ld s
e s u lt s in T a b le P 3 .1 4 - 2 .
T a b le P 3 .1 4 - 1
N o . o f P o in ts
V a lu e
N N2 − 1
0
2
2 / 9
N − 2
3 / 9
4
4 / 9
3 N − 8
6 / 9
1
( N − 2 ) N2 − 2
C H A P T E R 3 . P R O B L E M S O L U T IO N S
3 4
T a b le P 3 .1 4 - 2
N o . o f P o in ts
V a lu e
2
N
2
2
N
2
− 1 4 N
2 8
1 4 N −
1 2 8
9 8
1 6 N −
− 1 6 N
+ 9 8
0
2 / 9
3 / 9
4 / 9
5 / 9
6 / 9
1
2 2 4
2 5 6
+ 1 2 8
Problem 3.15
(a
th
in
th
th
) C o n s id e r a
e 1 / 9 s c a le fa
te n s ity v a lu e
e re s p o n s e o
e r ig h t, it p ic
3 × 3 m a s k
c to r), th e
s o f p ix e ls
f th e m a s k
k s u p o n ly
fi rs t. B
n e t e ffe
u n d e r t
. H o w e
o n e n e
e c a u s e
c t o f th
h e m a s
v e r, w h
w c o lu
a ll
e lo
k . I
e n
m n
th e c o e ffi c ie n ts a re 1 (w e
w p a s s fi lte r o p e r a tio n is t
n itia lly , it ta k e s 8 a d d itio n
th e m a s k m o v e s o n e p ix e
. T h e n e w re s p o n s e c a n b
a re
o a
s to
l lo
e c
ig n o r in
d d a ll th
p ro d u c
c a tio n t
o m p u te
g
e
e
o
d
a s
R
n e w
= R
o ld
− C
1
+ C
3
w h e r e C 1 is th e s u m o f p ix e ls u n d e r th e fi r s t c o lu m n o f th e m a s k b e fo r e it w a s
m o v e d , a n d C 3 is th e s im ila r s u m in th e c o lu m n it p ic k e d u p a fte r it m o v e d . T h is
is th e b a s ic b o x - fi lte r o r m o v in g - a v e r a g e e q u a tio n . F o r a 3 × 3 m a s k it ta k e s 2
a d d itio n s to g e t C 3 (C 1 w a s a lr e a d y c o m p u te d ). T o th is w e a d d o n e s u b tr a c tio n
a n d o n e a d d itio n to g e t R n ew . T h u s , a to ta l o f 4 a r ith m e tic o p e r a tio n s a re n e e d e d
to u p d a te th e re s p o n s e a fte r o n e m o v e . T h is is a re c u r s iv e p r o c e d u re fo r m o v in g
fr o m le ft to r ig h t a lo n g o n e r o w o f th e im a g e . W h e n w e g e t to th e e n d o f a r o w , w e
m o v e d o w n o n e p ix e l (th e n a tu re o f th e c o m p u ta tio n is th e s a m e ) a n d c o n tin u e
th e s c a n in th e o p p o s ite d ire c tio n .
F o r a m a s k o f s iz e n × n , (n − 1 ) a d d itio n s a r e n e e d e d to o b ta in C 3 , p lu s th e
s in g le s u b tr a c tio n a n d a d d itio n n e e d e d to o b ta in R n e w , w h ic h g iv e s a to ta l o f
(n + 1 ) a r ith m e tic o p e r a tio n s a fte r e a c h m o v e . A b r u te - fo r c e im p le m e n ta tio n
w o u ld r e q u ir e n 2 − 1 a d d itio n s a fte r e a c h m o v e .
(b ) T h e c o m p u ta tio n a l a d v a n ta g e is
A =
n
2
− 1
=
n + 1
(n + 1 )(n − 1 )
= n − 1 .
(n + 1 )
T h e p lo t o f A a s a fu n c tio n o f n is a s im p le lin e a r fu n c tio n s ta r tin g a t A =
n = 2 .
1 fo r
3 5
Problem 3.16
( a ) T h e k e y to s o lv in g th is p r o b le m is to r e c o g n iz e (1 ) th a t th e c o n v o lu tio n r e s u lt a t a n y lo c a tio n (x , y ) c o n s is ts o f c e n te r in g th e m a s k a t th a t p o in t a n d th e n
fo r m in g th e s u m o f th e p r o d u c ts o f th e m a s k c o e ffi c ie n ts w ith th e c o r re s p o n d in g
p ix e ls in th e im a g e ; a n d (2 ) th a t c o n v o lu tio n o f th e m a s k w ith th e e n tir e im a g e
r e s u lts in e v e r y p ix e l in th e im a g e b e in g v is ite d o n ly o n c e b y e v e r y e le m e n t o f
t h e m a s k ( i.e ., e v e r y p ix e l is m u lt ip lie d o n c e b y e v e r y c o e ffi c ie n t o f t h e m a s k ) .
B e c a u s e th e c o e ffi c ie n ts o f th e m a s k s u m to z e r o , th is m e a n s th a t th e s u m o f th e
p r o d u c ts o f th e c o e ffi c ie n ts w ith th e s a m e p ix e l a ls o s u m to z e r o . C a r r y in g o u t
th is a r g u m e n t fo r e v e r y p ix e l in th e im a g e le a d s to th e c o n c lu s io n th a t th e s u m
o f th e e le m e n ts o f th e c o n v o lu tio n a r r a y a ls o s u m to z e r o .
(b
is
re
c o
) T h e o
ro ta te d
la tin g a
r r e la tio
n ly d
b y 1
n im
n im
iffe
8 0 ◦
a g e
a g e
re n c e b e tw e
. T h is d o e s
w ith a m a s
w h o s e e le m
e n
n o t
k w
e n t
c o
a f
h o
s a
n v
fe c
s e
ls o
o lu tio n a
t th e c o n
c o e ffi c ie
s u m to z
n d
c lu
n ts
e ro
c o r r e la tio n is th a t th e m a s k
s io n s re a c h e d in (a ), s o c o rs u m to z e r o w ill p r o d u c e a
.
Problem 3.17
O n e o f th e e a s ie s t w a y s to lo o
s u p e r p o s itio n . L e t f (x , y ) a n
r e s p e c tiv e ly . A s s u m in g s q u a
e x p re s s f (x , y ) a s th e s u m o
n o n z e r o p ix e l (in itia lly , w e a s
r u n n in g h (x , y ) o v e r f (x , y ) c
h (x ,y )
k a t re p e a te d
d h (x , y ) d e n
re im a g e s o f
f a t m o s t N 2
s u m e th a t N
a n b e e x p re s
f (x ,y ) = h (x ,y )
a p p lic a tio n s o
o te th e im a g e
s iz e N × N fo
im a g e s , e a c h
c a n b e in fi n ite
s e d a s th e fo llo
f 1 (x ,y ) +
f a s p a tia l fi lte r is to u s e
a n d th e fi lte r fu n c tio n ,
r c o n v e n ie n c e , w e c a n
o f w h ic h h a s o n ly o n e
). T h e n , th e p ro c e s s o f
w in g c o n v o lu tio n :
f 2 (x ,y ) + ··· +
f
N
2
(x ,y ) .
S u p p o s e fo r illu s tr a tiv e p u r p o s e s th a t f i (x , y ) h a s v a lu e 1 a t its c e n te r, w h ile th e
o t h e r p ix e ls a r e v a lu e d 0 , a s d is c u s s e d a b o v e ( s e e F ig . P 3 .1 7 a ) . If h (x , y ) is a
3 × 3 m a s k o f 1 / 9 ’s ( F i g . P 3 . 1 7 b ) , t h e n c o n v o l v i n g h ( x , y ) w i t h f i ( x , y ) w i l l p r o d u c e a n i m a g e w i t h a 3 × 3 a r r a y o f 1 / 9 ’s a t i t s c e n t e r a n d 0 s e l s e w h e r e , a s F i g .
P 3 .1 7 ( c ) s h o w s . If h (x , y ) is n o w a p p lie d t o t h is im a g e , t h e r e s u lt in g im a g e w ill
b e a s s h o w n in F ig . P 3 .1 7 ( d ) . N o t e t h a t t h e s u m o f t h e n o n z e r o p ix e ls in b o t h
F ig s . P 3 .1 7 ( c ) a n d ( d ) is t h e s a m e , a n d e q u a l t o t h e v a lu e o f t h e o r ig in a l p ix e l.
T h u s , it is in tu itiv e ly e v id e n t th a t s u c c e s s iv e a p p lic a tio n s o f h (x , y ) w ill ” d iffu s e ” th e n o n z e r o v a lu e o f f i (x , y ) (n o t a n u n e x p e c te d r e s u lt, b e c a u s e h (x , y )
is a b lu r r in g fi lte r ). S in c e th e s u m r e m a in s c o n s ta n t, th e v a lu e s o f th e n o n z e r o
e le m e n ts w ill b e c o m e s m a lle r a n d s m a lle r, a s th e n u m b e r o f a p p lic a tio n s o f th e
fi lte r in c r e a s e s . T h e o v e r a ll r e s u lt is g iv e n b y a d d in g a ll th e c o n v o lv e d f k (x , y ),
fo r k = 1 , 2 , ..., N 2 .
3 6
C H A P T E R 3 . P R O B L E M S O L U T IO N S
F ig u r e P 3 .1 7
It is n o te d th a t e v e r y ite r a tio n o f b lu r r in g fu r th e r d iffu s e s th e v a lu e s o u tw a r d ly fr o m th e s ta r tin g p o in t. In th e lim it, th e v a lu e s w o u ld g e t in fi n ite ly s m a ll,
b u t, b e c a u s e th e a v e r a g e v a lu e r e m a in s c o n s ta n t, th is w o u ld r e q u ir e a n im a g e o f
in fi n ite s p a tia l p r o p o r tio n s . It is a t th is ju n c tio n th a t b o r d e r c o n d itio n s b e c o m e
im p o r ta n t. A lth o u g h it is n o t r e q u ir e d in th e p r o b le m s ta te m e n t, it is in s tr u c tiv e
to d is c u s s in c la s s th e e ffe c t o f s u c c e s s iv e a p p lic a tio n s o f h (x , y ) to a n im a g e o f
fi n ite p r o p o r tio n s . T h e n e t e ffe c t is th a t, b e c a u s e th e v a lu e s c a n n o t d iffu s e o u tw a r d p a s t th e b o u n d a r y o f th e im a g e , th e d e n o m in a to r in th e s u c c e s s iv e a p p lic a tio n s o f a v e r a g in g e v e n tu a lly o v e r p o w e r s th e p ix e l v a lu e s , d r iv in g th e im a g e to
z e r o in t h e lim it . A s im p le e x a m p le o f t h is is g iv e n in F ig . P 3 .1 7 ( e ) , w h ic h s h o w s
a n a r r a y o f s iz e 1 × 7 th a t is b lu r r e d b y s u c c e s s iv e a p p lic a tio n s o f th e 1 × 3 m a s k
h ( y ) = 13 [ 1 , 1 , 1 ] . W e s e e t h a t , a s l o n g a s t h e v a l u e s o f t h e b l u r r e d 1 c a n d i f f u s e
o u t, th e s u m , S , o f th e r e s u ltin g p ix e ls is 1 . H o w e v e r, w h e n th e b o u n d a r y is m e t,
3 7
a n a
tre a
d ia t
b o u
g in s
1 / (3
s s u m p tio n m u s t b e m a d e re g a rd in g h o
te d . H e r e , w e u s e d th e c o m m o n ly m a d
e ly p a s t th e b o u n d a r y a r e 0 . T h e m a s
n d a r y , h o w e v e r. In th is e x a m p le , w e s e
to d e c r e a s e w ith s u c c e s s iv e a p p lic a tio
)n w o u ld o v e r p o w e r th e s u m o f th e p ix
w m a s k o p e
e a s s u m p tio
k o p e r a tio n
e th a t th e s u
n s o f th e m a
e l v a lu e s , y ie
r a tio n s o n th e b o rd e r a re
n th a t p ix e l v a lu e im m e d o e s n o t g o b e y o n d th e
m o f th e p ix e l v a lu e s b e s k . In th e lim it, th e te r m
ld in g a n a r r a y o f 0 s .
Problem 3.18
( a ) T h e r e a r e n 2 p o in ts in a n n × n m e d ia n fi lte r m a s k . B e c a u s e n is o d d , th e
m e d ia n v a lu e , ζ , is s u c h th a t th e r e a r e (n 2 − 1 )/ 2 p o in ts w ith v a lu e s le s s th a n o r
e q u a l to ζ a n d th e s a m e n u m b e r w ith v a lu e s g r e a te r th a n o r e q u a l to ζ . H o w e v e r, b e c a u s e th e a r e a A (n u m b e r o f p o in ts ) in th e c lu s te r is le s s th a n o n e h a lf
n 2 , a n d A a n d n a r e in te g e r s , it fo llo w s th a t A is a lw a y s le s s th a n o r e q u a l to
(n 2 − 1 )/ 2 . T h u s , e v e n in th e e x tr e m e c a s e w h e n a ll c lu s te r p o in ts a r e e n c o m p a s s e d b y th e fi lte r m a s k , th e r e a r e n o t e n o u g h p o in ts in th e c lu s te r fo r a n y o f
th e m to b e e q u a l to th e v a lu e o f th e m e d ia n (r e m e m b e r, w e a r e a s s u m in g th a t
a ll c lu s te r p o in ts a r e lig h te r o r d a r k e r th a n th e b a c k g r o u n d p o in ts ). T h e r e fo r e , if
th e c e n te r p o in t in th e m a s k is a c lu s te r p o in t, it w ill b e s e t to th e m e d ia n v a lu e ,
w h ic h is a b a c k g r o u n d s h a d e , a n d th u s it w ill b e “ e lim in a te d ” fr o m th e c lu s te r.
T h is c o n c lu s io n o b v io u s ly a p p lie s to th e le s s e x tr e m e c a s e w h e n th e n u m b e r o f
c lu s te r p o in ts e n c o m p a s s e d b y th e m a s k is le s s th a n th e m a x im u m s iz e o f th e
c lu s te r.
( b ) F o r th e c o n c lu s io n r e a c h e d in (a ) to h o ld , th e n u m b e r o f p o in ts th a t w e c o n s id e r c lu s te r (o b je c t) p o in ts c a n n e v e r e x c e e d (n 2 − 1 )/ 2 . T h u s , tw o o r m o r e d iffe r e n t c lu s te r s c a n n o t b e in c lo s e e n o u g h p r o x im ity fo r th e fi lte r m a s k to e n c o m p a s s p o in ts fr o m m o r e th a n o n e c lu s te r a t a n y m a s k p o s itio n . It th e n fo llo w s th a t
n o tw o p o in ts fr o m d iffe r e n t c lu s te r s c a n b e c lo s e r th a n th e d ia g o n a l d im e n s io n
o f th e m a s k m in u s o n e c e ll (w h ic h c a n b e o c c u p ie d b y a p o in t fr o m o n e o f th e
c lu s te r s ). A s s u m in g a g r id s p a c in g o f 1 u n it, th e m in im u m d is ta n c e b e tw e e n
p
2 (n − 1 ). In o th e r
a n y tw o p o in ts o f d iffe r e n t c lu s te r s th e n m u s t g r e a te r th a n
w o r d s , th e s e p o in ts m u s t b e s e p a r a te d b y a t le a s t th e d is ta n c e s p a n n e d b y n − 1
c e lls a lo n g th e m a s k d ia g o n a l.
Problem 3.19
( a ) N u m e r ic a lly s o r t th e n
2
v a lu e s . T h e m e d ia n is
ζ = [(n
2
+ 1 )/ 2 ]- th la r g e s t v a lu e .
C H A P T E R 3 . P R O B L E M S O L U T IO N S
3 8
( b ) O n c e th e v a lu e s h a v e b e e n s o r te d o n e tim e , w e s im p ly d e le te th e v a lu e s in
th e tr a ilin g e d g e o f th e n e ig h b o r h o o d a n d in s e r t th e v a lu e s in th e le a d in g e d g e
in th e a p p r o p r ia te lo c a tio n s in th e s o r te d a r r a y .
Problem 3.20
(a ) T h e m o s t e x t
a 3 - p ix e l g a p , a lo
a c o m p le te ly b la
m a s k s iz e u p is
T h u s , th e s m a lle
(b ) T h e s m a
o n ly tw o p ix
n a r y , lik e th
A m in , a n d t h
le s s th a n B .
c re a te o n e b
re m
n g a
n k fi
g u a
s t m
e c a s e is w
th in s e g m
e ld . S in c e
ra n te e d to
a s k th a t w
h e n th e m a s k
e n t, in w h ic h
th is is k n o w n
e n c o m p a s s
ill d o th e jo b
is p o
c a s e
to b e
s o m e
is a 5
s itio
a 3 ×
th e
o f t
× 5 a
n e d o n th
3 m a s k w
la r g e s t g a
h e p ix e ls
v e r a g in g
lle s t a v e r a g e v a lu e p r o d u c e d b y th e m a s k is
e ls o f th e s e g m e n t. T h is a v e r a g e v a lu e is a g r
e r e s t o f th e s e g m e n t p ix e ls . D e n o te th e s m a
e b in a r y v a lu e s o f p ix e ls in th e th in s e g m e n t
T h e n , s e ttin g th e b in a r iz in g th r e s h o ld s lig h tly
in a r y p ix e l o f v a lu e B in th e c e n te r o f th e m a s
e c e n te
o u ld e n
p , th e n
in th e
m a s k .
w h e n it
a y - s c a le
lle s t a v e
b y B . C
s m a lle r
k .
r p ix e l o f
c o m p a s s
e x t (o d d )
s e g m e n t.
e n c o m p a s s e s
v a lu e , n o t b ir a g e v a lu e b y
le a r ly , A m in i s
t h a n A m in w i ll
Problem 3.21
F
a
h
e
q
a
r o m F ig . 3 .3 3 w e k n o w t h a t t h e v e r t ic a l b a r s a r e 5 p ix e ls w id e , 1 0 0 p ix e ls h ig h ,
n d th e ir s e p a r a tio n is 2 0 p ix e ls . T h e p h e n o m e n o n in q u e s tio n is r e la te d to th e
o r iz o n ta l s e p a r a tio n b e tw e e n b a r s , s o w e c a n s im p lify th e p r o b le m b y c o n s id r in g a s in g le s c a n lin e th r o u g h th e b a r s in th e im a g e . T h e k e y to a n s w e r in g th is
u e s tio n lie s in th e fa c t th a t th e d is ta n c e (in p ix e ls ) b e tw e e n th e o n s e t o f o n e b a r
n d th e o n s e t o f th e n e x t o n e (s a y , to its r ig h t) is 2 5 p ix e ls .
C o n s id e r t h e s c a n lin e s h o w n in F ig . P 3 .2 1 . A ls o s h o w n is a c r o s s s e c t io n
o f a 2 5 × 2 5 m a s k . T h e r e s p o n s e o f th e m a s k is th e a v e r a g e o f th e p ix e ls th a t it
e n c o m p a s s e s . W e n o te th a t w h e n th e m a s k m o v e s o n e p ix e l to th e r ig h t, it lo s e s
o n e v a lu e o f th e v e r tic a l b a r o n th e le ft, b u t it p ic k s u p a n id e n tic a l o n e o n th e
r i g h t , s o t h e r e s p o n s e d o e s n ’t c h a n g e . I n f a c t , t h e n u m b e r o f p i x e l s b e l o n g i n g
to th e v e r tic a l b a r s a n d c o n ta in e d w ith in th e m a s k d o e s n o t c h a n g e , r e g a r d le s s
o f w h e r e th e m a s k is lo c a te d (a s lo n g a s it is c o n ta in e d w ith in th e b a r s , a n d n o t
n e a r th e e d g e s o f th e s e t o f b a rs ).
T h e fa c t th a t th e n u m b e r o f b a r p ix e ls u n d e r th e m a s k d o e s n o t c h a n g e is d u e
to th e p e c u lia r s e p a r a tio n b e tw e e n b a r s a n d th e w id th o f th e lin e s in r e la tio n
to th e 2 5 - p ix e l w id th o f th e m a s k T h is c o n s ta n t re s p o n s e is th e re a s o n w h y n o
w h ite g a p s a r e s e e n in th e im a g e s h o w n in th e p r o b le m s ta te m e n t. N o te th a t th is
c o n s ta n t re s p o n s e d o e s n o t h a p p e n w ith th e 2 3 × 2 3 o r th e 4 5 × 4 5 m a s k s b e c a u s e
th e y a re n o t ” s y n c h r o n iz e d ” w ith th e w id th o f th e b a r s a n d th e ir s e p a r a tio n .
3 9
F ig u r e P 3 .2 1
Problem 3.22
T h e re a re a t m o s t q 2 p o in ts in th e a re a in w h ic h w e w a n t to re d u c e th e in te n s ity le v e l o f e a c h p ix e l to o n e - te n th its o r ig in a l v a lu e . C o n s id e r a n a v e r a g in g
m a s k o f s iz e n × n e n c o m p a s s in g th e q × q n e ig h b o r h o o d . T h e a v e r a g in g m a s k
h a s n 2 p o in ts o f w h ic h w e a r e a s s u m in g th a t q 2 p o in ts a r e fr o m th e o b je c t a n d
th e r e s t fr o m th e b a c k g r o u n d . N o te th a t th is a s s u m p tio n im p lie s s e p a r a tio n b e tw e e n o b je c ts th a t, a t a m in im u m , is e q u a l to th e a r e a o f th e m a s k a ll a r o u n d
e a c h o b je c t. T h e p r o b le m b e c o m e s in tr a c ta b le u n le s s th is a s s u m p tio n is m a d e .
T h is c o n d itio n w a s n o t g iv e n in th e p r o b le m s ta te m e n t o n p u r p o s e in o r d e r to
fo r c e th e s tu d e n t to a r r iv e a t th a t c o n c lu s io n . If th e in s tr u c to r w is h e s to s im p lify
th e p r o b le m , th is s h o u ld th e n b e m e n tio n e d w h e n th e p r o b le m is a s s ig n e d . A
fu r th e r s im p lifi c a tio n is to te ll th e s tu d e n ts th a t th e in te n s ity le v e l o f th e b a c k g r o u n d is 0 .
L e t B r e p r e s e n t th e in te n s ity le v e l o f b a c k g r o u n d p ix e ls , le t a i d e n o te th e in te n s ity le v e ls o f p o in ts in s id e th e m a s k a n d o i th e le v e ls o f th e o b je c ts . In a d d itio n , le t S a d e n o te th e s e t o f p o in ts in th e a v e r a g in g m a s k , S o th e s e t o f p o in ts in
th e o b je c t, a n d S b th e s e t o f p o in ts in th e m a s k th a t a r e n o t o b je c t p o in ts . T h e n ,
th e re s p o n s e o f th e a v e r a g in g m a s k a t a n y p o in t o n th e im a g e c a n b e w r itte n a s
R =
1
a
2
n
a
⎡
2
n
⎡
2
n
=
q
n
⎤
o
o
⎢q
⎣
q
1
=
i
a
⎢
⎣
1
=
2 S
i
j
2 S
Q +
a
a
o
2
o
2
o
2
2
+
j
j
1
n
2
2 S
j
k
⎤
2 S
b
2
⎥
⎦+
− q
⎥
⎦
⎡
1
2
n
o
(n
k
2
⎤
⎢
⎣
a
a
)B
k
2 S
b
k
⎥
⎦
C H A P T E R 3 . P R O B L E M S O L U T IO N S
4 0
w h e
a v e r
o f th
th a n
re
a g
e
o
Q
e
m
n
d e n o te
v a lu e o
a s k a t a
e -te n th
s th e a v
f o b je c t
n y p o in
Q max , o
q
n
e r a g e v a lu e o f o b je c t p o in ts . L e t th e m a x im u m e x p e c te d
p o in ts b e d e n o te d b y Q m a x . T h e n w e w a n t th e re s p o n s e
t o n th e o b je c t u n d e r th is m a x im u m c o n d itio n to b e le s s
r
2
2
Q
m a x
1
+
n
2
(n
2
− q
2
m a x
− B )
1
<
)B
1 0
Q
m a x
fr o m w h ic h w e g e t th e re q u ire m e n t
1 0 (Q
n > q
fo r th e
s ity is
b y th e
th is a n
m in im
0 , th e n
in s tr u c
s w e r fo
u m s iz e o f
th e m in im
to r, o r th e s
llo w s a lm o
th e
u m
tu d
s t b
a v e
m a
e n t
y in
(Q
ra
s k
m
s p
1 / 2
− 1 0 B )
m a x
g in g m a s k . N o te th a t if th e b a c k g r o u n d in te n p
1 0 q . If th is w a s a fa c t s p e c ifi e d
s iz e is n <
a d e th is a s s u m p tio n fr o m th e b e g in n in g , th e n
e c tio n .
Problem 3.23
T h e s tu d e n t s h o u ld r e a liz e th a t b o th th e L a p la c ia n a n d th e a v e r a g in g p r o c e s s
a r e lin e a r o p e r a tio n s , s o it m a k e s n o d iffe r e n c e w h ic h o n e is a p p lie d fi r s t.
Problem 3.24
T h e L a p la c ia n o p e r a to r is d e fi n e d a s
2
r
2
@
f =
f
2
@ x
2
@
+
f
2
@ y
fo r th e u n r o ta te d c o o rd in a te s , a n d a s
2
r
f =
2
@
@ x
f
02
+
@
@ y
2
f
02
.
fo r r o ta te d c o o rd in a te s . It is g iv e n th a t
x = x
0
c o s θ − y
0
s in θ
a n d
y = x
0
s in θ + y
0
c o s θ
w h e r e θ is th e a n g le o f r o ta tio n . W e w a n t to s h o w th a t th e r ig h t s id e s o f th e fi r s t
tw o e q u a tio n s a r e e q u a l. W e s ta r t w ith
@ f
=
@ x 0
=
@ f @ x
@ f @ y
+
@ x @ x 0
@ y @ x 0
@ f
@ f
c o s θ +
s in θ .
@ x
@ y
4 1
0
T a k in g th e p a r tia l d e r iv a tiv e o f th is e x p re s s io n a g a in w ith re s p e c t to x
2
@
f
@ x
02
2
@
=
f
2
c o s 2 θ +
@ x
@
@ x
@ f
@ y
@ f
@ x
@
s in θ c o s θ +
@ y
2
@
c o s θ s in θ +
y ie ld s
f
2
@ y
s in
2
θ .
N e x t, w e c o m p u te
@ f
@ y
@ x
@ y
@ f @ y
@ y @ y 0
@ f
f
s in θ +
c o s θ .
x
@ y
@ f
@ x
@
= −
@
=
0
+
0
T a k in g th e d e r iv a tiv e o f th is e x p re s s io n a g a in w ith re s p e c t to y
@
@ y
2
f
02
=
2
@
@ x
f
2
s in
2
θ −
@
@ x
@ f
@ y
@
c o s θ s in θ −
@ f
@ x
@ y
0
g iv e s
s in θ c o s θ +
2
@
@ y
f
2
c o s
2
θ .
A d d in g th e tw o e x p r e s s io n s fo r th e s e c o n d d e r iv a tiv e s y ie ld s
@
@ x
2
f
02
+
@
@ y
2
f
02
=
2
@
@ x
f
2
+
2
@
@ y
f
2
w h ic h p r o v e s th a t th e L a p la c ia n o p e r a to r is in d e p e n d e n t o f r o ta tio n .
Problem 3.25
T h e L a p la c ia n m a s k w ith a − 4 in th e c e n te r p e r fo r m s a n o p e r a tio n p r o p o r tio n a l
to d iffe re n tia tio n in th e h o r iz o n ta l a n d v e r tic a l d ire c tio n s . C o n s id e r fo r a m o m e n t a 3 × 3 “ L a p la c ia n ” m a s k w ith a − 2 in th e c e n te r a n d 1 s a b o v e a n d b e lo w
th e c e n te r. A ll o th e r e le m e n ts a r e 0 . T h is m a s k w ill p e r fo r m d iffe r e n tia tio n in
o n ly o n e d ir e c tio n , a n d w ill ig n o r e in te n s ity tr a n s itio n s in th e o r th o g o n a l d ir e c tio n . A n im a g e p r o c e s s e d w ith s u c h a m a s k w ill e x h ib it s h a r p e n in g in o n ly o n e
d ir e c tio n . A L a p la c ia n m a s k w ith a - 4 in th e c e n te r a n d 1 s in th e v e r tic a l a n d
h o r iz o n ta l d ir e c tio n s w ill o b v io u s ly p r o d u c e a n im a g e w ith s h a r p e n in g in b o th
d ir e c tio n s a n d in g e n e r a l w ill a p p e a r s h a r p e r th a n w ith th e p r e v io u s m a s k . S im ila r ly , a n d m a s k w ith a − 8 in th e c e n te r a n d 1 s in th e h o r iz o n ta l, v e r tic a l, a n d
d ia g o n a l d ir e c tio n s w ill d e te c t th e s a m e in te n s ity c h a n g e s a s th e m a s k w ith th e
− 4 in th e c e n te r b u t, in a d d itio n , it w ill a ls o b e a b le to d e te c t c h a n g e s a lo n g th e
d ia g o n a ls , th u s g e n e r a lly p r o d u c in g s h a r p e r - lo o k in g r e s u lts .
C H A P T E R 3 . P R O B L E M S O L U T IO N S
4 2
F ig u r e P 3 .2 6
Problem 3.26
(a ) T h e s iz e a n d c o e ffi c ie n ts o f th e 3 ×
d e fi n itio n o f th e L a p la c ia n g iv e n in E
o f c o e ffi c ie n ts (a n d th u s s iz e o f th e m
th e s e c o n d d e r iv a tiv e . A la r g e r “ L a p la
p le m e n tin g a s e c o n d d e r iv a tiv e , s o w e
s h a r p e r r e s u lts . In fa c t, a s e x p la in e d in
3 L a p la c ia n m a s
q . ( 3 .6 - 6 ) . In o
a s k ) is a d ire c t r
c ia n - lik e ” m a s k
c a n n o t e x p e c t t
p a r t (b ), ju s t th e
k c o m e d ir e c tly fr o m th
th e r w o rd s , th e n u m b e
e s u lt o f th e d e fi n itio n o
w o u ld n o lo n g e r b e im
h a t in g e n e r a l it w ill g iv
o p p o s ite o c c u r s .
e
r
f
e
( b ) In g e n e r a l, in c r e a s in g th e s iz e o f th e “ L a p la c ia n - lik e ” m a s k p r o d u c e s b lu r r in g . T o s e e w h y th is is s o , c o n s id e r a n im a g e c o n s is tin g o f tw o v e r tic a l b a n d s ,
a b la c k b a n d o n th e le ft a n d a w h ite b a n d o n th e r ig h t, w ith th e tr a n s itio n b e tw e e n th e b a n d s o c c u r r in g th r o u g h th e c e n te r o f th e im a g e . T h a t is , th e im a g e
h a s a s h a r p v e r t ic a l e d g e t h r o u g h it s c e n t e r. F r o m F ig . 3 .3 6 w e k n o w t h a t a s e c o n d d e r iv a tiv e s h o u ld p r o d u c e a d o u b le e d g e in th e r e g io n o f th e v e r tic a l e d g e
w h e n a 3 × 3 L a p la c ia n m a s k is c e n te r e d o n th e e d g e . A s th e c e n te r o f th e m a s k
m o v e s m o r e th a n tw o p ix e ls o n e ith e r s id e o f th e e d g e th e e n tir e m a s k w ill e n c o m p a s s a c o n s ta n t a r e a a n d its r e s p o n s e w o u ld b e z e r o , a s it s h o u ld . H o w e v e r,
s u p p o s e th a t th e m a s k is m u c h la r g e r. A s its c e n te r m o v e s th r o u g h , s a y , th e b la c k
(0 ) a r e a , o n e h a lf o f th e m a s k w ill b e to ta lly c o n ta in e d in th a t a r e a . H o w e v e r, d e p e n d in g o n its s iz e , p a r t o f th e m a s k w ill b e c o n ta in e d in th e w h ite (2 5 5 ) a r e a .
T h e s u m o f p r o d u c ts w ill th e r e fo r e b e d iffe r e n t fr o m 0 . T h is m e a n s th a t th e r e
w ill b e a r e s p o n s e in a n a r e a w h e r e th e r e s p o n s e s h o u ld h a v e b e e n 0 b e c a u s e
t h e m a s k is c e n t e r e d o n a c o n s t a n t a r e a . F ig u r e P 3 .2 6 s h o w s t h e s e e ffe c t s . T h e
4 3
0
0
0
0
1
0
0
0
0
1
9
17
(b)
(a)
F ig u r e P 3 .2 7
fi g u re o n
r e s u lt o f u
s h o w s th e
a n d 1 2 5 ×
m a s k s iz e
th e to p le ft is th e b a n d ju
s in g a 3 × 3 L a p la c ia n m a
r e s u lts o f u s in g “ L a p la c ia
1 2 5 , r e s p e c tiv e ly . T h e p r
is e v id e n t in th e s e r e s u lts
s t
s k
n o g
.
m
w
lik
re
e n tio n
ith a n
e ” m a s
s s iv e ly
e d a n
8 in t
k s o f
in c re
d
h e
s iz
a s
th e
c e
e s
in g
fi
n t
1 5
b
g u re n
e r. T h e
× 1 5 , 3
lu r r in g
e x t
o t
5 ×
a s
to
h e r
3 5 ,
a r
it is th e
fi g u re s
7 5 × 7 5 ,
e s u lt o f
Problem 3.27
W it h r e fe r e n c e t o E q s . ( 3 .6 - 8 ) a n d ( 3 .6 - 9 ) , a n d u s in g k
fo llo w in g e q u a tio n fo r u n s h a r p m a s k in g :
g (x ,y )
1 w e c a n w r ite th e
f (x , y ) − f¯ (x , y )
2 f ( x , y ) − f¯ ( x , y )
=
f (x ,y ) +
=
C o n v o lv in g f (x , y ) w ith th e
in g f (x ,y ) w ith th e m a s k in
o p e r a tio n s a r e lin e a r, w e c a n
e q u a tio n th a t u s in g tw o m a s
3 .3 2 ( a ) p r o d u c e s t h e c o m p o
w ith f (x ,y ) p r o d u c e s g (x ,y )
=
m a s
F ig .
u s e
k s o
s ite
, th e
k in
3 .3 2
s u p
f th e
m a s
u n s
F ig . P 3 .2 7 ( a ) p r o d u c e
( a ) p r o d u c e s f¯ ( x , y ) . T
e r p o s itio n , a n d w e s e e
fo r m in F ig . P 3 .2 7 ( a ) a
k in F ig . P 3 .2 7 ( b ) . C o
h a r p r e s u lt.
s f
h e
fro
n d
n v
(x
n ,
m
th
o lv
,y
b e
th
e
in
). C o n v o lv c a u s e th e s e
e p re c e d in g
m a s k in F ig .
g th is m a s k
Problem 3.28
C o n s id e r th e fo llo w in g e q u a tio n :
f (x ,y ) − r
2
f (x ,y )
=
f (x ,y ) −
f (x + 1 ,y ) +
f (x − 1 ,y ) +
f (x ,y + 1 )
=
+ f (x ,y − 1 ) − 4 f (x ,y )
6 f (x ,y ) −
f (x + 1 ,y ) +
+ f (x ,y − 1 ) +
f (x ,y )
f (x − 1 ,y ) +
f (x ,y + 1 )
C H A P T E R 3 . P R O B L E M S O L U T IO N S
4 4
=
=
w h
te r
T re
fa c
e r
e d
a t
to
e f
a t
in g
rs ,
(x ,y ) d
(x ,y ) a
th e c o
w e m a y
e n
n d
n s
w
1 .2 f (x , y )−
1
f (x + 1 ,y ) + f (x − 1 ,y ) +
5
+ f (x ,y − 1 ) + f (x ,y )
5
1 .2 f (x , y ) −
5
r ig h
a lity
n in
th e L
t s
fa
E q
a p
f (x ,y )
o te s th e a v e r a g e o f f (x ,y ) in a p re d e fi n e d n e ig h b o r h o o d c e n in c lu d in g th e c e n te r p ix e l a n d its fo u r im m e d ia te n e ig h b o r s .
ta n ts in th e la s t lin e o f th e a b o v e e q u a tio n a s p r o p o r tio n a lity
r ite
2
f (x ,y ) − r
T h e
tio n
g iv e
in g
f (x ,y + 1 )
id e o f th is
c to rs to b
s . ( 3 .6 - 8 )
la c ia n fr o
e q u a tio
e o f th e
a n d ( 3 .6
m a n im
f (x ,y )
n is re
s a m e
-9 ). T
a g e is
c o
fo
h u
p r
f (x ,y ) −
f (x ,y ).
g n iz e d w ith in
r m a s th e d e
s , it h a s b e e n
o p o r tio n a l to
th
fi n
d e
u n
e ju s t
itio n
m o n s
s h a rp
-m
o f
tra
m
e n t
u n s
te d
a s k
2
1 / 2
io n e d p r o p o rh a r p m a s k in g
th a t s u b tra c tin g .
Problem 3.29
( a ) F r o m P r o b le m 3 .2 4 , f (x , y )
@ f
=
@ x 0
a n d
@ f
@ y
0
@ f
c o s θ +
@ x
@ f
s in θ
@ y
@ f
s in θ +
@ x
= −
@ f
c o s θ
@ y
fr o m w h ic h it fo llo w s th a t
2
@ f
@ x
@ f
@ y
+
0
o r
@ f
@ x
0
2
@ f
@ y
+
2
2
0
1 / 2
@ f
@ y
+
@ f
@ x
=
0
2
@ f
@ x
=
2
+
@ f
@ y
2
.
T h u s , w e s e e th a t th e m a g n itu d e o f th e g r a d ie n t is a n is o tr o p ic o p e r a to r.
( b ) F r o m E q . ( 3 .6 - 1 0 ) , ( 3 .6 - 1 2 ) , a n d t h e p r e c e d in g r e s u lt s ,
jG
x
j =
@ f
@ x
G
y
=
@ f
@ y
,
4 5
jG
0
x
@ f
@ x
j =
a n d
G
C l e a r l y , jG
x
0
j+
G
y
0
y
6= j G
0
x
=
@ f
@ y
j+
G
=
0
@ f
s in θ
@ y
@ f
c o s θ +
@ x
=
0
@ f
c o s θ
@ y
@ f
s in θ +
@ x
−
,
.
.
y
Problem 3.30
It is g iv e n th a t th e r a n g e o f illu m in a tio n s ta y s in th e lin e a r p o r tio n o f th e c a m e r a
r e s p o n s e r a n g e , b u t n o v a lu e s fo r th e r a n g e a r e g iv e n . T h e fa c t th a t im a g e s s ta y
in th e lin e a r r a n g e im p lie s th a t im a g e s w ill n o t b e s a tu r a te d a t th e h ig h e n d o r
b e d r iv e n in th e lo w e n d to s u c h a n e x te n t th a t th e c a m e r a w ill n o t b e a b le to
r e s p o n d , th u s lo s in g im a g e in fo r m a tio n ir r e tr ie v a b ly . T h e o n ly w a y to e s ta b lis h
a b e n c h m a r k v a lu e fo r illu m in a tio n is w h e n th e v a r ia b le (d a y lig h t) illu m in a tio n
is n o t p r e s e n t. L e t f 0 (x , y ) d e n o te a n im a g e ta k e n u n d e r a r tifi c ia l illu m in a tio n
o n ly , w it h n o m o v in g o b je c t s ( e .g ., p e o p le o r v e h ic le s ) in t h e s c e n e . T h is b e c o m e s th e s ta n d a r d b y w h ic h a ll o th e r im a g e s w ill b e n o r m a liz e d . T h e r e a r e n u m e r o u s w a y s to s o lv e th is p r o b le m , b u t th e s tu d e n t m u s t s h o w a w a r e n e s s th a t
a r e a s in th e im a g e lik e ly to c h a n g e d u e to m o v in g o b je c ts s h o u ld b e e x c lu d e d
fr o m th e illu m in a tio n - c o r r e c tio n a p p r o a c h .
O n e w a y is to s e le c t v a r io u s r e p r e s e n ta tiv e s u b a r e a s o f f 0 (x , y ) n o t lik e ly to
b e o b s c u r e d b y m o v in g o b je c ts a n d c o m p u te th e ir a v e r a g e in te n s itie s . W e th e n
s e le c t th e m in im u m a n d m a x im u m o f a ll th e in d iv id u a l a v e r a g e v a lu e s , d e n o te d
b y , f m in a n d f m a x . T h e o b je c t i v e t h e n i s t o p r o c e s s a n y i n p u t i m a g e , f ( x , y ) , s o
t h a t i t s m i n i m u m a n d m a x i m u m w i ll b e e q u a l t o f m in a n d f m a x , r e s p e c t i v e ly .
T h e e a s ie s t w a y to d o th is is w ith a lin e a r tr a n s fo r m a tio n fu n c tio n o f th e fo r m
f
(x ,y ) = a f (x ,y ) + b
o u t
w h e r e f o u t is th e s c a le d o u tp u t im a g e . It is e a s ily v e r ifi e d th a t th e o u tp u t im a g e
w ill h a v e th e r e q u ir e d m in im u m a n d m a x im u m v a lu e s if w e c h o o s e
a =
f
f
−
m a x
−
f
m a x
f
m in
m in
a n d
b =
w h e re f
m a x
a n d f
m in
f
m in
f
f
m a x
m a x
f
−
−
f
m a x
f
m in
m in
a r e th e m a x im u m a n d m in im u m v a lu e s o f th e in p u t im a g e .
C H A P T E R 3 . P R O B L E M S O L U T IO N S
4 6
N o te th a t th e k e y a s s u m
th e lin e a r o p e r a tin g r a n g e
tie s a re n o t a n is s u e . A n o t
p r is e a r e la tiv e ly s m a ll a r e
o b je c ts w o u ld o v e r p o w e r t
n o t m a k e s e n s e . If th e s tu
to g r a m e q u a liz a tio n ), h e / s
tio n s .
p tio n b e h in d th is m e th o d is th a t a ll im a
o f th e c a m e r a , th u s s a tu r a tio n a n d o th
h e r im p lic it a s s u m p tio n is th a t m o v in g
a in th e fi e ld o f v ie w o f th e c a m e r a , o t
h e s c e n e a n d th e v a lu e s o b ta in e d fr o m
d e n t s e le c ts a n o th e r a u to m a te d a p p r o
h e m u s t d is c u s s th e s a m e o r s im ila r ty p
g e s s ta y w ith in
e r n o n lin e a r io b je c ts c o m h e r w is e th e s e
f 0 (x , y ) w o u ld
a c h ( e .g ., h is e s o f a s s u m p -
Problem 3.31
S u b s t it u t in g d ir e c t ly in t o E q . ( 3 .8 - 9 )
2
b −
c −
b
c
2
a
1
=
a
− a
− a
2
=
b
=
1
2
a + c
.
2
Problem 3.32
A ll fi g u r e s c a n b e c o n s tr u c te d u s in g v a r io u s m e m b e r s h ip fu n c tio n s fr o m F ig .
3 .4 6 a n d t h e d e fi n it io n o f fu z z y u n io n ( t h e p o in t w is e m a x im a b e t w e e n t h e c u r v e s ) ,
a s fo llo w s .
( a ) T h e fi g u r e in p a r t (a ) o f th e p r o b le m s ta te m e n t c a n b e c o n s tr u c te d u s in g tw o
t r a p e z o id a l fu n c t io n s [F ig s . 3 .4 6 ( b ) ] a n d o n e t r ia n g u la r fu n c t io n [F ig . 3 .4 6 ( a ) ],
a s F ig . P 3 .3 2 ( a ) s h o w s .
( b ) T h e fi g u r e in p a r t (b ) o f th e p r o b le m s ta te m e n t c a n b e c o n s tr u c te d u s in g tw o
tr a p e z o id a l fu n c tio n s in w h ic h th e o n e o n th e r ig h t te r m in a te s v e r tic a lly , a s F ig .
P 3 .3 2 ( b ) s h o w s .
( c ) T h e fi g u r e in p a r t (c ) o f th e p r o b le m s ta te m e n t c a n b e c o n s tr u c te d u s in g tw o
t r ia n g u la r fu n c t io n s , a s F ig . P 3 .3 2 ( c ) s h o w s .
Problem 3.33
T
b
o
a
h e
o r h
n e 3 ×
th ic k n e
o o d in c
p ix e l- th
3 n e ig h
s s o f th
re a s e s .
ic k s tr a
b o r h o o
e b o
W e
ig h t
d is
u n d a r ie s
s u p p o r t t
b la c k lin e
u s e d , a n y
in c
h is
r u
n e
re a
c o
n n
ig h
s e s a s a th e s iz
n c lu s io n w ith a
in g v e r tic a lly th
b o r h o o d s w h o s
e o f th e fi lte r in g n e ig h n e x a m p le . C o n s id e r a
r o u g h a w h ite im a g e . If
e c e n te rs a re m o re th a n
4 7
1
1
1
.5
.5
.5
0
a - c
a - b
a
a + b
a + c
1
z
0
a
a +c
b
(a)
b +c
d
1
z 0
(b)
a - b
a =.5 a +b
1
z
(c)
F ig u r e P 3 .3 2
tw o p ix e ls a w a y fr o m
c e n te r p ix e l w ill b e d
lo c a tio n , if w e in c r e a
b e e n c o m p a s s e d a n d
d e s ig n a te d a b o u n d a
th e s iz e o f th e n e ig h b
fr o m th e lin e b e fo r e t
is , th e th ic k n e s s o f th
h o o d in c re a s e s .
th e lin e w ill p r o d u c e d iffe r e n c e s w ith v a lu e s o f z e r o a n d th e
e s ig n a te d a r e g io n p ix e l. L e a v in g th e c e n te r p ix e l a t s a m e
s e th e s iz e o f th e n e ig h b o r h o o d to , s a y , 5 × 5 , th e lin e w ill
n o t a ll d iffe r e n c e s b e z e r o , s o th e c e n te r p ix e l w ill n o w b e
r y p o in t, th u s in c re a s in g th e th ic k n e s s o f th e b o u n d a r y . A s
o r h o o d in c r e a s e s , w e w o u ld h a v e to b e fu r th e r a n d fu r th e r
h e c e n te r p o in t c e a s e s to b e c a lle d a b o u n d a r y p o in t. T h a t
e b o u n d a r y d e te c te d in c re a s e s a s th e s iz e o f th e n e ig h b o r-
Problem 3.34
( a ) If th e in te n s ity o f th e c e n te r p ix e l o f a 3 × 3 r e g io n is la r g e r th a n th e in te n s ity
o f a ll its n e ig h b o r s , th e n d e c r e a s e it. If th e in te n s ity is s m a lle r th a n th e in te n s ity
o f a ll its n e ig h b o r s , th e n in c r e a s e it. E ls e , d o n o t n o th in g .
( b ) R u le s
IF d 2 is P O A N D d
IF d 2 is N E A N D d
E L S E v is Z R .
N o te : In r u le 1 , a ll
(z 5 ) is le s s th a n th
z 50 m o r e p o s i t i v e s
te r p ix e l c lo s e r to
d iffe re n c e s a re n e
n o a c tio n b e c a u s e
s h o u ld b e z e r o (k e
(c
re
o v
fu
) T
s p
e r
n c
h e
e c t
la p
tio
4
4
is P O A N D d
is N E A N D d
6
6
is P O A N D d
is N E A N D d
8
8
is P O T H E N v is P O
is N E T H E N v is N E
p o s itiv e d iffe r e n c e s m e a n th a t th e in te n s ity o f th e n o is e p u ls e
a t o f a ll it s 4 - n e ig h b o r s . T h e n w e ’ll w a n t t o m a k e t h e o u t p u t
o th a t w h e n it is a d d e d to z 5 it w ill b r in g th e v a lu e o f th e c e n th e v a lu e s o f its n e ig h b o r s . T h e c o n v e r s e is tr u e w h e n a ll th e
g a tiv e . A m ix tu r e o f p o s itiv e a n d n e g a tiv e d iffe r e n c e s c a lls fo r
th e c e n te r p ix e l is n o t a c le a r s p ik e . In th is c a s e th e c o r r e c tio n
e p in m in d th a t z e r o is a fu z z y s e t to o ).
m e m b e r s h ip fu n c tio n s P O a n d N E a r e tr ia n g le s fr o m − 1 to 0 a n d 0 to 1 ,
iv e ly . M e m b e r s h ip fu n c tio n Z R is a ls o a tr ia n g le . It is c e n te r e d o n 0 a n d
s t h e o t h e r tw o s lig h t ly . F ig u r e P 3 .3 4 ( a ) s h o w s t h e s e t h r e e m e m b e r s h ip
n s .
C H A P T E R 3 . P R O B L E M S O L U T IO N S
4 8
(d )
o ffi =
o b t
F ig
c e n
2 ,4
a in
u re
te r
,6 ,8
th e
P 3
e le
. T
n e
.3 4 ( b ) s h o w s in g
m e n ts in th e 3 × 3
h e v a lu e v in th e c
w i n t e n s i t y , z 50 , i n
r a p h ic f
n e ig h b
e n te r is
th a t lo c
o r
o r
th
a t
m
h o
e c
io n
th e th r e e r u le s fr o m p a r t (b ). T h e
o d s a re d iffe re n c e s d i = z i − z 5 , fo r
o r r e c tio n th a t w ill b e a d d e d to z 5 to
.
( e ) F ig u r e P 3 .3 4 ( c ) d e s c r ib e s t h e fu z z y s y s t e m r e s p o n s ib le fo r g e n e r a t in g t h e c o r r e c t io n fa c t o r, v . T h is d ia g r a m is s im ila r t o F ig . 3 .5 2 , w it h t w o m a in d iffe r e n c e s :
(1 ) th e r u le s a r e c o m p o s e d o f A N D s , m e a n in g th a t w e h a v e to u s e th e m in o p e r a t io n w h e n a p p ly in g t h e lo g ic a l o p e r a t io n s ( s t e p 2 in F ig . 3 .5 2 ) ; a n d ( 2 ) w e s h o w
h o w to im p le m e n t a n E L S E r u le . T h is r u le is n o th in g m o r e th a t c o m p u tin g 1
m in u s th e m in im u m v a lu e o f th e o u tp u ts o f s te p 2 , a n d u s in g th e r e s u lt to c lip
th e Z R m e m b e r s h ip fu n c tio n . It is im p o r ta n t to u n d e r s ta n d th a t th e o u tp u t o f
th e fu z z y s y s te m is th e c e n te r o f g r a v ity o f th e r e s u lt o f a g g r e g a tio n (s te p 4 in F ig .
3 .5 2 ) . T h u s , fo r e x a m p le , if b o t h t h e fi r s t a n d s e c o n d r u le s w e r e t o p r o d u c e a
m in im u m o f 0 , th e “ s tr e n g th ” o f th e E L S E r u le w o u ld b e 1 . T h is w o u ld p r o d u c e
th e c o m p le te Z R m e m b e r s h ip fu n c tio n in th e im p lic a tio n s te p (s te p 3 in F ig .
3 .5 2 ) . T h e o t h e r t w o r e s u lt s w o u ld b e z e r o , s o t h e r e s u lt o f a g g r e g a t io n w o u ld b e
th e Z R fu n c tio n . T h e c e n te r o f g r a v ity o f th is fu n c tio n is 0 , s o th e o u tp u t o f th e
s y s te m w o u ld b e v = 0 , in d ic a tin g n o c o r r e c tio n , a s it s h o u ld b e . F o r th e p a r tic u la r v a lu e s o f t h e d i 0s s h o w n in F ig . P 3 .3 4 ( c ) , t h e a g g r e g a t e d fu n c t io n is b ia s e d t o
th e r ig h t, s o th e v a lu e o f th e c o r r e c tio n (c e n te r o f g r a v ity ) w ill b e p o s itiv e . T h is is
a s it s h o u ld b e b e c a u s e th e d iffe r e n c e s a r e a ll p o s itiv e , in d ic a tin g th a t th e v a lu e
o f z 5 is le s s th a n th e v a lu e o f its 4 - n e ig h b o r s .
4 9
1
NE
PO
ZE
0
0
L - 1
0
L - 1
- L +1
- L +1
0
L - 1
- L +1
(a)
IF
d
d
IF
d
2 is PO
d
v
4 is PO
d
THEN
6 is PO
d
v is PO
2 is NE
d
8 is PO
ELSE
d
v
4 is NE
6 is NE
THEN
v is NE
8 is NE
v is ZR
(b)
1. Fuzzify inputs.
2. Apply fuzzy logical
operation(s) (AND =min).
3. Apply implication
method (min).
l1
d
IF
2 is PO
d
AND
d
6 is PO
AND
4 is PO
d
AND
THEN
8 is PO
v is PO
l
IF
d
2 is NE
AND
d
AND
4 is NE
d
6 is NE
AND
d
8 is NE
2
v is NE
THEN
l E =min{1 - l 1, 1 - l 2}=1
d
ELSE v is ZR
d
2
4
d
d
6
4. Apply
aggregation
method (max).
8
v
Neighborhood
differences
Input z
5. Defuzzify
(center of
gravity.)
Correction from
fuzzy system
5
v
+
(c)
F ig u r e 3 .3 4
Output, z
5
Chapter 4
Problem Solutions
Problem 4.1
F (μ )
1
=
f (t )e
− j 2 π μ t
d t
− 1
T
=
A e
− j 2 π μ t
d t
0
=
=
=
=
If w
4 .1
fu n
e x p
− A
− A
T
e − j2π μ t 0 =
j 2 π μ
j 2 π μ
A
e jπ μ T − e − jπ μ T e
j 2 π μ
A
s in π μ T e − jπ μ T
π μ
s in π μ T
A T
e − jπ μ T .
π μ T
e
− j 2 π μ t
− 1
− j π μ T
e le t T = W , th e o n ly d iffe r e n c e b e tw e e n th is r e s u lt a n d th e r e s u lt in E x a m p le
is th e e x p o n e n tia l te r m . It is a p h a s e te r m th a t a c c o u n ts fo r th e s h ift in th e
c tio n . T h e m a g n itu d e o f th e F o u r ie r tr a n s fo r m is th e s a m e in b o th c a s e s , a s
e c te d .
5 1
C H A P T E R 4 . P R O B L E M S O L U T IO N S
5 2
Problem 4.2
(a ) T o p r o v e in fi n ite p e r io d ic ity in b o th d ire c tio n s w ith p e r io d 1 / Δ T , w e h a v e to
s h o w t h a t F˜ μ + k [ 1 / Δ T ] = F˜ ( μ ) f o r k = 0 , ± 1 , ± 2 , . . . . F r o m E q . ( 4 .3 - 5 ) ,
F˜
μ + k [1 / Δ T ]
1
1
=
4 T
n = − 1
1
1
=
4 T
n = − 1
1
1
4 T
=
F˜
=
k
F
μ +
F
μ +
k − n
4 T
μ −
m
4 T
F
−
4 T
m = − 1
n
4 T
μ
w h e r e th e th ir d lin e fo llo w s fr o m th e fa c t th a t k a n d n a r e in te g e r s a n d th e lim its
o f s u m m a tio n a r e s y m m e tr ic a b o u t th e o r ig in . T h e la s t s te p fo llo w s fr o m E q .
( 4 .3 - 5 ) .
( b ) A g a i n , w e n e e d t o s h o w t h a t w e h a v e t o s h o w t h a t F˜
k = 0 , ± 1 , ± 2 , . . . . F r o m E q . ( 4 .4 - 2 ) ,
F˜
μ + k / Δ T
1
=
n
e
− j 2 π
f
n
e
− j 2 π μ n Δ T
f
n
e
− j 2 π μ n Δ T
n = − 1
1
=
=
n = − 1
1
(
f
μ + k / Δ T
μ + k / Δ T
e
)
=
F˜ ( μ ) f o r
n Δ T
− j 2 π k n
n = − 1
=
F˜
μ
w h e r e th e th ir d lin e fo llo w s fr o m th e fa c t th a t e − j 2 π k n = 1 b e c a u s e b o th k a n d n
a r e i n t e g e r s ( s e e E u l e r ’s f o r m u l a ) , a n d t h e l a s t l i n e f o l l o w s f r o m E q . ( 4 . 4 - 2 ) .
Problem 4.3
F r o m t h e d e fi n it io n o f t h e 1 - D F o u r ie r t r a n s fo r m in E q . ( 4 .2 - 1 6 ) ,
5 3
F
μ
1
=
− 1
1
=
− 1
− j
2
=
=
− j
2
− j 2 π μ t
f (t ) e
d t
− j 2 π μ t
s in (2 π n t ) e
1
e
j 2 π n t
− e
e
j 2 π n t
e
− 1
1
d t
− j 2 π n t
− j 2 π μ t
e
− j 2 π μ t
d t
1
− j
2
d t −
− 1
e
− j 2 π n t
e
− j 2 π μ t
d t .
− 1
F r o m th e tr a n s la tio n p r o p e r ty in T a b le 4 .3 w e k n o w th a t
j 2 π μ
f (t )e
0
t
, F
μ − μ
0
a n d w e k n o w fr o m th e s ta te m e n t o f th e p r o b le m th a t th e F o u r ie r tr a n s fo r m o f a
c o n s ta n t f (t ) = 1 is a n im p u ls e . T h u s ,
j 2 π μ
(1 ) e
0
t
, δ
μ − μ
0
.
T h u s , w e s e e th a t th e le ftm o s t in te g r a l in th e th e la s t lin e a b o v e is th e F o u r ie r
tr a n s fo r m o f (1 ) e j 2 π n t , w h ic h is δ μ − n , a n d s im ila r ly , th e s e c o n d in te g r a l is
th e tr a n s fo r m o f (1 ) e − j 2 π n t , o r δ μ + n . C o m b in in g a ll r e s u lts y ie ld s
F
μ
=
j
2
δ
μ + n
− δ
μ − n
a s d e s ire d .
Problem 4.4
(a ) T h e p e r io d is s u c h th a t 2 π n t = 2 π , o r t = 1 / n .
(b ) T h e fre q
fo r m o f th e
tra n s fo r m o
lu s tr a te d in
s tr u c tio n if
s a tis fi e d ).
u e n c y is 1 d iv id e d b y
g iv e n s in e w a v e lo o k
f th e s a m p le d d a ta (s
F ig . P 4 .4 ( b ) ( th e d a s h
th e s in e fu n c tio n w e
th e p e r io d , o r n . T h e c
s a s in F ig . P 4 .4 ( a ) ( s
h o w in g a fe w p e r io d s
e d b o x is a n id e a l fi lte
r e s a m p le d , w ith th e
o n tin u o u s F o u r ie r tr a n s e e P r o b le m 4 .3 ) , a n d t h e
) h a s th e g e n e r a l fo r m ilr th a t w o u ld a llo w r e c o n s a m p lin g th e o r e m b e in g
( c ) T h e N y q u is t s a m p lin g r a te is e x a c tly tw ic e th e h ig h e s t fr e q u e n c y , o r 2 n . T h a t
is , (1 / Δ T ) = 2 n , o r Δ T = 1 / 2 n . T a k in g s a m p le s a t t = ± Δ T , ± 2 Δ T , . . . w o u ld
y ie ld th e s a m p le d fu n c tio n s in (2 π n Δ T ) w h o s e v a lu e s a r e a ll 0 s b e c a u s e Δ T =
C H A P T E R 4 . P R O B L E M S O L U T IO N S
5 4
F (m)
-n
n
m
(a)
F (m)
....
1 - n
DT
1
DT
-n
n
-1
DT
1
DT +n
m
....
(b)
F (m)
....
1 - n
DT
n
1
DT
-n
-1
DT
1
DT +n
m
....
(c)
F ig u r e P 4 .4
1 / 2 n a n d n is a n in t e g e r. In t e r m s o f F ig . P 4 .4 ( b ) , w e s e e t h a t w h e n Δ T = 1 / 2 n
a ll th e p o s itiv e a n d n e g a tiv e im p u ls e s w o u ld c o in c id e , th u s c a n c e lin g e a c h o th e r
a n d g iv in g a r e s u lt o f 0 fo r th e s a m p le d d a ta .
(d ) W h
s u c h a
in th is
fo r m a
th e b o
a n y s h
e n th e s a m p lin g r a te is le s s
s th e o n e illu s tr a te d in F ig .
c a s e . F o r s o m e v a lu e s o f s a
s in g le s in e w a v e a n d a p lo t
o k . O th e r v a lu e s w o u ld r e s
a p e o b ta in a b le b y s a m p lin g
th a n th e N y q u is t r a te , w e c a
P 4 .4 ( c ) , w h ic h is t h e s u m
m p lin g , th e s u m o f th e tw o
o f th e s a m p le s w o u ld a p p e
u lt in fu n c tio n s w h o s e s a m
th e s u m o f tw o s in e s .
n h a v e a s itu a tio n
o f tw o s in e w a v e s
s in e s c o m b in e to
a r a s in F ig . 4 .8 o f
p le s c a n d e s c r ib e
5 5
Problem 4.5
S t a r t in g fr o m E q . ( 4 .2 - 2 0 ) ,
f (t )
1
g (t ) =
f (τ )g (t − τ )d τ .
− 1
T h e F o u r ie r tr a n s fo r m o f th is e x p re s s io n is
=
f (t )
g (t )
1
=
− 1
1
=
⎡
⎣
⎤
1
f (τ )g (t − τ )d τ ⎦e
− 1
⎡
f (τ ) ⎣
− 1
1
− j 2 π μ t
d t
⎤
g (t − τ )e
− j 2 π μ t
d t ⎦d τ .
− 1
T h e te r m in s id e th e in n e r b r a c k e ts is th e F o u r ie r tr a n s fo r m o f g (t − τ ). B u t, w e
k n o w fr o m t h e t r a n s la t io n p r o p e r t y ( T a b le 4 .3 ) t h a t
=
g (t − τ ) = G (μ )e
− j 2 π μ τ
s o
=
f (t )
g (t )
1
=
− 1
G (μ )
=
=
f (τ ) G (μ )e
1
f (τ )e
− j 2 π μ τ
− j 2 π μ τ
d τ
d τ
− 1
G (μ )F (μ ).
T h is p r o v e s th a t m u ltip lic a tio n in th e fr e q u e n c y d o m a in is e q u a l to c o n v o lu tio n
in th e s p a tia l d o m a in . T h e p r o o f th a t m u ltip lic a tio n in th e s p a tia l d o m a in is
e q u a l to c o n v o lu tio n in th e s p a tia l d o m a in is d o n e in a s im ila r w a y .
C H A P T E R 4 . P R O B L E M S O L U T IO N S
5 6
Problem 4.6
f (t )
=
f˜ ( t )
h (t )
1
=
− 1
1
=
− 1
1
=
h ( z ) f˜ ( t − z ) d z
1
=
s in [π (t − n Δ T ) / Δ T ]
[π (t − n Δ T ) / Δ T ]
f (n Δ T )
n = − 1
1
f (t − z ) δ (t − n Δ T − z )d z
n = − 1
s in (π z / Δ T )
f (t − z ) δ (t − n Δ T − z )d z
(π z / Δ T )
− 1
n = − 1
1
=
1
s in (π z / Δ T )
(π z / Δ T )
f (n Δ T ) s in c [(t − n Δ T ) / Δ T ] .
n = − 1
Problem 4.7
T h e te n t fu n c tio n is o b
fo r m o f a b o x is a s in
F o u r ie r tr a n s fo r m o f th
tr a n s fo r m s , it fo llo w s t
tio n s q u a re d .
ta in e
c fu n
e s p a
h a t th
d b y c
c tio n
tia l c o
e F o u
.
o n v
B
n v o
r ie r
o lv
e c a
lu t
tra
in
u
io
n
g tw o b o
s e , b y th
n o f tw o
s fo r m o f
x e s
e c
fu n
a te
, a n
o n v
c tio
n t f
d r e c a ll th a t
o lu tio n th e
n s is th e p r o
u n c tio n is a
th e tra n s o re m , th e
d u c t th e ir
s in c fu n c -
Problem 4.8
( a ) W e s o lv e th is p r o b le m b y d ir e c t s u b s titu tio n u s in g o r th o g o n a lity . S u b s titu tin g E q . ( 4 .4 - 5 ) in t o ( 4 .4 - 4 ) y ie ld s
M − 1
F
=
m
n = 0
=
F
− j 2 π r n / M
F r e
⎦e
− j 2 π m n / M
r = 0
⎡
F
r = 0
⎤
M − 1
1
⎣
M
M − 1
1
M
=
⎡
r
⎣
⎤
M − 1
e
− j 2 π r n / M
e
− j 2 π m n / M
⎦
n = 0
m
w h e r e th e la s t s te p fo llo w s fr o m th e o r th o g o n a lity c o n d itio n g iv e n in th e p r o b le m s t a t e m e n t . S u b s t it u t in g E q . ( 4 .4 - 4 ) in t o ( 4 .6 - 5 ) a n d u s in g t h e s a m e b a s ic
p r o c e d u r e y ie ld s a s im ila r id e n tity fo r f n .
5 7
( b ) W e s o lv e th is p r o b le m a s a b o v e , b y d ir e c t s u b s titu tio n a n d u s in g o r th o g o n a lit y . S u b s t it u t in g E q . ( 4 .4 - 7 ) in t o ( 4 .4 - 6 ) y ie ld s
M − 1
F (u )
=
x = 0
=
1
M
=
⎡
1
⎣
M
M − 1
⎤
M − 1
r = 0
⎡
F (r ) ⎣
r = 0
⎦e
− j 2 π r x / M
F (u )e
− j 2 π u x / M
⎤
M − 1
− j 2 π r x / M
e
e
− j 2 π u x / M
⎦
x = 0
F (u )
w h e r e th e la s t s te p fo llo w s fr o m th e o r th o g o n a lity c o n d itio n g iv e n in th e p r o b le m s t a t e m e n t . S u b s t it u t in g E q . ( 4 .4 - 6 ) in t o ( 4 .6 - 7 ) a n d u s in g t h e s a m e b a s ic
p r o c e d u r e y ie ld s a s im ila r id e n tity fo r f (x ).
Problem 4.9
T o p r o v e in fi n ite p e r io d ic ity w e h a v e to s h o w th a t F (u + k M ) =
0 , ± 1 , ± 2 , . . . . W e d o t h is b y d ir e c t s u b s t it u t io n in t o E q . ( 4 .4 - 6 ) :
F (u ) fo r k
=
M − 1
F (u + k M )
=
f (x )e
− j 2 π [u + k M ]x / M
x = 0
⎡
⎤
M − 1
=
⎣
=
F (u )
f (x )e
− j 2 π u x / M
⎦e
− j 2 π k x
x = 0
w h e r e th e la s t s te p fo llo w s fr o m th e fa c t th a t e − j2 π k x = 1 b e c a u s e k a n d x a r e
in te g e r s . N o te th a t th is h o ld s fo r p o s itiv e a n d n e g a tiv e v a lu e s o f k . W e p r o v e th e
v a lid it y o f E q . ( 4 .4 - 9 ) in a s im ila r w a y .
Problem 4.10
W it h r e fe r e n c e t o t h e s t a t e m e n t o f t h e c o n v o lu t io n t h e o r e m g iv e n in E q s . ( 4 .2 2 1 ) a n d ( 4 .2 - 2 2 ) , w e n e e d t o s h o w t h a t
f (x )
h (x ) , F (u )H (u )
a n d th a t
f (x )h (x ) , F (u )
H (u ).
C H A P T E R 4 . P R O B L E M S O L U T IO N S
5 8
F r o m E q . ( 4 .4 - 1 0 ) a n d t h e d e fi n it io n o f t h e D F T in E q . ( 4 .4 - 6 ) ,
⎡
M − 1
=
f (x )
h (x )
M − 1
⎣
=
x = 0
m = 0
M − 1
⎤
f (m )h (x − m )⎦e
⎡
f (m ) ⎣
=
m = 0
− j 2 π u x / M
⎤
M − 1
h (x − m )e
− j 2 π u x / M
⎦
x = 0
M − 1
=
f (m )H (u )e
− j 2 π u m / M
m = 0
M − 1
=
H (u )
f (m )e
=
H (u )F (u ).
− j 2 π u m / M
m = 0
T h e o th e r h a lf o f th e d is c r e te c o n v o lu tio n th e o r e m is p r o v e d in a s im ila r m a n n e r.
Problem 4.11
W it h r e fe r e n c e t o E q . ( 4 .2 - 2 0 ) ,
f (t ,z )
h (t ,z ) =
1
1
− 1
− 1
f (α ,β )h (t − α ,z − β )d α d β .
Problem 4.12
T h e lim it o f th e im a g in g s y s te m is 1 p ix e l p e r s q u a r e , s o e a c h r o w
in th e r e s u ltin g im a g e w o u ld b e o f th e fo r m . . . 1 0 1 0 1 0 1 0 1 . . . . T
t h is w a v e fo r m is P = 2 m m , s o t h e m a x fr e q u e n c y is μ = 1 / P = 0 .5
T o a v o id a lia s in g w e h a v e to s a m p le a t a r a te th a t e x c e e d s tw ic e th
o r 2 (0 .5 ) = 1 s a m p le / m m . T h a t is , t h e s a m p lin g r a t e w o u ld h a v e
s a m p le / m m . S o , e a c h s q u a r e h a s to c o r r e s p o n d to s lig h tly m o r e th
in th e im a g in g s y s te m .
a n d c o lu m n
h e p e r io d o f
c y c le s / m m .
is fre q u e n c y
to e x c e e d 1
a n o n e p ix e l
Problem 4.13
S h r in k in g c
is n o t th e c
n e w d e ta il
ra te , s o z o o
a n
a s
in
m
c a u s e a lia s in
e in z o o m in g
in tr o d u c e d b
in g c a n n o t re
g b e
, w h
y z o
s u lt
c a u s e
ic h in
o m in g
in a lia
th e e ffe c tiv e s a m p lin g r a te is r e d u c e d . T h is
tr o d u c e s a d d itio n a l s a m p le s . A lth o u g h n o
, it c e r ta in ly d o e s n o t r e d u c e th e s a m p lin g
s in g .
5 9
Problem 4.14
F r o m E q . ( 4 .5 - 7 ) ,
F (μ ,ν ) = =
f (t ,z )
1
1
− 1
− 1
=
− j 2 π (μ t + ν z )
f (t ,z )e
R e c a ll th a t in
u s e (t ,z ) a n d
c o n tin u o u s v
(x ,y ) a n d (u ,
v a r ia b le s .
d t d z .
F r o m E q . ( 2 .6 - 2 ) , t h e F o u r ie r t r a n s fo r m o p e r a t io n is lin e a r if
=
a
f 1 (t ,z ) + a
1
f 2 (t ,z )
2
= a
=
1
f 1 (t ,z ) + a
=
2
f 2 (t ,z ) .
S u b s titu tin g in to th e d e fi n itio n o f th e F o u r ie r tr a n s fo r m y ie ld s
=
a
1
f 1 (t ,z ) + a
2
f 2 (t ,z )
1
1
− 1
− 1
=
a
=
a
1
− 1
+ a
=
a
1
=
1
1
− 1
− 1
2
f 1 (t ,z ) + a
=
2
2
f 2 (t ,z )
d t d z
− j 2 π (μ t + ν z )
f (t ,z )e
1
− 1
f 1 (t ,z ) + a
− j 2 π (μ t + ν z )
× e
1
1
f 2 (t ,z )e
d t d z
− j 2 π (μ t + ν z )
d t d z
f 2 (t ,z ) .
w h e r e th e s e c o n d s te p fo llo w s fr o m th e d is tr ib u tiv e p r o p e r ty o f th e in te g r a l.
S im ila r ly , fo r th e d is c r e te c a s e ,
M − 1 N − 1
=
a
1
f 1 (x ,y ) + a
2
f 2 (x ,y )
=
a
1
f 1 (x ,y ) + a
2
f 2 (x ,y ) e
− j 2 π (u x / M + v y / N )
x = 0 y = 0
M − 1 N − 1
=
− j 2 π (u x / M + v y / N )
f 1 (x ,y )e
a
1
x = 0
y = 0
M − 1 N − 1
+ a
f 2 (x ,y )e
2
x = 0
=
a
1
=
− j 2 π (u x / M + v y / N )
y = 0
f 1 (x ,y ) + a
2
=
f 2 (x ,y ) .
T h e lin e a r ity o f th e in v e r s e tr a n s fo r m s is p r o v e d in e x a c tly th e s a m e w a y .
Problem 4.15
S e le c t a n im a g e o f y o u r c h o ic e a n d c o m p u te its a v e r a g e v a lu e :
f¯ ( x , y ) =
1
M N
M − 1 N − 1
f (x ,y ).
x = 0
y = 0
th is
(μ ,ν
a r ia b
v ) fo
c h a p te r w e
) fo r
le s , a n d
r d is c re te
C H A P T E R 4 . P R O B L E M S O L U T IO N S
6 0
C o m p u te th
1 / M N w a s
2 1 )]. S im ila
D F T . F in a lly
m u la tio n o f
o f f (x ,y ) a n d
d in fr o n t o f th
( 0 , 0 ) = f¯ ( x , y )
p
M N f (x ,y
0 )=
e D F T a n d ID F
e D F T
in c lu d e
r ly , if F
, if F (0 ,
b o th th
o b
e
th
),
T .
ta in F (0
ID F T [s e
e 1 / M N
th e te r m
,0 ).
e E q
te r m
p
1 /
F (0 ,0
( 4 .5 - 1
a s in c
N w a
If
s .
w
M
) =
5 ),
lu d
s in
M N f¯ ( x , y ) ,
( 4 .5 - 1 6 ) a n d
e d in fr o n t o
c lu d e d in th e
th e n
( 4 .6 f th e
fo r-
Problem 4.16
( a ) F r o m E q . ( 4 .5 - 1 5 ) ,
M − 1 N − 1
=
f (x ,y )e
j 2 π (u
0
x + v
0
y )
=
j 2 π (u
f (x ,y )e
x = 0
0
x + v
0
y )
e
− j 2 π (u x / M + v y / N )
y = 0
M − 1 N − 1
=
f (x ,y )e
x = 0
=
− j 2 π
[
(u − u
0
)x / M + (v − v
0
)y / N
]
y = 0
F (u − u
,v − v 0 ).
0
( b ) F r o m E q . ( 4 .5 - 1 6 ) ,
=
− 1
F (u ,v )e
− j 2 π (u x
0
+ u y 0 )
=
M − 1 N − 1
1
M N
0
+ u y 0 )
j 2 π (u x / M + v y / N )
M − 1 N − 1
M N
F (u ,v )
u = 0 v = 0
× e
=
− j 2 π (u x
u = 0 v = 0
× e
1
=
F (u ,v )e
f (x − x
0
j 2 π [(u (x − x
0
)/ M + v (y − y 0 )/ N )]
,y − y 0 ).
Problem 4.17
F r o m E q . ( 4 .5 - 7 ) ,
F
μ ,ν
=
=
1
1
− 1
1
− 1
1
− 1
− 1
f (t ,z ) e
s in
2 π μ
− j 2 π (μ t + ν z )
0
t + 2 π ν 0 z
d t d z
e
− j 2 π (μ t + ν z )
d t d z .
6 1
E x p r e s s in g th e s in e fu n c tio n in te r m s o f e x p o n e n tia ls y ie ld s
F
μ ,ν
− j
2
=
1
j 2 π (μ
e
⎡
− j
⎣
2
=
1
− 1
− 1
1
j 2 π (μ
− 1
⎡
− j
⎣
−
2
− e
− j 2 π (μ
t + ν 0 z )
0
− j 2 π (μ t + ν z )
e
d t d z
⎤
1
e
− 1
t + ν 0 z )
0
0
1
t + ν 0 z )
− j 2 π (μ t + ν z )
d t d z ⎦
⎤
1
− j 2 π (μ
e
− 1
e
0
t + ν 0 z )
e
− j 2 π (μ t + ν z )
d t d z ⎦.
− 1
W e re c o g n iz e th e te r m s in s id e th e b r a c k e ts a s th e F o u r ie r tr a n s fo r m s o f
j 2 π (μ
(1 )e
t + ν 0 z )
0
a n d
(1 )e
F ro m
4 .3 w
fo r m
tra n s
th e
e k n o
. T h e
fo r m
p r o b le
w th a t
re fo re ,
. P lu g g
=
m
th
=
in
s in
− j 2 π (μ
s ta te m e n t w e k n o w
e e x p o n e n tia l ju s t in
( 1 ) e j 2 π (μ 0 t + ν 0 z ) = δ
g th e s e r e s u lts in th e
2 π μ
0
t + 2 π ν 0 z
0
t + ν 0 z )
th a
tro d
(μ −
p re
j
=
.
t th e =
u c e s a
μ 0 ,ν −
c e d in g
[1 ] = δ (μ
s h ift in th
ν 0 ), a n d s
e q u a tio n
δ (μ + μ
2
0
,ν ), a
e o r ig
im ila r
y ie ld
n d fr o m T a b le
in o f th e tr a n s ly fo r th e o th e r
s ,
) − δ (ν − ν 0 ) .
Problem 4.18
W e c o n s id e r t h e 1 - D c a s e fi r s t . F r o m E q . ( 4 .4 - 6 ) ,
M − 1
F (u ) =
f (x )e
− j 2 π u x / M
.
x = 0
W h e n f ( x ) = 1 a n d u = 0 , F ( u ) = 1 . W h e n f ( x ) = 1 a n d u 6= 0 ,
M − 1
F (u ) =
e
− j 2 π u x / M
.
x = 0
T h is e x p r e s s io n is 0 fo r a n y in te g e r v a lu e o f u in th e r a n g e [0 , N − 1 ]. T h e r e a r e
v a r i o u s w a y s o f p r o v i n g t h i s . O n e o f t h e m o s t i n t u i t i v e i s b a s e d o n u s i n g E u l e r ’s
fo r m u la to e x p r e s s th e e x p o n e n tia l te r m a s
e
− j 2 π u x / M
= c o s (2 π u x / M ) − j s in (2 π u x / M ).
C H A P T E R 4 . P R O B L E M S O L U T IO N S
6 2
T h is e x p re s s io n d e s c r ib e s a u n it v e c to r in th e c o
te re d a t th e o r ig in a n d its d ire c tio n d e p e n d s o n
a n y in te g e r v a lu e o f u in th e r a n g e [0 , N − 1 ] th
v a lu e s o f x in th e s a m e r a n g e . T h is m e a n s th a t th
b e r o f r e v o lu tio n s a b o u t th e o r ig in in e q u a l in c
v a lu e o f c o s (2 π u x / M ), th e r e w ill b e a c o r r e s p o n
T h is p r o d u c e s a z e r o s u m fo r th e re a l p a r t o f th
a p p ly th e im a g in a r y p a r t. T h e r e fo r e , w h e n f (x
F (u ) = 0 . T h u s , w e h a v e s h o w n th a t fo r d is c re te
if u = 0
i f u =6 0 .
1
= [1 ] = δ (u ) =
m p le x p la n e . T h e v e c to r is c e n th e v a lu e o f th e a r g u m e n t. F o r
e a r g u m e n t r a n g e s o v e r in te g e r
e v e c to r m a k e s a n in te g e r n u m re m e n ts . T h u s , fo r a n y p o s itiv e
d in g n e g a tiv e v a lu e o f th is te r m .
e e x p o n e n t. S im ila r c o m m e n ts
) = 0 a n d u 6= 0 , i t f o l l o w s t h a t
q u a n titie s ,
0
A s im ila r p r o c e d u r e a p p lie s in th e c a s e o f tw o v a r ia b le s , a n d w e h a v e th a t
if u = 0 a n d v = 0
o th e r w is e .
1
= [1 ] = δ (u ,v ) =
0
Problem 4.19
F r o m E q . ( 4 .5 - 1 5 ) , a n d u s in g t h e e x p o n e n t ia l r e p r e s e n t a t io n o f t h e s in e fu n c t io n ,
w e h a v e
M − 1 N − 1
F (u ,v )
=
2 π u
s in
x = 0
− j
2
=
− j
2
=
=
x + 2 π v 0 y
e
− j 2 π (u x / M + v y / N )
y = 0
M − 1 N − 1
e
j 2 π (u
x + v
0
0
y )
− e
− j 2 π (u
0
x + v
y )
0
e
− j 2 π (u x / M + v y / N )
x = 0 y = 0
M − 1 N − 1
e
j 2 π (M u
0
x / M + N v
0
y / N )
e
− j 2 π (u x / M + v y / N )
x = 0 y = 0
− j
−
2
=
0
M − 1 N − 1
e
− j 2 π (M u
0
x / M + N v 0 y / N )
e
− j 2 π (u x / M + v y / N )
x = 0 y = 0
− j
j
= ( 1 ) e j 2 π (M u 0 x / M + N v 0 y / N ) +
= ( 1 ) e − j 2 π (u 0 x
2
2
j
[δ (u + M u 0 ,v + N v 0 ) − δ (u − M u 0 ,v − N v 0 )]
2
+ v
0
y )
w h e r e t h e fo u r t h s t e p fo llo w s fr o m t h e d is c u s s io n in P r o b le m 4 .1 7 , a n d t h e la s t
lin e fo llo w s fr o m T a b le 4 .3 .
6 3
Problem 4.20
T
o
a
w
F
h e fo llo w in g a r e p r o o
th e r p r o p e r tie s a re g iv
s im a g in a r y , its re a l p a
h o s e re a l a n d im a g in a
o u r ie r tr a n s fo r m p a ir s
fs
e n
r t
r y
. S
o f
in
is z
p a
im
s o m e o f th e p r o p e r tie s in T a b le
C h a p te r 4 . R e c a ll th a t w h e n w e
e r o . W e u s e th e te r m c o m p le x to
r ts a r e n o t z e r o . W e p r o v e o n ly th
ila r te c h n iq u e s a r e u s e d to p r o v e
4 .1 .
re fe
d e n
e fo
th e
P r
r to
o te
r w a
in v
o o fs
a fu
a fu
rd p
e rs e
o f th
n c tio
n c tio
a r t th
p a r t.
e
n
n
e
(a ) P r o p e r ty 2 : If f (x ,y ) is im a g in a r y , f (x ,y ) , F ∗ (− u ,− v ) = − F (u ,v ). P r o o f:
B e c a u s e f (x ,y ) is im a g in a r y , w e c a n e x p re s s it a s j g (x ,y ), w h e re g (x ,y ) is a re a l
fu n c tio n . T h e n th e p r o o f is a s fo llo w s :
⎤∗
⎡
F
∗
(− u − v )
⎢
⎣
=
M − 1 N − 1
j 2 π (u x / M + v y / N )
j g (x ,y )e
x = 0
⎥
⎦
y = 0
M − 1 N − 1
=
− j 2 π (u x / M + v y / N )
− j g (x ,y )e
x = 0 y = 0
M − 1 N − 1
=
j g (x ,y ) e
−
x = 0
− j 2 π (u x / M + v y / N )
y = 0
M − 1 N − 1
=
f (x ,y )e
−
x = 0
=
− j 2 π (u x / M + v y / N )
y = 0
− F (u ,v ).
(b ) P r o p e r ty 4 : If f (x ,y ) is im a g in a r y , th e n R (u ,v ) is o d d a n d I (u ,v ) is e v e n .
P r o o f: F is c o m p le x , s o it c a n b e e x p r e s s e d a s
F (u ,v )
=
re a l [F (u ,v )] + j im a g [F (u ,v )]
=
T h e n , −
B u t, b e
th e n fo
is o d d )
F (u ,
c a u s e
llo w s
a n d I
v )
f
fro
(u
= − R (u ,v ) −
(x ,y ) is im a g
m th e p re v io
,v ) = I (− u ,−
R (u ,v ) + j I (u ,v ).
j I
in
u s
v )
(u ,v
a r y ,
tw o
(I is
) a n
F ∗(−
e q u
e v e
d F ∗(− u ,− v ) = R (− u ,− v ) − j I (− u ,− v ).
u ,− v ) = − F (u ,v ) (s e e P ro p e r ty 2 ). It
a t io n s t h a t R (u , v ) = − R (− u , − v ) ( i.e ., R
n ).
(c ) P r o p e r ty 5 : f (− x ,− y ) , F ∗ (u ,v ). T h a t is , if f (x ,y ) is re a l a n d its tr a n s fo r m is
F (u , v ), th e n th e tr a n s fo r m o f f (− x , − y ) is F ∗ (u , v ). A n d c o n v e r s e ly . P r o o f: F r o m
E x a m p le 4 .1 2 ,
M − 1 N − 1
=
f (− x ,− y )
f (m ,n )e
=
m = 0 n = 0
j 2 π (u m / M + v n / N )
.
C H A P T E R 4 . P R O B L E M S O L U T IO N S
6 4
T o s e e w h a t h a p p e n s w h e n f (x , y ) is r e a l, w e w r ite th e r ig h t s id e o f th is e q u a tio n
a s
⎡
M − 1 N − 1
f (m ,n )e
⎣
j 2 π (u m / M + v n / N )
=
m = 0 n = 0
⎡
=
⎣
=
F
⎤∗
M − 1 N − 1
∗
f (m ,n )e
− j 2 π (u m / M + v n / N )
m = 0 n = 0
⎦
⎤∗
M − 1 N − 1
f (m ,n )e
− j 2 π (u m / M + v n / N )
⎦
m = 0 n = 0
∗
(u ,v )
w h e r e th e s e c o n d s te p fo llo w s fr o m th e fa c t th a t f (x , y ) is r e a l. T h u s , w e h a v e
s h o w n th a t
=
f (− x ,− y )
= F
∗
(u ,v ).
( d ) P r o p e r ty 7 : W h e n f (x , y ) is c o m p le x , f ∗ (x , y ) , F
∗
(− u ,− v ). P ro o f:
M − 1 N − 1
=
f ∗ (x ,y )
f ∗ (x ,y )e
=
x = 0
y = 0
⎡
=
⎢
⎣
M − 1 N − 1
F
∗
(e )
c o n
is e
th e
P ro
v e r
v e n
n w
p e r ty 9 : If f (x ,y ) is re
s e ly . P r o o f: B e c a u s e
a n d its im a g in a r y p a
e w ill h a v e c o m p le te d
⎤∗
f (x ,y )e
x = 0
=
− j 2 π (u x / M + v y / N )
j 2 π (u x / M + v y / N )
⎥
⎦
y = 0
(− u ,− v ).
a l a n d
f (x ,y )
r t is o d
th e p r
o d
is
d .
o o
d , th e n F (u ,v ) is im a g in a r y a n d o d d , a n d
r e a l, w e k n o w th a t th e r e a l p a r t o f F (u , v )
If w e c a n s h o w th a t F is p u r e ly im a g in a r y ,
f.
M − 1 N − 1
F (u ,v )
=
f (x ,y )e
x = 0
− j 2 π (u x / M + v y / N )
y = 0
M − 1 N − 1
=
f (x ,y )
x = 0
e
− j 2 π (u x / M )
e
− j 2 π (v y / N )
y = 0
M − 1 N − 1
=
[o d d ] e v e n − j o d d
x = 0
y = 0
e v e n − j o d d
6 5
M − 1 N − 1
[o d d ] (e v e n )(e v e n ) − 2 j (e v e n )(o d d ) − (o d d )(o d d )
=
x = 0
y = 0
M − 1 N − 1
=
M − 1 N − 1
[(o d d )(e v e n )] − 2 j
x = 0
y = 0
[(e v e n )(e v e n )]
x = 0
y = 0
M − 1 N − 1
[(o d d )(e v e n )]
−
x = 0
=
y = 0
im a g in a r y
w h e r e t h e la s t s t e p fo llo w s fr o m E q . ( 4 .6 - 1 3 ) .
(f ) P
e v e n
re a l
re a l
n a r y
ro p e r ty
, a n d c o
p a r t o f F
p a r t is 0
, w e c a n
1 0 : If f (x ,y ) is
n v e r s e ly . P r o o f
(u ,v ) is o d d a n
, th e n w e w ill h
e x p re s s it a s j g
im a g in a r y a n d e v e n
: W e k n o w th a t w h e
d its im a g in a r y p a r t
a v e p r o v e d th is p r o
(x ,y ), w h e re g is a r
h e n F (u ,v ) is im a g in a r y a n
f (x ,y ) is im a g in a r y , th e n th
e v e n . If w e c a n s h o w th a t th
r ty . B e c a u s e f (x ,y ) is im a g
l fu n c tio n . T h e n ,
, t
n
is
p e
e a
d
e
e
i-
M − 1 N − 1
F (u ,v )
=
f (x ,y )e
x = 0
− j 2 π (u x / M + v y / N )
y = 0
M − 1 N − 1
=
j g (x ,y )
x = 0
− j 2 π (u x / M )
e
e
− j 2 π (v y / N )
y = 0
M − 1 N − 1
=
[j e v e n ][e v e n − j o d d ][e v e n − j o d d ]
x = 0
y = 0
M − 1 N − 1
[j e v e n ] (e v e n )(e v e n ) − 2 j (e v e n )(o d d ) − (o d d )(o d d )
=
x = 0
y = 0
M − 1 N − 1
=
M − 1 N − 1
[(e v e n )(e v e n )] + 2
j
x = 0
y = 0
[(e v e n )(o d d )]
x = 0 y = 0
M − 1 N − 1
[(e v e n )(e v e n )]
− j
x = 0 y = 0
=
w h e re th
(g ) P ro p e
c o n v e rs e
is o d d a n
im a g in a r y
e la s t s te p
r ty 1 1 : If f
ly . P r o o f:
d its im a g
fo
(x
If
in
llo w
,y )
f (x
a r y
s fro m E q . (4
is im a g in a r y
,y ) is im a g in
p a r t is e v e n .
.6 - 1
a n d
a r y ,
If w
3 ).
o d d , th e n F (u ,v ) is re a l a n d o d d , a n d
w e k n o w th a t th e re a l p a r t o f F (u ,v )
e c a n s h o w th a t th e im a g in a r y p a r t is
C H A P T E R 4 . P R O B L E M S O L U T IO N S
6 6
z e r o , th e n w e w ill h a v e th e p r o o f fo r th is p r o p e r ty . A s a b o v e ,
M − 1 N − 1
F (u ,v )
=
[j o d d ] (e v e n )(e v e n ) − 2 j (e v e n )(o d d ) − (o d d )(o d d )
x = 0
y = 0
M − 1 N − 1
=
[j o d d ][e v e n − j o d d ][e v e n − j o d d ]
x = 0
y = 0
M − 1 N − 1
=
M − 1 N − 1
[(o d d )(e v e n )] + 2
j
x = 0
y = 0
[(e v e n )(e v e n )]
x = 0
y = 0
M − 1 N − 1
[(o d d )(e v e n )]
− j
x = 0 y = 0
=
re a l
w h e r e t h e la s t s t e p fo llo w s fr o m E q . ( 4 .6 - 1 3 ) .
(h )
e v e
im a
b o t
f (x
P ro p e r
n , a n d
g in a r y
h its re
,y ) = f
f (x ,y )
ly . P r o
F (u ,v )
a g in a r y
j f ie (x ,
If
rs e
o f
im
) +
ty 1 2 :
c o n v e
p a r ts
a l a n d
r e ( x , y
is c
o f:
a re
p a
y ).
o m p le
H e re ,
e v e n .
r ts a re
T h e n ,
x a n d e
w e h a v
R e c a ll
e v e n . T
v e n , th
e to p r
th a t if
h u s , w
e n F
o v e t
f (x ,y
e c a n
(u ,
h a t
) is
w r
v )
b o
a n
ite
is
th
e
th
c o m
th e
v e n
is fu
p le x
re a l
fu n c
n c tio
a n d
a n d
tio n ,
n a s
M − 1 N − 1
F (u ,v )
=
[ f
x = 0
r e
(x ,y ) + j f
i e
(x ,y )]e
− j 2 π (u x / M + v y / N )
y = 0
M − 1 N − 1
=
f
x = 0
r e
(x ,y )e
− j 2 π (u x / M + v y / N )
y = 0
M − 1 N − 1
=
+
j f
x = 0
T h e
w h ic
im a g
th a t
e v e n
p ro o
fi rs t t
h w e
in a r y
th e th
im a g
f.
e r m o
k n o w
e v e n
e tra n
in a r y
f th is
is a re
fu n c t
s fo r m
p a r t,
re s u
a l, e
io n ,
o f a
a n d
lt is r e c o g n
v e n fu n c tio
w h ic h w e k
c o m p le x , e
is th u s a c o
i e
(x ,y )e
− j 2 π (u x / M + v y / N )
.
y = 0
iz e d a s th e D F T o f a
n . T h e s e c o n d te r m is
n o w is im a g in a r y a n d
v e n fu n c tio n , h a s a n e
m p le x e v e n fu n c tio n .
r e a l, e v e n
th e D F T o
e v e n . T h u
v e n re a l p a
T h is c o n c
fu n c tio n ,
f a p u r e ly
s , w e s e e
r t a n d a n
lu d e s th e
( i) P r o p e r ty 1 3 : If f (x , y ) is c o m p le x a n d o d d , th e n F (u , v ) is c o m p le x a n d o d d ,
6 7
a n d c o n v e r s e ly . P r o o f: T h e p r o o f p a r a lle ls th e p r o o f in (h ).
M − 1 N − 1
F (u ,v )
=
x = 0
[ f
r o
(x ,y ) + j f
f
r o
(x ,y )e
− j 2 π (u x / M + v y / N )
(x ,y )]e
i o
y = 0
M − 1 N − 1
=
x = 0
− j 2 π (u x / M + v y / N )
y = 0
M − 1 N − 1
+
j f
x = 0
T
o
k
w
h e fi rs t t
d d . T h e
n o w is re
a n te d to
e r m is
s e c o n
a l a n d
p ro v e
i o
(x ,y )e
− j 2 π (u x / M + v y / N )
.
y = 0
th e D F T o f a n o d d , re a l fu n c tio n , w h ic h k n o w is im a g in a r y a n d
d te r m is th e D F T o f p u r e ly im a g in a r y o d d fu n c tio n , w h ic h w e
o d d . T h u s , th e s u m o f th e tw o is a c o m p le x , o d d fu n c tio n , a s w e
.
Problem 4.21
R e c a ll th a t th e r e a s o n fo r p a d d in g is to e s ta b lis h a “ b u ffe r ” b e tw e e n th e p e r io d s
th a t a r e im p lic it in th e D F T . Im a g in e th e im a g e o n th e le ft b e in g d u p lic a te d in fi n ite ly m a n y tim e s to c o v e r th e x y - p la n e . T h e r e s u lt w o u ld b e a c h e c k e r b o a r d ,
w ith e a c h s q u a r e b e in g in th e c h e c k e r b o a r d b e in g th e im a g e (a n d th e b la c k e x te n s io n s ). N o w im a g in e d o in g th e s a m e th in g to th e im a g e o n th e r ig h t. T h e
r e s u lts w o u ld b e id e n tic a l. T h u s , e ith e r fo r m o f p a d d in g a c c o m p lis h e s th e s a m e
s e p a r a tio n b e tw e e n im a g e s , a s d e s ire d .
Problem 4.22
U n le s s a ll
d u c e s s ig n
T h e s e c a n
th e s p a tia l
h o r iz o n ta l
b o rd e rs
ifi c a n t d
b e s tro n
d o m a in
a x e s o f t
o n o f a n im a g
is c o n tin u itie s
g h o r iz o n ta l a
in tr o d u c e h ig h
h e s p e c tr u m .
e a r e b la c k , p a d d
(e d g e s ) a t o n e o r
n d v e r tic a l e d g e s
-fre q u e n c y c o m p
in g th
m o re
. T h e
o n e n t
e im a g e w it
b o rd e rs o f
s e s h a rp tra
s a lo n g th e v
Problem 4.23
( a ) T h e a v e r a g e s o f th e tw o im a g e s a r e c o m p u te d a s fo llo w s :
f¯ ( x , y ) =
1
M N
M − 1 N − 1
f (x ,y )
x = 0
y = 0
h
th
n s
e r
0 s in
e im
itio n
tic a l
tro a g e .
s in
a n d
C H A P T E R 4 . P R O B L E M S O L U T IO N S
6 8
a n d
f ¯p ( x , y )
P − 1 Q − 1
1
=
f
P Q
f (x ,y )
P Q
x = 0
M N
P Q
=
(x ,y )
M − 1 N − 1
1
=
p
x = 0 y = 0
y = 0
f¯ ( x , y )
w h e r e th e s e c o n d s te p is r e s u lt o f th e fa c t th a t th e im a g e is p a d d e d w ith 0 s . T h u s ,
th e r a tio o f th e a v e r a g e v a lu e s is
r =
P Q
M N
T h u s , w e s e e th a t th e r a tio in c re a s e s a s a fu n c tio n o f P Q , in d ic a tin g th a t th e
a v e r a g e v a lu e o f th e p a d d e d im a g e d e c r e a s e s a s a fu n c tio n o f P Q . T h is is a s
e x p e c te d ; p a d d in g a n im a g e w ith z e r o s d e c r e a s e s its a v e r a g e v a lu e .
( b ) Y e s , t h e y a r e e q u a l . W e k n o w t h a t F ( 0 , 0 ) = M N f¯ ( x , y ) a n d F
A n d , f r o m p a r t ( a ) , f¯p ( x , y ) = M N f¯ ( x , y ) / P Q . T h e n ,
F
p
(0 ,0 )
P Q
F p (0 ,0 )
=
=
p
( 0 , 0 ) = P Q f¯p ( x , y ) .
M N F (0 ,0 )
P Q
M N
F (0 ,0 ).
Problem 4.24
W e s o lv e th e p r o b le m b y d ir e c t s u b s titu tio n in to th e d e fi n itio n o f th e 2 - D D F T :
M − 1 N − 1
F (u + k
1
M ,v + k
2
N )
=
x = 0
f (x ,y )e
− j 2 π ([u + k
f (x ,y )e
− j 2 π (u x / M + v y / N )
1
M ]x / M + [v + k
2
N ]y / N )
y = 0
M − 1 N − 1
=
x = 0
=
e
− j 2 π k
1
x
e
− j 2 π k
2
y
y = 0
F (u ,v )
w ith k 1 a n d k 2 h a v in g v a lu e s 0 , ± 1 , ± 2 , . . . . T h e la s t s te p fo llo w s fr o m th e fa c t th a t
k 1 x a n d k 2 y a r e in te g e r s , w h ic h m a k e s th e tw o r ig h tm o s t e x p o n e n tia ls e q u a l to
1 .
6 9
Problem 4.25
( a ) F r o m E q . ( 4 .4 - 1 0 ) a n d t h e d e fi n it io n o f t h e 1 - D D F T ,
M − 1
=
f (x )
h (x )
=
f (x )
− j 2 π u x / M
h (x )e
x = 0
M − 1 M − 1
=
− j 2 π u x / M
f (m )h (x − m )e
x = 0 m = 0
M − 1
=
M − 1
f (m )
m = 0
h (x − m )e
− j 2 π u x / M
x = 0
b u t
M − 1
h (x − m )e
− j 2 π u x / M
x = 0
= = [h (x − m )] = H (u )e
− j 2 π m u / M
w h e r e t h e la s t s t e p fo llo w s fr o m E q . ( 4 .6 - 4 ) . S u b s t it u t in g t h is r e s u lt in t o t h e
p r e v io u s e q u a tio n y ie ld s
M − 1
=
f (x )
h (x )
=
− j 2 π m u / M
f (m )e
H (u )
m = 0
=
F (u )H (u ).
T h e o th e r p a r t o f th e c o n v o lu tio n th e o r e m is d o n e in a s im ila r m a n n e r.
(b ) A s in (a ),
M − 1 N − 1
=
f (x ,y )
h (x ,y )
=
f (x ,y )
x = 0
y = 0
M − 1 N − 1
=
x = 0
y = 0
=
M − 1 N − 1
=
⎡
⎣
h (x ,y )e
M − 1 N − 1
⎤
f (m ,n )h (x − m ,y − n )⎦
m = 0 n = 0
× e
− j 2 π (u x / M + v y / N )
M − 1 N − 1
f (m ,n )
m = 0 n = 0
M − 1 N − 1
=
h (x − m ,y − n )
x = 0
× e
y = 0
− j 2 π (u x / M + v y / N )
f (m ,n )e
m = 0 n = 0
=
− j 2 π (u x / M + v y / N )
F (u ,v )H (u ,v ).
− j 2 π (u m / M + v n / N )
H (u ,v )
C H A P T E R 4 . P R O B L E M S O L U T IO N S
7 0
( c ) C o r r e la tio n is d o n e in th e s a m e w a y , b u t b e c a u s e o f th e d iffe r e n c e in s ig n in
th e a r g u m e n t o f h th e r e s u lt w ill b e a c o n ju g a te :
M − 1 N − 1
=
f (x ,y )
h (x ,y )
=
f (x ,y )
x = 0
y = 0
M − 1 N − 1
=
x = 0
⎡
⎣
y = 0
M − 1 N − 1
⎤
f (m ,n )h (x + m ,y + n )⎦
m = 0 n = 0
− j 2 π (u x / M + v y / N )
× e
M − 1 N − 1
=
M − 1 N − 1
f (m ,n )
m = 0 n = 0
h (x + m ,y + n )
x = 0
y = 0
− j 2 π (u x / M + v y / N )
× e
M − 1 N − 1
=
− j 2 π (u x / M + v y / N )
h (x ,y )e
f (m ,n )e
j 2 π (u m / M + v n / N )
H (u ,v )
m = 0 n = 0
∗
=
(u ,v )H (u ,v ).
F
( d ) W e b e g in w ith o n e v a r ia b le :
=
d f (z )
d z
1
=
d f (z )
e
d z
− 1
− j 2 π ν z
d z
In te g r a tio n b y p a r ts h a s th e fo llo w in g g e n e r a l fo r m ,
s d w = s w −
L e t s = e
− j 2 π ν z
a n d w =
d w =
w d s .
f (z ). T h e n , d w / d z = d f (z )/ d z o r
d f (z )
d z
d z
d s = (− j 2 π ν )e
a n d
− j 2 π ν z
d z
s o it fo llo w s th a t
=
d f (z )
d z
1
=
− 1
=
=
=
d f (z )
e
d z
f (z ) e
(j 2 π ν )
− j 2 π ν z
− j 2 π ν z
1
1
1
− 1
(j 2 π ν )F (ν )
d z
1
−
f (z )e
f (z )(− j 2 π ν )e
− 1
− j 2 π ν z
d z
− j 2 π ν z
d z
7 1
b e c a u s e f (± 1 ) = 0 b y a s s u m p tio n (s e e T a b le 4 .3 ). C o n s id e r n e x t th e s e c o n d
d e r iv a tiv e . D e fi n e g (z ) = d f (z )/ d z . T h e n
d g (z )
d z
=
= ( j 2 π ν )G (ν )
w h e re G (ν ) is th e F o u r ie r tr a n s fo r m o f g (z ). B u t g (z ) =
(j 2 π ν )F (ν ), a n d
d 2 f (z )
=
= (j 2 π ν )2 F (ν ).
d z 2
d f (z )/ d z , s o G (ν ) =
C o n tin u in g in th is m a n n e r w o u ld r e s u lt in th e e x p r e s s io n
d
=
If
s a
to
w
w e n o
m e re
in d ic
o u ld h
w g o to
s u lt a s
a te th e
a v e F (
2
in
v
μ
-D a n d
th e p r
a r ia b le
,ν ). T h
n
f (z )
d z
n
= (j 2 π ν )n F (ν ).
ta k e th e d e r iv a tiv e o f o n ly o n e v a r ia b le , w e w o u ld g e t th e
e c e d in g e x p re s s io n , b u t w e h a v e to u s e p a r tia l d e r iv a tiv e s
to w h ic h d iffe r e n tia tio n a p p lie s a n d , in s te a d o f F (μ ), w e
u s ,
n
@
=
D e fi n e g (t ,z ) = @
n
f (t ,z )/ @ t
=
m
@
f (t ,z )
@ z n
n
= (j 2 π ν )n F (μ ,ν ).
, th e n
g (t ,z )
@ t m
= (j 2 π μ )
m
G (μ ,ν ).
B u t G (μ ,ν ) is th e tr a n s fo r m o f g (t ,z ) = @ n f (t ,z )/ @ t n , w h ic h w e k n o w is e q u a l
to ( j 2 π μ )n F (μ , ν ). T h e r e fo r e , w e h a v e e s ta b lis h e d th a t
=
@
@ t
m
@
@ z
n
f (t ,z )
= (j 2 π μ )
m
(j 2 π ν )n F (μ ,ν ).
B e c a u s e th e F o u r ie r tr a n s fo r m is u n iq u e , w e k n o w th a t th e in v e r s e tr a n s fo r m
o f th e r ig h t o f th is e q u a tio n w o u ld g iv e th e le ft, s o th e e q u a tio n c o n s titu te s a
F o u r ie r tr a n s fo r m p a ir (k e e p in m in d th a t w e a r e d e a lin g w ith c o n tin u o u s v a r ia b le s ).
Problem 4.26
( a ) F r o m C h a p te r 3 , th e L a p la c ia n o f a fu n c tio n f (t , z ) o f tw o c o n tin u o u s v a r ia b le s is d e fi n e d a s
@ 2 f (t ,z )
@ 2 f (t ,z )
r 2 f (t ,z ) =
+
.
@ t 2
@ z 2
C H A P T E R 4 . P R O B L E M S O L U T IO N S
7 2
W e o b ta in th e F o u r ie r tr a n s fo r m o f th e L a p la c ia n u s in g th e r e s u lt fr o m P r o b le m
4 .2 5 ( e n t r y 1 2 in T a b le 4 .3 ) :
2
r
=
f (t ,z )
2
@
=
=
f (t ,z )
@ t 2
+ =
@
2
f (t ,z )
@ z 2
(j 2 π μ )2 F (μ ,ν ) + (j 2 π ν )2 F (μ ,ν )
=
− 4 π
=
2
(μ
2
+ ν
2
)F (μ ,ν ).
W e r e c o g n iz e th is a s th e fa m ilia r fi lte r in g e x p r e s s io n G (μ , ν ) = H (μ , ν )F (μ , ν ), in
w h ic h H (μ ,ν ) = − 4 π 2 (μ 2 + ν 2 ).
( b ) A s th e p r e c e d in g d e r iv a tio n s h o w s , th e L a p la c ia n fi lte r a p p lie s to c o n tin u o u s
v a r ia b le s . W e c a n g e n e r a te a fi lte r fo r u s in g w ith th e D F T s im p ly b y s a m p lin g
th is fu n c tio n :
H (u ,v ) = − 4 π 2 (u 2 + v 2 )
fo r u = 0 ,1 ,2 , . . . , M − 1 a n d v = 0 , 1 , 2 , . . . , N − 1 . W h e n w o r k in g w it h c e n t e r e d
tr a n s fo r m s , th e L a p la c ia n fi lte r fu n c tio n in th e fr e q u e n c y d o m a in is e x p r e s s e d
a s
H (u ,v ) = − 4 π 2 ([u − M / 2 ]2 + [v − N / 2 ]2 ).
In s u m m a r y , w e h a v e th e fo llo w in g F o u r ie r tr a n s fo r m p a ir r e la tin g th e L a p la c ia n
in th e s p a tia l a n d fre q u e n c y d o m a in s :
r
2
f (x ,y ) , − 4 π
2
([u − M / 2 ]2 + [v − N / 2 ]2 )F (u ,v )
w h e r e it is u n d e r s to o d th a t th e fi lte r is a s a m p le d v e r s io n o f a c o n tin u o u s fu n c tio n .
( c ) T h e L a p la c ia n fi lte r is is o tr o p ic , s o its s y m m e tr y is a p p r o x im a te d m u c h c lo s e r
b y a L a p la c ia n m a s k h a v in g th e a d d itio n a l d ia g o n a l te r m s , w h ic h r e q u ir e s a − 8
in th e c e n te r s o th a t its re s p o n s e is 0 in a re a s o f c o n s ta n t in te n s ity .
Problem 4.27
( a ) T h e s p a tia l a v e r a g e (e x c lu d in g th e c e n te r te r m ) is
1
g (x ,y ) =
4
f (x ,y + 1 ) +
f (x + 1 ,y ) +
f (x − 1 ,y ) +
f (x ,y − 1 ) .
F r o m p r o p e r t y 3 in T a b le 4 .3 ,
G (u ,v )
=
=
1
e j2π v /N + e
4
H (u ,v )F (u ,v )
j 2 π u / M
+ e
− j 2 π u / M
+ e
− j 2 π v / N
F (u ,v )
7 3
w h e re
1
H (u ,v ) =
[c o s (2 π u / M ) + c o s (2 π v / N )]
2
is th e fi lte r tr a n s fe r fu n c tio n in th e fr e q u e n c y d o m a in .
( b ) T o s e e th a t th is is a lo w p a s s fi lte r, it h e lp s to e x p r e s s th
in th e fo r m o f o u r fa m ilia r c e n te r e d fu n c tio n s :
1
[c o s (2 π [u − M / 2 ])/ M ) + c o s (2 π [v −
H (u ,v ) =
2
C o n s id e r o n e v a r ia b le fo r c o n v e n ie n c e . A s u r a n g e s fr o m
o f c o s (2 π [u − M / 2 ]/ M ) s ta r ts a t − 1 , p e a k s a t 1 w h e n u
th e fi lte r ) a n d th e n d e c r e a s e s to − 1 a g a in w h e n u = M .
a m p litu d e o f th e fi lte r d e c r e a s e s a s a fu n c tio n o f d is ta n c e
c e n te r e d fi lte r, w h ic h is th e c h a r a c te r is tic o f a lo w p a s s fi lte
is e a s ily c a r r ie d o u t w h e n c o n s id e r in g b o th v a r ia b le s s im u
e p re c e d in g e q u a tio n
=
N / 2 ]/ N )] .
0 to M
M / 2
T h u s , w
fro m th
r. A s im
lta n e o u
− 1 , th e
(th e c e n
e s e e th
e o r ig in
ila r a r g u
s ly .
v a
te
a t
o f
m
lu
r o
th
th
e n
e
Problem 4.28
( a ) A s in P r o b le m 4 .2 7 , t h e fi lt e r e d im a g e is g iv e n b y :
g (x ,y ) =
f (x + 1 ,y ) −
f (x ,y ) +
f (x ,y + 1 ) −
f (x ,y ).
F r o m p r o p e r t y 3 in T a b le 4 .3 ,
G (u ,v )
=
j 2 π u / M
F (u ,v )e
=
[e
=
j 2 π u / M
− F (u ,v ) + F (u ,v )e
j 2 π v / N
− 1 ]F (u ,v ) + [e
j 2 π v / N
− F (u ,v )
− 1 ]F (u ,v )
H (u ,v )F (u ,v )
w h e r e H (u , v ) is th e fi lte r fu n c tio n :
H (u ,v )
(e
=
=
2 j
j 2 π u / M
− 1 ) + (e
s in (π u / M )e
j 2 π v / N
j π u / M
− 1 )
+ s in (π v / N )e
j π v / N
.
( b ) T o s e e th a t th is is a h ig h p a s s fi lte r, it h e lp s to e x p r e s s th e fi lte r fu n c tio n in th e
fo r m o f o u r fa m ilia r c e n te r e d fu n c tio n s :
H (u ,v ) = 2 j
s in (π [u − M / 2 ]/ M )e
T h e fu n c tio n is 0 a t th e c e n te r
v a lu e o f th e fi lte r d e c r e a s e s , r e a
M − 1 a n d v = M − 1 . T h e n e g a t
d e r iv a tiv e s a re ta k e n . If, in s te a
f (x + 1 ,y ) a n d f (x ,y ) − f (x ,y +
lim itin g v a lu e . T h e im p o r ta n t
h ig h e r fre q u e n c ie s a re p a s s e d ,
j π u / M
+ s in (π [v − N / 2 ]/ N )e
o f th e fi lte r u = M / 2 ).
c h in g its lim itin g v a lu e
iv e lim itin g v a lu e is d u e
d w e h a d ta k e n d iffe re n
1 ), th e fi lte r w o u ld h a v e
p o in t h e re is th a t th e d
w h ic h is th e c h a r a c te r is
A s
o f
to
c e
te
j π v / N
.
u a n d v in c re a s e , th e
c lo s e to − 4 j w h e n u =
th e o rd e r in w h ic h th e
s o f th e fo r m f (x ,y ) −
n d e d to w a rd a p o s itiv e
c te r m is e lim in a te d a n d
tic o f a h ig h p a s s fi lte r.
f
e
e
t
C H A P T E R 4 . P R O B L E M S O L U T IO N S
7 4
Problem 4.29
T h e fi lte r e d fu n c tio n is g iv e n b y
g (x ,y ) = [ f (x + 1 ,y ) +
f (x − 1 ,y ) +
f (x ,y + 1 ) +
f (x ,y − 1 )] − 4 f (x ,y ).
A s in P r o b le m 4 .2 8 ,
G (u ,v ) = H (u ,v )F (u ,v )
w h e re
H (u ,v )
=
e
j 2 π u / M
+ e
− j 2 π u / M
j 2 π v / N
+ e
+ e
− j 2 π v / N
− 4
2 [c o s (2 π u / M ) + c o s (2 π v / N ) − 2 ] .
=
S h iftin g th e fi lte r to th e c e n te r o f th e fr e q u e n c y r e c ta n g le g iv e s
H (u ,v ) = 2 [c o s (2 π [u − M / 2 ] / M ) + c o s (2 π [v − N / 2 ] / N ) − 2 ] .
W h e n (u ,v ) =
u e s a w a y fro m
o rd e r in w h ic h
e lim in a te d a n d
a h ig h p a s s fi lte
(M
th e
d e
th
r.
/ 2 ,N
c e n
r iv a t
e h ig
/
te
iv
h
2 )
r,
e s
e r
(t
H
a
fr
h e
(u
re
e q
c e n te r
,v ) d e c
ta k e n .
u e n c ie
o f th e s h ifte
re a s e s (a s in
T h e im p o r ta
s a re p a s s e d ,
d fi lte r ),
P r o b le m
n t p o in t
w h ic h is
H (u
4 .2
is t
th e
,v ) =
8 ) b e c
h e th e
c h a ra
0 .
a u s
d c
c te
F o r v a le o f th e
te r m is
r is tic o f
Problem 4.30
T h e a
a n d s
T h e F
(a s in
c o m p
n s w e
q u a re
o u r ie
P ro b
u te d
r is
ro o
r tra
le m
d ire
n o . T
ts in v
n s fo r
4 .2 8 )
c tly in
h e F o u r ie r tr
o lv e d in c o m
m c o u ld b e u
, b u t th e s q u
th e s p a tia l d
a n s fo r m is
p u tin g th e
s e d to c o m
a re s , s q u a r
o m a in .
a lin
g ra d
p u te
e ro o
e a r p ro c e
ie n t a re n
th e d e r iv
t, o r a b s o
s s , w
o n lin
a tiv e
lu te
h ile
e a r
s a s
v a lu
th e
o p e r
d iffe
e s m
s q
a t
re
u
u a re
io n s .
n c e s
s t b e
Problem 4.31
W e w a n t to s h o w th a t
=
− 1
A e
− (μ
2
+ ν
2
)/ 2 σ
2
= A 2 π σ
2
e
− 2 π
2
σ
2
(t 2 + z
2
)
.
T h e e x p la n a tio n w ill b e c le a r e r if w e s ta r t w ith o n e v a r ia b le . W e w a n t to s h o w
th a t, if
H (μ ) = e
− μ
2
/ 2 σ
2
7 5
th e n
h (t )
− 1
=
=
H (μ )
1
=
− 1
p
=
2
− μ
e
2 π σ
2
/ 2 σ
2
− 2 π
σ
j 2 t μ t
e
2
2
t
d μ
.
W e c a n e x p re s s th e in te g r a l in th e p re c e d in g e q u a tio n s a s
1
h (t ) =
1
2 σ 2
−
e
2
[μ
2
− j 4 π σ
μ t ]
d μ .
− 1
M a k in g u s e o f th e id e n tity
e
(2 π )2 σ 2 t 2
2
−
(2 π )2 σ 2 t 2
2
e
= 1
in th e p r e c e d in g in te g r a l y ie ld s
h (t )
=
=
e
e
1
(2 π )2 σ 2 t 2
2
−
−
e
−
1
2 σ 2
[μ
e
−
1
2 σ 2
[μ − j 2 π σ
− 1
1
(2 π )2 σ 2 t 2
2
2
2
− j 4 π σ
2
μ t − (2 π )2 σ
t ]
2
4
t 2 ]
d μ .
d μ .
− 1
2
N e x t, w e m a k e th e c h a n g e o f v a r ia b le s r = μ − j 2 π σ
p re c e d in g in te g r a l b e c o m e s
−
h (t ) = e
1
(2 π )2 σ 2 t 2
2
r 2
2 σ 2
−
e
t . T h e n , d r = d μ a n d th e
d r .
− 1
F in a lly , w e m u ltip ly a n d d iv id e th e r ig h t s id e o f th is e q u a tio n b y
ta in
⎤
⎡
1
p
r 2
(2 π )2 σ 2 t 2
1
−
−
⎣p
2
h (t ) =
2 π σ e
e 2σ2 d r ⎦.
2 π σ − 1
p
2 π σ a n d o b -
T h e e x p r e s s io n in s id e th e b r a c k e ts is r e c o g n iz e d a s th e G a u s s ia n p r o b a b ility
d e n s ity fu n c tio n w h o s e v a lu e fr o m - 1 to 1 is 1 . T h e r e fo r e ,
h (t ) =
p
− 2 π
2 π σ e
2
2
σ
2
t
.
W ith th e p r e c e d in g r e s u lts a s b a c k g r o u n d , w e a r e n o w r e a d y to s h o w th a t
h (t ,z )
− 1
A e
A 2 π σ
2
=
=
=
e
− (μ
2
− 2 π
2
+ ν
2
σ
2
)/ 2 σ
(t 2 + z
2
2
)
.
C H A P T E R 4 . P R O B L E M S O L U T IO N S
7 6
B y s u b s titu tin g d ir e c tly in to th e d e fi n itio n o f th e in v e r s e F o u r ie r tr a n s fo r m w e
h a v e :
h (t ,z )
1
=
1
− 1
1
=
⎡
− 1
2
− (μ
A e
1
⎣
)/ 2 σ
2
j 2 π (μ t + ν z )
e
⎤
!
μ 2
2 σ 2
−
A e
− 1
2
+ ν
+ j 2 π μ t
d μ d ν
"
d μ ⎦e
−
#
ν 2
2 σ 2
+ j 2 π ν z
d ν .
− 1
T h e in te g r a l in s id e th e b r a c k e ts is re c o g n iz e d fr o m th e p re v io u s d is c u s s io n to b e
p
2
2 2
e q u a l to A 2 π σ e − 2 π σ t . T h e n , th e p re c e d in g in te g r a l b e c o m e s
h (t ,z ) = A
p
− 2 π
2 π σ e
2
2
σ
"
1
2
t
−
e
ν 2
2 σ 2
#
+ j 2 π ν z
− 1
W e n o w re c o g n iz e th e re m a in in g in te g r a l to b e e q u a l to
w h ic h w e h a v e th e fi n a l r e s u lt:
p
p
2
2 2
2 π σ e −
h (t ,z ) =
A 2 π σ e − 2π σ t
=
A 2 π σ
2
e
2
− 2 π
σ
2
(t 2 + z
2
)
d ν .
p
2 π
2 π σ e
2
2
σ
z
− 2 π
2
σ
2
z
2
, fro m
2
.
Problem 4.32
T h e s p a tia l fi lte r is o b ta in e d b y ta k in g th e in v e r s e F o u r ie r tr a n s fo r m o f th e fr e q u e n c y d o m a in fi lte r :
h
H P
(t ,z )
=
=
=
=
=
− 1
1 − H
L P
− 1
[1 ] − =
− 1
δ (0 ) − A 2 π σ
(μ ,ν )
H
2
e
L P ( μ
− 2 π 2 σ
,ν )
2
(t 2 + z
2
)
.
T h is r e s u lt is fo r c o n tin u o u s fu n c tio n s . T o u s e th e m w ith d is c r e te v a r ia b le s w e
s im p ly s a m p le th e fu n c tio n in to its d e s ir e d d im e n s io n s .
Problem 4.33
T h e c o m p le x c o n ju g a te s im p ly c h a n g e s j to − j in th e in v e r s e tr a n s fo r m , s o th e
im a g e o n th e r ig h t is g iv e n b y
M − 1 N − 1
=
− 1
F
∗
(u ,v )
=
x = 0
F (u ,v )e
− j 2 π (u x / M + v y / N )
F (u ,v )e
j 2 π (u (− x )/ M + v (− y )/ N )
y = 0
M − 1 N − 1
=
x = 0
=
y = 0
f (− x ,− y )
7 7
w h ic h s im p ly m ir r o r s f (x , y ) a b o u t th e o r ig in , th u s p r o d u c in g th e im a g e o n th e
r ig h t.
Problem 4.34
T h e e q u a lly - s p a c e d , v e r tic a l b a r s o n th e le ft, lo w e r th ir d o f th e im a g e .
Problem 4.35
W ith
c a n b
k n o w
g iv e s
re fe re n c e to
e e x p re s s e d
d o n o t h a v e
a n im p u ls e a
E q .
a 1
a n
t th
(4
m
im
e o
.9 - 1 ),
in u s
p u ls e
r ig in
a ll th e
th e tra
a t th e
in th e
h ig h p a
n s fe r fu
o r ig in ).
h ig h p a s
s s fi
n c t
T h
s s p
lt
io
e
a
e rs
n o
in v
tia l
in d is c u s s e d in S e c t io n 4 .9
f lo w p a s s fi lte r (w h ic h w e
e r s e F o u r ie r tr a n s fo r m o f 1
fi lte r s .
Problem 4.36
(
o
t
r
a
o
a ) T h e r in g in fa
n ly (th e fo llo w in
h e d a rk c e n te r a
e s u lt lo o k s s o b r
re m u c h h ig h e r
f th e r e s u lt.
c t
g
re
ig
th
h a s a d a rk c e
im a g e s h o w s
a is a v e r a g e d
h t is th a t th e
a n a n y w h e re
n te r a re a a s a re s u
th e r e s u lt o f h ig h p
o u t b y th e lo w p a s
d is c o n tin u ity (e d g
e ls e in th e im a g e ,
lt o f th e h ig h p a s s o p e r a tio n
a s s fi lte r in g o n ly ). H o w e v e r,
s fi lte r. T h e r e a s o n th e fi n a l
e ) o n b o u n d a r ie s o f th e r in g
th u s d o m in a tin g th e d is p la y
( b ) F ilte r in g w ith th e F o u r ie r tr a n s fo r m is a lin e a r p r o c e s s . T h e o r d e r d o e s n o t
m a tte r.
Problem 4.37
( a ) O n e a p p lic a tio n o f th e fi lte r g iv e s :
G (u ,v )
=
H (u ,v )F (u ,v )
=
e
2
− D
(u ,v )/ 2 D
0
2
F (u ,v ).
S im ila r ly , K a p p lic a tio n s o f th e fi lte r w o u ld g iv e
G
T h e in v e
th e G a u s
p a s s fi lte
v a lu e o f
lo w p a s s
rs e D F T
s ia n fi lte
r, p a s s in
th e im a g
fi lte r in g
o
r.
g
e
w
K
(u ,v ) = e
− K D
f G K (u , v ) w o u ld g iv
If K is “ la r g e e n o u g h
o n ly F (0 , 0 ). W e k n o
. S o , th e r e is a v a lu e
ill s im p ly p r o d u c e a
2
(u ,v )/ 2 D
e th e
,” t h e
w th a
o f K
c o n s t
0
2
F (u ,v ).
im a g e r e s u ltin g
G a u s s ia n L P F w
t th is te r m is e q
a fte r w h ic h th e
a n t im a g e . T h e
fro m K
ill b e c o m
u a l to th
r e s u lt o f
v a lu e o f
p a s
e a
e a v
re p
a ll
s e s o f
n o tc h
e ra g e
e a te d
p ix e ls
C H A P T E R 4 . P R O B L E M S O L U T IO N S
7 8
F ig u r e P 4 .3 6 .
o n
th
p r
o f
th is
e a n s
o a c h
fi lte r
im
w e
a n
in g
a g e w ill b e e q u a l to th e a v e r a g e v a lu e o f th e o r ig in a l im a g e . N o te th a t
r a p p lie s e v e n a s K a p p r o a c h e s in fi n ity . In th is c a s e th e fi lte r w ill a p im p u ls e a t th e o r ig in , a n d th is w o u ld s till g iv e u s F (0 , 0 ) a s th e r e s u lt
.
( b ) T o g u a r a n te e th e r e s u lt in (a ), K h a s to b e c h o s e n la r g e e n o u g h s o th a t th e
fi lte r b e c o m e s a n o tc h p a s s fi lte r (a t th e o r ig in ) fo r a ll v a lu e s o f D (u , v ). K e e p in g
in m in d th a t in c r e m e n ts o f fr e q u e n c ie s a r e in u n it v a lu e s , th is m e a n s
H
K
(u ,v ) = e
− K D
2
(u ,v )/ 2 D
0
2
=
0
1
if (u ,v ) = (0 ,0 )
O th e r w is e .
B e c a u s e u a n d v a r e in te g e r s , th e c o n d itio n s o n th e s e c o n d lin
a r e s a tis fi e d fo r a ll u > 1 a n d / o r v > 1 . W h e n u = v = 0 ,
H K (u , v ) = 1 , a s d e s ir e d . W e w a n t a ll v a lu e s o f th e fi lte r to b e
o f t h e d is t a n c e fr o m t h e o r ig in t h a t a r e g r e a t e r t h a n 0 ( i.e ., fo r v
v g r e a te r th a n 0 ). H o w e v e r, th e fi lte r is a G a u s s ia n fu n c tio n , s o
g r e a te r th a n 0 fo r a ll fi n ite v a lu e s o f D (u , v ). B u t, w e a r e d e
n u m b e r s , w h ic h w ill b e d e s ig n a te d a s z e r o w h e n e v e r th e v a lu e
th a n o n e - h a lf th e s m a lle s t p o s itiv e n u m b e r r e p r e s e n ta b le in th
u s e d . A s g iv e n in th e p r o b le m s ta te m e n t, th e v a lu e o f th is n u
v a lu e s o f K fo r w h ic h fo r w h ic h th e fi lte r fu n c tio n is g r e a te r t
e in th is e q u a tio n
D (u ,v ) = 0 , a n d
z e r o fo r a ll v a lu e s
a lu e s o f u a n d / o r
its v a lu e is a lw a y s
a lin g w ith d ig ita l
o f th e fi lte r is le s s
e c o m p u te r b e in g
m b e r i s c m in . S o ,
h a n 0 .5 × c m in w i ll
7 9
s u ffi c e . T h a t is , w e w a n t th e m in im u m v a lu e o f K fo r w h ic h
e
− K D
2
(u ,v )/ 2 D
0
2
< 0 .5 c
m in
o r
>
K
>
A s n o
B e c a u
o r ig in
th e re
ln (0 .5 c
−
2
D
−
2 D
)
m in
(u ,v )/ 2 D
0
2
ln (0 .5 c
2
D
0
2
)
m in
.
(u ,v )
e q u a tio n fo r h o ld fo r a ll v a lu e s o f D 2 (u , v ) > 0 .
a s e s a s a fu n c tio n o f in c re a s in g d is ta n c e fr o m th e
p o s s ib le v a lu e o f D 2 (u , v ), w h ic h is 1 . T h is g iv e s
te d a b o v e , w e w a n t th is
s e th e e x p o n e n tia l d e c re
, w e c h o o s e th e s m a lle s t
s u lt
K
> − 2 D
2
0
ln (0 .5 c
m in
)
w h i c h y i e ld s a p o s i t i v e n u m b e r b e c a u s e c m in < < 1 . T h i s r e s u lt g u a r a n t e e s t h a t
th e lo w p a s s fi lte r w ill a c t a s a n o tc h p a s s fi lte r, le a v in g o n ly th e v a lu e o f th e tr a n s fo r m a t th e o r ig in . T h e im a g e w ill n o t c h a n g e p a s t th is v a lu e o f K .
Problem 4.38
( a ) T h e k e y fo r th e s tu d e n t to b e a b le to s o lv e th is p r o b le m is to tr e a t th e n u m b e r
o f a p p lic a tio n s (d e n o te d b y K ) o f th e h ig h p a s s fi lte r a s 1 m in u s K a p p lic a tio n s
o f th e c o r r e s p o n d in g lo w p a s s fi lte r, s o th a t
H
K
(u ,v )
=
=
H K (u ,v )F (u ,v )
$
2
1 − e − K D (u ,v )/ 2
w h e r e th e G a u s s ia n lo w p a s s fi lte r
r e c tly w ith th e e x p r e s s io n o f th e
a n d a tte m p t to r a is e it to th e K th p
T h e s o lu tio n to th e p r o b le m p
h o w e v e r, th e fi lte r w ill a p p r o a c h a
w ill p r o d u c e a n im a g e w ith z e r o a v
th e r e is a v a lu e o f K a fte r w h ic h
s im p ly p r o d u c e a c o n s ta n t im a g e .
D
0
2
%
H (u ,v )
is
G
o
a
n
e
fr o m P r o b le m 4 .3 7 . S t u d e n t s w h o s t a r t d i2
2
a u s s i a n h i g h p a s s fi l t e r 1 − e − K D (u ,v )/ 2 D 0
w e r w ill r u n in to a d e a d e n d .
r a lle ls t h e s o lu t io n o f P r o b le m 4 .3 7 . H e r e ,
o tc h fi lte r th a t w ill ta k e o u t F (0 , 0 ) a n d th u s
r a g e v a lu e (th is im p lie s n e g a tiv e p ix e ls ). S o ,
th e r e s u lt o f r e p e a te d h ig h p a s s fi lte r in g w ill
( b ) T h e p r o b le m h e r e is to d e te r m in e th e v a lu e o f K fo r w h ic h
H
K
(u ,v ) = 1 − e
− K D
2
(u ,v )/ 2 D
0
2
=
1
0
if (u ,v ) = (0 ,0 )
o th e r w is e .
C H A P T E R 4 . P R O B L E M S O L U T IO N S
8 0
B e c a u s e u a n d v a r e in te g e r s , th e c o n d itio n s o n th e s e c o n d lin e in th is e q
h a v e to b e s a tis fi e d fo r a ll u ≥ 1 a n d / o r v ≥ 1 . W h e n u = v = 0 , D (u , v ) =
H K (u , v ) = 0 , a s d e s ir e d . W e w a n t a ll v a lu e s o f th e fi lte r to b e 1 fo r a ll v a
t h e d is t a n c e fr o m t h e o r ig in t h a t a r e g r e a t e r t h a n 0 ( i.e ., fo r v a lu e s o f u a n
g re a te r th a n 0 ). F o r H K (u ,v ) to b e c o m e 1 , th e e x p o n e n tia l te r m h a s to b
0 fo r v a lu e s o f u a n d / o r v g r e a te r th a n 0 . T h is is th e s a m e r e q u ir e m e n
P r o b le m 4 .3 7 , s o th e s o lu tio n o f t h a t p r o b le m a p p lie s h e r e a s w e ll.
u a tio n
0 , a n d
lu e s o f
d / o r v
e c o m e
t a s in
Problem 4.39
( a ) E x p r e s s fi lte r in g a s c o n v o lu tio n to r e d u c e a ll p r o c e s s e s to th e s p a tia l d o m a in .
T h e n , th e fi lte r e d im a g e is g iv e n b y
g (x ,y ) = h (x ,y )
f (x ,y )
w h e r e h is th e s p a tia l fi lte r (in v e r s e F o u r ie r tr a n s fo r m o f th e fr e q u e n c y - d o m a in
fi lte r ) a n d f is th e in p u t im a g e . H is to g r a m p r o c e s s in g th is r e s u lt y ie ld s
g 0(x ,y )
=
=
T
g (x ,y )
T
h (x ,y )
f (x ,y ) ,
w h e r e T d e n o te s th e h is to g r a m e q u a liz a tio n tr a n s fo r m a tio n . If w e h is to g r a m e q u a liz e fi r s t, th e n
g (x ,y ) = T f (x ,y )
a n d
g 0(x ,y ) = h (x ,y )
f (x ,y ) .
T
In g e n e r a l, T is a n o n lin e a r fu n c tio n d e te r m in e d b y th e n a tu r e o f th e p ix e ls in
t h e i m a g e f r o m w h i c h i t i s c o m p u t e d . T h u s , i n g e n e r a l , T h ( x , y ) f ( x , y ) =6
h (x ,y ) T f (x ,y ) a n d th e o rd e r d o e s m a tte r.
(b ) A
tra s t
m e n
e q u a
te r in
e q u a
s in
o f a
t is
liz e
g p r
liz e
d ic a te d in S e c
n im a g e . A lth o
u s u a lly n o t d r
d fi r s t, th e g a in
o c e s s . T h e re fo
th e im a g e a fte
t io n 4 .9 , h ig h p a s s fi lt e r in g s e v e r e ly d im in is
u g h h ig h - fr e q u e n c y e m p h a s is h e lp s s o m e ,
a m a t ic ( s e e F ig . 4 .5 9 ) . T h u s , if a n im a g e
in c o n tr a s t im p r o v e m e n t w ill e s s e n tia lly b e
r e , th e p r o c e d u r e in g e n e r a l is to fi lte r fi r s t a n
r th a t.
h e s th e c o n th e im p r o v e is h is to g r a m
lo s t in th e fi ld h is to g r a m -
Problem 4.40
F r o m E q . ( 4 .9 - 3 ) , t h e t r a n s fe r fu n c t io n o f a B u t t e r w o r t h h ig h p a s s fi lt e r is
H (u ,v ) =
$
1 +
1
D
0
D (u ,v )
%2 n .
8 1
W e w a n t th e fi lte r to h a v e a v a lu e o f γ L w h e n D (u
h ig h v a lu e s o f D (u , v ). T h e p r e c e d in g e q u a tio n is
th is :
(γ H − γ
H (u ,v ) = γ L +
$
D 0
1 +
D (u ,v
,v ) = 0 , a n d a p p ro a c h γ H fo r
e a s ily m o d ifi e d to a c c o m p lis h
)
%2 n .
L
)
T h e v a lu e o f n c o n tr o ls th e s h a r p n e s s o f th e tr a n s itio n b e tw e e n γ
L
a n d γ
H
.
Problem 4.41
B e c a u s e M = 2
n
, w e c a n w r it e E q s . ( 4 .1 1 - 1 6 ) a n d ( 4 .1 1 - 1 7 ) a s
1
m (n ) =
M n
2
a n d
a (n ) = M n .
P r o o f b y in d u c tio n b e g in s b y s h o w in g th a t b o th e q u a tio n s h o ld fo r n = 1 :
m (1 ) =
2
1
(2 )(1 ) = 1
a (1 ) = (2 )(1 ) = 2 .
a n d
W e k n o w t h e s e r e s u lt s t o b e c o r r e c t fr o m t h e d is c u s s io n in S e c t io n 4 .1 1 .3 . N e x t ,
w e a s s u m e th a t th e e q u a tio n s h o ld fo r n . T h e n , w e a r e r e q u ir e d to p r o v e th a t
t h e y a ls o a r e t r u e fo r n + 1 . F r o m E q . ( 4 .1 1 - 1 4 ) ,
m (n + 1 ) = 2 m (n ) + 2
n
.
S u b s titu tin g m (n ) fr o m a b o v e ,
m (n + 1 )
=
1
+ 2
2 n n
2
(n + 1 )
+ 2
2
2
=
2
1
=
=
M n
2
n
1
2
2
n + 1
n
(n + 1 ).
T h e r e fo r e , E q . ( 4 .1 1 - 1 6 ) is v a lid fo r a ll n .
F r o m E q . ( 4 .1 1 - 1 7 ) ,
a (n + 1 ) = 2 a (n ) + 2
n
n + 1
.
C H A P T E R 4 . P R O B L E M S O L U T IO N S
8 2
S u b s titu tin g th e a b o v e e x p r e s s io n fo r a (n ) y ie ld s
a (n + 1 )
=
2 M n + 2
=
2 (2
=
2
n
n + 1
n + 1
n ) + 2
n + 1
(n + 1 )
w h ic h c o m p le te s th e p r o o f.
Problem 4.42
C o n s id e r a s in g le s ta r, m o d e le d a s a n im p u ls e δ (x − x
f (x ,y ) = K δ (x − x
0
,y − y 0 ). T h e n ,
0
,y − y 0 )
fr o m w h ic h
z (x ,y )
=
=
ln f (x , y ) = ln K + ln δ (x − x
0
K
0
+ δ (x − x
0
0
,y − y 0 )
,y − y 0 ).
T a k in g th e F o u r ie r tr a n s fo r m o f b o th s id e s y ie ld s
=
z (x ,y )
=
=
F r o m th is r e s u lt, it is e v id e n t th a
a t th e o r ig in o f th e fr e q u e n c y p la
p o n e n t w ill ta k e c a r e o f th e p r o b
im p u ls e s (s ta r s ) is im p le m e n te d
th e fi lte r w ill b e th e s a m e . A t th e
c o m b in e d b y a d d itio n , fo llo w e d
n e s s b e tw e e n th e s ta r s is p re s e r v
=
K
0
+ =
δ (0 ,0 ) + e
δ 0(x − x
− 2 π (u x
0
0
,y − y 0 )
+ v y 0 )
.
t th e c o n tr ib u tio n o f illu m in a tio n is a n im p u ls e
n e . A n o tc h fi lte r th a t a tte n u a te s o n ly th is c o m le m . E x te n s io n o f th is d e v e lo p m e n t to m u ltip le
b y c o n s id e r in g o n e s ta r a t a tim e . T h e fo r m o f
e n d o f th e p r o c e d u r e , a ll in d iv id u a l im a g e s a r e
b y in te n s ity s c a lin g s o th a t th e r e la tiv e b r ig h te d .
Problem 4.43
T h e p r o b le m c a n b e s o lv e d b y c a r r y in g o u t th e fo llo w in g s te p s :
1 . P e r fo r m a m e d ia n fi lte r in g o p e r a tio n .
2 . F o llo w (1 ) b y h ig h - fr e q u e n c y e m p h a s is .
3 . H is to g r a m - e q u a liz e th is r e s u lt.
4 . C o m p u te th e a v e r a g e g r a y le v e l, K
0
. A d d th e q u a n tity (K − K
0
) to a ll p ix e ls .
8 3
Output
I
I
1
Input
r
2
F ig u r e P 4 .4 3
5 . P e r fo r m t h e t r a n s fo r m a t io n s s h o w n in F ig . P 4 .4 3 , w h e r e r is t h e in p u t g r a y
le v e l, a n d R , G , a n d B a r e fe d in to a n R G B c o lo r m o n ito r.
to
to
to
to
to
to
to
to
to
p
p
p
p
p
p
p
p
p
w
w
w
w
w
w
w
w
w
h
h
h
h
h
h
h
h
h
ite
ite
ite
ite
ite
ite
ite
ite
ite
c h
c h
c h
c h
c h
c h
c h
c h
c h
a r
a r
a r
a r
a r
a r
a r
a r
a r
a c
a c
a c
a c
a c
a c
a c
a c
a c
te
te
te
te
te
te
te
te
te
r to
r to
r to
r to
r to
r to
r to
r to
r to
fo
fo
fo
fo
fo
fo
fo
fo
fo
rc
rc
rc
rc
rc
rc
rc
rc
rc
e fi
e fi
e fi
e fi
e fi
e fi
e fi
e fi
e fi
g u
g u
g u
g u
g u
g u
g u
g u
g u
re
re
re
re
re
re
re
re
re
to
to
to
to
to
to
to
to
to
th
th
th
th
th
th
th
th
th
e t
e t
e t
e t
e t
e t
e t
e t
e t
o p
o p
o p
o p
o p
o p
o p
o p
o p
w
w
w
w
w
w
w
w
w
h
h
h
h
h
h
h
h
h
it
it
it
it
it
it
it
it
it
e c
e c
e c
e c
e c
e c
e c
e c
e c
h
h
h
h
h
h
h
h
h
a r
a r
a r
a r
a r
a r
a r
a r
a r
a c
a c
a c
a c
a c
a c
a c
a c
a c
te
te
te
te
te
te
te
te
te
r t
r t
r t
r t
r t
r t
r t
r t
r t
o
o
o
o
o
o
o
o
o
fo
fo
fo
fo
fo
fo
fo
fo
fo
rc
rc
rc
rc
rc
rc
rc
rc
rc
e fi
e fi
e fi
e fi
e fi
e fi
e fi
e fi
e fi
g u
g u
g u
g u
g u
g u
g u
g u
g u
re
re
re
re
re
re
re
re
re
to
to
to
to
to
to
to
to
to
th
th
th
th
th
th
th
th
th
e
e
e
e
e
e
e
e
e
Chapter 5
Problem Solutions
Problem 5.1
T h e s o lu t io n s a r e s h o w n in F ig . P 5 .1 , fr o m le ft t o r ig h t .
F ig u r e P 5 .1
Problem 5.2
T h e s o lu tio n s a r e s h o w n in th e fo llo w in g fi g u r e , fr o m le ft to r ig h t.
F ig u r e P 5 .2
8 5
8 6
C H A P T E R 5 . P R O B L E M S O L U T IO N S
Problem 5.3
T h e s o lu t io n s a r e s h o w n in F ig . P 5 .3 , fr o m le ft t o r ig h t .
F ig u r e P 5 .3
Problem 5.4
T h e s o lu t io n s a r e s h o w n in F ig . P 5 .4 , fr o m le ft t o r ig h t .
F ig u r e P 5 .4
Problem 5.5
T h e s o lu t io n s a r e s h o w n in F ig . P 5 .5 , fr o m le ft t o r ig h t .
F ig u r e P 5 .5
8 7
Problem 5.6
T h e s o lu t io n s a r e s h o w n in F ig . P 5 .6 , fr o m le ft t o r ig h t .
F ig u r e P 5 .6
Problem 5.7
T h e s o lu t io n s a r e s h o w n in F ig . P 5 .7 , fr o m le ft t o r ig h t .
F ig u r e P 5 .7
Problem 5.8
T h e s o lu t io n s a r e s h o w n in F ig . P 5 .8 , fr o m le ft t o r ig h t .
F ig u r e P 5 .8
C H A P T E R 5 . P R O B L E M S O L U T IO N S
8 8
F ig u r e P 5 .9
Problem 5.9
T h e s o lu t io n s a r e s h o w n in F ig . P 5 .9 , fr o m le ft t o r ig h t .
Problem 5.10
(a ) T h e
p ix e l is z
p o in ts v
th e a r ith
k e y to
e ro . D
a lu e d
m e tic
th is p r o b le m is th a
r a w a p r o fi le o f a n id
1 . T h e g e o m e tr ic m
m e a n w ill g iv e in te r
t th e g e o m e tr
e a l e d g e w ith
e a n w ill g iv e o
m e d ia te v a lu e
ic m e a n is z e r o w h e n e v e r a n y
a fe w p o in ts v a lu e d 0 a n d a fe w
n ly v a lu e s o f 0 a n d 1 , w h e r e a s
s (b lu r ).
( b ) B la c k is 0 , s o th e g e o m e tr ic m e a n w ill r e tu r n v a lu e s o f 0 a s lo n g a s a t le a s t
o n e p ix e l in th e w in d o w is b la c k . B e c a u s e th e c e n te r o f th e m a s k c a n b e o u ts id e
th e o r ig in a l b la c k a r e a w h e n th is h a p p e n s , th e fi g u r e w ill b e th ic k e n e d .
Problem 5.11
T h e k e y to u n d e r s ta n d in g th
o f th e p ix e ls in th e n e ig h b o
s ta n t, w ith th e im p u ls e n o is e
th e n o is e s p ik e to b e v is ib le ,
o f its n e ig h b o r s . A ls o k e e p in
p o w e r in th e d e n o m in a to r.
e b e h a
r h o o d
p o in t
its v a lu
m in d
v io r o f th e c o n tr a - h a r m
s u r r o u n d in g a n o is e im
b e in g in th e c e n te r o f th
e m u s t b e c o n s id e r a b ly
th a t th e p o w e r in th e n u
( a ) B y d e fi n itio n , p e p p e r n o is e is a lo w v a lu e (r e a
s u r r o u n d e d b y lig h t v a lu e s . T h e c e n te r p ix e l (th e
in fl u e n c e in th e s u m s . If th e a re a s p a n n e d b y th
s ta n t, th e r a tio w ill a p p r o a c h th e v a lu e o f th e p ix e
r e d u c in g th e e ffe c t o f th e lo w - v a lu e p ix e l. F o r e x a
th e fi lte r fo r a d a r k p o in t o f v a lu e 1 in a 3 × 3 r e g io
Q = 0 .5 , fi lt e r = 9 8 .7 8 ; fo r Q = 1 , fi lt e r = 9 9 .8 8 , fo
Q = 5 , fi lt e r = 1 0 0 .0 0 .
o n ic fi lte r is to th
p u ls e a s b e in g c
e n e ig h b o r h o o d .
la r g e r th a n th e v a
m e r a to r is 1 p lu s
lly 0 ). It is m o s
p e p p e r n o is e ),
e fi lte r is a p p r o
ls in th e n e ig h b
m p le , h e r e a r e s
n w ith p ix e ls o f
r Q = 2 , fi lte r =
t
w
x
o
o
v
9
in
o n
F o
lu
th
k
r
e
e
v is ib le w h e n
ill h a v e little
im a te ly c o n r h o o d — th u s
m e v a lu e s o f
a lu e 1 0 0 : F o r
9 .9 9 ; a n d fo r
8 9
(b ) T h e
s m a ll. T
n e g a tiv e
th e n b e
a s a th e
p ix e ls in
(c ) W h e
n o is e w
F o r s a lt
n o is e —
re v e rs e h a p p e n s
h e c e n te r p ix e l w
, s o th e s m a ll n u
th o u g h t o f a c o n
s a m e c o n s ta n t ra
th e n e ig h b o r h o o
n th e
ill b e r
n o is e
th e im
(d ) W h e n
n u m b e r o
n o m in a to
e x a m p le ,
9 / [1 / p 1 +
T h u s , lo w
If, fo r e x a
re s p o n s e
Q =
f p ix
r a re
fo r a
1 / p 2
p ix e
m p le
w ill b
w ro
a is e
th e
a g e
n g p o
d to a
im a g e
w ill b
w h e n t
ill n o w
m b e rs
s ta n t ra
is e d to
d . S o th
la r ity is
p o s itiv e
w ill b e c
e c o m e d
h e c e n te r p o in t is la r g e a
b e th e la r g e s t. H o w e v e r,
w ill d o m in a te th e r e s u lt.
is e d to th e p o w e r Q + 1 a
th e p o w e r Q . T h a t c o n s ta
e r a tio is ju s t th a t v a lu e .
n d its n e ig
th e e x p o n e
T h e n u m e
n d th e d e n
n t is th e v a
h b o rs a re
n t is n o w
ra to r c a n
o m in a to r
lu e o f th e
u s e d , th e la r g e n u m b e r s in th e c a s e o f th e s a lt
p o w e r, th u s th e n o is e w ill o v e r p o w e r th e r e s u lt.
o m e v e r y lig h t. T h e o p p o s ite is tr u e fo r p e p p e r
a rk .
− 1 , th e v a lu e o f th e n u m e r a to r a t a n y lo c a tio n is e q u a l to th e
e ls in th e n e ig h b o r h o o d (m n ). T h e te r m s o f th e s u m in th e d e 1 d iv id e d b y in d iv id u a l p ix e l v a lu e s in th e n e ig h b o r h o o d . F o r
3 × 3 e n ig h b o r h o o o d , th e r e s p o n s e o f th e fi lte r w h e n Q = − 1 is :
+ · · · 1 / p 9 ] w h e r e t h e p ’s a r e t h e p i x e l v a l u e s i n t h e n e i g h b o r h o o d .
l v a lu e s w ill te n d to p r o d u c e lo w fi lte r r e s p o n s e s , a n d v ic e v e r s a .
, th e fi lte r is c e n te r e d o n a la r g e s p ik e s u r r o u n d e d b y z e r o s , th e
e a lo w o u tp u t, th u s r e d u c in g th e e ffe c t o f th e s p ik e .
( e ) In a c o n s ta n t a r e a , th e fi lte r r e tu r n s th e v a lu e o f th e p ix e ls in th e a r e a , in d e p e n d e n tly o f th e v a lu e o f Q .
Problem 5.12
A b a n d p a s s fi lte r is o b ta in e d b y s u b tr a c tin g th e c o r r e s p o n d in g b a n d r e je c t fi lte r
fro m 1 :
H
(u ,v ) = 1 − H
B P
B R
(u ,v ).
T h e n :
( a ) Id e a l b a n d p a s s fi lte r :
⎧
⎨ 0
H
IB P
(u ,v ) =
⎩
1
0
i f D ( u , v ) < D 0 − W2
i f D 0 − W2 ≤ D ( u , v ) ≤ D
D ( u , v ) > D 0 + W2
0
+
W
2
.
C H A P T E R 5 . P R O B L E M S O L U T IO N S
9 0
( b ) B u tte r w o r th b a n d p a s s fi lte r :
H
(u ,v )
B B P
=
1 −
1
&
&
D (u ,v )W
2 ( u ,v ) − D
D
=
&
0
'2
0
n
2
n
2
D (u ,v )W
D 2 (u ,v )− D
1 +
'2
D (u ,v )W
D 2 (u ,v )− D
1 +
'2
0
.
n
2
( c ) G a u s s ia n b a n d p a s s fi lte r :
⎡
H G B P (u ,v )
⎢
1 − ⎣1 − e
=
=
2
e
2
D 2 (u ,v )− D 2
0
D (u ,v )W
1
−
−
⎤
2
D 2 (u ,v )− D 2
0
D (u ,v )W
1
⎥
⎦
2
Problem 5.13
W e u s e h ig h p a s s fi lt e r s t o c o n s t r u c t n o t c h fi lt e r s , a s in d ic a t e d in E q . ( 4 .1 0 .2 ) .
T h e id e a l n o tc h r e je c t fi tte r is g iv e n b y
(3
H IN R (u ,v ) =
H
(u ,v )H
k
− k
(u ,v )
k = 1
w h e re
H
k
(u ,v ) =
0
if D
if D
1
k
k
(u ,v ) ≤ D
(u ,v ) > D
0
0
w ith
D
in w h ic h (u
k
,v
k
k
(u − M / 2 − u
(u ,v ) =
k
)2 + (v − N / 2 − v
k
)
2
) a r e th e c e n te r s o f th e n o tc h e s . F o r th e G a u s s ia n fi lte r,
H
k
(u ,v ) = 1 − e
− D
2
k
(u ,v )/ 2 D
0
2
a n d th e n o tc h r e je c t fi lte r is g iv e n b y
(3
H G N R (u ,v ) =
k = 1
$
1 − e
− D
2
k
(u ,v )/ 2 D
0
2
%$
1 − e
− D
2
− k
(u ,v )/ 2 D
0
2
%
.
9 1
Problem 5.14
W e p r o c e e d a s fo llo w s :
F (u ,v )
1
=
− 1
1
=
− j 2 π (u x + v y )
f (x ,y )e
A s in (u
− j 2 π (u x + v y )
x + v 0 y )e
0
d x d y
d x d y .
− 1
U s in g th e e x p o n e n tia l d e fi n itio n o f th e s in e fu n c tio n ,
1
s in θ =
e
j θ
y )
− e
2 j
− e
− j θ
g iv e s u s
F (u ,v )
1
− j A
2
=
⎡
− j A
⎣
2
⎡
j A
⎣
2
=
j (u
e
− 1
x + v
0
− j (u
x + v
0
0
y )
− j 2 π (u x + v y )
e
d x d y
⎤
1
e
1
0
j 2 π (u
0
x / 2 π + v
0
y / 2 π )
e
d x d y ⎦−
− j 2 π (u x + v y )
− 1
e
⎤
− j 2 π (u
0
x / 2 π + v
0
y / 2 π )
e
− j 2 π (u x + v y )
d x d y ⎦.
− 1
T h e s e a re th e F o u r ie r tr a n s fo r m s o f th e fu n c tio n s
j 2 π (u
1 × e
x / 2 π + v 0 y / 2 π )
0
a n d
1 × e
r e s p e c tiv e ly . T h e F o u r ie r tr a n s f
th e e x p o n e n tia ls s h ift th e o r ig in
T a b le 4 .3 . T h u s ,
&
− j A
F (u ,v ) =
δ u −
2
− j 2 π (u
0
x / 2 π + v 0 y / 2 π )
o r m o f th e 1 g iv e s a n im p u ls e a t th e o r ig in , a n d
o f t h e im p u ls e , a s d is c u s s e d in S e c t io n 4 .6 .3 a n d
u
2 π
0
,v −
v
2 π
0
!
− δ
u +
u
2 π
0
,v +
v
0
!'
.
2 π
Problem 5.15
F r o m E q . ( 5 .4 - 1 9 )
σ
2
=
1
(2 a + 1 ) (2 b + 1 )
g (γ ) − w η (γ ) −
g − w η
2
C H A P T E R 5 . P R O B L E M S O L U T IO N S
9 2
w h e re “ γ ” in d ic a te s te r m s a ffe c te d b y th e s u m m a tio n s . L e ttin g K = 1 / (2 a +
1 )(2 b + 1 ), ta k in g th e p a r tia l d e r iv a tiv e o f σ 2 w ith re s p e c t to w a n d s e ttin g th e
r e s u lt e q u a l to z e r o g iv e s
@ σ
@ w
2
=
=
g (γ ) − w η (γ ) − g + w η
2
K
− g (γ )η (γ ) + g (γ )η + w η
K
2
− η (γ ) + η
(γ ) − w η (γ )η +
g η (γ ) − g η − w η η (γ ) + w η
=
− g η + g η + w η
=
− g η + g η + w
=
2
2
− w η
"
2
η
2
− η
= 0
2
+ g η − g η − w η
#
2
+ w η
2
0
w h e r e , fo r e x a m p le , w e u s e d th e fa c t th a t
1
g (γ )η (γ ) = g η .
(2 a + 1 ) (2 b + 1 )
S o lv in g fo r w g iv e s u s
g η − g η
w =
η
2
− η
2
.
F in a lly , in s e r tin g th e v a r ia b le s x a n d y ,
g (x ,y )η (x ,y ) − g (x ,y )η (x ,y )
w (x ,y ) =
η
2
(x ,y ) − η
2
(x ,y )
w h ic h a g r e e s w it h E q . ( 5 .4 - 2 1 ) .
Problem 5.16
F r o m E q . ( 5 .5 - 1 3 ) ,
1
g (x ,y ) =
f (α ,β )h (x − α ,y − β ) d α d β .
− 1
It is g iv e n th a t f (x , y ) = δ (x − a ), s o f (α , β ) = δ (α − a ). T h e n , u s in g th e im p u ls e
r e s p o n s e g iv e n in th e p r o b le m s ta te m e n t,
g (x ,y ) =
1
− 1
δ (α − a )e
−
$
(x − α )2 +
(
y − β
)
2
%
d α d β
9 3
$
1
=
1
=
δ (α − a )e
=
e
2
(x − α )
] e
(
−
[
2
(x − a )
[
−
δ (α − a )e
2
e
(
−
y − β
2
)
d α d β
$
1
] d α
$
1
]
(x − α )
%
2
)
y − β
− 1
− 1
−
[
−
(
−
e
y − β
)
2
%
d β
− 1
%
d β
− 1
w h e r e w e u s e d th e fa c t th a t th e in te g r a l o f th e im p u ls e is n o n z e r o o n ly w h e n
α = a . N e x t, w e n o te th a t
$
1
(
−
e
)
y − β
2
%
d β =
− 1
$
1
e
(
−
)
β − y
2
%
d β
− 1
w h ic h is in th e fo r m o f a c o n s ta n t tim e s a G a u s s ia n d e n s ity w ith v a r ia n c e σ
p
1 / 2 o r s ta n d a rd d e v ia tio n σ = 1 / 2 . In o th e r w o rd s ,
$
e
−
(
β − y
)
2
⎡
%
2 π (1 / 2 ) ⎣
=
1
− (1 / 2 )
e
(
β − y ) 2
(1 / 2 )
2
=
⎤
⎦.
2 π (1 / 2 )
T h e in te g r a l fr o m m in u s to p lu s in fi n ity o f th e q u a n tity in s id e th e b r a c k e ts is 1 ,
s o
p
2
π e − [ (x − a ) ]
g (x ,y ) =
w h ic h is a b lu r r e d v e r s io n o f th e o r ig in a l im a g e .
Problem 5.17
F o llo w in g th e im a g e c o o r d in a te c o n v e n tio n in th e b o o k , v e r tic a l m o tio n is in th e
x - d ire c tio n a n d h o r iz o n ta l m o tio n is in th e y - d ire c tio n . T h e n , th e c o m p o n e n ts
o f m o tio n a r e a s fo llo w s :
x
0
a t
T 1
(t ) =
a
0 ≤ t ≤ T 1
T 1 < t ≤ T
1
+ T
2
a n d
y 0 (t ) =
0
b (t − T 1 )
T 1
0 ≤ t ≤ T 1
T 1 < t ≤ T
1
+ T 2 .
C H A P T E R 5 . P R O B L E M S O L U T IO N S
9 4
T h e n , s u b s t it u t in g t h e s e c o m p o n e n t s o f m o t io n in t o E q . ( 5 .6 - 8 ) y ie ld s
T
H (u ,v )
T 1 + T
1
=
− j 2 π [u a t / T 1 ]
e
e
0
T
1
π u a
T
− j π u a
s in (π u a )e
2
− j 2 π u a
+ e
e
T
1
d t
1
2
d t
2
− j 2 π u a
+ e
− j 2 π v b (t − T 1 )/ T
1
T
T
=
− j π u a
s in (π u a )e
π u a
=
− j 2 π [u a + v b (t − T 1 )/ T 2 )
T 1 + T
T
=
2
d t +
− j 2 π v b τ / T
e
2
d τ
0
1
− j π u a
s in (π u a )e
π u a
T
− j 2 π u a
+ e
2
π v b
s in (π v b )e
− j π v b
w h e r e in th e th ir d lin e w e m a d e th e c h a n g e o f v a r ia b le s τ = t − T 1 . T h e b lu r r e d
im a g e is th e n
g (x ,y ) = = − 1 [H (u ,v )F (u ,v )]
w h e re F (u ,v ) is th e F o u r ie r tr a n s fo r m o f th e in p u t im a g e .
Problem 5.18
F o llo w in g t h e p r o c e d u r e in S e c t io n 5 .6 .3 ,
T
H (u ,v )
e
− j 2 π u x
e
− j 2 π u
e
− j π u a t
=
(t )
0
d t
0
T
=
[
(1 / 2 )a t
2
]d t
0
T
=
2
d t
0
T
c o s (π u a t 2 ) − j s in (π u a t 2 ) d t
=
)
=
0
T
2
2 π u a T
w h e re
*
C (z ) =
a n d
*
S (z ) =
p
C (
2
z
2 π
T
2
π
π u a T ) − j S (
c o s t 2 d t
0
z
0
s in t 2 d t .
p
π u a T )
9 5
T h e s e a r e F r e s n e l c o s in e a n d s in e in te g r a ls . T h e y c a n b e fo u n d , fo r e x a m p le ,
th e H a n d b o o k o f M a th e m a tic a l F u n c tio n s , b y A b r a m o w itz , o r o th e r s im ila r r e fe re n c e .
Problem 5.19
A b a s ic a p p r o a c h fo r re s to r in g a r o
a g e fr o m r e c ta n g u la r to p o la r c o o
d im e n s io n a l u n ifo r m m o tio n b lu r
c u s s e d in th is c h a p te r fo r h a n d lin g
b e a p p lie d to th e p r o b le m . T h e im
o rd in a te s a fte r re s to r a tio n . T h e m
w ith r e c ta n g u la r c o o r d in a te s (x , y )
c o o rd in a te s (r ,θ ), w h e re
ta tio n a lly b lu r r e d im a g e is to c o n v e r t th e
r d in a te s . T h e b lu r w ill th e n a p p e a r a s o
a lo n g th e θ - a x is . A n y o f th e te c h n iq u e s
u n ifo r m b lu r a lo n g o n e d im e n s io n c a n t
a g e is th e n c o n v e r te d b a c k to r e c ta n g u la r
a th e m a tic a l s o lu tio n is s im p le . F o r a n y p
w e g e n e r a te a c o r re s p o n d in g p ix e l w ith p
r =
x
2
a n d
θ = ta n
A
a
B
w
w
S
d is p la y
x is a n d
e c a u s e
e c a n u
ith x =
e c tio n s
o f th e r e s u ltin g im a g
w o u ld , in a d d itio n , a
th e e x te n t o f th e ro t
s e th e s a m e s o lu tio n
θ a n d y = r to o b t
5 .7 t h r o u g h 5 .9 t h e n
e w
p p e
a tio
w e
a in
b e c
− 1
+ y
im n e d is h e n
c o ix e l
o la r
2
!
y
x
.
o u ld s h o w a n im a g e th a
a r d is to r te d d u e to th e
n a l b lu r is k n o w n (it is
u s e d fo r u n ifo r m lin e a r
th e tr a n s fe r fu n c tio n .
o m e a p p lic a b le .
t is b
c o o r
g iv e
m o t
A n y
lu r r e d
d in a te
n a s π
io n (S e
o f th e
a lo n g
c o n v
/ 8 ra
c tio n
m e th
th e θ e r s io n .
d ia n s ),
5 .6 .3 ) ,
o d s in
Problem 5.20
M e a s u r e th e a v e r a g e v a lu e o f th e b a c k g r o u n d . S e t a ll p ix e ls in th e im a g e , e x c e p t th e c r o s s h a ir s , to th a t in te n s ity v a lu e . D e n o te th e F o u r ie r tr a n s fo r m o f
th is im a g e b y G (u ,v ). B e c a u s e th e c h a r a c te r is tic s o f th e c r o s s h a ir s a re g iv e n
w ith a h ig h d e g re e o f a c c u r a c y , w e c a n c o n s tr u c t a n im a g e o f th e b a c k g r o u n d
(o f th e s a m e s iz e ) u s in g th e b a c k g r o u n d in te n s ity le v e ls d e te r m in e d p r e v io u s ly .
W e th e n c o n s tr u c t a m o d e l o f th e c r o s s h a ir s in th e c o r r e c t lo c a tio n (d e te r m in e d
fr o m th e g iv e n im a g e ) u s in g th e d im e n s io n s p r o v id e d a n d in te n s ity le v e l o f th e
c r o s s h a ir s . D e n o te b y F (u ,v ) th e F o u r ie r tr a n s fo r m o f th is n e w im a g e . T h e
r a tio G (u , v )/ F (u , v ) is a n e s tim a te o f th e b lu r r in g fu n c tio n H (u , v ). In th e lik e ly
e v e n t o f v a n is h in g v a lu e s in F (u , v ), w e c a n c o n s tr u c t a r a d ia lly - lim ite d fi lte r u s in g t h e m e t h o d d is c u s s e d in c o n n e c t io n w it h F ig . 5 .2 7 . B e c a u s e w e k n o w F (u , v )
a n d G (u , v ), a n d a n e s tim a te o f H (u , v ), w e c a n r e fi n e o u r e s tim a te o f th e b lu r r in g fu n c t io n b y s u b s t it u t in g G a n d H in E q . ( 5 .8 - 3 ) a n d a d ju s t in g K t o g e t a s
C H A P T E R 5 . P R O B L E M S O L U T IO N S
9 6
c lo s e a s p o s s ib le to a g o o d r e s u lt fo r F (u , v ) (th e r e s u lt c a n b e e v a lu a te d v is u a lly
b y ta k in g th e in v e r s e F o u r ie r tr a n s fo r m ). T h e r e s u ltin g fi lte r in e ith e r c a s e c a n
th e n b e u s e d to d e b lu r th e im a g e o f th e h e a r t, if d e s ir e d .
Problem 5.21
T h e k e y to s o lv in g th is p r o b le m is to r e c o g n iz e th a t th e g iv e n fu n c tio n ,
x
h (x ,y ) =
2
2
+ y
2
− 2 σ
4
σ
x 2 + y 2
2 σ 2
−
e
is t h e t h e s e c o n d d e r iv a t iv e ( L a p la c ia n ) o f t h e fu n c t io n ( s e e S e c t io n 3 .6 .2 r e g a r d in g th e L a p la c ia n )
x 2 + y 2
2 σ 2
−
s (x ,y ) = e
.
T h a t is ,
r
2
[s (x ,y )]
2
@
=
2
x
=
@ 2 s (x ,y )
s (x ,y )
+
2
@ x
@ y 2
+ y 2 − 2 σ 2 − x 2 + 2y 2
2 σ
e
.
σ 4
( T h is r e s u lt is d e r iv e d in S e c t io n 1 0 .2 .6 ) . S o , it fo llo w s t h a t
H (u ,v )
=
=
h (x ,y )
=
=
2
r
s (x ,y ) .
B u t, w e k n o w fr o m th e s ta te m e n t o f P r o b le m 4 .2 6 (a ) th a t
=
r
2
s (x ,y )
2
= − 4 π
2
(u
2
+ v
)F (u ,v )
w h e re
F (u ,v )
=
=
T h
a G
g iv
a re
e re
a u
e n
th
fo re
s s ia
in e
e re
, w e
n fu n
n tr y
v e rs e
h a v e
c tio n
1 3 o f
o f th
re
.
T
e
d u c e d t
F ro m th
a b le 4 .3
e n tr y in
=
e
−
h e
e b
(n
th
p ro
a s ic
o te
e ta
x 2 + y 2
2 σ 2
=
s (x ,y )
=
e
b le m to
fo r m o f
th a t (x ,y
b le ), w e
= 2 π σ
2
e
−
c o
th
)
h a
x 2 + y 2
2 σ 2
.
m p u tin g th e F o u r ie r tr a n s fo r m o f
e G a u s s ia n F o u r ie r tr a n s fo r m p a ir
a n d (u , v ) in th e p r e s e n t p r o b le m
v e
− 2 π
2
σ
2
(u
2
+ v
2
)
9 7
s o w e h a v e th e fi n a l r e s u lt
H (u ,v )
2
− 4 π
=
3
− 8 π
=
2
+ v
2
(u
2
σ
2
(u
− 4 π
=
2
(u
)F (u ,v )
2
+ v
2
2
+ v
2
2 π σ
)
2
− 2 π
)e
2
σ
2
− 2 π
e
2
(u
2
σ
2
+ v
2
(u
2
+ v
)
)
a s d e s ire d . K e e p in m in d th a t th e p re c e d in g d e r iv a tio n s a re b a s e d o n a s s u m in g
c o n tin u o u s v a r ia b le s . A d is c r e te fi lte r is o b ta in e d b y s a m p lin g th e c o n tin u o u s
fu n c tio n .
Problem 5.22
T h is is a s im p le p lu g in p r o b le m . Its p u r p o s e is to g a in fa m ilia r ity w ith th e v a r io u s t e r m s o f t h e W ie n e r fi lt e r. F r o m E q . ( 5 .8 - 3 ) ,
jH ( u , v ) j2
1
H W (u ,v ) =
H ( u , v ) jH ( u , v ) j2 + K
w h e re
jH (u , v )j2
=
=
∗
(u ,v )H (u ,v )
H
2
(u ,v )
6
6 4 π
=
T h e n ,
H
⎡
H W (u ,v ) = − ⎣
3
− 8 π
6
6 4 π
2
+ v
2
(u
2
(u
2
(u
2
σ
4
σ
4
σ
+ v
2
+ v
2
)2 e
− 2 π
)e
2
− 4 π
)2 e
2
− 4 π
2
σ
2
(u
2
2
(u
2
+ v
σ
2
σ
2
(u
2
+ v
2
+ v
2
)
.
⎤
)
)
+ K
⎦.
Problem 5.23
T h is a ls o is a s im p le p lu g in p r o b le m , w h o s e p u r p o s e is th e s a m e a s th e p r e v io u s
p r o b le m . F r o m E q . ( 5 .9 - 4 )
H
C
(u ,v )
∗
H
=
(u ,v )
jH (u , v )j + γ jP (u , v )j2
=
2
−
2
8 π
6 4 π
4
σ
4
(u
2
2
(u
2
+ v
2
)2 e
σ
w h e re P (u ,v ) is th e F o u r ie r tr a n s fo r m o f th
T h is is a s fa r a s w e c a n r e a s o n a b ly c a r r y th is
o u t to s tu d e n ts th a t a fi lte r in th e fr e q u e n c y
is d is c u s s e d in S e c t io n 4 .9 .4 [s e e E q . ( 4 .9 - 5 ) ].
fo r P (u , v ) h e r e w o u ld o n ly in c r e a s e th e n u m
n o t h e lp in s im p lify in g th e e x p r e s s io n .
+ v
− 4 π
2
− 2 π σ
2
)e
σ
2
(u
2
+ v
2
2
(u
)
2
+ v
2
)
+ γ jP ( u , v ) j2
e L a p la c ia n o p e r a to r [E q . (5
p r o b le m . It is w o r th w h ile p o
d o m a in fo r th e L a p la c ia n o p
H o w e v e r, s u b s titu tin g th a t s o
b e r o f te r m s in th e fi lte r a n d
.9 - 5 ) ].
in tin g
e ra to r
lu tio n
w o u ld
C H A P T E R 5 . P R O B L E M S O L U T IO N S
9 8
Problem 5.24
B e c a u
(5 .5 - 1
o f th e
a n d n
th e c o
s e th e s
7 ) h o ld s
s y s te m
o is e a re
m p le te
y s te
. F u
fi rs t
u n c
re s p
m is
r th e
to F
o rre
o n s
a s s u m e d
r m o re , w e
(u ,v ) a n d
la te d . T h e
e . F ir s t, u s
G
1
lin e a
c a n
th e n
s u m
in g o
r a n d p
u s e s u p
to N (u
o f th e
n ly F (u
o s
e r
,v
tw
,v
itio
p o s
) b e
o in
),
n in v a
itio n a
c a u s e
d iv id u
r ia
n d
w e
a l
n t,
o b
k n
re s
it fo llo w s
ta in th e r
o w th a t th
p o n s e s th
th a t E
e s p o n
e im a
e n g iv
q .
s e
g e
e s
(u ,v ) = H (u ,v )F (u ,v )
a n d
jG
1
(u , v )j2 = jH (u , v )j2 jF (u , v )j2 .
T h e n , u s in g o n ly N (u , v ),
G
(u ,v ) = N (u ,v )
2
a n d ,
jG
2
( u , v ) j2 = jN ( u , v ) j2
s o th a t,
jG ( u , v ) j2
jG
=
=
1
( u , v ) j 2 + jG
2
(u , v )j2
jH ( u , v ) j2 jF ( u , v ) j2 + jN ( u , v ) j2 .
Problem 5.25
(a ) It is g iv e n th a t
F^ ( u , v )
2
= jR ( u , v ) j2 jG ( u , v ) j2 .
F r o m P r o b le m 5 .2 4 ( r e c a ll t h a t t h e im a g e a n d n o is e a r e a s s u m e d t o b e u n c o r r e la te d ),
2
F^ ( u , v ) = j R ( u , v ) j 2 j H ( u , v ) j 2 j F ( u , v ) j 2 + j N ( u , v ) j 2 .
F o rc in g
F^ ( u , v )
2
t o e q u a l jF (u , v )j2 g iv e s
R (u ,v ) =
jF (u , v )j2
1 / 2
.
jH ( u , v ) j2 jF ( u , v ) j2 + jN ( u , v ) j2
(b )
F^ ( u , v )
=
R (u , v )G (u , v )
jF (u , v )j2
=
=
jH ( u , v ) j2 jF ( u , v ) j2 + jN ( u , v ) j2
⎤1 / 2
⎡
1
⎥
⎢
⎦ G (u ,v )
⎣
jN ( u ,v ) j2
2
j H ( u , v ) j + jF ( u ,v ) j2
1 / 2
G (u ,v )
9 9
a n d , b e c a u s e jF (u , v )j2 = S
f
(u , v ) a n d jN (u , v )j2 = S
η
⎤1
/ 2
⎡
⎢
F^ ( u , v ) = ⎣
1
jH ( u , v ) j2 +
S
η
S
f
(u ,v )
(u ,v )
⎥
⎦
(u ,v ),
G (u ,v ).
Problem 5.26
O n
a b
a s
a s
(4 )
e p
lu r
s m
c lo
U s
o s s ib le s o lu tio n : (1 ) P e r fo
re d im a g e o f a b r ig h t, s in g
a ll a s p o s s ib le in th e fi e ld
s e ly a s p o s s ib le . (3 ) T h e F
e a W ie n e r fi lte r a n d v a r y
r
le
o
o
K
m im a g e a v e
s ta r to s im u
f v ie w o f th e
u r ie r tr a n s fo
u n til th e s h
r a g in g
la te a n
te le s c
r m o f
a rp e s t
to re d u c e n o is
im p u ls e (th e s
o p e to s im u la t
th is im a g e w ill
im a g e p o s s ib le
e . (2
ta r s
e a n
g iv e
is o
) O b ta in
h o u ld b e
im p u ls e
H (u ,v ).
b ta in e d .
Problem 5.27
T h e b a s ic id e a b e h in d th is p r o b le m is to u s e th e c a m e r a a n d r e p r e s e n ta tiv e c o in s
to m o d e l th e d e g r a d a tio n p r o c e s s a n d th e n u tiliz e th e r e s u lts in a n in v e r s e fi lte r
o p e r a tio n . T h e p r in c ip a l s te p s a r e a s fo llo w s :
1 . S e le c t c o in s a s c lo s e a s p o s s ib le in s iz e a n d c o n te n t a s th e lo s t c o in s . S e le c t
a b a c k g r o u n d th a t a p p r o x im a te s th e te x tu re a n d b r ig h tn e s s o f th e p h o to s
o f th e lo s t c o in s .
2 . S e t u p th e m u s e u m p h o to g r a p h ic c a m e r a in
b le to g iv e im a g e s th a t r e s e m b le th e im a g e s
p a y in g a tte n tio n to illu m in a tio n ). O b ta in a
e x p e r im e n ta tio n , o b ta in a T V c a m e r a c a p a
s e m b le th e te s t p h o to s . T h is c a n b e d o n e
a n im a g e p r o c e s s in g s y s te m a n d g e n e r a tin g
u s e d in th e e x p e r im e n t.
a g e o m e tr y a s c lo s e a s p o s s io f th e lo s t c o in s (th is in c lu d e s
fe w te s t p h o to s . T o s im p lify
b le o f g iv in g im a g e s th a t r e b y c o n n e c tin g th e c a m e r a to
d ig ita l im a g e s , w h ic h w ill b e
3 . O b ta in s e ts o f im a g e s o f e a c h c o in w ith d iffe r e n t le n s s e ttin g s . T h e r e s u ltin g im a g e s s h o u ld a p p r o x im a te th e a s p e c t a n g le , s iz e (in r e la tio n to
th e a r e a o c c u p ie d b y th e b a c k g r o u n d ), a n d b lu r o f th e p h o to s o f th e lo s t
c o in s .
4 . T h e le
fo r th e
th e c o
u n ifo r
n s s e
c o rr
in a n
m b a
ttin
e s p
d b
c k g
g fo r e a c h
o n d in g im
a c k g ro u n d
ro u n d , o r
im
a g e
a n
o th
a g
o
d
e r
e in
f a lo
r e p la
m e c
(3 )
s t c
c e
h a n
is a m o
o in . F o
th e m w
is m to
d e l
r e a
ith
a p p
o f th e b
c h s u c h
a s m a ll,
r o x im a t
lu r r in g p r o c e
s e ttin g , re m o
b r ig h t d o t o n
e a n im p u ls e
s s
v e
a
o f
C H A P T E R 5 . P R O B L E M S O L U T IO N S
1 0 0
(a)
(b)
(c)
F ig u r e P 5 .2 8
lig h t. D ig itiz e th e im p u ls e . Its F o u r ie r tr a n s fo r m is th e tr a n s fe r fu n c tio n o f
th e b lu r r in g p r o c e s s .
5 . D ig itiz e e a c h (b lu r r e d ) p h o to o f a lo s t c o in , a n d o b ta in its F o u r ie r tr a n s fo r m . A t th is p o in t, w e h a v e H (u ,v ) a n d G (u ,v ) fo r e a c h c o in .
6 . O b ta in a n a p p r o x im a tio n to F (u , v ) b y u s in g a W ie n e r fi lte r. E q u a tio n
(5 .8 - 3 ) is p a r tic u la r ly a ttr a c tiv e b e c a u s e it g iv e s a n a d d itio n a l d e g r e e o f
fre e d o m (K ) fo r e x p e r im e n tin g .
7 . T
s
b
o
h e in v e r s e
to re d im a g
a s ic s te p s
b ta in a c c e
F o u
e fo
w ith
p ta b
r ie
r a
v a
le
r tra n
c o in .
r io u s
r e s u lt
s fo r m
In g e
d iffe r
s in a
o f e a c h a p p ro
n e r a l, s e v e r a l
e n t s e ttin g s a n
p r o b le m s u c h
x im
e x p
d p
a s t
a t i o n F^ ( u , v ) g i v e s t h e r e e r im e n ta l p a s s e s o f th e s e
a r a m e te r s a re re q u ire d to
h is .
Problem 5.28
T h
a x
1 8
1 3
e
is
0 ◦
5 ◦
s o lu
is ρ
. In
. T h
tio n s a re s h o w n in th e
a n d th e v e r tic a l a x is is
(b ) th e fa t lo b e s o c c u r
e in te n s ity a t th a t p o in t
fo
θ
a t
is
llo w in g
, w ith
4 5 ◦ a n
d o u b le
θ
fi g u re s . In e a c
= 0 ◦ a t th e b
d th e s in g le p o
th e in te n s ity o
h
o t
in
f a
fi g u
to m
t o f
ll o t
re th
a n d
in te r
h e r p
e h o r iz o n ta l
g o in g u p to
s e c tio n is a t
o in ts .
Problem 5.29
B e c a u s e f (x , y ) is r o ta tio n a lly s y m m e tr ic , its p r o je c tio n s a r e th e s a m e fo r a ll a n g le s , s o a ll w e h a v e t o d o is o b t a in t h e p r o je c t io n fo r θ = 0 ◦ . E q u a t io n ( 5 .1 1 - 3 )
1 0 1
th e n b e c o m e s
<
= g (ρ ,θ )
f (x ,y )
1
1
− 1
1
− 1
=
=
f (ρ ,y )d y
− 1
=
f (x ,y )δ (x − ρ )d x d y
1
A
e
− 1
=
A e
− y
1
2
− ρ
2
(− ρ
2
)
− y
e
2
d y
d y .
− 1
B e c a u s e
1
1
2
2 π σ
it fo llo w s b y le ttin g σ
2
2
− z
e
2
/ 2 σ
d z = 1
− 1
= 1 / 2 in th is e q u a tio n th a t,
p
T h e n ,
1
1
2
− z
e
π
d z = 1 .
− 1
1
e
− y
2
d y =
p
π
− 1
a n d
g (ρ ,θ ) = A
p
π e
− ρ
2
.
Problem 5.30
( a ) F r o m E q . ( 5 .1 1 - 3 ) ,
<
f (x ,y )
= g (ρ ,θ )
=
=
=
=
1
1
− 1
1
− 1
1
− 1
1
− 1
1
− 1
− 1
1
0
f (x ,y )δ (x c o s θ + y s in θ − ρ )d x d y
δ (x ,y )δ (x c o s θ + y s in θ − ρ )d x d y
1 × δ (0 − ρ )d x d y
if ρ = 0
o th e r w is e .
C H A P T E R 5 . P R O B L E M S O L U T IO N S
1 0 2
w h e r e th e th ir d s te p fo llo w s fr o m th e fa c t th a t δ (x , y ) is z e r o if x a n d / o r y a r e n o t
z e ro .
( b ) S im ila r ly , s u b s t it u t in g in t o E q . ( 5 .1 1 - 3 ) ,
<
= g (ρ ,θ )
f (x ,y )
1
1
− 1
1
− 1
1
− 1
1
− 1
1
− 1
− 1
=
=
=
f (x ,y )δ (x c o s θ + y s in θ − ρ )d x d y
δ (x − x
1 × δ (x
0
0
,y − y 0 )δ (x c o s θ + y s in θ − ρ )d x d y
c o s θ + y
0
s in θ − ρ )d x d y .
F r o m th e d e fi n itio n o f th e im p u ls e , th is r e s u lt is 0 u n le s s
ρ = x
c o s θ + y
0
0
s in θ
w h ic h is th e e q u a tio n o f a s in u s o id a l c u r v e in th e ρ θ - p la n e .
Problem 5.31
( a ) F r o m S e c tio n 2 .6 , w e k n o w th a t a n o p e r a to r, O , is lin e a r if O (a f 1 + b f 2 ) =
a O ( f 1 ) + b O ( f 2 ). F r o m t h e d e fi n it io n o f t h e R a d o n t r a n s fo r m in E q . ( 5 .1 1 - 3 ) ,
O (a f
+ b f 2 )
1
=
=
1
1
− 1
− 1
1
(a f
1
a
− 1
1
f 1 δ (x c o s θ + y s in θ − ρ )d x d y
− 1
+ b
=
+ b f 2 )δ (x c o s θ + y s in θ − ρ )d x d y
1
1
− 1
− 1
a O ( f 1 ) + b O ( f 2 )
f 2 δ (x c o s θ + y s in θ − ρ )d x d y
th u s s h o w in g th a t th e R a d o n tr a n s fo r m is a lin e a r o p e r a tio n .
( b ) L e t p = x − x 0 a n d q = y − y 0 . T h e n d p = d x a n d d q = d y . F r o m E q . ( 5 .1 1 - 3 ) ,
th e R a d o n tr a n s fo r m o f f (x − x 0 ,y − y 0 ) is
g (ρ ,θ )
=
=
=
=
1
1
− 1
1
− 1
1
− 1
1
− 1
1
− 1
− 1
g (ρ − x
0
f (x − x
0
,y − y 0 )δ (x c o s θ + y s in θ − ρ )d x d y
f (p ,q ) δ
(p + x
f (p ,q ) δ
p c o s θ + q s in θ − (ρ − x
c o s θ − y
0
0
) c o s θ + (q + y 0 ) s in θ − ρ ) d p d q
s in θ ,θ ).
0
c o s θ − y
0
s in θ ) d p d q
1 0 3
( c ) F r o m C h a p t e r 4 ( P r o b le m 4 .1 1 ) , w e k n o w t h a t t h e c o n v o lu t io n o f t w o fu n c tio n f a n d h is d e fi n e d a s
c (x ,y )
=
f (x ,y )
=
h (x ,y )
1
1
− 1
− 1
f (α ,β )h (x − α ,y − β )d α d β .
W e w a n t to s h o w th a t < f c g = < f
< f h g ,w h e r e < d e n o t e s t h e R a d o n t r a n s fo r m . W e d o t h is b y s u b s t it u t in g t h e c o n v o lu t io n e x p r e s s io n in t o E q . ( 5 .1 1 - 3 ) .
T h a t is ,
⎤
⎡ 1
1
1
1
< f c g
f (α ,β )h (x − α ,y − β )d α d β ⎦
⎣
=
− 1
− 1
− 1
− 1
× δ (x c o s θ + y s in θ − ρ )d x d y
f (α ,β )
=
α
β
⎡
⎤
× ⎣
w h e re w e u s e d
a n d th e ir v a r ia b
in g w ith th e in t
++
h (x
x +y +
= x 0 y 0h
h (x − α ,y − β )δ (x c o s θ + y s in θ − ρ )d x d y ⎦d α d β
x
y
th e s u b s c r ip ts in th e in te g r a ls fo r c la r ity b e tw e e n th e in te g r a ls
le s . A ll in te g r a ls a r e u n d e r s to o d to b e b e tw e e n − 1 a n d 1 . W o r k e g r a ls in s id e th e b r a c k e ts w ith x 0 = x − α a n d y 0 = y − β w e h a v e
− α ,y − β )δ (x c o s θ + y s in θ − ρ )d x d y
(x 0,y 0)δ (x
0
c o s θ + y
0
s in θ − [ρ − α c o s θ − β s in θ ])d x 0d y
0
= < f h g (ρ − α c o s θ − β s in θ ,θ ).
W e r
b e in
T h e
fu n c
e c o g n iz e
g w ith re
n o ta tio n
tio n o f ρ
< f c g
th e
s p e c
in th
− α c
s e c o n
t to ρ
e la s t
o s θ −
d in te g
a n d θ
lin e is
β s in θ
ra l
, it
u s e
a n
a s th e
is a fu
d to in
d θ .” T
R a d o n tr a n s fo r m o f h , b u t in s te a d o f
n c tio n o f ρ − α c o s θ − β s in θ a n d θ .
d ic a te “ th e R a d o n tr a n s fo r m o f h a s a
h e n ,
f (α ,β )
=
α
β
⎡
× ⎣
=
α
β
⎤
h (x − α ,y − β )δ (x c o s θ + y s in θ − ρ )d x d y ⎦d α d β
x
y
f (α , β )< f h g (ρ − ρ 0, θ )d α d β
C H A P T E R 5 . P R O B L E M S O L U T IO N S
1 0 4
w h e re ρ 0 =
c a n w r ite
α c o s θ + β s in θ . T h e n , b a s e d o n th e p r o p e r tie s o f th e im p u ls e , w e
< f h g (ρ − ρ 0, θ ) =
< f h g (ρ − ρ 0, θ )δ (α c o s θ + β s in θ − ρ 0)d ρ 0.
0
ρ
T h e n ,
< f c g =
β
f (α ,β )
=
α
β
⎡
× ⎣
⎤
< f h g (ρ − ρ ,θ )δ (α c o s θ + β s in θ − ρ )d ρ ⎦d α d β
0
ρ
0
0
< f h g (ρ − ρ ,θ ) ⎣
ρ
0
ρ
0
<
⎤
f (α ,β )δ (α c o s θ + β s in θ − ρ )d α d β ⎦d ρ
0
α
< f h g (ρ − ρ 0, θ )<
=
0
⎡
0
=
=
α
f (α , β ) < f h g (ρ − ρ 0, θ ) d α d β
0
β
(ρ 0, θ )d ρ
f
0
< f h g
f
w h e r e th e fo u r th s te p fo llo w s fr o m th e d e fi n itio n o f th e R a d o n tr a n s fo r m a n d th e
fi fth s te p fo llo w s fr o m th e d e fi n itio n o f c o n v o lu tio n . T h is c o m p le te s th e p r o o f.
Problem 5.32
T h e s o lu tio n is a s fo llo w s :
1
2 π
f (x ,y )
=
0
G (! ,θ )e
j 2 π ! (x c o s θ + y s in θ )
! d ! d θ
0
1
π
=
0
G (! ,θ )e
j 2 π ! (x c o s θ + y s in θ )
! d ! d θ
0
1
π
+
0
G (! ,θ )e
j 2 π ! (x c o s [θ + 1 8 0 ◦ ]+ y s in [θ + 1 8 0 ◦ )
! d ! d θ .
0
B u t G (θ + 1 8 0 ◦ ,θ ) = G (− ! ,θ ), s o th e p re c e d in g e q u a tio n c a n b e e x p re s s e d a s
π
f (x ,y )
=
0
1
− 1
w h ic h a g r e e s w it h E q . ( 5 .1 1 - 1 5 ) .
j! j G ( ! , θ ) e
j 2 π ! (x c o s θ + y s in θ )
d ! d θ
1 0 5
Problem 5.33
T h e a r g u m e n t o f fu n c t io n s in E q .( 5 .1 1 - 2 4 ) m a y b e w r it t e n a s :
r c o s (β + α − ' ) − D s in α = r c o s (β − ' ) c o s α − [r s in (β − ' ) + D ] s in α .
F r o m F ig . 5 .4 7 ,
R c o s α
0
=
R + r s in (β − ' )
R s in α
0
=
r c o s (β − ' ).
T h e n , s u b s titu tin g in th e e a r lie r e x p r e s s io n ,
r c o s (β + α − ' ) − R s in α
R s in α 0 c o s α − R c o s α 0 s in α
=
=
=
R (s in α 0 c o s α − c o s α 0 s in α )
0
R s in (α
− α )
w h ic h a g r e e s w it h E q . ( 5 .1 1 - 2 5 ) .
Problem 5.34
F r o m t h e e x p la n a t io n o f E q . ( 5 .1 1 - 1 8 ) ,
1
s (ρ ) =
j! j e
− 1
j 2 π ! ρ
d ! .
L e t ρ = R s in α , a n d k e e p in m in d th a t α / R s in α is a lw a y s p o s itiv e . T h e n ,
1
s (R s in α ) =
− 1
j! j e
j 2 π ! R s in α
d ! .
N e x t, d e fi n e th e tr a n s fo r m a tio n
!
0
! R s in α
.
α
=
T h e n ,
α
d !
R s in α
d ! =
0
a n d w e c a n w r ite
&
s (R s in α )
=
=
a s d e s ire d .
α
'2
1
R s in α
− 1
'2
&
α
s (α )
R s in α
!
0
e
j 2 π ! 0α
d !
0
Chapter 6
Problem Solutions
Problem 6.1
F r o m F ig . 6 .5 in t h e b o o k , x = 0 .4 3 a n d y = 0 .4 . S
z = 0 .1 7 . T h e s e a r e t h e t r ic h r o m a t ic c o e ffi c ie n t s .
v a lu e s X , Y , a n d Z , w h ic h a r e r e la te d to th e tr
(6 .1 - 1 ) th r o u g h (6 .1 - 3 ). N o te h o w e v e r, th a t a ll
d iv id e d b y th e s a m e c o n s ta n t, s o th e ir p e rc e n ta
c o e ffi c ie n ts a re th e s a m e a s th o s e o f th e c o e ffi c
X = 0 .4 3 , Y = 0 .4 0 , a n d Z = 0 .1 7 .
in c e x + y + z = 1 , it fo llo w s th a t
W e a r e in te r e s te d in tr is tim u lu s
ic h r o m a tic c o e ffi c ie n ts b y E q s .
th e tr is tim u lu s c o e ffi c ie n ts a r e
g e s r e la tiv e to th e tr ic h r o m a tic
ie n ts . T h e re fo re , th e a n s w e r is
Problem 6.2
D e n o te b y c th e g iv e n c o lo r, a n d le t its c o o r d in a te s b e d e n o te d b y (x
d is ta n c e b e tw e e n c a n d c 1 is
d (c ,c 1 ) =
S im ila r ly th e d is ta n c e b e tw e e n c
d (c 1 ,c 2 ) =
T h e p e rc e n ta g e p
1
o f c
1
e n ta
x a m
0 % .
lu e s
0
1
a n d c
(x
1
− x
)2 +
1
− x
y
− y
0
1 / 2
2
1
,y 0 ). T h e
.
2
2
)2 +
y
1
− y
2
2
1 / 2
.
in c is
p
T h e p e rc
s e e , fo r e
a n d p 2 =
1 0 0 % . V a
(x
0
1
=
g e p 2 o f c 2 is s im
p le , th a t w h e n c =
S im ila r ly , w h e n d
in b e tw e e n a re e
d (c 1 ,c 2 ) − d (c ,c 1 )
× 1 0 0 .
d (c 1 ,c 2 )
p ly p 2 = 1 0 0
c 1 , th e n d (c
(c ,c 1 ) = d (c
a s ily s e e n to
1 0 7
− p 1 . In
,c 1 ) = 0 a
1 , c 2 ) , it f o
fo llo w fr o
th e p re c e d in g e
n d it fo llo w s th a
llo w s th a t p 1 = 0
m th e s e s im p le
q u a
t p 1
% a
r e la
tio n w e
= 1 0 0 %
n d p 2 =
tio n s .
C H A P T E R 6 . P R O B L E M S O L U T IO N S
1 0 8
F ig u r e P 6 .3
Problem 6.3
C o n s id e r F
tr ia n g le a n
b o u n d a r y .
b o rd e r o f t
th e lin e s e g
b y th e c o lo
ig . P 6 .3 , in w h ic h c 1 , c 2 , a n d c 3 a r e t h e g iv e n v e
d c is a n a r b itr a r y c o lo r p o in t c o n ta in e d w ith in th
T h e k e y to s o lv in g th is p r o b le m is to r e a liz e th a
h e tr ia n g le is m a d e u p o f p r o p o r tio n s fr o m th e tw
m e n t th a t c o n ta in s th e p o in t. T h e c o n tr ib u tio n to
r v e r te x o p p o s ite th is lin e is 0 % .
r tic e s o f t
e tr ia n g le
t a n y c o lo
o v e r tic e s
a p o in t o n
h e c o lo r
o r o n its
r o n th e
d e fi n in g
th e lin e
T h e lin e s e g m e n t c o n n e c tin g p o in ts c 3 a n d c is s h o w n e x te n d e d (d a s h e d s e g m e n t) u n til it in te r s e c ts th e lin e s e g m e n t c o n n e c tin g c 1 a n d c 2 . T h e p o in t o f in te r s e c tio n is d e n o te d c 0 . B e c a u s e w e h a v e th e v a lu e s o f c 1 a n d c 2 , if w e k n e w
c 0 , w e c o u ld c o m p u te th e p e r c e n ta g e s o f c 1 a n d c 2 c o n ta in e d in c 0 b y u s in g th e
m e t h o d d e s c r ib e d in P r o b le m 6 .2 . L e t t h e r a t io o f t h e c o n t e n t o f c 1 a n d c 2 in c 0
b e d e n o t e d b y R 1 2 . If w e n o w a d d c o lo r c 3 t o c 0 , w e k n o w fr o m P r o b le m 6 .2 t h a t
th e p o in t w ill s ta r t to m o v e to w a r d c 3 a lo n g th e lin e s h o w n . F o r a n y p o s itio n o f
a p o in t a lo n g th is lin e w e c o u ld d e te r m in e th e p e r c e n ta g e o f c 3 a n d c 0 , a g a in , b y
u s in g t h e m e t h o d d e s c r ib e d in P r o b le m 6 .2 . W h a t is im p o r t a n t t o k e e p in m in d
th a t th e r a tio R 1 2 w ill r e m a in th e s a m e fo r a n y p o in t a lo n g th e s e g m e n t c o n n e c tin g c 3 a n d c 0 . T h e c o lo r o f th e p o in ts a lo n g th is lin e is d iffe r e n t fo r e a c h p o s itio n ,
b u t th e r a tio o f c 1 to c 2 w ill r e m a in c o n s ta n t.
S o , if w e c a n o b ta in c 0 , w e c a n th e n d e te r m in e th e r a tio R 1 2 , a n d th e p e rc e n ta g e o f c 3 , in c o lo r c . T h e p o in t c 0 is n o t d iffi c u lt to o b ta in . L e t y = a 1 2 x + b 1 2 b e
1 0 9
th e s tr a ig h t lin e c o n ta in in g p o in ts c 1 a n d c 2 , a n d y = a 3 c x + b 3 c th e lin e c o n ta in in g c 3 a n d c . T h e in te r s e c tio n o f th e s e tw o lin e s g iv e s th e c o o r d in a te s o f c 0 . T h e
lin e s c a n b e d e te r m in e d u n iq u e ly b e c a u s e w e k n o w th e c o o r d in a te s o f th e tw o
p o in t p a ir s n e e d e d to d e te r m in e th e lin e c o e ffi c ie n ts . S o lv in g fo r th e in te r s e c tio n in te r m s o f th e s e c o o rd in a te s is s tr a ig h tfo r w a rd , b u t te d io u s . O u r in te re s t
h e r e is in th e fu n d a m e n ta l m e th o d , n o t th e m e c h a n ic s o f m a n ip u la tin g s im p le
e q u a tio n s s o w e d o n o t g iv e th e d e ta ils .
A t th is ju n c tu r e w e h a v e th e p e r c e n ta g e o f c 3 a n d th e r a tio b e tw e e n c 1 a n d
c 2 . L e t th e p e r c e n ta g e s o f th e s e th r e e c o lo r s c o m p o s in g c b e d e n o te d b y p 1 , p 2 ,
a n d p 3 r e s p e c tiv e ly . W e k n o w th a t p 1 + p 2 = 1 0 0 − p 3 , a n d th a t p 1 / p 2 = R 1 2 ,
s o w e c a n s o lv e fo r p 1 a n d p 2 . F in a lly , n o te th a t th is p r o b le m c o u ld h a v e b e e n
s o lv e d th e s a m e w a y b y in te r s e c tin g o n e o f th e o th e r tw o s id e s o f th e tr ia n g le .
G o in g to a n o th e r s id e w o u ld b e n e c e s s a r y , fo r e x a m p le , if th e lin e w e u s e d in
th e p r e c e d in g d is c u s s io n h a d a n in fi n ite s lo p e . A s im p le te s t to d e te r m in e if th e
c o lo r o f c is e q u a l to a n y o f th e v e r tic e s s h o u ld b e th e fi r s t s te p in th e p r o c e d u r e ;
in th is c a s e n o a d d itio n a l c a lc u la tio n s w o u ld b e r e q u ir e d .
Problem 6.4
U s e c o lo r fi lte r s th a t
th r e e o b je c ts . W ith a
re s p o n d s to th a t w a v
c h ro m e c a m e ra . A m
fro m a c o m p u te r. If
fi lte r s w ill b e a p p r o x
re s p o n s e o f th e th re e
a r e s h a r p ly tu n e d to th e w a v e le n g th s
s p e c ifi c fi lte r in p la c e , o n ly th e o b je c
e le n g th w ill p r o d u c e a s ig n ifi c a n t r e s p
o to r iz e d fi lte r w h e e l c a n b e u s e d to c o
o n e o f th e c o lo r s is w h ite , th e n th e r e
im a te ly e q u a l a n d h ig h . If o n e o f th e
fi lte r s w ill b e a p p r o x im a te ly e q u a l a n d
o f th e
ts w h o
o n s e o
n tro l fi
s p o n s e
c o lo r s
lo w .
c o lo r s o f th e
s e c o lo r c o r n th e m o n o lte r p o s itio n
o f th e th re e
is b la c k , th e
Problem 6.5
A t th e c e n te r p o in t w e h a v e
1
2
R +
2
1
B + G =
1
2
(R + G + B ) +
1
2
G = m id g r a y +
2
1
G
w h ic h lo o k s to a v ie w e r lik e p u r e g r e e n w ith a b o o s t in in te n s ity d u e to th e a d d itiv e g r a y c o m p o n e n t.
C H A P T E R 6 . P R O B L E M S O L U T IO N S
1 1 0
C o lo r
B la c k
R e d
Y e llo w
G re e n
C y a n
B lu e
M a g e n ta
W h ite
R
1
1
G ra y
0 .5
0 .5
G
0
0
1
0
1
0
1
1
0
1
0
0
1
0
T a b le P
B
M o
0
0
2
0
2
0
1
1
1
2
1
2
0 .5
6 .6
n o R
0
5 5
5 5
0
0
0
5 5
5 5
1 2 8
M o n o G
0
0
2 5 5
2 5 5
2 5 5
0
0
2 5 5
M o n o B
0
0
0
0
2 5 5
2 5 5
2 5 5
2 5 5
1 2 8
1 2 8
Problem 6.6
F o r th e im a g e g
th a t th e R G B c o
2 5 5 re p re s e n tin
d is p la y s s h o w n
iv e n , t
m p o n
g b la c
in F ig
h e m a x im u m in te n s ity a n d s a tu r a tio n re q u ire m e n t m e a n s
e n t v a lu e s a r e 0 o r 1 . W e c a n c r e a t e T a b le P 6 .6 w it h 0 a n d
k a n d w h ite , r e s p e c tiv e ly . T h u s , w e g e t th e m o n o c h r o m e
. P 6 .6 .
Problem 6.7
T h e r e a r e 2 8 = 2 5 6 p o s s ib le v a lu e s in e a c h 8 - b it im a g e . F o r a c o lo r to b e g r a y , a ll
R G B c o m p o n e n ts h a v e to b e e q u a l, s o th e r e a r e 2 5 6 s h a d e s o f g r a y .
Problem 6.8
( a ) A ll p ix e l v a lu e s in th e R e d im a g e a r e 2 5 5 . In th e G r e e n im a g e , th e fi r s t c o lu m n
i s a l l 0 ’s ; t h e s e c o n d c o l u m n a l l 1 ’s ; a n d s o o n u n t i l t h e l a s t c o l u m n , w h i c h i s
F ig u r e P 6 .6
1 1 1
c o m p o s e d o f a l l 2 5 5 ’s . I n t h e B l u e i m a g e , t h e fi r s t r o w i s a l l 2 5 5 ’s ; t h e s e c o n d r o w
a l l 2 5 4 ’s , a n d s o o n u n t i l t h e l a s t r o w w h i c h i s c o m p o s e d o f a l l 0 ’s .
( b ) L e t t h e a x is n u m b e r in g b e t h e s a m e a s in F ig . 6 .7 in t h e b o o k . T h e n : (0 , 0 , 0 ) =
w h ite , (1 , 1 , 1 , ) = b la c k , (1 , 0 , 0 ) = c y a n , (1 , 1 , 0 ) = b lu e , (1 , 0 , 1 ) = g r e e n , (0 , 1 , 1 ) =
r e d , (0 , 0 , 1 ) = y e llo w , (0 , 1 , 0 ) = m a g e n ta .
( c ) T h e o n e s th a t d o n o t c o n ta in th e b la c k o r w h ite p o in t a r e fu lly s a tu r a te d . T h e
o th e r s d e c r e a s e in s a tu r a tio n fr o m th e c o r n e r s to w a r d th e b la c k o r w h ite p o in t.
C o lo r
B la c k
R e d
Y e llo w
G re e n
C y a n
B lu e
M a g e n ta
W h ite
R
1
G ra y
0 .5
G
C
1
B
0
0
0
0
1
1
1
1
0 .5
0 .5
0 .5
0
0
1
0
1
0
1
1
0
1
0
0
1
0
1
0
0
1
1
1
0
0
T a b le P 6 .9
M
Y
1
1
1
1
0
1
0
1
0
0
1
0
1
0
0
0
0 .5
0 .5
M o n
2 5
0
0
2 5
2 5
2 5
o C
0
M o n o M
2 5 5
2 5 5
0
0
0
2 5 5
2 5 5
0
1 2 8
1 2 8
5
5
5
5
0
M o n
2 5
2 5
2 5
2 5
o Y
5
5
5
5
0
0
0
0
1 2 8
Problem 6.9
(a ) F o r th e im a g e g iv e n , th e m a x im u m in te n s ity a n d s a tu r a tio n re q u ire m e n t
m e a n s t h a t t h e R G B c o m p o n e n t v a lu e s a r e 0 o r 1 . W e c a n c r e a t e T a b le P 6 .9 u s in g
E q . ( 6 .2 - 1 ) . T h u s , w e g e t t h e m o n o c h r o m e d is p la y s s h o w n in F ig . P 6 .9 ( a ) .
( b ) T h e r e s u ltin g d is p la y is th e c o m p le m e n t o f th e s ta r tin g R G B im a g e . F r o m le ft
t o r ig h t , t h e c o lo r b a r s a r e ( in a c c o r d a n c e w it h F ig . 6 .3 2 ) w h it e , c y a n , b lu e , m a g e n ta , r e d , y e llo w , g r e e n , a n d b la c k . T h e m id d le g r a y b a c k g r o u n d is u n c h a n g e d .
F ig u r e P 6 .9
C H A P T E R 6 . P R O B L E M S O L U T IO N S
1 1 2
Problem 6.10
E q u a tio n ( 6 .2
a s in g le c o m p
G , a n d Y o f B
F r o m E q . ( 6 .5
-1
o
.
-6
) re
n e n
F o r
), w
v e a ls th
t o f th e
c la r ity ,
e k n o w
a t
c o
w
th
e a c h c o m p o n e n t o f th e C M Y im a g e is a fu n c tio n o f
r re s p o n d in g R G B im a g e — C is a fu n c tio n o f R , M o f
e w ill u s e a p r im e to d e n o te th e C M Y c o m p o n e n ts .
a t
s
i
= k r
i
fo r i = 1 , 2 , 3 ( fo r t h e R , G , a n d B c o m p o n e n t s ) . A n d fr o m E q . ( 6 .2 - 1 ) , w e k n o w
th a t th e C M Y c o m p o n e n ts c o r re s p o n d in g to th e r i a n d s i (w h ic h w e a re d e n o tin g
w ith p r im e s ) a re
r
i
0
= 1 − r
i
a n d
s
i
0
= 1 − s i.
T h u s ,
r i = 1 − r
i
0
a n d
s
i
0
= 1 − s
= 1 − k r i = 1 − k
i
1 − r
i
0
s o th a t
s
i
0
= k r i0 + ( 1 − k ) .
Problem 6.11
( a ) T h e p u r e s t g r e e n is 0 0 F F 0 0 , w h ic h c o r r e s p o n d s to c e ll (7 , 1 8 ).
( b ) T h e p u r e s t b lu e is 0 0 0 0 F F , w h ic h c o r r e s p o n d s to c e ll (1 2 , 1 3 ).
Problem 6.12
U s in g E q s
th a t, in a c
c o s − 1 (0 / 0
( 6 .2 - 3 ) y ie
d is p la y s h
. ( 6 .2
c o rd
). In
ld s S
o w n
-2
a n
a d
=
in
) t h r o u g h ( 6 .2 - 4 ) , w e
c e w it h E q . ( 6 .2 - 2 ) , h
d itio n , s a tu r a tio n is
1 − 3 m in (0 ) / (3 × 0 )
F ig . P 6 .1 2 .
g e
u e
u n
= 1
t th e r e s u lts
is u n d e fi n e
d e fi n e d w h
− (0 / 0 ). T h
s h
d w
e n
u s ,
o w n
h e n
R =
w e g
in
R
G
e t
T a b
= G
= B
th e
le P 6 .1 2 .
= B s in c
= 0 s in c
m o n o c h
N o te
e θ =
e E q .
ro m e
1 1 3
C o lo r
B la c k
R e d
Y e llo w
G re e n
C y a n
B lu e
M a g e n ta
W h ite
R
1
1
1
H
–
0
0 .1 7
0 .3 3
0 .5
0 .6 7
0 .8 3
–
G ra y
0 .5
0 .5
0 .5
–
G
0
B
0
0
0
1
0
1
1
0
0
1
0
0
1
0
1
1
0
0
1
1
T a b le P 6 .1 2
S
I
0
0
1
0 .3 3
1
0 .6 7
1
0 .3 3
1
0 .6 7
1
0 .3 3
1
0 .6 7
0
1
0
0 .5
M o n o H
–
0
4 3
8 5
1 2 8
1 7 0
2 1 3
–
–
M o n
–
2 5
2 5
2 5
2 5
2 5
2 5
0
0
o S
5
5
5
5
5
5
M o n o I
0
8 5
1 7 0
8 5
1 7 0
8 5
1 7 0
2 5 5
1 2 8
Problem 6.13
W it h r e fe r e n c e t o t h e H S I c o lo r c ir c le in F ig . 6 .1 4 ( b ) , d e e p p u r p le is fo u n d a t
a p p r o x im a te ly 2 7 0 ◦ . T o g e n e r a te a c o lo r r e c ta n g le w ith th e p r o p e r tie s r e q u ir e d in
th e p r o b le m s ta te m e n t, w e c h o o s e a fi x e d in te n s ity I , a n d m a x im u m s a tu r a tio n
(th e s e a r e s p e c tr u m c o lo r s , w h ic h a r e s u p p o s e d to b e fu lly s a tu r a te d ), S . T h e
fi r s t c o lu m n in th e r e c ta n g le u s e s th e s e tw o v a lu e s a n d a h u e o f 2 7 0 ◦ . T h e n e x t
c o lu m n (a n d a ll s u b s e q u e n t c o lu m n s ) w o u ld u s e th e s a m e v a lu e s o f I a n d S , b u t
th e h u e w o u ld b e d e c r e a s e d to 2 6 9 ◦ , a n d s o o n a ll th e w a y d o w n to a h u e o f 0 ◦ ,
w h ic h c o r r e s p o n d s to r e d . If th e im a g e is lim ite d to 8 b its , th e n w e c a n o n ly h a v e
2 5 6 v a r ia tio n s in h u e in th e r a n g e fr o m 2 7 0 ◦ d o w n to 0 ◦ , w h ic h w ill r e q u ir e a
d iffe r e n t u n ifo r m s p a c in g th a n o n e d e g r e e in c r e m e n ts o r, a lte r n a tiv e ly , s ta r tin g
a t a 2 5 5 ◦ a n d p r o c e e d in in c r e m e n ts o f 1 , b u t th is w o u ld le a v e o u t m o s t o f th e
p u r p le . If w e h a v e m o r e th a n e ig h t b its , th e n th e in c r e m e n ts c a n b e s m a lle r.
L o n g e r s tr ip s a ls o c a n b e m a d e b y d u p lic a tin g c o lu m n v a lu e s .
F ig u r e P 6 .1 2
1 1 4
C H A P T E R 6 . P R O B L E M S O L U T IO N S
Problem 6.14
T h e re a re tw o im p o r ta n t a s p
H S I s p a c e a n d th e o th e r is to
v a lu e s g r o w a s a fu n c tio n o f a
e v e r im a g e a re a is u s e d . T h e
a r a d iu s w h e n th e a n g le is 0
b y , s a y , o n e d e g r e e , a n d a ll th
V a lu e s o f th e s a tu r a tio n im a g
o r ig in . T h e in te n s ity im a g e
m in d it is n o t d iffi c u lt to w r it
e c ts to th is p r o b le m . O n e is to a p p r o a c h it in th e
u s e p o la r c o o r d in a te s to c r e a te a h u e im a g e w h o s e
n g le . T h e c e n te r o f th e im a g e is th e m id d le o f w h a tn , fo r e x a m p le , th e v a lu e s o f th e h u e im a g e a lo n g
◦ w o u l d b e a l l 0 ’s . T h e n t h e a n g l e i s i n c r e m e n t e d
e v a l u e s a l o n g t h a t r a d i u s w o u l d b e 1 ’s , a n d s o o n .
e d e c r e a s e lin e a r ly in a ll r a d ia l d ir e c tio n s fr o m th e
is ju s t a s p e c ifi e d c o n s ta n t. W ith th e s e b a s ic s in
e a p r o g r a m th a t g e n e r a te s th e d e s ir e d r e s u lt.
Problem 6.15
T h e h u e , s a t u r a t io n , a n d in t e n s it y im a g e s a r e s h o w n in F ig . P 6 .1 5 , fr o m le ft t o
r ig h t.
Problem 6.16
( a ) It is g iv e n t h a t t h e c o lo r s in F ig . 6 .1 6 ( a ) a r e p
g iv e n th a t th e g r a y - le v e l im a g e s in th e p r o b le m
la tte r c o n d itio n m e a n s th a t h u e (a n g le ) c a n o
n u m b e r o f 2 5 6 v a lu e s . B e c a u s e h u e v a lu e s a r e
0 ◦ to 3 6 0 ◦ th is m e a n s th a t fo r a n 8 - b it im a g e th e
h u e v a lu e s a r e n o w 3 6 0 / 2 5 5 . A n o th e r w a y o f lo
3 6 0 ] h u e s c a le is c o m p r e s s e d to th e r a n g e [0 ,
(th e fi r s t p r im a r y c o lo r w e e n c o u n te r ), w h ic h is
in te g e r ) in th e in te g e r s c a le o f th e 8 - b it im a g e s
S im ila r ly , g r e e n , w h ic h is 1 2 0 ◦ b e c o m e s 8 5 in
c o m p u te th e v a lu e s o f th e o th e r tw o r e g io n s a s
F ig u r e P 6 .1 5
r im a r y s p e c tr u m c o lo r s . It a ls o is
s ta te m e n t a re 8 - b it im a g e s . T h e
n ly b e d iv id e d in to a m a x im u m
re p re s e n te d in th e in te r v a l fr o m
in c re m e n ts b e tw e e n c o n tig u o u s
o k in g a t th is is th a t th e e n tire [0 ,
2 5 5 ]. T h u s , fo r e x a m p le , y e llo w
6 0 ◦ n o w b e c o m e s 4 3 (th e c lo s e s t
h o w n in th e p r o b le m s ta te m e n t.
th is im a g e . F r o m th is w e e a s ily
b e in g 1 7 0 a n d 2 1 3 . T h e re g io n in
1 1 5
t h e m id d le is p u r e w h it e [e q u a l p r o p o r t io n s o f r e d g r e e n a n d b lu e in F ig . 6 .6 1 ( a ) ]
s o its h u e b y d e fi n itio n is 0 . T h is a ls o is tr u e o f th e b la c k b a c k g r o u n d .
( b ) T h e c o lo r s a r e s p e c tr u m c o lo r s , s o th e y a r e fu lly s a tu r a te d . T h e r e fo r e , th e
v a lu e s 2 5 5 s h o w n a p p ly ‘ to a ll c ir c le r e g io n s . T h e r e g io n in th e c e n te r o f th e
c o lo r im a g e is w h ite , s o its s a tu r a tio n is 0 .
( c ) T h e k e y to g e ttin g th e v a lu e s in th is fi g u r e is to r e a liz e th a t th e c e n te r p o r tio n o f th e c o lo r im a g e is w h ite , w h ic h m e a n s e q u a l in te n s itie s o f fu lly s a tu r a te d
r e d , g r e e n , a n d b lu e . T h e r e fo r e , th e v a lu e o f b o th d a r k e r g r a y r e g io n s in th e
in t e n s it y im a g e h a v e v a lu e 8 5 ( i.e ., t h e s a m e v a lu e a s t h e o t h e r c o r r e s p o n d in g
r e g io n ). S im ila r ly , e q u a l p r o p o r tio n s o f th e s e c o n d a r ie s y e llo w , c y a n , a n d m a g e n ta p r o d u c e w h ite , s o th e tw o lig h te r g r a y r e g io n s h a v e th e s a m e v a lu e (1 7 0 )
a s th e r e g io n s h o w n in th e fi g u r e . T h e c e n te r o f th e im a g e is w h ite , s o its v a lu e is
2 5 5 .
Problem 6.17
( a ) B e c a u s e th e in fr a r e d im a g e w h ic h w a s u s e d in p la c e o f th e r e d c o m p o n e n t
im a g e h a s v e r y h ig h g r a y - le v e l v a lu e s .
(b )
T h r
s h o
s h a
T
e
w
d
h e w a te r
s h o ld th e
n in F ig .
e o f b lu e
a p p e a rs a s s
im a g e w ith
P 6 .1 7 . It is
p re s e n ts n o
(c ) N o te th a t th e
W e a lr e a d y k n o w
a c tu a lly r e m o v e s
th e o th e r m a n -m
m o v a l o f th e re d
c a n b e d o n e b y u
o lid
a th
c le a
d iffi
b la c k
re s h o
r th a t
c u ltie
p re d o m in a n t c
h o w to ta k e o
th e “ b a c k g ro u
a d e s tr u c tu re s
[a n d th e b la c k
s in g th e te c h n
( 0 ) in t h e n e a r in fr a r e d im a g e [F ig . 6 .2 7 ( d ) ].
ld v a lu e s lig h tly la r g e r th a n 0 . T h e r e s u lt is
c o lo r in g a ll th e b la c k p o in ts in th e d e s ir e d
s .
o lo r o f n a tu r a l te r r a in is in
u t th e w a te r fro m (b ). T h e
n d ” o f r e d a n d b la c k w o u ld
, w h ic h a p p e a r m o s tly in a
if y o u d o n o t w a n t to u s e
iq u e d is c u s s e d in S e c tio n 6
v a r io u
re fo re
le a v e
b lu is h
th e m
.7 .2 .
s s h a d e s o f re d .
, a m e th o d th a t
p r e d o m in a n tly
lig h t c o lo r. R e e th o d a s in (b )]
Problem 6.18
U s
h a
p u
is ,
o f
in g E q . ( 6 .2 - 3 ) , w e s e e t h a t t h e b a s ic p r o b le m is t h a t m a n y
v e th e s a m e s a tu r a tio n v a lu e . T h is w a s d e m o n s tr a te d in P r o b
r e r e d , y e llo w , g r e e n , c y a n , b lu e , a n d m a g e n ta a ll h a d a s a tu r
a s lo n g a s a n y o n e o f t h e R G B c o m p o n e n t s is 0 , E q . ( 6 .2 - 3 ) y ie
1 .
C o n s id e r R G B c o lo r s (1 , 0 , 0 ) a n d (0 , 0 .5 9 , 0 ), w h ic h r e p r e s e n
a n d g r e e n . T h e H S I t r ip le t s fo r t h e s e c o lo r s [p e r E q . ( 6 .4 - 2 ) t h r o
(0 , 1 , 0 .3 3 ) a n d (0 .3 3 , 1 , 0 .2 ), r e s p e c t iv e ly . N o w , t h e c o m p le m e n
d iff
le m
a tio
ld s
e re n t
6 .1 2 ,
n o f 1
a s a tu
c o lo r s
w h e re
. T h a t
r a tio n
t s h a d e s o f re d
u g h ( 6 .4 - 4 ) ] a r e
ts o f th e b e g in -
C H A P T E R 6 . P R O B L E M S O L U T IO N S
1 1 6
F ig u r e P 6 .1 7
n in g R G B v a lu
c o r re s p o n d in g
t h r o u g h ( 6 .4 - 4
re d , a s ta r tin g
1 , w h ile fo r th
m e n te d ” s a tu r
d iffe re n t “c o m
tio n to c o m p u
e s ( s e e S e c t io n 6 .5 .2 ) a r e (0 , 1 , 1 ) a n d (1 , 0 .4 1 , 1 ), r e s p e c t
c o lo r s a r e c y a n a n d m a g e n ta . T h e ir H S I v a lu e s [p e r E q
) ] a r e (0 .5 , 1 , 0 .6 6 ) a n d (0 .8 3 , 0 .4 8 , 0 .8 ), r e s p e c t iv e ly . T h u
s a tu r a tio n o f 1 y ie ld e d th e c y a n “ c o m p le m e n te d ” s a tu
e g r e e n , a s ta r tin g s a tu r a tio n o f 1 y ie ld e d th e m a g e n ta
a t io n o f 0 .4 8 . T h a t is , t h e s a m e s t a r t in g s a t u r a t io n r e s u lt
p le m e n te d ” s a tu r a tio n s . S a tu r a tio n a lo n e is n o t e n o u g h
te th e s a tu r a tio n o f th e c o m p le m e n te d c o lo r.
iv e ly ; th e
s . ( 6 .4 - 2 )
s , fo r th e
r a tio n o f
“ c o m p le e d in tw o
in fo r m a -
Problem 6.19
T h
6 .3
n o
0 .5
h a
e c o
2 . T
r m a
), th
lf o f
m p
h e
liz e
e h
th e
le m
h u e
d b y
u e o
c irc
e n t o f a c o lo r is th
c o m p o n e n t is th e a
3 6 0 d e g re e s . F o r a
f th e c o m p le m e n ta
le ( i.e ., fo r 0 .5 ≤ H ≤
e c o lo r
n g le fr o
c o lo r o n
r y c o lo r
1 ), th e
o p
m
th
is
h u
p o s ite
re d in
e to p
H + 0
e o f th
it o
a c o
h a lf
.5 . F
e c o
n th e c o lo r c ir c le
u n te r c lo c k w is e d ir
o f t h e c ir c le ( i.e ., 0
o r a c o lo r o n th e b
m p le m e n t is H − 0
o f F ig .
e c tio n
≤ H ≤
o tto m
.5 .
Problem 6.20
T h e R G B t r a n s fo r m a t io n s fo r a c o m p le m e n t [fr o m F ig . 6 .3 3 ( b ) ] a r e :
s
i
= 1 − r
i
w h e re i = 1 ,2 ,3 (fo r th e R , G , a n d B c o m p o n e n ts ). B u t fr o m th e d e fi n itio n o f th e
C M Y s p a c e in E q . ( 6 .2 - 1 ) , w e k n o w t h a t t h e C M Y c o m p o n e n t s c o r r e s p o n d in g t o
r i a n d s i , w h ic h w e w ill d e n o te u s in g p r im e s , a r e
r
0
s
i
i
0
= 1 − r i
= 1 − s i.
1 1 7
F ig u r e P 6 .2 1
T h u s ,
r i = 1 − r
i
0
a n d
s
i
0
= 1 − s
i
= 1 − (1 − r i) = 1 −
1 −
1 − r
i
0
s o th a t
s 0 = 1 − r i0 .
Problem 6.21
T h e R G B tra n
a re a s , e ffe c tiv
a n d b lu e c o m
s o th a t th e c o
s h o w n in F ig .
s fo r m a tio n s h o
e ly c o m p r e s s in
p o n e n ts s h o u ld
lo r s d o n o t c h a
P 6 .2 1 .
u ld d
g a ll
b e t
n g e .
a rk e n
v a lu e s
ra n s fo
T h e g
th
to
r m
e n
e h
w a
e d
e ra
ig
rd
w
l
h lig h ts a n d lig h te
th e m id to n e s . T
ith th e s a m e m a p
s h a p e o f th e c u r v
n th e s h
h e re d , g
p in g fu n
e w o u ld
a d
re
c t
b e
o w
e n ,
io n
a s
Problem 6.22
B a s e d o n t h e d is c u s s io n is S e c t io n 6 .5 .4 a n d w it h r e fe r e n c e t o t h e c o lo r w h e e l
in F ig . 6 .3 2 , w e c a n d e c r e a s e t h e p r o p o r t io n o f y e llo w b y ( 1 ) d e c r e a s in g y e llo w ,
(2 ) in c r e a s in g b lu e , (3 ) in c r e a s in g c y a n a n d m a g e n ta , o r (4 ) d e c r e a s in g r e d a n d
g re e n .
Problem 6.23
T h e L ∗ a ∗ b ∗ c o m p o n e n t s a r e c o m p u te d u s in g E q s . ( 6 .5 - 9 ) t h r o u g h ( 6 .5 - 1 2 ) . R e fe re n c e w h ite is R = G = B = 1 . T h e c o m p u ta tio n s a re b e s t d o n e in a s p re a d s h e e t,
a s s h o w n in T a b le P 6 .2 3 .
C H A P T E R 6 . P R O B L E M S O L U T IO N S
1 1 8
Problem 6.24
T h e s im p le
c o lo r s p a c e
o n th e in te
r e s u ltin g in
b a c k to th e
s t a p p ro a c h c o n c e p tu a
, p e r fo r m h is to g r a m s p
n s ity (I ) c o m p o n e n t o
te n s ity c o m p o n e n t w it
s ta r tin g c o lo r s p a c e .
lly is
e c ifi
n ly (
h th e
to tra n s fo r
c a tio n p e r
le a v in g H
o r ig in a l h
m
th
a n
u e
e v e r y
e d is c
d S a
a n d s
in p u
u s s io
lo n e )
a tu ra
t im a
n in
, a n d
tio n
g e
S e
c
c o
to th
c tio n
o n v e
m p o
e H S I
3 .3 .2
r t th e
n e n ts
Problem 6.25
T h e g iv e n c o lo r im a g e is s h o w n in F ig . P 6 .2 5 ( a ) . A s s u m e t h a t t h e c o m p o n e n t
im a g e v a lu e s o f th e H S I im a g e a r e in th e r a n g e [0 , 1 ]. C a ll th e c o m p o n e n t im a g e s
H (h u e ), S (s a tu r a tio n ), a n d I (in te n s ity ).
( a ) It is g iv e n th a t th e im a g e is fu lly s a tu r a te d , s o im a g e H w ill b e c o n s ta n t w ith
v a lu e 1 . S im ila r ly , a ll t h e s q u a r e s a r e a t t h e ir m a x im u m v a lu e s o , fr o m E q . ( 6 .2 4 ), th e in te n s ity im a g e a ls o w ill b e c o n s ta n t, w ith v a lu e 1 / 3 [th e m a x im u m v a lu e
o f a n y (n o r m a liz e d ) p ix e l in th e R G B im a g e is 1 , a n d it is g iv e n th a t n o n e o f th e
s q u a r e s o v e r la p ]. T h e h u e c o m p o n e n t im a g e , H , is s h o w n in F ig . P 6 .2 5 ( b ) . R e c a ll
fr o m F ig . 6 .1 4 t h a t t h e v a lu e o f h u e is a n a n g le . B e c a u s e t h e r a n g e o f v a lu e s o f H
is n o r m a liz e d to [0 , 1 ], w e s e e fr o m th a t fi g u r e , fo r e x a m p le , th a t a s w e g o a r o u n d
th e c ir c le in th e c o u n te r c lo c k w is e d ir e c tio n a h u e v a lu e o f 0 c o r r e s p o n d s to r e d ,
a v a lu e o f 1 / 3 to g r e e n , a n d a v a lu e o f 2 / 3 to b lu e . T h u s , th e im p o r ta n t p o in t to
b e m a d e in F ig . P 6 .2 5 (b ) is th a t th e g r a y v a lu e in th e im a g e c o r r e s p o n d in g to th e
r e d s q u a r e s h o u ld b e b la c k , th e v a lu e c o r r e s p o n d in g to th e g r e e n g r e e n s q u a r e
s h o u ld b e lo w e r - m id g r a y , a n d th e v a lu e o f th e th e s q u a r e c o r r e s p o n d in g to b lu e
s h o u ld b e a lig h te r s h a d e o f g r a y th a n th e s q u a r e c o r r e s p o n d in g to g r e e n . A s y o u
c a n s e e , t h is in d e e d is th e c a s e fo r th e s q u a r e s in F ig . P 6 .2 5 ( b ) . F o r th e s h a d e s o f
r e d , g r e e n , a n d b lu e in F ig . P 6 .2 5 ( a ) , t h e e x a c t v a lu e s a r e H = 0 , H = 0 .3 3 , a n d
H = 0 .6 7 , r e s p e c t iv e ly .
1 1 9
F ig u r e P 6 .2 5
( b ) T h e s a tu r a tio n im a g e is c o n s ta n t, s o s m o o th in g it w ill p r o d u c e th e s a m e c o n s ta n t v a lu e .
( c ) F ig u r e P 6 .2 5 ( c ) s h o w s t h e r e s u lt o f b lu r r in g th e h u e im a g e . W h e n t h e a v e r a g in g m a s k is fu lly c o n ta in e d in a s q u a r e , th e r e is n o b lu r r in g b e c a u s e th e v a lu e
o f e a c h s q u a re is c o n s ta n t. W h e n th e m a s k c o n ta in s p o r tio n s o f tw o o r m o re
s q u a r e s th e v a lu e p r o d u c e d a t th e c e n te r o f th e m a s k w ill b e b e tw e e n th e v a lu e s
o f th e tw o s q u a r e s , a n d w ill d e p e n d th e r e la tiv e p r o p o r tio n s o f th e s q u a r e s o c c u p ie d b y th e m a s k . T o s e e e x a c tly w h a t th e v a lu e s a r e , c o n s id e r a p o in t in th e
c e n t e r o f r e d m a s k in F ig . P 6 .2 5 ( c ) a n d a p o in t in t h e c e n t e r o f t h e g r e e n m a s k
o n th e to p le ft. W e k n o w fr o m (a ) a b o v e th a t th e v a lu e o f th e r e d p o in t is 0 a n d
th e v a lu e o f th e g r e e n p o in t is 0 .3 3 . T h u s , th e v a lu e s in th e b lu r r e d b a n d b e t w e e n r e d a n d g r e e n v a r y fr o m 0 t o 0 .3 3 b e c a u s e a v e r a g in g is a lin e a r o p e r a tio n .
F ig u r e P 6 .2 5 (d ) s h o w s th e r e s u lt o f g e n e r a tin g a n R G B im a g e w ith th e b lu r r e d
C H A P T E R 6 . P R O B L E M S O L U T IO N S
1 2 0
h u e c o m p o n e n t im a g e s a n d th e o r ig in a l s a tu r a tio n a n d in te n s ity
v a lu e s a lo n g th e lin e ju s t d is c u s s e d a r e tr a n s itio n s fr o m g r e e n to r
6 .1 4 w e s e e t h a t t h o s e t r a n s it io n s e n c o m p a s s t h e s p e c t r u m fr o m
th a t in c lu d e s c o lo r s s u c h a s y e llo w [a ll th o s e c o lo r s a r e p r e s e n t in
a lth o u g h th e y a r e s o m e w h a t d iffi c u lt to s e e ]. T h e r e a s o n fo r th e d
lin e in th is fi g u r e is th a t th e a v e r a g e v a lu e s a lo n g th a t r e g io n a r e n
b e t w e e n r e d a n d b lu e , w h ic h w e k n o w fr o m fr o m F ig . 6 .1 4 is g r e e n
im a g e s . T h e
e d . F r o m F ig .
g re e n to re d
F ig . P 6 .2 5 ( d ) ,
ia g o n a l g re e n
e a r ly m id w a y
.
Problem 6.26
T h is is a s im p le p r o b le m to e n c o u r a g e th e s tu d e n t to th in k a b o u t th e m e a n in g
o f t h e e le m e n t s in E q . ( 6 .7 - 2 ) . W h e n C = I , it fo llo w s t h a t C − 1 = I a n d E q . ( 6 .7 - 2 )
b e c o m e s
1 / 2
D (z ,a ) = (z − a )T (z − a )
.
B u t th e te r m in s id e th e b r a c k e ts is re c o g n iz e d a s th e in n e r p r o d u c t o f th e v e c to r
(z − a ) w it h it s e lf, w h ic h , b y d e fi n it io n , is e q u a l t o t h e r ig h t s id e o f E q . ( 6 .7 - 1 ) .
Problem 6.27
( a ) T h e c u b e is c o m p o s e d o f s ix in te r s e c tin g p la n e s in R G B s p a c e . T h e g e n e r a l
e q u a tio n fo r s u c h p la n e s is
a z
R
+ b z
G
+ c z
B
+ d = 0
w h e r e a , b , c , a n d d a r e p a r a m e t e r s a n d t h e z ’s a r e t h e c o m p o n e n t s o f a n y p o i n t
(v e c to r ) z in R G B s p a c e ly in g o n th e p la n e . If a n R G B p o in t z d o e s n o t lie o n th e
p la n e , a n d its c o o r d in a te s a r e s u b s titu te d in th e p r e c e d in g e q u a tio n , th e e q u a tio n w ill g iv e e ith e r a p o s itiv e o r a n e g a tiv e v a lu e ; it w ill n o t y ie ld z e r o . W e s a y
th a t z lie s o n th e p o s itiv e o r n e g a tiv e s id e o f th e p la n e , d e p e n d in g o n w h e th e r
th e r e s u lt is p o s itiv e o r n e g a tiv e . W e c a n c h a n g e th e p o s itiv e s id e o f a p la n e b y
m u ltip ly in g its c o e ffi c ie n ts (e x c e p t d ) b y − 1 . S u p p o s e th a t w e te s t th e p o in t a
g iv e n in th e p r o b le m s ta te m e n t to s e e w h e th e r it is o n th e p o s itiv e o r n e g a tiv e
s id e e a c h o f th e s ix p la n e s c o m p o s in g th e b o x , a n d c h a n g e th e c o e ffi c ie n ts o f
a n y p la n e fo r w h ic h th e r e s u lt is n e g a tiv e . T h e n , a w ill lie o n th e p o s itiv e s id e o f
a ll p la n e s c o m p o s in g th e b o u n d in g b o x . In fa c t a ll p o in ts in s id e th e b o u n d in g
b o x w ill y ie ld p o s itiv e v a lu e s w h e n th e ir c o o r d in a te s a r e s u b s titu te d in th e e q u a tio n s o f th e p la n e s . P o in ts o u ts id e th e b o x w ill g iv e a t le a s t o n e n e g a tiv e (o r z e r o
if it is o n a p la n e ) v a lu e . T h u s , th e m e th o d c o n s is ts o f s u b s titu tin g a n u n k n o w n
c o lo r p o in t in th e e q u a tio n s o f a ll s ix p la n e s . If a ll th e r e s u lts a r e p o s itiv e , th e
p o in t is in s id e th e b o x ; o th e r w is e it is o u ts id e th e b o x . A fl o w d ia g r a m is a s k e d
1 2 1
F ig u r e P 6 .2 9
f o r i n t h e p r o b l e m s t a t e m e n t t o m a k e i t s i m p l e r t o e v a l u a t e t h e s t u d e n t ’s l i n e o f
re a s o n in g .
( b ) If th e b o x is lin e d u p w ith th e R G B c o o r d in a te a x e s , th e n th e p la n e s in te r s e c t
th e R G B c o o r d in a te p la n e s p e r p e n d ic u la r ly . T h e in te r s e c tio n s o f p a ir s o f p a r a lle l p la n e s e s ta b lis h a r a n g e o f v a lu e s a lo n g e a c h o f th e R G B a x is th a t m u s t b e
c h e c k e d to s e e if th e if a n u n k n o w n p o in t lie s in s id e th e b o x o r n o t. T h is c a n b e
d o n e o n a n im a g e p e r im a g e b a s is ( i.e ., t h e t h r e e c o m p o n e n t im a g e s o f a n R G B
im a g e ), d e s ig n a tin g b y 1 a c o o rd in a te th a t is w ith in its c o r re s p o n d in g r a n g e a n d
0 o th e r w is e . T h e s e w ill p r o d u c e th r e e b in a r y im a g e s w h ic h , w h e n A N D e d , w ill
g iv e a ll th e p o in ts in s id e th e b o x .
Problem 6.28
T h e s k e tc h is a n e lo n g a te d e llip s o id a l fi g u r e in w h ic h th e le n g th lin e d u p w ith
th e R e d (R ) a x is is 8 tim e s lo n g e r th a t th e o th e r tw o d im e n s io n s . In o th e r w o r d s ,
th e fi g u r e lo o k s lik e a b lim p a lig n e d w ith th e R - a x is .
Problem 6.29
S e t o n e o f th e th r e e p r im a r y im a g e s to a c o n s ta n t v a lu e (s a y , 0 ), th e n c o n s id e r
t h e t w o im a g e s s h o w n in F ig . P 6 .2 9 . If w e fo r m e d a n R G B c o m p o s it e im a g e b y
1 2 2
C H A P T E R 6 . P R O B L E M S O L U T IO N S
le ttin g th e im a g e o n th e le ft b e th e r e d c o m p o n e n t a n d th e im a g e o n th e r ig h t
th e g r e e n c o m p o n e n t, th e n th e r e s u lt w o u ld b e a n im a g e w ith a g r e e n r e g io n o n
th e le ft s e p a r a te d b y a v e r tic a l e d g e fr o m a r e d r e g io n o n th e r ig h t. T o c o m p u te
th e g r a d ie n t o f e a c h c o m p o n e n t im a g e w e ta k e s e c o n d - o rd e r p a r tia l d e r iv a tiv e s .
In th is c a s e , o n ly th e c o m p o n e n t o f th e d e r iv a tiv e in th e h o r iz o n ta l d ir e c tio n is
n o n z e r o . If w e m o d e l th e e d g e a s a r a m p e d g e th e n a p r o fi le o f th e d e r iv a tiv e
im a g e w o u ld a p p e a r a s s h o w n in F ig . P 6 .2 9 ( s e e S e c t io n 1 0 .2 .4 fo r m o r e d e t a il
o n th e p r o fi le o f e d g e s ). T h e m a g n ifi e d v ie w s h o w s c le a r ly th a t th e d e r iv a tiv e s
o f th e tw o im a g e s a re m ir r o r s o f e a c h o th e r. T h u s , if w e c o m p u te d th e g r a d ie n t
v e c to r o f e a c h im a g e a n d a d d e d th e r e s u lts a s s u g g e s te d in th e p r o b le m s ta te m e n t, th e c o m p o n e n ts o f th e g r a d ie n t w o u ld c a n c e l o u t, g iv in g a z e r o g r a d ie n t
fo r a c o lo r im a g e th a t h a s a c le a r ly d e fi n e d e d g e b e tw e e n tw o d iffe r e n t c o lo r r e g io n s . T h is s im p le e x a m p le illu s tr a te s th a t th e g r a d ie n t v e c to r o f a c o lo r im a g e
is n o t e q u iv a le n t to th e r e s u lt o f fo r m in g a c o lo r g r a d ie n t v e c to r fr o m th e s u m o f
th e g r a d ie n t v e c to r s o f th e in d iv id u a l c o m p o n e n t im a g e s .
Chapter 7
Problem Solutions
Problem 7.1
F o llo w in g th e e x p la n a tio n in E x a m p le 7 .1 , th e d e c o d e r is a s s h o w n in F ig . P 7 .1 .
Problem 7.2
A m e a n a p
th e s ta r tin
m e a n a p p
a v e ra g e s o
p r o x im a tio
g im a g e is
r o x im a tio n
v e r f x ,y
n p y ra m
o f s iz e 4
p y r a m id
a n d s u b
id
×
. T
s a
is
4
h
m
c re a
, J =
e le v
p lin
te d b y fo r m in g 2 × 2 b lo c k a v e r a g e s . S in c e
2 a n d f x , y is p la c e d in le v e l 2 o f th e
e l 1 a p p r o x im a tio n is (b y ta k in g 2 × 2 b lo c k
g )
3 .5
1 1 .5
5 .5
1 3 .5
Level j-1
approximation
2
Upsampler
Interpolation
filter
Prediction
Level j
prediction
residual
+
F ig u r e P 7 .1
1 2 3
Level j
approximation
C H A P T E R 7 . P R O B L E M S O L U T IO N S
1 2 4
a n d t h e le v e l 0 a p p r o x im a t io n is s im ila r ly [8 .5 ]. T h e c o m p le t e d m e a n a p p r o x im a tio n p y r a m id is
⎡
⎤
1
2
3
4
⎢
⎥
6
7
8 ⎥
3 .5
5 .5
⎢ 5
[8 .5 ] .
⎢
⎥
1 0
1 1
1 2 ⎦ 1 1 .5
1 3 .5
⎣ 9
1 3
1 4
1 5
1 6
P ix e
u a l
tio n
p lin
tio n
l r e p lic a tio n is u s
p y r a m id . L e v e l
a p p r o x im a tio n ,
g th e le v e l 1 a p p
(o r ig in a l im a g e )
⎡
1
2
⎢
6
⎢ 5
⎢
9
1
0
⎣
1 3
1 4
e d in th e g
0 o f th e p
[8 .5 ]. T h e
r o x im a tio
. T h u s , w e
3
4
7
8
1 1
1 5
1 2
1 6
e n e r a tio n
re d ic tio n
le v e l 2 p r e
n a n d s u b
g e t
⎤ ⎡
3 .5
⎥ ⎢
⎥ ⎢ 3 .5
⎥− ⎢
⎦ ⎣ 1 1 .5
1 1 .5
⎡
− 2 .5
⎢
⎢ 1 .5
⎢
⎣ − 2 .5
1 .5
o f th e c
re s id u a
d ic tio n
tr a c tin g
3 .5
3 .5
1 1 .5
1 1 .5
o m p le m e n
l p y r a m id
re s id u a l is
it fr o m th
− 1 .5
2 .5
− 1 .5
2 .5
5 .5
5 .5
1 3 .5
1 3 .5
ta r y p re
is th e lo
o b ta in e
e le v e l 2
tio n
s t re
y u p
p ro x
re s id s o lu s a m im a -
⎤
5 .5
5 .5
1 3 .5
1 3 .5
− 2 .5
1 .5
− 2 .5
1 .5
d ic
w e
d b
a p
⎥
⎥
⎥
⎦
=
⎤
− 1 .5
2 .5
− 1 .5
2 .5
⎥
⎥
⎥.
⎦
S im ila r ly , th e le v e l 1 p r e d ic tio n r e s id u a l is o b ta in e d b y u p s a m p lin g th e le v e l 0
a p p r o x im a tio n a n d s u b tr a c tin g it fr o m th e le v e l 1 a p p r o x im a tio n to y ie ld
3 .5
1 1 .5
5 .5
1 3 .5
−
8 .5
8 .5
8 .5
8 .5
T h e p re d ic tio n re s id u a l p y r a m id is th e re fo r
⎡
− 2 .5
− 1 .5
− 2 .5
− 1 .5
⎢
2 .5
1 .5
2 .5
⎢ 1 .5
⎢
− 1 .5
− 2 .5
− 1 .5
⎣ − 2 .5
1 .5
2 .5
1 .5
2 .5
− 5
3
=
− 3
5
.
e
⎤
⎥
⎥
⎥
⎦
− 5
3
− 3
5
[8 .5 ] .
Problem 7.3
T h e n u m b e r o f e le m e n ts in a J + 1 le v e l p y r a m id w h e r e N =
2
4
N 2 o r 43 2 J
= 43 2 2 J ( s e e S e c t i o n 7 . 1 . 1 ) :
3
,
1
4 2 J
1
1
2 J
1 +
≤
2
+
+ ... +
2
1
2
J
3
(4 )
(4 )
(4 )
2
J
is b o u n d e d b y
1 2 5
fo r J > 0 . W e c a n g e n e r a te th e fo llo w in g ta b le :
P y r a m id E le m e n ts
1
5
2 1
8 5
..
.
J
0
1
2
3
..
.
1
C o m p re s s
1
5 / 4 =
2 1 / 1 6 =
8 5 / 6 4 =
io n R a tio
1 .2 5
1 .3 1 2 5
1 .3 2 8
..
.
4 / 3 = 1 .3 3
A ll b u t th e tr iv ia l c a s e , J = 0 , a r e e x p a n s io n s . T h e e x p a n s io n fa c to r is a fu n c tio n
o f J a n d b o u n d e d b y 4 / 3 o r 1 .3 3 .
Problem 7.4
F i r s t c h e c k f o r o r t h o n o r m a l i t y . I f t h e fi l t e r s a r e o r t h o n o r m a l ., t h e y a r /e b i o r t h o g o p1 , p1
n a l b y d e fi n itio n . B e c a u s e K e v e n = 2 , n = 0 ,1 , a n d g 0 (n ) =
fo r n = 0 ,1 ,
2
2
E q . ( 7 .1 - 1 4 ) y ie ld s :
.
/
p 1 , p− 1
(− 1 )n g 0 (1 − n ) = (− 1 )n g 0 (1 − n ) = g 0 (0 ) ,− g 0 (1 ) =
= g 1 (n )
2
2
.
/
1p
1p
g 0 (1 − n ) =
, 2 = h 0 (n )
. 2
/
p− 1 , p 1
= h 1 (n ) .
g 1 (1 − n ) =
2
2
T h u s , t h e fi lt e r s a r e o r t h o n o r m a l a n d w ill a ls o s a t is fy E q . ( 7 .1 - 1 3 ) . In a d d it io n ,
t h e y w ill s a t is fy t h e b io r t h o g o n a lit y c o n d it io n s s t a t e d in E q s . ( 7 .1 - 1 2 ) a n d ( 7 .1 1 1 ) , b u t n o t ( 7 .1 - 1 0 ) . T h e fi lt e r s a r e b o t h o r t h o n o r m a l a n d b io r t h o g o n a l.
Problem 7.5
(a ) f
(b ) f
(c ) f
f (− n ) = f 1 , 0 .2 5 , 0 .5 , 0 g .
(n ) =
b
(n ) = (− 1 )
c
(d ) f
(e ) f
(n ) = − f (n ) = f 0 , − 0 .5 , − 0 .2 5 , − 1 g .
a
e
d
(n ) =
f
c
e
f (n ) = f 0 , − 0 .5 , 0 .2 5 , − 1 g .
(− n ) = f − 1 , 0 .2 5 , − 0 .5 , 0 g .
(n ) = (− 1 )
( f ) F ilte r f
n
n
f
b
(n ) = f 1 , − 0 .2 5 , 0 .5 , 0 g .
(n ) in ( e ) c o r r e s p o n d s t o E q . ( 7 .1 - 9 ) .
Problem 7.6
T a b le 7 .1 in t h e b o o k d e fi n e s g 0 (n ) fo r n = 0 , 1 , 2 , ..., 7 t o b e a b o u t { 0 .2 3 , 0 .7 2 ,
0 .6 3 , - 0 .0 3 , - 0 .1 9 , 0 .0 3 , 0 .0 3 , - 0 .0 1 } . U s in g E q , ( 7 .1 - 1 4 ) w it h K e v e n = 8 , w e c a n
C H A P T E R 7 . P R O B L E M S O L U T IO N S
1 2 6
w r ite
g
n
(n ) = (− 1 )
1
g
0
(7 − n ) .
T h u s g 1 (n ) is a n o r d e r - r e v e r s e d a n d m o d u la te d c o p y o f g
- 0 .0 3 , 0 .0 3 , 0 .1 9 , - 0 .0 3 , - 0 .6 3 , 0 .7 2 , - 0 .2 3 } .
0
(n )— t h a t is , { – 0 .0 1 ,
T o n u m e r ic a lly p r o v e t h e o r t h o n o r m a lit y o f t h e fi lt e r s , le t m = 0 in E q . ( 7 .1 1 3 ):
0
g
(n ) g
i
j
(n )
1
= δ
fo r i , j = f 0 ,1 g .
i − j
Ite r a tin g o v e r i a n d j , w e g e t
2
g
2n
2
g
0
2
(n ) =
0
n
(n ) g
1
g
1
2
(n ) = 1
(n ) = 0 .
n
S u b s titu tio n o f th e fi lte r c o e ffi c ie n t v a lu e s in to th e s e e q u a tio n s y ie ld s :
g
0
(n ) g
(n )
1
=
(0 .2 3 ) (− 0 .0 1 ) + (0 .7 2 ) (− 0 .0 3 ) + (0 .6 3 ) (0 .0 3 )
n
+ (− 0 .0 3 ) (0 .1 9 ) + (− 0 .1 9 ) (− 0 .0 3 ) + (0 .0 3 ) (− 0 .6 3 )
+ (0 .0 3 ) (0 .7 2 ) + (− 0 .0 1 ) (− 0 .2 3 )
=
g
2
0
(n )
0
=
g
n
2
1
(n )
n
=
(± 0 .2 3 ) 2 + (0 .7 2 ) 2 + (± 0 .6 3 )2 + (− 0 .0 3 )
2
+ (± 0 .1 9 )2 + (0 .0 3 ) 2 + (± 0 .0 3 )2 + (− 0 .0 1 )
=
2
1 .
Problem 7.7
R e c o n s tr u c tio n is p e r fo r m e d b y re v e r s in g th e d e c o m p o s itio n p r o c e s s — th a t is ,
b y r e p la c in g th e d o w n s a m p le r s w ith u p s a m p le r s a n d th e a n a ly s is fi lte r s b y th e ir
s y n t h e s is fi lt e r c o u n t e r p a r t s , a s F ig . P 7 .7 s h o w s .
1 2 7
g0(n)
2
a(m,n)
Columns
(along n)
dV(m,n)
g0(m)
2
Rows
(along m)
g1(n)
2
Columns
dH(m,n)
f(m,n)
g0(n)
2
Columns
g1(m)
2
Rows
dD(m,n)
g1(n)
2
Columns
F ig u r e P 7 .7
Problem 7.8
T h e H a a r tr a n s fo r m m a tr ix fo r N = 8 is
⎡
1
1
1
1
⎢
1
1
1
⎢ p1
p
p
p
⎢
2
2
−
2
−
2
⎢
⎢
1 ⎢ 0
0
0
0
H 8 = p ⎢
− 2
0
0
8 ⎢ 2
⎢
0
2
− 2
⎢ 0
⎢
0
0
0
⎣ 0
0
0
0
0
⎤
1
1
1
1
− 1
0
p
− 1
0
p
− 1
0
p
− 1
0
p
2
0
0
0
− 2
0
0
−
2
−
0
0
2
2
⎥
⎥
⎥
⎥
⎥
⎥
⎥.
⎥
⎥
⎥
⎥
⎦
2
0
0
0
0
0
− 2
2
Problem 7.9
( a ) E q u a tio n ( 7 .1 - 1 8 ) d e fi n e s th e 2 × 2 H a a r t r a n s fo r m a t io n m a t r ix a s
H
=
2
p
1
2
1
1
1
− 1
.
T h u s , u s in g E q . ( 7 .1 - 1 7 ) , w e g e t
T = H F H
=
5
− 3
T
p
=
4
0
2
1
2
1
1
1
− 1
6
3
− 1
2
1
1
1
− 1
.
C H A P T E R 7 . P R O B L E M S O L U T IO N S
1 2 8
(b ) F ir s t, c o m p u te
− 1
H
a
=
2
b
d
c
s u c h th a t
a
b
d
c
1
p
1
1
− 1
1
2
=
1
0
0
1
.
S o lv in g th is m a tr ix e q u a tio n y ie ld s
− 1
H
p
=
2
1
1
1
− 1
1
2
= H
2
T
= H
2
.
T h u s ,
F = H
T
T H
2
1
p
=
1
− 1
1
2
3
=
1
− 1
2
6
5
− 3
4
1
0
1
− 1
1
.
Problem 7.10
(a ) T h e b a s is is o r th o n o r m a l a n d th e c o e ffi c ie n ts a re c o m p u te d b y th e v e c to r
e q u iv a le n t o f E q . ( 7 .2 - 5 ) :
α
p1
=
0
5
=
α
1
p
p1
2
3
2
2
2
2
p1
=
=
p
p1
−
2
3
2
2
2
2
s o ,
5
p
2
2
'
0
+
p
2
2
'
1
=
5
p
p1
2
p1
=
2
2
3
.
2
2
+
p
p1
2
2
−
2
p1
2
1 2 9
(b ) T h e b a s is is b io r th o n o r m a l a n d th e c o e ffi c ie n ts a re c o m p u te d b y th e v e c to r
e q u iv a le n t o f E q . ( 7 .2 - 3 ) :
α
1
− 1
0
1
=
0
3
2
= 1
α
1
=
3
2
= 2
s o ,
'
+ 2 '
0
1
=
1
3
=
1
+ 2
0
1
.
2
( c ) T h e b a s is is o v e r c o m p le te a n d th e c o e ffi c ie n ts a r e c o m p u te d b y th e v e c to r
e q u iv a le n t o f E q . ( 7 .2 - 3 ) :
α
2
=
0
= 2
$
α
2
p
1
=
−
1
3
0
3
3
2
p
=
−
= − 1 −
3
3
$
2
2
3
= − 1 +
α
% 3
3
1
p
−
3
2
p
3
3
3
3
% 3
2
C H A P T E R 7 . P R O B L E M S O L U T IO N S
1 3 0
F ig u r e P 7 .1 1
s o ,
2 '
0
+
− 1 +
2
p
3
3
'
1
+
− 1 −
2
p
3
3
'
2
1
= 2
p
2
− 1 +
2
− 1 −
=
+
0
3
p
3
2
3
2
+
2
3
3
3
1
−p
−
−
1
p2
3
2
.
Problem 7.11
A s c a n b e s e e n i n F i g . P 7 . 1 1 , s c a l i n g f u n c t i o n ' 0 ,0 ( x ) c a n n o t b e w r i t t e n a s a s u m
o f d o u b l e r e s o l u t i o n c o p i e s o f i t s e l f . N o t e t h e g a p b e t w e e n ' 1 ,0 ( x ) a n d ' 1 ,1 ( x ) .
1 3 1
F ig u r e P 7 .1 2
Problem 7.12
S u b s t it u t in g j = 3 in t o E q . ( 7 .2 - 1 3 ) , w e g e t
'
= S p a n
V
3
k
3 ,k
(x )
3
/
3
2 2 ' 2 3 x − k
k
4
3 p
= S p a n 2 2 ' (8 x − k ) .
.
V
3
= S p a n
V
k
U s in g t h e H a a r s c a lin g fu n c t io n [E q . ( 7 .2 - 1 4 ) ], w e t h e n g e t t h e r e s u lt s h o w n in
F ig . P 7 .1 2 .
Problem 7.13
F r o m E q . ( 7 .2 - 1 9 ) , w e fi n d t h a t
3 ,3
(x ) = 2
= 2
a n
in
E q
1 /
d u s in
F ig . P
. ( 7 .2 p
2 a n
g th e H a
7 .1 3 . T o
2 8 ) a n d t
d h (1 ) =
a r w a v e le t fu n c
e x p re s s
3 ,3 ( x
h e H a a r w a v e le
p
− 1 / 2 . T h u s
tio
) a
t v
w e
3 / 2
p
(2 3 x − 3 )
(8 x − 3 )
2
n d e fi n it io n fr o m E q . ( 7 .2 - 3 0 ) , o b t a in t h e p lo t
s a fu n c tio n o f s c a lin g fu n c tio n s , w e e m p lo y
e c t o r d e fi n e d in E x a m p le 7 .6 — t h a t is , h (0 ) =
g e t
(x ) =
h
(n )
p
2 ' (2 x − n )
n
s o th a t
(8 x − 3 ) =
(n )
h
p
2 ' (2 [8 x − 3 ] − n )
n
=
p
p
1
2
2 ' (1 6 x − 6 ) +
− 1
p
p
2
= ' (1 6 x − 6 ) − ' (1 6 x − 7 ).
2 ' (1 6 x − 7 )
C H A P T E R 7 . P R O B L E M S O L U T IO N S
1 3 2
T h e n , s in c e
3 ,3
(x ) = 2
p
3 ,3
(8 x − 3 ) fr o m a b o v e , s u b s titu tio n g iv e s
2
= 2
= 2
p
(8 x − 3 )
2
p
p
2 ' (1 6 x − 6 ) − 2
2 ' (1 6 x − 7 ).
Problem 7.14
U s in g E q . ( 7 .2 - 2 2 ) ,
V
3
= V
2
⊕ W
2
= V
1
⊕ W
1
⊕ W
2
= V
0
⊕ W
0
⊕ W
1
⊕ W
2
T h e s c a lin g a n d w a v e le t fu n c t io n s a r e p lo t t e d in F ig . P 7 .1 4 .
Problem 7.15
W ith j
0
= 1 th e a p p r o x im a tio n c o e ffi c ie n ts a re c 1 (0 ) a n d c 1 (1 ):
1 / 2
x
2
c 1 (0 ) =
x
p
2
2 d x =
p
2
2 4
0
1
c 1 (1 ) =
p
2 d x =
7
p
2 4
2
.
1 / 2
ψ3, 3(x) =2 2 ψ(8x-3)
22
0
-2 2
0
3/8
F ig u r e P 7 .1 3
1
1 3 3
V0
V1
V2
V3
W0
W1
W2
F ig u r e P 7 .1 4
T h e re fo re , th e V
1
a p p r o x im a tio n is
p
2
2 4
'
1 ,0
(x ) +
7
p
2 4
2
'
1 ,1
(x )
w h ic h , w h e n p lo t t e d , is id e n t ic a l t o t h e V 1 a p p r o x im a t io n in F ig . 7 .1 5 ( d ) . T h e
la s t tw o c o e ffi c ie n ts a r e d 1 (0 ) a n d d 1 (1 ), w h ic h a r e c o m p u te d a s in th e e x a m p le .
T h u s , th e e x p a n s io n is
y =
p
2 4
2
'
1 ,0
(x ) +
7
p
2 4
2
'
1 ,1
(x ) +
−
p
3 2
2
1 ,0
(x ) −
3
p
3 2
2
1 ,1
(x )
+ ···
Problem 7.16
( a ) B e c a u s e M = 4 , J = 2 , a n d j 0 = 1 , t h e s u m m a t io n s in E q s . ( 7 .3 - 5 ) t h r o u g h
( 7 .3 - 7 ) a r e p e r fo r m e d o v e r n = 0 , 1 , 2 , 3 , j = 1 , a n d k = 0 , 1 . U s in g H a a r fu n c t io n s
a n d a s s u m in g th a t th e y a re d is tr ib u te d o v e r th e r a n g e o f th e in p u t s e q u e n c e , w e
C H A P T E R 7 . P R O B L E M S O L U T IO N S
1 3 4
g e t
W
'
1
(1 ,0 ) =
1
=
W
'
1
(1 ,1 ) =
1
1
1
(1 )(
2
=
2 ) + (4 )(
1 ,1
f (0 )
2
1
(1 ,1 ) =
p
f (1 )'
(0 ) +
p
p
f (0 )
2
1
2
s o th a t th e D W T is f 5
(0 ) +
f (1 )
2 ) + (4 )(−
p
(0 ) +
f (1 )
1 ,0
1 ,1
(1 ) +
1 ,1
p
1 ,0
=
1 ,1
(2 ) +
p
2 )
=
(1 ) +
f (2 )
1 ,0
1 ,1
(1 ) +
p
2 / 2 ,− 3
2 / 2 ,− 3
p
f (2 )
2 ) + (0 )(−
p
1 ,1
p
f (3 )'
p
5 2
2
1 ,0
(3 )
f (3 )'
p
− 3 2
2
1 ,1
(3 )
(2 ) +
2 ) + (− 3 )(0 ) + (0 )(0 )
p
p
(2 ) +
1 ,0
f (2 )'
2 ) + (0 )(
(1 )(0 ) + (4 )(0 ) + (− 3 )(
2 / 2 ,− 3
f (2 )'
2 ) + (− 3 )(0 ) + (0 )(0 )
f (1 )'
(0 ) +
(1 ) +
1 ,0
(1 )(0 ) + (4 )(0 ) + (− 3 )(
2
(1 ,0 ) =
1 ,0
f (0 )'
2
=
W
(1 )(
2
=
W
f (0 )'
2
=
(2 ) +
2 )
=
f (3 )
p
− 3 2
2
1 ,0
(3 )
f (3 )
p
− 3 2
2
1 ,1
(3 )
2 / 2 g .
( b ) U s in g E q . ( 7 .3 - 7 ) ,
f (n ) =
W
2
1
[W
'
(1 ,0 )
(1 ,0 )'
1 ,0
1 ,0
(n ) + W
(n ) + W
'
(1 ,1 )
(1 ,1 )'
1 ,1
1 ,1
(n )+
(n )]
w h ic h , w ith n = 1 , b e c o m e s
p
p
p
2
f (1 ) =
(5 )( 2 ) + (− 3 )(0 ) + (− 3 )( 2 ) + (− 3 )(0 )
4
p
2 ( 2 )2
= 1 .
=
4
Problem 7.17
In
in
W
fu
ila
a c
is
tu itiv e ly , th e c o n tin u o u s w a v e le t tr a n s fo r m (C
d e x ” b e tw e e n th e s ig n a l a n d th e w a v e le t a t v
h e n th e in d e x is la r g e , th e r e s e m b la n c e is s tr
n c tio n is s im ila r to its e lf a t d iffe r e n t s c a le s , th e
r a t d iffe r e n t s c a le s . T h e C W T c o e ffi c ie n t v a lu
te r is tic p a tte r n . A s a r e s u lt, w e c a n s a y th a t th e
s e lf- s im ila r — lik e a fr a c ta l s ig n a l.
W T ) c a lc u la te s a “ r e s e m b la n c e
a r io u s s c a le s a n d tr a n s la tio n s .
o n g ; e ls e it is w e a k . T h u s , if a
r e s e m b la n c e in d e x w ill b e s im e s (th e in d e x ) w ill h a v e a c h a r fu n c tio n w h o s e C W T is s h o w n
1 3 5
{-1/ 2 , -3/ 2, 7/ 2, -3/ 2, 0}
{-1/ 2, 1/ 2}
2
Wψ(1, n) = {-3/ 2, -3/ 2}
{1/ 2, 1/ 2}
2
Wϕ(1, n) = {5/ 2, -3/ 2}
Wϕ(2, n) = f(n)
= {1, 4, -3, 0}
{1/ 2 , 5/ 2, 1/ 2, -3/ 2, 0}
F ig u r e P 7 .1 9
Problem 7.18
( a ) T h e s c a le a n d tr a n s la tio n p a r a m e te r s a r e c o n tin u o u s , w h ic h le a d s to th e
o v e r c o m p le te n e s s o f th e tr a n s fo r m .
(b
is
o f
o f
) T h e D W T is a
s u ffi c ie n t fo r
te n e a s ie r to in
th e fu n c tio n o
b
re
te
r
e t
c o
rp
im
te r c h o ic e w h e
n s tr u c tio n o f
re t b e c a u s e th
a g e . F o r e x a m
n w e n e e d
th e o r ig in a
e b u ilt- in r e
p le , s e e th e
a s p a c
l fu n c
d u n d
s e lf- s
e s a v
tio n
a n c y
im ila
in g re
o r im
te n d s
r ity o
p r
a g
to
f P
e s
e .
re
ro
e n ta t
T h e
in fo r
b le m
io n th a t
C W T is
c e tr a its
7 .1 7 .
Problem 7.19
T h e fi lt e r b a n k is t h e fi r s t b a n k in F ig . 7 .1 9 , a s s h o w n in F ig . P 7 .1 9 :
Problem 7.20
T h e c o m p le x ity is d e te r m in e d b y th e n u m b e r o f c o e ffi c ie n ts in th e s c a lin g a n d
w a v e le t v e c t o r s — t h a t is , b y n in E q s . ( 7 .2 - 1 8 ) a n d ( 7 .2 - 2 8 ) . T h is d e fi n e s t h e
n u m b e r o f ta p s in fi lte r s h (− n ), h ' (− n ), h (n ), a n d h ' (n ).
Problem 7.21
( a ) I n p u t ' ( n ) = f 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 g = ' 0 ,0 ( n ) f o r a t h r e e - s c a l e w a v e l e t t r a n s f o r m
w ith H a a r s c a lin g a n d w a v e le t fu n c tio n s . S in c e w a v e le t tr a n s fo r m c o e ffi c ie n ts
m e a s u r e th e s im ila r ity o f th e in p u t to th e b a s is fu n c tio n s , th e r e s u ltin g tr a n s fo r m
is
f W
'
(0 ,0 ),W
(0 ,0 ),W
W
(1 ,0 ),W
(2 ,3 )g = f 2
(1 ,1 ),W
p
(2 ,0 ),W
2 ,0 ,0 ,0 ,0 ,0 ,0 ,0 g .
(2 ,1 ),W
(2 ,2 )
C H A P T E R 7 . P R O B L E M S O L U T IO N S
1 3 6
= k = 0 .
p
(b ) U s in g th e s a m e re a s o n in g a s in p a r t (a ), th e tr a n s fo r m is f 0 ,2 2 ,0 ,0 ,0 ,0 ,0 ,0 g .
T h e W
'
(0 , 0 ) t e r m c a n b e c o m p u t e d u s in g E q . ( 7 .3 - 5 ) w it h j
0
( c ) F o r th e g iv e n tr a n s fo r m , W (2 , 2 ) = B a n d a ll o th e r tr a n s fo r m c o e ffi c ie n ts a r e
0 . T h u s , th e in p u t m u s t b e p r o p o r tio n a l to
2 ,2 ( x ) . T h e i n p u t s e q u e n c e m u s t b e
o f t h e fo r m f 0 , 0 , 0 , 0 , C , − C , 0 , 0 g fo r s o m e C . T o d e t e r m in e C , u s e E q . ( 7 .3 - 6 ) t o
w r ite
W
(2 ,2 ) =
1
p
f f (0 )
8
f (4 )
=
2 ,2
p
1
(4 ) +
2 ,2
(0 ) +
f (5 )
p
2 ,2
(5 ) +
2 ,2
(1 ) +
f (6 )
f (2 )
2 ,2
(6 ) +
2 ,2
(2 ) +
f (7 )
f (3 )
2 ,2
2 ,2
(3 )+
(7 )g
f (0 )(0 ) + (0 )(0 ) + (0 )(0 ) + (0 )(0 ) + (C )(2 ) + (− C )(− 2 )+
8
(0 )(0 ) + (0 )(0 )g
=
f (1 )
1
8
f 2 C + 2 C g =
4 C
p
8
=
p
2 C .
p
B e c a u s e th is c o e ffi c ie n t is k n o w n to h a v e th e v a lu e B , w e h a v e th a t 2 C = B o r
p
2
B .
C =
2
p
p
T h u s , th e in p u t s e q u e n c e is f 0 ,0 ,0 ,0 , 2 B / 2 ,−
2 B / 2 , 0 , 0 g . T o c h e c k th e r e s u lt
s u b s t it u t e t h e s e v a lu e s in t o E q . ( 7 .3 - 6 ) :
p
1
2
W (2 ,2 ) = p f (0 )(0 ) + (0 )(0 ) + (0 )(0 ) + (0 )(0 ) + (
B )(2 )+
2
8
p
2
B )(− 2 ) + (0 )(0 ) + (0 )(0 )g
(−
2
p
1 p
2 B g
= p f 2 B +
8
= B .
Problem 7.22
T h e y a re b
r e s o lu tio n
F W T , th e s e
m id , th e y a
T o c o n s
in F ig . 7 .1 0
p la c e th e 6
o th m u lti- r e s o lu tio n r e p r e s e n ta tio n s th a t e m p lo
a p p r o x im a tio n im a g e a n d a s e r ie s o f “d iffe re n c
“d iffe re n c e ” im a g e s a re th e tr a n s fo r m d e ta il c o e ffi
r e th e p r e d ic tio n r e s id u a ls .
tr u c t th e a p p r o x im a tio n p y r a m id th a t c o r re s p o n
(a ), w e w ill u s e th e F W T − 1 2 - d s y n th e s is b a n k o
4 × 6 4 a p p r o x im a t io n “ c o e ffi c ie n t s ” fr o m F ig . 7 .1 0
y a s in g le r e d u c e d e ” im a g e s . F o r th e
c ie n ts ; fo r th e p y r a d s to th e tra n s fo r m
f F ig . 7 .2 4 ( c ) . F ir s t ,
(a ) a t th e to p o f th e
1 3 7
D
22
22
{-1/ 2, 1/ 2}
2
Each Row Columns
(along n)
Wϕ(1, m, n)
= 3 -1
62
2
42
{1/ 2, 1/ 2}
2
Wψ(0, 0, 0)
=[0]
{-1/ 2, 1/ 2}
2
Each
Column
Rows
(along m)
{1/ 2, 1/ 2}
2
Each
Column
Rows
{-1/ 2, 1/ 2}
2
Each
Column
Rows
{1/ 2, 1/ 2}
2
Each
Column
Rows
V
Wψ(0, 0, 0)
=[4]
5 -3
40
H
Wψ(0, 0, 0)
=[-3]
Each Row Columns
Ordered per
Fig. 7.24(b)
Wϕ(0, 0, 0)
=[5]
F ig u r e P 7 .2 3
p y r a m id b e in g c o n s tr u c te d . T h e n u s e it, a lo n g w ith 6 4 × 6 4 h o r iz o n ta l,
a n d d ia g o n a l d e t a il c o e ffi c ie n t s fr o m t h e u p p e r - le ft o f F ig . 7 .1 0 ( a ) , t o d
fi lt e r b a n k in p u t s in F ig . 7 .2 4 ( c ) . T h e o u t p u t w ill b e a 1 2 8 × 1 2 8 a p p r o x
o f th e o r ig in a l im a g e a n d s h o u ld b e u s e d a s th e n e x t le v e l o f th e a p p r o x
p y r a m id . T h e 1 2 8 × 1 2 8 a p p r o x im a tio n is th e n u s e d w ith th e th re e 1 2 8 ×
t a il c o e ffi c ie n t im a g e s in t h e u p p e r 1 / 4 o f t h e t r a n s fo r m in F ig . 7 .1 0 ( a )
t h e s y n t h e s is fi lt e r b a n k in F ig . 7 .2 4 ( c ) a s e c o n d t im e — p r o d u c in g a 2 5 6 ×
p r o x im a tio n th a t is p la c e d a s th e n e x t le v e l o f th e a p p r o x im a tio n p y r a m
p r o c e s s is th e n re p e a te d a th ird tim e to re c o v e r th e 5 1 2 × 5 1 2 o r ig in a
w h ic h is p la c e d a t th e b o tto m o f th e a p p r o x im a tio n p y r a m id . T h u s , th e
im a tio n p y r a m id w o u ld h a v e 4 le v e ls .
v e r tic a l,
r iv e th e
im a tio n
im a tio n
1 2 8 d e to d r iv e
2 5 6 a p id . T h is
l im a g e ,
a p p ro x -
Problem 7.23
O n e p a s s t h r o u g h t h e F W T 2 - d fi lt e r b a n k o f F ig . 7 .2 4 ( a ) is a ll t h a t is r e q u ir e d
( s e e F ig . P 7 .2 3 ) .
Problem 7.24
A s c a n b e
in v a r ia n t.
in th e p r o
c o m p u ta t
s e e n
If th e
b le m
io n p
in t
in p
a re
ro c e
h e s e q u e n c e o
u t is s h ifte d , th
1 2 8 × 1 2 8 , th e
s s . T h e fi lte r b
f im a g e s t
e tra n s fo r
y b e c o m e
a n k o f F ig
h a
m
th
. 7
t a re
c h a n
e W '
.2 4 ( a
s h o w
g e s . S
(7 ,m
) c a n
n , th e
in c e a
,n ) in
b e u s
D W T is
ll o r ig in
p u ts fo r
e d w ith
n o t
a l im
th e
j + 1
s h ift
a g e s
F W T
= 7 .
C H A P T E R 7 . P R O B L E M S O L U T IO N S
1 3 8
F o r a s in g le s c a le tr a n s fo r m , tr a n s fo r m c o e ffi c ie n ts W ' (6 , m , n ) a n d W i (6 , m , n )
fo r i = H , V , D a r e g e n e r a te d . W ith H a a r w a v e le ts , th e tr a n s fo r m a tio n p r o c e s s
s u b d iv id e s th e im a g e in to n o n - o v e r la p p in g 2 × 2 b lo c k s a n d c o m p u te s 2 - p o in t
a v e r a g e s a n d d iffe r e n c e s (p e r th e s c a lin g a n d w a v e le t v e c to r s ). T h u s , th e r e a r e
n o h o r iz o n ta l, v e r tic a l, o r d ia g o n a l d e ta il c o e ffi c ie n ts in th e fi r s t tw o tr a n s fo r m s
s h o w n ; th e in p u t im a g e s a r e c o n s ta n t in a ll 2 × 2 b lo c k s (s o a ll d iffe r e n c e s a r e
0 ). If th e o r ig in a l im a g e is s h ifte d b y o n e p ix e l, d e ta il c o e ffi c ie n ts a r e g e n e r a te d
s in c e th e re a re th e n 2 × 2 a re a s th a t a re n o t c o n s ta n t. T h is is th e c a s e in th e th ird
tra n s fo r m s h o w n .
Problem 7.25
T h e t a b le is c o m p le t e d a s s h o w n in F ig . P 7 .2 5 . T h e fu n c t io n s a r e d e t e r m in e d
u s in g E q s . ( 7 .2 - 1 8 ) a n d ( 7 .2 - 2 8 ) w it h t h e H a a r s c a lin g a n d w a v e le t v e c t o r s fr o m
E x a m p le s 7 .5 a n d 7 .6 :
' (x ) = ' (2 x ) + ' (2 x − 1 )
(x ) = ' (2 x ) − ' (2 x − 1 ).
T o o r d e r th e w a v e le t fu n c tio n s in fr e q u e n c y , c o u n t th e n u m b e r
th a t a r e m a d e b y e a c h fu n c tio n . F o r e x a m p le , V 0 h a s th e fe w e s t tr a
2 ) a n d lo w e s t fr e q u e n c y c o n te n t, w h ile W 2 ,A A h a s th e m o s t (9 ) tr
c o r r e s p o n d in g ly h ig h e s t fr e q u e n c y c o n te n t. F r o m to p to b o tto m
th e r e a r e 2 , 3 , 5 , 4 , 9 , 8 , 6 , a n d 7 tr a n s itio n s , r e s p e c tiv e ly . T h e r e fo r e ,
o r d e r e d s u b s p a c e s a r e (fr o m lo w to h ig h fr e q u e n c y ) V 0 , W 0 , W 1 ,D
W 2 ,D D , W 2 ,A D , a n d W 2 ,A A .
o f tr a n s itio n s
n s itio n s (o n ly
a n s itio n s a n d
in th e fi g u re ,
th e fre q u e n c y
, W 1 ,A , W 2 ,D A ,
Problem 7.26
( a ) T h e a n a ly s is t r e e is s h o w n in F ig . P 7 .2 6 ( a ) .
( b ) T h e c o r r e s p o n d in g fr e q u e n c y s p e c t r u m is s h o w n in F ig . P 7 .2 6 ( b ) .
Problem 7.27
F ir s t u s e th e e n tr o p y m e a s u r e to fi n d th e s ta r tin g v a lu e fo r th e in p u t s e q u e n c e ,
w h ic h is
7
E f f (n )g =
f
2
(n ) ln
f
2
(n )
= 2 .7 7 2 6 .
n = 0
T h e n p e r fo r m a n ite r a tio n o f th e F W T a n d c o m p u te th e e n tr o p y o f th e g e n e r a te d
a p p r o x im a t io n a n d d e t a il c o e ffi c ie n t s . T h e y a r e 2 .0 7 9 4 a n d 0 , r e s p e c t iv e ly . S in c e
1 3 9
V0
V1
W0
W1,A
V2
W1,D
W1
W2,AA
W2,A
W2,AD
W2,DA
V3
W2
W2,DD
W2,D
F ig u r e P 7 .2 5
t h e ir s u m is le s s t h a n t h e s t a r t in g e n t r o p y o f 2 .7 7 2 6 , w e w ill u s e t h e d e c o m p o s itio n .
B e c
ra n te d
it s h o u
d e c o m
a u
. T
ld
p o
s e
h u
b e
s it
th
s ,
d
io
e d e ta
w e p e
e c o m
n s a re
il e n tr
r fo r m
p o s e d
c a lle d
o p
a n
a g
fo
y is
o th
a in
r. T
0 , n o f
e r F W
. T h is
h e re s
u r th e r d e c o m
T ite r a tio n o n
p r o c e s s is th e
u ltin g o p tim a
p o s itio
th e a p
n re p e
l tre e is
n o
p ro
a te
s h
f th e
x im
d u n
o w n
d e ta il is
a tio n to
til n o fu
in F ig . P
VJ
H
VJ -1
VJ -2
H
WJ -2
V
WJ -2
WJ -1
V
WJ -1
D
WJ -1
D
D
D
D
D
WJ -1,A WJ -1,H WJ -1,V WJ -1,D
WJ -2
D
D
D
D
WJ -1,HA WJ -1,HH WJ -1,HV WJ -1,HD
F ig u r e P 7 .2 6 ( a )
w a
s e e
r th
7 .2
rif
e r
7 .
C H A P T E R 7 . P R O B L E M S O L U T IO N S
1 4 0
ωVERT
a = VJ -2
π
w z
o n
ml x
w
n o
x l m
WJ -1
c
d
H
b =WJ -2
z
V
c =WJV-2
D
d =WJ -2
d
D
x =WJ -1,A
H
−π
WJ -1
a
b
d
w
z
WJ -1
π
d
c
ml
o n x
H
b
z = WJD-1,V
D
w =WJ -1,D
D
V
WJ -1
x
l m
n o
z
w
l = WJ -1,HA
D
m =WJ -1,HH
D
n =WJ -1,HV
−π
D
o =WJ -1,HD
F ig u r e P 7 .2 6 ( b )
2.7726
2.0794
1.3863
0.6931
0
ωHORIZ
0
0
F ig u r e P 7 .2 7
D u m m y w h it e t e x t t o m o v e fi g u r e ..
0
0
Chapter 8
Problem Solutions
Problem 8.1
( a ) A h is to g r a m e q u a liz e d im a g e (in th e o r y ) h a s a n in te n s ity d is tr ib u tio n w h ic h
is u n ifo r m . T h a t is , a ll in t e n s it ie s a r e e q u a lly p r o b a b le . E q . ( 8 .1 - 4 ) t h u s b e c o m e s
L
w h e re
a re e q
fe w e r
to c o v
1 / 2 n is t
u a lly p r o
b its th a n
e r th e 2 n
h e p
b a b
a n y
le v e
ro b a b
le , th e
o th e r
ls . T h
a v g
ility o f
re is n o
. T h u s ,
is , o f c o
L
o c c
a d
w e
u rs
a v g
2
n
2
1
=
− 1
l (r k )
n
k = 0
u r re n c e o f a n y in te n
v a n ta g e to a s s ig n in g
a s s ig n e a c h th e fe w e
e is n b its a n d L a v g
2
1
=
=
S in c
p a r t
s s ib
m e s
e a ll in
ic u la r
le b its
n b its
te n s itie s
in te n s ity
re q u ire d
a ls o :
− 1
(n )
n
2
=
n
s ity .
a n y
s t p o
b e c o
k = 0
1
2
n
2
n
n
n .
(b ) S in c e s p a tia l re d u n d a n c y is a s s o c ia te d w ith th e g e o m e tr ic a r r a n g e m e n t o f
th e in te n s ite s s in th e im a g e , it is p o s s ib le fo r a h is to g r a m e q u a liz e d im a g e to
c o n ta in a h ig h le v e l o f s p a tia l r e d u n d a n c y - o r n o n e a t a ll.
Problem 8.2
( a ) A s in g le lin e o f r a w d a ta c o n ta in s n 1 = 2 n b its . T h e m a x im u m r u n le n g th
w o u ld b e 2 n a n d th u s r e q u ir e n b its fo r r e p r e s e n ta tio n . T h e s ta r tin g c o o r d in a te
1 4 1
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 4 2
o f e a c h
p ix e l lin
u s e d to
lin e . T h
r u n a
e . S in
s ig n a
u s , th
ls o
c e
l th
e to
re q u ire s n
a r u n le n g
e s ta r t o f e
ta l n u m b e
b its s in c
th o f 0 c a
a c h n e w
r o f b its r
n
=
2
e it
n n
lin e
e q u
m a y
o t o c
- a n
ire d
2 n + N
=
a v g
1 + N
2 n
b e
c u
a d
to
a r b itr a
r a n d th
d itio n a
c o d e a n
r ily lo c a te
e r u n - le n
l 2 n b its a
y s c a n lin
d w ith in th e 2 n
g th p a ir (0 ,0 ) is
re re q u ire d p e r
e is
(n + n )
a v g
w h e r e N a v g is th e a v e r a g e n u m b e r o f r u n - le n g th p a ir s o n a lin e . T o a c h ie v e s o m e
le v e l o f c o m p r e s s io n , C m u s t b e g r e a te r th a n 1 . S o ,
n
n
=
C
1
2
1 + N
2 n
a n d
N
(b ) F o r n = 1 0 , N
n
2
=
a v g
<
2
> 1
a v g
n − 1
− 1 .
n
m u s t b e le s s t h a n 5 0 .2 r u n - le n g t h p a ir s p e r lin e .
a v g
Problem 8.3
T h e o r ig
e n c e s b e
m u s t b e
s ig n a l- to
in a l p ix e l in t
tw e e n th e m
m u ltip lie d b
- n o is e c a lc u
f
B a s e 1 0
1 0 8
1 3 9
1 3 5
2 4 4
1 7 2
1 7 3
5 6
9 9
e n s itie
a re s h o
y 1 6 to
la tio n s
s , th e ir 4 - b it q u a n tiz e d c o u n te r p a r ts , a n d th e d iffe rw n in T a b le P 8 .3 . N o t e t h a t t h e q u a n t iz e d in t e n s it ie s
d e c o d e o r d e c o m p re s s th e m fo r th e r m s e rro r a n d
.
x ,y
0
1
1
1
1
1
0
0
B a s e
1 1 0 1
0 0 0 1
0 0 0 0
1 1 1 0
0 1 0 1
0 1 0 1
0 1 1 1
1 1 0 0
2
1 0 0
0 1 1
1 1 1
1 0 0
1 0 0
1 0 1
0 0 0
0 1 1
T a b le P
f^ x , y
B a s e 2
B
0 1 1 0
1 0 0 0
1 0 0 0
1 1 1 1
1 0 1 0
1 0 1 0
0 0 1 1
0 1 1 0
8 .3
a s e 1 0
6
8
8
1 5
1 0
1 0
3
6
1 6 f^ x , y − f
B a s e 1 0
-1 2
-1 1
-7
-4
-1 2
-1 3
-8
-3
x ,y
1 4 3
U s in g E q . ( 8 .1 - 1 0 ) , t h e r m s e r r o r is
e
5
6
6 1
7
=
r m s
8
*
1
8
*
=
7
1 6 f^ x , y
f
−
x ,y
2
x = 0 y = 0
(− 1 2 )2 + (− 1 1 )2 + (− 7 )2 + (− 4 )2 + (− 1 2 )2 + (− 1 3 )2 + (− 8 )2 + (− 3 )
=
=
0
1
8
2
(7 1 6 )
9 .4 6
o r a b o u t 9 .5 in t e n s it y le v e ls . F r o m E q . ( 8 .1 - 1 1 ) , t h e s ig n a l- t o - n o is e r a t io is
2
S N R
=
2
m s
9 6
=
=
'
0
x = 0
2 +
0
x = 0
2
7
1 2
y = 0
8 2 +
2
7
1 6 f^ x , y
1 2 8
2
1 6 f^ x , y
y = 0
2
+ 2 4 0
f
−
2
2
x ,y
+ 1 6 0
7 1 6
2
+ 1 6 0
2
+ 4 8
2
+ 9 6
2
1 6 2 3 0 4
7 1 6
2 2 7 .
Problem 8.4
(a
u s
d e
d e
) T a b
e d in
c im a
c o d e
le P 8 .4 s
e a c h s t
l e q u iv a
d IG S in
h o w s th e
e p , th e 4 le n t o f th
te n s itie s
s ta r
b it I
e IG
a n d
tin
G S
S c
th
g in te n s ity v a lu e s ,
c o d e a n d its e q u iv
o d e m u ltip lie d b y
e in p u t v a lu e s , a n d
th e ir 8 - b it c o
a le n t d e c o d e
1 6 ), th e e rro r
th e s q u a re d
d e s , th e IG S s u m
d v a lu e (th e
b e tw e e n th e
e rro r.
( b ) U s in g E q . ( 8 .1 - 1 0 ) a n d t h e s q u a r e d e r r o r v a lu e s fr o m T a b le P 8 .4 , t h e r m s e r r o r
is
*
1
e r m s =
(1 4 4 + 2 5 + 4 9 + 1 6 + 1 6 + 1 6 9 + 6 4 + 9 )
8
*
1
(4 9 2 )
=
8
=
7 .8 4
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 4 4
o r a b o u t 7 .8 in t e n s it y le v e ls . F r o m E q . ( 8 .1 - 1 1 ) , t h e s ig n a l- t o - n o is e r a t io is
S N R
9 6
=
m s
'
1 0 8
1 3 9
1 3 5
2 4 4
1 7 2
1 7 3
5 6
9 9
8 - b it C o d e
0 1
1 0
1 0
1 1
1 0
1 0
0 0
0 1
1 0
0 0
0 0
1 1
1 0
1 0
1 1
1 0
2
+ 1 4 4
+ 1 2 8
2
+ 2 4 0
2
+ 1 7 6
4 9 2
2
+ 1 6 0
2
+ 6 4
2
+ 9 6
2
1 7 3 8 2 4
4 9 2
3 5 3 .
=
In te n s ity
2
1 1
1 0
0 1
0 1
1 1
1 1
1 0
0 0
0 0
1 1
1 1
0 0
0 0
0 1
0 0
1 1
0 0
0 1
1 0
1 0
1 1
1 0
1 0
0 1
0 1
S u
0 0
1 0
0 1
0 0
1 1
1 1
1 0
0 0
1 0
m
0 0
1 1
0 1
1 1
0 1
0 0
1 1
0 1
1 0
T a b le P 8 .4
IG S C o d e
D e c o d e d IG S
E rro r
S q u a re E rro r
-1 2
5
-7
-4
4
-1 3
8
-3
1 4 4
2 5
4 9
1 6
1 6
1 6 9
6 4
9
0 0
0 0
1 1
1 0
0 0
0 0
0 1
0 1
0 0
0 1
1 0
1 0
1 1
1 0
1 0
0 1
0 1
1 0
0 1
0 0
1 1
1 1
1 0
0 0
1 0
9 6
1 4 4
1 2 8
2 4 0
1 7 6
1 6 0
6 4
9 6
Problem 8.5
(a ) T h e m a x im u m c o m p re s s io n w ith H u ffm a n c o d in g is C =
(b ) N o ,
p e r in te
a tim e .
(c ) O n e
m a n e n
a n d H u
8
5 .3
= 1 .5 0 9 .
b u t H u ffm a n c o d in g d o e s p r o v id e th e s m a lle s t p o s s ib le n u m b e r o f b its
n s ity v a lu e s u b je c t to th e c o n s tr a in t th a t th e in te n s itie s a r e c o d e d o n e a t
p o s s ib ility is to e lim in a te s p a tia l r e d u n d a n c ie s in th e im a g e b e fo r e H u ffc o d in g . F o r in s ta n c e , c o m p u te th e d iffe r e n c e s b e tw e e n a d ja c e n t p ix e ls
ffm a n c o d e th e m .
Problem 8.6
T h e c o n v e r s io n fa c to r s a r e c o m p u te d u s in g th e lo g a r ith m ic r e la tio n s h ip
lo g
a
x =
1
lo g
b
a
lo g
b
x .
T h u s , 1 H a r t le y = 3 .3 2 1 9 b it s a n d 1 n a t = 1 .4 4 2 7 b it s .
1 4 5
Problem 8.7
3
4
L e t t h e s e t o f s o u r c e s y m b o ls b e a 1 , a 2 , ..., a q w it h p r o b a b ilit ie s P (a 1 ) , P (a
T h e n , u s in g E q . ( 8 .1 - 6 ) a n d t h e fa c t t h a t t h e s u m o f a ll P (a i ) is 1 , w e g e t
⎤
⎡
q
⎢
lo g q ⎣
=
lo g q − H
P
a
j
) , ..., P
q
⎥
⎦+
P
j = 1
a
j
lo g P
a
j
j = 1
q
=
2
q
P
lo g q +
a
j
P
j = 1
a
lo g P
j
a
j
j = 1
q
=
P
a
lo g q P
j
a
j
.
j = 1
U s in g t h e lo g r e la t io n s h ip fr o m P r o b le m 8 .6 , t h is b e c o m e s
q
= lo g e
P
a
j
ln q P
a
j
.
j = 1
T h e n , m u ltip ly in g th e in e q u a lity ln x ≤ x − 1 b y - 1 to g e t ln 1 / x ≥
p ly in g it to th is la s t r e s u lt,
⎡
⎤
q
1
⎦
P a j ⎣1 −
lo g q − H
≥
lo g e
q
P
a
j
i j = 1
⎤
⎡
q
q
P
a
1
j
⎥
⎢
≥
lo g e ⎣
P a j −
⎦
q
a j
j = 1
j = 1 P
≥
lo g e [1 − 1 ]
≥
0
1 − x a n d a p -
s o th a t
lo g q ≥ H .
T h e re fo r
e q u a lity
p o in t in
1 / q fo r a
e , H is a lw a y s le s s th a n , o r e q u a l to , lo g q . F u r th e r m o r e , in v ie w o f th e
c o n d itio n (x = 1 ) fo r ln 1 / x ≥ 1 − x , w h ic h w a s in tr o d u c e d a t o n ly o n e
th e a b o v e d e r iv a tio n , w e w ill h a v e s tr ic t e q u a lity if a n d o n ly if P (a j ) =
ll j .
Problem 8.8
(a ) T h e re a re tw o u n iq u e c o d e s .
a
T
q
.
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 4 6
21
243
95
169
3/8
3/8
1/8
1/8
3/8
3/8
2/8
5/8
3/8
21
243
95
169
1
00
000
001
1
00
01
0
1
Code assignments
Source reductions
F ig u r e P 8 .9
( b ) T h e c o d e s a r e : (1 ) 0 , 1 1 , 1 0 a n d (2 ) 1 , 0 0 , 0 1 . T h e c o d e s a r e c o m p le m e n ts
o f o n e a n o th e r. T h e y a r e c o n s tr u c te d b y fo llo w in g th e H u ffm a n p r o c e d u r e fo r
th r e e s y m b o ls o f a r b itr a r y p r o b a b ility .
Problem 8.9
( a ) T h e e n t r o p y o f t h e im a g e is e s t im a t e d u s in g E q . ( 8 .1 - 7 ) t o b e
2 5 5
H˜
=
p
−
r
(r k ) lo g
2
p
r
(r k )
k = 0
=
−
=
=
4
4
1 2
1 2
4
4
lo g 2
+
lo g 2
+
lo g 2
+
3 2
3 2
3 2
3 2
3 2
3 2
− [− 0 .5 3 0 6 − 0 .3 7 5 − 0 .3 7 5 − 0 .5 3 0 6 ]
1 2
lo g
3 2
1 2
2
3 2
1 .8 1 1 b it s / p ix e l.
T h e p r o b a b ilit ie s u s e d in t h e c o m p u t a t io n a r e g iv e n in T a b le P 8 .9 - 1 .
In te n s ity
2 1
9 5
1 6 9
2 4 3
T a b le P 8 .9 - 1
C o u n t
P ro b
1 2
3
4
1
4
1
1 2
3
a b ility
/ 8
/ 8
/ 8
/ 8
( b ) F ig u r e P 8 .9 s h o w s o n e p o s s ib le H u ffm a n s o u r c e r e d u c t io n a n d c o d e a s s ig n m e n t . U s e t h e p r o c e d u r e s d e s c r ib e d in S e c t io n 8 .2 .1 . T h e in t e n s it ie s a r e fi r s t
a r r a n g e d in o r d e r o f p r o b a b ility fr o m th e to p to th e b o tto m (a t th e le ft o f th e
s o u r c e r e d u c tio n d ia g r a m ). T h e le a s t p r o b a b le s y m b o ls a r e th e m c o m b in e d to
c r e a te a r e d u c e d s o u r c e a n d th e p r o c e s s is r e p e a te d fr o m le ft to r ig h t in th e d ia g r a m . C o d e w o r d s a r e th e n a s s ig n e d to th e r e d u c e d s o u r c e s y m b o ls fr o m r ig h t
to le ft. T h e c o d e s a s s ig n e d to e a c h in te n s ity v a lu e a r e r e a d fr o m th e le ft s id e o f
th e c o d e a s s ig n m e n t d ia g r a m .
1 4 7
( c ) U s in g E q . ( 8 .1 - 4 ) , t h e a v e r a g e n u m b e r o f b it s r e q u ir e d t o r e p r e s e n t e a c h p ix e l
in th e H u ffm a n c o d e d im a g e (ig n o r in g th e s to r a g e o f th e c o d e its e lf ) is
L
a v g
3
= 1
3
8
+ 2
1
+ 3
8
1
+ 3
8
=
8
1 5
= 1 .8 7 5 b it s / p ix e l.
8
T h u s , th e c o m p re s s io n a c h ie v e d is
8
C =
B e c a u s e th
c o d in g re d
4 .2 7
× 1 0 0
4 .4 1 7
re m o v a l o f
e t
u n
o r
c o
h
d
9
d
1 .8 7 5
e o re tic a l c o m p re s s io
a n c y i s 1 . 88 1 1 = 4 . 4 1 7 ,
6 .6 7 % o f t h e m a x im u
in g r e d u n d a n c y a lo n
= 4 .2 7 .
n r e s u ltin g fr o m th e e lim in a tio n o f a ll
th e H u ffm a n c o d e d im a g e a c h ie v e s
m c o m p r e s s io n p o s s ib le th r o u g h th e
e .
( d ) W e c a n c o m p u te th e r e la tiv e fr e q u e n c y o f p a ir s o f p ix e ls b y a s s u m in g th a t
th e im a g e is c o n n e c te d fr o m lin e to lin e a n d e n d to b e g in n in g . T h e r e s u ltin g
p r o b a b ilit ie s a r e lis t e d in T a b le P 8 .9 - 2 .
T a b le
In te n s ity p a ir
C
(2 1 , 2 1 )
(2 1 , 9 5 )
(9 5 , 1 6 9 )
(1 6 9 , 2 4 3 )
(2 4 3 , 2 4 3 )
(2 4 3 , 2 1 )
P 8 .9 - 2
o u n t
P ro b a
8
1 /
4
1 /
4
1 /
4
1 /
8
1 /
4
1 /
b ility
4
8
8
8
4
8
T h e e n t r o p y o f t h e in t e n s it y p a ir s is e s t im a t e d u s in g E q . ( 8 .1 - 7 ) a n d d iv id in g b y
2 (b e c a u s e th e p ix e ls a r e c o n s id e r e d in p a ir s ):
1
2
H˜
=
−
=
=
1
1
lo g
1
2
+
1
2 4
4
8
2 .5
2
1 .2 5 b it s / p ix e l.
lo g
1
2
8
+
1
8
lo g
1
2
8
+
1
8
lo g
1
2
8
+
1
4
lo g
1
2
4
+
1
8
lo g
1
2
8
T h e d iffe r e n c e b e tw e e n th is v a lu e a n d th e e n tr o p y in (a ) te lls u s th a t a m a p p in g
c a n b e c r e a t e d t o e lim in a t e (1 .8 1 1 − 1 .2 5 ) = 0 .5 6 b it s / p ix e l o f s p a t ia l
re d u n d a n c y .
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 4 8
( e ) C o n s tr u c t a d iffe r e n c e im a g e b y r e p lic a tin g th e fi r s t c o lu m n o f th e o r ig in a l
im a g e a n d u s in g th e a r ith m e tic d iffe r e n c e b e tw e e n a d ja c e n t c o lu m n s fo r th e
r e m a in in g e le m e n ts . T h e d iffe r e n c e im a g e is
2 1
0
0
7 4
7 4
7 4
0
0
2 1
0
0
7 4
7 4
7 4
0
0
2 1
0
0
7 4
7 4
7 4
0
0
2 1
0
0
7 4
7 4
7 4
0
0
T h e p r o b a b ilitie s o f its v a r io u s e le m e n ts a r e g iv e n in T a b le 8 .9 - 3 .
T a b le P 8 .9 - 3
In te n s ity d iffe re n c e
C o u n t
2 1
4
0
1 6
7 4
1 2
P ro b a
1 /
1 /
3 /
b ility
8
2
8
T h e e n t r o p y o f t h e d iffe r e n c e im a g e is e s t im a t e d u s in g E q . ( 8 .1 - 7 ) t o b e
H˜ = −
1
8
lo g
1
2
1
+
8
1
lo g
2
( f ) T h e e n tr o p y c a lc u la te d in (a ) is
d e p e n d e n t p ix e ls . T h e e n tr o p y (o
s m a lle r th a t th e v a lu e fo u n d in (a
in d e p e n d e n t . T h e r e is a t le a s t (1 .8
d a n c y in th e im a g e . T h e d iffe re n c e
th a t s p a tia l r e d u n d a n c y , le a v in g o n
2
3
+
2
3
lo g
8
2
= 1 .4 1 b it s / p ix e l.
8
b a s e d o n th e a s s u m p tio n
f th e p ix e l p a ir s ) c o m p u te
), r e v e a ls th a t th e p ix e ls a
1 1 − 1 .2 5 ) = 0 .5 6 b it s / p ix e
im a g e m a p p in g u s e d in (e
ly (1 .4 1 − 1 .2 5 ) = 0 .1 6 b it s /
o f s ta tis tic a lly in d in (d ), w h ic h is
r e n o t s ta tis tic a lly
l o f s p a tia l re d u n ) re m o v e s m o s t o f
p ix e l.
Problem 8.10
T h e d e c o d e d m e s s a g e is a
3
a
6
a
6
a
2
a
5
a
2
a
2
a
2
a
4
.
Problem 8.11
T h e G o lo m b c o d e is s h o w n in T a b le P 8 .1 1 . It is c o m p u t e d b y t h e p r o c e d u r e o u t lin e d in S e c t io n 8 .2 .2 in c o n ju n c t io n w it h E q . ( 8 .2 - 1 ) .
1 4 9
In te g e r n
U n a r y c o d e o f
0
8n 9
m
T a b le P 8 .1 1
T r u n c a te d n m o d m
0
0
0 0
0 1 0
0 1 1
1 0 0
1 0 1
1 0 1
1 1 0
1 1 0
1 1 0
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1
0
1 0
2
0
1 1
3
1 0
0
4
1 0
1 0
5
1 0
1 1
6
1 1 0
1 1 0
1 1 0
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
0
7
8
9
1 0
1 1
1 2
1 3
1 4
1 5
1 0
1 0 1
0
0
0
1 0
1 0 1
0
1 0
1 0
1 0
1 1 0
G o lo m b c o d e G
0
1 0
1 1
0
3
(n )
0
1
0
1 0
1 1
0 0
0 1
0 1
1 0
1 0
1 0
1 1
0
1
0
1 0
1 1
0 0
Problem 8.12
T o d e c o d e G
m
(n ), le t k =
:
lo g
2
m
;
a n d c = 2
k
− m . T h e n :
1 . C o u n t th e n u m b e r o f 1 s in a le ft- to - r ig h t s c a n o f a c o n c a te n a te d G m (n ) b it
s e q u e n c e b e fo r e r e a c h in g th e fi r s t 0 , a n d m u ltip ly th e n u m b e r o f 1 s b y m .
2 . If th e d e c im a l e q u iv a le n t o f th e n e x t k − 1 b its is le s s th a n c , a d d it to r e s u lt
fr o m s te p 1 ; e ls e a d d th e d e c im a l e q u iv a le n t o f th e n e x t k b its a n d s u b tr a c t
c .
F o r e x a m p le , to d e c o d e th e fi r s t G 3 (n ) c o d e
:
;
1 1 1 1 0 1 0 0 1 1 ..., le t k =
lo g 2 3 = 2 , c = 2 2 −
in a le ft- to - r ig h t s c a n o f th e b it s tr e a m b e fo
th e n u m b e r o f 1 s b y 3 y ie ld s 1 2 (th e r e s u lt
id e n tifi e d in s te p 1 is a 1 , w h o s e d e c im a l e q u
S o a d d th e d e c im a l e q u iv a le n t o f th e 2 b its
a n d s u b tr a c t 1 . T h u s , th e fi r s t in te g e r is 1 2 +
th e n e x t c o d e w o rd , w h ic h b e g in s w ith b its 0
in th e G o lo m b c o d e d b it s e q u e n c e
3 = 1 , a n d n o te th a t th e re a re 4 1 s
r e r e a c h in g th e fi r s t 0 . M u ltip ly in g
o f s te p 1 ). T h e b it fo llo w in g th e 0
iv a le n t is n o t le s s t h a n c , i.e ., 1 ≮ 1 .
fo llo w in g th e 0 id e n tifi e d in s te p 1
2 − 1 = 1 3 . R e p e a t th e p ro c e s s fo r
1 1 ...
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 5 0
Problem 8.13
T h e p r o b a b ilit y m a s s fu n c t io n d e fi n e d b y E q . ( 8 .2 - 2 ) is fo r a n in fi n it e s e t o f in te g e r s . It is n o t p o s s ib le to lis t a n in fi n ite s e t o f s o u r c e s y m b o ls a n d p e r fo r m th e
s o u r c e r e d u c t i o n s r e q u i r e d b y H u f f m a n ’s a p p r o a c h .
Problem 8.14
T h e e x p o n e n tia l G o lo m b c o d e is s h o w n in T a b le P 8 .1 4 . It is c o m p u te d b y th e
p r o c e d u r e o u t lin e d in S e c t io n 8 .2 .2 in c o n ju n c t io n w it h E q s . ( 8 .2 - 5 ) a n d ( 8 .2 - 6 ) .
In te g e r n
T a b le P 8 .1 4
P a ra m e te r i
G o lo m b c o d e G
0
0
1
0
2
0
3
0
4
1
5
1
6
1
7
1
8
1
9
1
1 0
1
1 1
1
1 2
2
1 3
2
1 4
2
1 5
2
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 0
1 0 0
1 0 0
1 0 1
1 0 1
1 0 1
1 0 1
1 1 0
1 1 0
1 1 0
1 1 0
2
e x p
(n )
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
0 0
0 0
0 0
0 0
0 0
0 1
1 0
1 1
Problem 8.15
T o d e c o d e G
k
e x p
(n ):
1 . C o u n t t h e n u m b e r o f 1 s i n a l e f t - t o - r i g h t s c a n o f a c o n c a t e n a t e d G ek x p ( n )
b it s e q u e n c e b e fo r e r e a c h in g th e fi r s t 0 , a n d le t i b e th e n u m b e r o f 1 s
c o u n te d .
2 . G e t th e k + i b its fo llo w in g th e 0 id e n tifi e d in s te p 1 a n d le t d b e its d e c im a l
e q u iv a le n t.
1 5 1
70
60
50
40
m
Equation (8.2-3)
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
ρ
F ig u r e P 8 .1 6
3 . T h e d e c o d e d in te g e r is th e n
i − 1
d +
2
j + k
.
j = 0
F o r e x a m p l e , t o d e c o d e t h e fi r s t G e2 x p ( n ) c o d e i n t h e b i t s t r e a m 1 0 1 1 1 0 1 1 . . . , l e t
i = 1 , th e n u m b e r o f 1 s in a le ft- to - r ig h t s c a n o f th e b it s tr e a m b e fo r e fi n d in g th e
fi r s t 0 . G e t th e 2 + 1 = 3 b its fo llo w in g th e 0 , th a t is , 1 1 1 s o d = 7 . T h e d e c o d e d
in te g e r is th e n
1 − 1
7 +
2
j + 2
= 7 + 2
2
= 1 1 .
j = 0
R e p e a t th e p r o c e s s fo r th e n e x t c o d e w o rd , w h ic h b e g in s w ith th e b it s e q u e n c e
0 1 1 ...
Problem 8.16
T h e g r a p h s h o w n in F ig u r e P 8 .1 6 w a s o b t a in e d u s in g M A T L A B . N o t e t h a t t h e
fu n c tio n is n o t d e fi n e d fo r ρ = f 0 ,1 g .
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 5 2
Encoding sequence
b
b
a
c
0.156
0.156
d
d
d
d
d
0.3
0.5
1
d
0.15536
c
c
c
c
c
0.1544
a
0
0.14
b
a
a
a
0.1
b
b
b
b
a
0.1528
0.14
F ig u r e P 8 .1 7
Problem 8.17
F ig u r e P 8 .1 7 s h o w s t h e a r it h m e t ic c o d in g p r o c e s s a s d e s c r ib e d in S e c t io n 8 .2 .3 .
A n y v a lu e in t h e in t e r v a l [0 .1 5 4 4 , 0 .1 5 5 3 6 ) a t t h e r ig h t s id e o f t h e fi g u r e c a n b e
u s e d t o c o d e t h e s e q u e n c e . F o r e x a m p le , t h e v a lu e 0 .1 5 5 .
Problem 8.18
T h e a r ith m e tic d e c o d in g p r o c e s s is th e re v e r s e o f th e e n c o d in g p
b y d iv id in g th e [0 , 1 ) in te r v a l a c c o rd in g to th e s y m b o l p r o b a b
s h o w n in T a b le P 8 .1 8 . T h e d e c o d e r im m e d ia t e ly k n o w s t h e m
b e g in s w ith a n “ e ”, s in c e th e c o d e d m e s s a g e lie s in th e in te r v a l
m a k e s it c le a r th a t th e s e c o n d s y m b o l is a n “ a ”, w h ic h n a r r o w s
[0 .2 , 0 .2 6 ) . T o fu r t h e r s e e t h is , d iv id e t h e in t e r v a l [0 .2 , 0 .5 ) a c c o r d
b o l p r o b a b ilitie s . P r o c e e d in g lik e th is , w h ic h is th e s a m e p r o c e d u
t h e m e s s a g e , w e g e t “ e a ii!” .
S y m b o l
a
e
i
o
u
!
T a b le P 8 .1 8
P r o b a b ility
0 .2
0 .3
0 .1
0 .2
0 .1
0 .1
[0
[0
[0
[0
[0
[0
R a
.0
.2
.5
.6
.8
.9
n g
, 0
, 0
, 0
, 0
, 0
, 1
e
.2
.5
.6
.8
.9
.0
)
)
)
)
)
)
ro c e d u re . S ta r t
ilitie s . T h is is
e s s a g e 0 .2 3 3 5 5
[0 .2 , 0 .5 ) . T h is
th e in te r v a l to
in g to th e s y m re u s e d to c o d e
1 5 3
Problem 8.19
A s
y o
c a
T a
s u m e
u a s s u
s e , th e
b le P 8
th a t th e fi rs t 2 5 6 c o d
m e 7 - b it A S C II, th e
A S C II ” a ” c o rre s p o n
.1 9 . T h e e n c o d e d o u
e s in th e s ta r tin g d ic tio n a r y a re th e A S C II c o d e s . If
fi r s t 1 2 8 lo c a tio n s a r e a ll th a t a r e n e e d e d . In e ith e r
d s to lo c a tio n 9 7 . T h e c o d in g p r o c e e d s a s s h o w n in
tp u t is
9 7 2 5 6 2 5 7 2 5 8 9 7 .
R e c o g n iz e d
a
a
a a
a
a a
a a a
a
a a
a a a
a a a a
a
C h a ra c te r
a
a
a
a
a
a
a
a
a
a
a
T a b le P 8 .1 9
O u tp u t
D ic t. A d d re s s
D ic t. E n tr y
9 7
2 5 6
a a
2 5 6
2 5 7
a a a
2 5 7
2 5 8
a a a a
2 5 8
9 7
2 5 9
a a a a a
Problem 8.20
T h e in p u t t o t h e L Z W d e c o d in g a lg o r it h m in E x a m p le 8 .7 is
3 9 3 9 1 2 6 1 2 6 2 5 6 2 5 8 2 6 0 2 5 9 2 5 7 1 2 6
T h e s ta r tin g d ic tio n a r y , to b e c o n s is te n t w ith th e c o d in g its e lf, c
lo c a tio n s – w ith th e fi r s t 2 5 6 c o r r e s p o n d in g to in te n s ity v a lu e s 0 t
T h e d e c o d in g a lg o r ith m b e g in s b y g e ttin g th e fi r s t e n c o d e d v a lu e , o u
c o r r e s p o n d in g v a lu e fr o m th e d ic tio n a r y , a n d s e ttin g th e “ r e c o g n iz e
to th e fi r s t v a lu e . F o r e a c h a d d itio n a l e n c o d e d v a lu e , w e (1 ) o u tp u
n a r y e n tr y fo r th e p ix e l v a lu e (s ), (2 ) a d d a n e w d ic tio n a r y e n tr y w h
is th e “ r e c o g n iz e d s e q u e n c e ” p lu s th e fi r s t e le m e n t o f th e e n c o d e d
p r o c e s s e d , a n d (3 ) s e t th e “ re c o g n iz e d s e q u e n c e ” to th e e n c o d e d
p r o c e s s e d . F o r t h e e n c o d e d o u t p u t in E x a m p le 8 .1 2 , t h e s e q u e n c e o
is a s s h o w n in T a b le P 8 .2 0 .
N o te , fo r e x a m p le , in r o w 5 o f th e ta b le th a t th e n e w d ic tio n a r y
c a tio n 2 5 9 is 1 2 6 - 3 9 , th e c o n c a te n a tio n o f th e c u r r e n tly r e c o g n iz e
o n ta in s 5 1 2
h ro u g h 2 5 5 .
tp u ttin g th e
d s e q u e n c e ”
t th e d ic tio o s e c o n te n t
v a lu e b e in g
v a lu e b e in g
f o p e r a tio n s
e n tr y fo r lo d s e q u e n c e ,
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 5 4
1 2 6 , a n d th e fi r s t e le m e n t o f th e e n c o d e d v a lu e b e in g p r o c e s s e d – th e 3 9 fr o m th e
3 9 - 3 9 e n tr y in d ic tio n a r y lo c a tio n 2 5 6 . T h e o u tp u t is th e n r e a d fr o m th e th ir d
c o lu m n o f th e ta b le to y ie ld
3 9
3 9
3 9
3 9
3 9
3 9
3 9
3 9
1 2 6
1 2 6
1 2 6
1 2 6
1 2 6
1 2 6
1 2 6
1 2 6
w h e re it is a s s u m e d th a t th e d e c o d e r k n o w s o r is g iv e n th e s iz e o f th e im a g e th a t
w a s re c e iv e d . N o te th a t th e d ic tio n a r y is g e n e r a te d a s th e d e c o d in g is c a r r ie d
o u t.
R e c o g n iz e d
3 9
3 9
1 2 6
1 2 6
2 5 6
2 5 8
2 6 0
2 5 9
2 5 7
E n c o d e d V a lu e
3 9
3 9
1 2 6
1 2 6
2 5 6
2 5 8
2 6 0
2 5 9
2 5 7
1 2 6
T a b le P 8 .2 0
P ix e ls
3 9
3 9
1 2 6
1 2 6
3 9 -3 9
1 2 6 -1 2 6
3 9 -3 9 -1 2 6
1 2 6 -3 9
3 9 -1 2 6
1 2 6
D ic t. A d d re s s
D ic t. E n tr y
2 5 6
2 5 7
2 5 8
2 5 9
2 6 0
2 6 1
2 6 2
2 6 3
2 6 4
3 9 -3 9
3 9 -1 2 6
1 2 6 -1 2 6
1 2 6 -3 9
3 9 -3 9 -1 2 6
1 2 6 -1 2 6 -3 9
3 9 -3 9 -1 2 6 -1 2 6
1 2 6 -3 9 -3 9
3 9 -1 2 6 -1 2 6
Problem 8.21
U s in g t h e B M P s p e c ifi c a t io n g iv e n in E x a m p le 8 .8 o f S e c t io n 8 .2 .5 , t h e fi r s t t w o
b y te s in d ic a te th a t th e u n c o m p r e s s e d d a ta b e g in s w ith a r u n o f 4 s w ith le n g th
3 . In a s im ila r m a n n e r, th e s e c o n d tw o b y te s c a ll fo r a r u n o f 6 s w ith le n g th 5 .
T h e fi rs t fo u r b y te s o f th e B M P e n c o d e d s e q u e n c e a re e n c o d e d m o d e . B e c a u s e
th e 5 th b y te is 0 a n d th e 6 th b y te is 3 , a b s o lu te m o d e is e n te r e d a n d th e n e x t
th r e e v a lu e s a r e ta k e n a s u n c o m p r e s s e d d a ta . B e c a u s e th e to ta l n u m b e r o f b y te s
in a b s o lu te m o d e m u s t b e a lig n e d o n a 1 6 - b it w o r d b o u n d a r y , th e 0 in th e 1 0 th
b y te o f th e e n c o d e d s e q u e n c e is p a d d in g a n d s h o u ld b e ig n o r e d . T h e fi n a l tw o
b y te s s p e c ify a n e n c o d e d m o d e r u n o f 4 7 s w ith le n g th 2 . T h u s , th e c o m p le te
u n c o m p re s s e d s e q u e n c e is { 4 , 4 , 4 , 6 , 6 , 6 , 6 , 6 , 1 0 3 , 1 2 5 , 6 7 , 4 7 , 4 7 } .
1 5 5
Problem 8.22
( a ) U s in g E q . ( 8 .2 - 9 ) , fo r m T a b le P 8 .2 2 .
B in
0 0
0 0
0 0
0 0
0 1
0 1
0 1
0 1
a r y
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
G ra y
0 0
0 0
0 0
0 0
0 1
0 1
0 1
0 1
T a b le P 8 .2 2
C o d e
B in
0 0
1 0
1 0
0 1
1 1
1 0
1 0
1 0
1 1
1 0
1 1
1 1
1 1
0 1
1 1
0 0
a r y
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
G ra y
1 1
1 1
1 1
1 1
1 0
1 0
1 0
1 0
C
0
0
1
1
1
1
0
0
o d e
0
1
1
0
0
1
1
0
( b ) T h e p r o c e d u r e is to w o r k fr o m th e m o s t s ig n ifi c a n t b it to th e le a s t s ig n ifi c a n t
b it u s in g th e e q u a tio n s :
a
a
= g m
= g i ⊕ a
m − 1
− 1
i
i + 1
0 ≤ i ≤ m − 2 .
T h e d e c o d e d b in a r y v a lu e is th u s 0 1 0 1 1 0 0 1 1 1 0 1 0 .
Problem 8.23
F o llo w in g th e p r o c e d u r e in th e fl o w c h a r t o f F ig . 8 .1 4 in th e b o o k , th e p r o p e r
c o d e is
0 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0 1 1 0 0 0 1
w h e r e th e s p a c e s h a v e b e e n in s e r te d fo r r e a d a b ility a lo n e . T h e c o d in g m o d e
s e q u e n c e is p a s s , v e r tic a l (1 le ft), v e r tic a l (d ir e c tly b e lo w ), h o r iz o n ta l (d is ta n c e s
3 a n d 4 ), a n d p a s s .
Problem 8.24
( a ) - ( b ) F o llo w in g th e p r o c e d u r e o u t lin e d in S e c t io n 8 .2 .8 , w e o b ta in t h e r e s u lt s
s h o w n in T a b le P 8 .2 4 .
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 5 6
D C C o e ffi c ie n t D iffe re n c e
-7
-6
-5
-4
4
5
6
7
T a b le P 8 .2 4
T w o ’s C o m p l e
1 ...1 0
1 ...1 0
1 ...1 0
1 ...1 1
0 ...0 1
0 ...0 1
0 ...0 1
0 ...0 1
m e n t V a lu e
0 1
1 0
1 1
0 0
0 0
0 1
1 0
1 1
C o
0 0 0
0 0 0
0 0 0
0 0 0
0 0 1
0 0 1
0 0 1
0 0 1
d e
0 0
0 1
1 0
1 1
0 0
0 1
1 0
1 1
Problem 8.25
In g e n e r a l, th e n u m b e r o f M A D c o m p u ta tio n s fo r s in g le - p ix e l p r e c is io n a n d d is p la c e m e n ts ± d x a n d ± d y is (2 d x + 1 ) 2 d y + 1 . W ith b o th d x a n d d y e q u a l to
± 8 , th e n u m b e r o f M A D c o m p u ta tio n s fo r th is p r o b le m is 1 7 2 o r 2 8 9 p e r m a c r o b lo c k . W ith 8 × 8 m a c r o b lo c k s , e a c h M A D c o m p u ta tio n in v o lv e s 8 2 (3 ) = 1 9 2
o p e r a tio n s (6 4 s u b tr a c tio n s , a b s o lu te v a lu e s , a n d a d d itio n s ). S o , th e to ta l n u m b e r o f m a th o p e r a tio n s fo r 8 × 8 m a c r o b lo c k s a n d s in g le p ix e l d is p la c e m e n ts
d x = d y = 8 is 2 8 9 × 1 9 2 = 5 5 ,4 8 8 .
F o r 14 p i x e l p r e c i s i o n , t h e n u m b e r o f M A D c o m p u t a t i o n s i s m u l t i p l i e d b y
1 6 , y ie ld in g 1 6 × 2 8 9 = 4 , 6 2 4 M A D c o m p u ta tio n s p e r m a c r o b lo c k . T h e n u m b e r o f m a th o p e r a tio n s fo r e a c h M A D c o m p u ta tio n is in c re a s e d fr o m 1 9 2 to
1 9 2 + 8 2 (4 ) = 4 4 8 , w h e r e th e a d d itio n a l o p e r a tio n s a r e fo r b ilin e a r in te r p o la tio n .
S o , t h e t o t a l n u m b e r o f m a t h o p e r a t i o n s f o r 8 × 8 m a c r o b l o c k s a n d 14 p i x e l d i s p la c e m e n ts d x = d y = 8 (u s in g b ilin e a r in te r p o la tio n ) is 4 6 2 4 × 4 4 8 = 2 , 0 7 1 , 5 5 2 .
Problem 8.26
T h e re a re s e v e r a l w a y s in w h ic h b a c k w a rd m o tio n - c o m p e n s a te d p re d ic tio n c a n
h e lp :
1 . It p r o v id e s a m e a n s o f p r e d ic tin g u n c o v e r e d b a c k g r o u n d w h e n a n o b je c t
is m o v in g .
2 . It p r o v id e s a m e a n s o f p r e d ic tin g p ix e ls o n th e e d g e s o f fr a m e s w h e n th e
c a m e r a is p a n n in g .
3 . B a c k w a rd a n d fo r w a rd p re d ic tio n c a n a v e r a g e th e n o is e in tw o re fe re n c e
fr a m e s to y ie ld a m o r e a c c u r a te p r e d ic tio n .
1 5 7
Encoded
Macroblock
Variable
Length
Decoder
Encoded
Motion
Vector
Variable
Length
Decoder
Inverse
Quantizer
Inverse Mapper
(e.g., DCT-1)
Decoded
Image
Macroblock
+
Motion Estimator
and Compensator
F ig u r e P 8 .2 7
Problem 8.27
T h e a p p r o p r ia t e M P E G d e c o d e r is s h o w n in F ig . P 8 .2 7 .
Problem 8.28
(a ) S u b s titu tin g ρ
o f R a n d r , w e g e t
= 0 in t o E q . ( 8 .2 - 4 9 ) a n d e v a lu a t in g it t o fo r m t h e e le m e n t s
h
ρ
1
2
R = σ
ρ
r = σ
1
ρ
2
ρ
.
2
(b ) F ir s t fo r m th e in v e r s e o f R ,
R
− 1
1
=
σ
2
− ρ
1
1 − ρ
2
− ρ
.
1
T h e n , p e r fo r m t h e m a t r ix m u lt ip lic a t io n o f E q . ( 8 .2 - 4 5 ) :
− 1
α = R
T h u s , α
1
= ρ a n d α
2
r =
σ
σ
2
2
ρ
1 − ρ
2
1 − ρ
0
2
=
ρ
0
.
= 0 .
( c ) T h e v a r ia n c e is c o m p u t e d u s in g E q . ( 8 .2 - 4 8 ) :
σ
e
2
= σ
2
− α
T
r =
ρ
0
ρ
ρ
2
= σ
2
1 − ρ
2
.
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 5 8
Problem 8.29
T h e d e r iv a tio n p r o c e e d s b y s u b s titu tin g th e u n ifo r m p r o b a b ility fu n c tio n in to
E q s . ( 8 .2 - 5 7 ) - ( 8 .2 - 5 9 ) a n d s o lv in g t h e r e s u lt in g s im u lt a n e o u s e q u a t io n s w it h
L = 4 . E q u a t io n ( 8 .2 - 5 8 ) y ie ld s
0
= 0
1
=
s
s
s
1
(t
2
= 1 .
2
+ t 2 )
1
S u b s titu tin g th e s e v a lu e s in to th e in te g r a ls d e fi n e d b y E q . (8 .2 - 5 7 ), w e g e t tw o
e q u a tio n s . T h e fi r s t is (a s s u m in g s 1 ≤ A )
s
1
(s − t 1 ) p (s ) d s = 0
s
1
1
(t 1 + t 2 )
2
2 A
0
1
+ t 2 )2 − 4 t 1 (t
(t
1
+ t 2 ) (t
2
− t 1 s
2
0
(t
1
2
s
(s − t 1 ) d s =
1
2
(t
1
+ t 2 )
0
+ t 2 )
=
0
− 3 t 1 )
=
0
= 0
s o
1
=
− t
2
=
3 t 1 .
t
t
2
T h e fi r s t o f th e s e r e la tio n s d o e s n o t m a k e s e n s e s in c e b o th t 1 a n d t 2 m u s t b e p o s itiv e . T h e s e c o n d r e la tio n s h ip is a v a lid o n e . T h e s e c o n d in te g r a l y ie ld s (n o tin g
th a t s 1 is le s s th a n A s o th e in te g r a l fr o m A to 1 is 0 b y th e d e fi n itio n o f p (s ))
s
2
(s − t 2 ) p (s ) d s = 0
s
1
A
1
(s − t 2 ) d s =
2 A
1
2
4 A
(t 1 + t 2 )
2
− 8 A t
2
− (t
1
2
s
2
A
− t 2 s
+ t 2 )2 − 4 t 2 (t
1
1
2
(t
1
+ t 2 )
+ t 2 ) = 0 .
= 0
1 5 9
S u b s titu tin g t
2
= 3 t
1
fr o m th e fi r s t in te g r a l s im p lifi c a tio n in to th is r e s u lt, w e g e t
8 t
t
1
−
2
− 6 A t
1
A
2
(8 t
2
+ A
1
=
1
− 2 A )
0
0
A
=
t
t
=
1
1
2
A
=
.
4
B a c k s u b s titu tin g th e s e v a lu e s o f t 1 , w e fi n d th e c o r r e s p o n d in g t
2
=
2
=
t
t
3 A
2
3 A
4
a n d
a n d
s
1
= A
s
1
=
A
2
fo r
fo r
1
=
A
1
=
A
t
t
2
a n d s
1
v a lu e s :
4
2
.
B e c a u s e s 1 = A is n o t a r e a l s o lu tio n (th e s e c o n d in te g r a l e q u a tio n w o u ld th e n
b e e v a lu a te d fr o m A to A , y ie ld in g 0 o r n o e q u a tio n ), th e s o lu tio n is g iv e n b y th e
s e c o n d . T h a t is ,
s 1 = A2
s 2 = 1
s 0 = 0
A
t 1 = 4
t 2 = 3 4A
.
Problem 8.30
T h e r e a r e m a n y w a y s to a p p r o a c h th is p r o b le m . F o r e x a m p le , JP E G - 2 0 0 0 c o m p r e s s i o n c o u l d b e u s e d . H e r e ’s a n o t h e r a l t e r n a t i v e . B e c a u s e t h e T 1 t r a n s f e r r a t e
is 1 .5 4 4 M b it / s e c , a 6 s e c o n d t r a n s fe r w ill p r o v id e
(1 .5 4 4 × 1 0 6 )(6 s e c ) = 9 .2 6 4 × 1 0
6
b its
o f d a ta . T h e in itia l a p p r o x im a tio n o f th e X - r a y m u s t c o n ta in n o m o re th a n th is
n u m b e r o f b its . T h e re q u ire d c o m p re s s io n r a tio is th u s
C =
4 0 9 6 × 4 0 9 6 × 1 2
= 2 1 .7 3
9 .2 6 4 × 1 0 6
T h e JP E G tr a n s fo r m c o d in g a p p r o a c h o f S e c tio n
c o m p r e s s io n a n d p r o v id e r e a s o n a b ly g o o d r e c o
c o d e r, th e X - r a y c a n b e JP E G c o m p re s s e d u s in g a
a b o u t a 2 5 :1 c o m p r e s s io n . W h ile it is b e in g t r a n
re m o te v ie w in g s ta tio n , th e e n c o d e r c a n d e c o d e
id e n tify th e “ d iffe r e n c e s ” b e tw e e n th e r e s u ltin g
8 .2 .8 c a n a c h
n s tr u c tio n s .
n o r m a liz a tio n
s m itte d o v e r t
th e c o m p re s s e
X -ra y a p p ro x
ie v e th is le v e l o f
A t th e X -ra y e n a r r a y th a t y ie ld s
h e T 1 lin e to th e
d JP E G d a ta a n d
im a tio n a n d th e
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 6 0
o r ig in a l X - r a y im a g e . S in c e w e w is h to tr a n s m it th e s e “d iffe re n c e s ” o v e r a s p a n
o f 1 m in u te w ith re fi n e m e n ts e v e r y 5 - 6 s e c o n d s , th e re c a n b e n o m o re th a n
6 0
6
o r 1 0
re fi n e
m e n t
n a l X m u s t
is
m
c o
ra
b e
e n ts
rre s
y a n
o b t
. If w
p o n d
d th e
a in e d
e a s s u m e th
s to th e “d if
JP E G re c o n s
p e r b it (to a
C =
w
tr
m
“d
a t
fe r
tr u
llo
1 2
e n
c t
w
to
6 0
5
1 2
re fi n e
c e s” b
e d a p p
a 6 s e c
m e n t
e tw e e
r o x im
o n d a
to
s a re
n o n
a tio
v e ra
m a d e
e o f th
n , th e n
g e tra n
a n d
e 1 2
th e
s fe r
th a t
b its
c o m p
tim e
e a
in
re
fo
c h
th
s s
r e
re fi n e e o r ig iio n th a t
a c h b it)
4 0 9 6 × 4 0 9 6 × 1
= 1 .8 1
9 .2 6 4 × 1 0 6
h e re , a s b e fo re , th e b o tto m o f th e fr a c tio n is th e n u m b e r o f b its th a t c a n b e
a n s m itte d o v e r a T 1 lin e in 6 s e c o n d s . T h u s , th e “ d iffe r e n c e ” d a ta fo r e a c h b it
u s t b e c o m p r e s s e d b y a fa c to r ju s t le s s th a n 2 . O n e s im p le w a y to g e n e r a te th e
iffe re n c e in fo r m a tio n ” is to X O R th e a c tu a l X - r a y w ith th e re c o n s tr u c te d JP E G
a p p r o x im a tio n . T h e r e s u ltin g b in a r y im a g e w ill c o n ta in a 1 in e v e r y b it p o s itio n
a t w h ic h th e a p p r o x im a tio n d iffe r s fr o m th e o r ig in a l. If th e X O R r e s u lt is tr a n s m itte d o n e b it a t a tim e b e g in n in g w ith th e M S B a n d e n d in g w ith th e L S B , a n d
e a c h b it is c o m p r e s s e d b y a n a v e r a g e fa c t o r o f 1 .8 1 :1 , w e w ill a c h ie v e t h e p e r fo r m a n c e th a t is r e q u ir e d in th e p r o b le m s ta te m e n t. T o a c h ie v e a n a v e r a g e e r r o r fr e e b it - p la n e c o m p r e s s io n o f 1 .8 1 :1 ( s e e S e c t io n 8 .2 .7 ) , t h e X O R d a t a c a n b e
G r a y c o d e d , r u n - le n g th c o d e d , a n d fi n a lly v a r ia b le - le n g th c o d e d . A c o n c e p tu a l
b lo c k d ia g r a m fo r b o th t h e e n c o d e r a n d d e c o d e r a r e g iv e n in F ig . P 8 .3 0 . N o te
th a t th e d e c o d e r c o m p u te s th e b it re fi n e m e n ts b y X O R in g th e d e c o d e d X O R d a ta
w ith th e re c o n s tr u c te d JP E G a p p r o x im a tio n .
Problem 8.31
T
F
a
o
o d e m o n
W T fi lte r
n d e v e n
u tp u t o f
s tra te
b a n k
o u tp u
s te p 4
Y
4
th e e
m e th
ts o f
o f th
(0 )
=
q u iv a le n c e o f th
o d , w e s im p ly d
th e liftin g a lg o r
e a lg o r ith m c a n
=
Y
2
(0 ) + δ [Y
X (0 ) + β [Y
1
3
e liftin
e r iv e g
ith m o
b e w r
g b a s e d a p p r o a c h a n d th e tr a d itio n a l
e n e r a l e x p re s s io n s fo r o n e o f th e o d d
f E q . (8 .2 - 6 2 ). F o r e x a m p le , th e Y (0 )
itte n a s
(− 1 ) + Y
(− 1 ) + Y
1
3
(1 )]
(1 )] + δ [Y
3
(− 1 ) + Y
3
(1 )]
1 6 1
X-ray
JPEG
Encoder
MUX
JPEG
Decoder
Control/
Sequencer
Bit
Selector
MUX
1-bit
XOR
Gray
Coder
Bit
Selector
MUX
Compressed
Data In
Compressed
T1
Transmitter Data Out
Variable
Run-length
Length
Coder
Coder
Encoder
JPEG
Decoder
Frame
Buffer
MUX
Control/
Sequencer
12-bit
XOR
Display
Generator
X-ray
to Display
Decoder
Zero Pad
Variable Run-length
Gray
and
Length
Decoder Decoder
Position Bit
Decoder
F ig u r e P 8 .3 0
w h e r e t h e s u b s c r i p t s o n t h e Y ’s h a v e b e e n a d d e d t o i d e n t i f y t h e s t e p o f t h e l i f t in g a lg o r ith m fr o m w h ic h th e v a lu e is g e n e r a te d . C o n tin u in g th is s u b s titu tio n
p a t t e r n f r o m e a r l i e r s t e p s o f t h e a l g o r i t h m u n t i l Y 4 ( 0 ) i s a f u n c t i o n o f X ’s o n l y ,
w e g e t
Y (0 )
1 + 2 α β + 2 α δ + 6 α β γ δ + 2 γ δ
=
β + 3 β γ δ + δ
+
β γ δ
+
α β γ δ
+
+
+
X (4 )
X (− 1 )
α β + 4 α β γ δ + α δ + γ δ
β γ δ
α β γ δ
X (2 )
X (3 )
β + 3 β γ δ + δ
+
+
X (1 )
α β + 4 α β γ δ + α δ + γ δ
+
X (0 )
X (− 3 )
X (− 4 ) .
X (− 2 )
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 6 2
T h u s , w e c a n fo r m t h e lo w p a s s a n a ly s is fi lt e r c o e ffi c ie n t s s h o w n in T a b le P 8 .3 1 1 .
C o e ffi c ie
±
±
±
±
T a b le P 8 .3 1 - 1
E x p re s s io n
α β γ δ / K
β γ δ / K
α β + 4 α β γ δ + α δ + γ δ / K
β + 3 β γ δ + δ / K
1 + 2 α β + 2 α δ + 6 α β γ δ + 2 γ δ / K
n t In d e x
4
3
2
1
0
H e re , th e c o e ffi c
a n d th e d iv is io n
c o e ffi c ie n t v a lu e
β , γ , δ , a n d K fr
d e r iv a tio n b e g in
ie n t e
b y K
s in c
o m th
n in g
x p re s s io
is in a c c
o lu m n 3
e te x t in
w ith
Y
3
n s a
o rd a
a re
to th
(1 ) = Y
re ta
n c e
d e te
e e x
k e n d ire c
w ith s te p
r m in e d b
p re s s io n s
(1 ) + γ [Y
1
2
tly fr o m o
6 o f E q . (
y s u b s titu
o f c o lu m
(0 ) + Y
2
u r
8 .2
tin
n 2
V a lu e
0 .0 2 6 7 4 8 7 5 7
- 0 .0 1 6 8 6 4 1 1 8
- 0 .0 7 8 2 2 3 2 6
0 .2 6 6 8 6 4 1 1
0 .6 0 2 9 4 9 0 1
e x p a n s io n o f Y (0 )
-6 2 ). T h e
g th e v a lu e s o f α ,
. A s im ila r
(2 )]
y ie ld s
Y (1 )
α + 3 α β γ + δ
=
1 + 2 β γ
+
X (1 )
α + 3 α β γ + δ
+
β γ
+
α β γ
+
β γ
+
+
α β γ
X (0 )
X (2 )
X (3 )
X (4 )
X (− 1 )
X (− 2 )
fr o m w h ic h w e c a n o b ta in th e h ig h p a s s a n a ly s is fi lte r c o e ffi c ie n ts s h o w n in
T a b le P 8 .3 1 - 2
C o e ffi c ie n t In d e x
-2
-1
0
1
2
3
4
T a b le
E x
−
−
− K α
− K
− K α
−
−
P 8 .3 1 - 2
p re s s io n
K α β γ
K β γ
+ 3 α β γ + δ
1 + 2 β γ
+ 3 α β γ + δ
K β γ
K α β γ
-0
0
0
-1
0
0
-0
.0
.0
.5
.1
.5
.0
.0
V a lu
9 1 2 7
5 7 5 4
9 1 2 7
1 5 0 8
9 1 2 7
5 7 5 4
9 1 2 7
e
1 7
3 5
1 7
7 0
1 7
3 5
1 7
6 2
2 5
6 6
5 3
6 6
2 5
6 2
1 6 3
Problem 8.32
F r o m E q . ( 8 .6 - 5 ) a n d t h e p r o b le m s t a t e m e n t , w e g e t t h a t
μ
ε
= μ 0 = 8
= ε 0 + 2 − 2 = ε
2 L L
2 L L
= 8 .
0
S u b s t it u t in g t h e s e v a lu e s in t o E q . ( 8 .2 - 6 4 ) , w e fi n d t h a t fo r t h e 2 L L s u b b a n d
Δ
2 L L
(8 + 0 )− 8
= 2
H e re , w e h a v e a s s u m e d a n 8 -b
6 5 ) , ( 8 .2 - 6 4 ) , a n d F ig . 8 .4 8 ( t o
g e t
Δ 2 H H = 2 (8 + 2
Δ 2H L = Δ 2 L H
Δ 1 H H = 2 (8 + 2
Δ 1H L = Δ 1 L H
8
1 +
2
= 1 .0 0 3 9 0 6 2 5 .
1 1
it im a g e s o t h a t R b = 8 . L ik e w is e , u s in g E q s . ( 8 .2 fi n d th e a n a ly s is g a in b its fo r e a c h s u b b a n d ), w e
)− 8
= 2
)− 8
= 2
8
1 +
1 1
2
(8 + 1 )− 8
1 +
8
2
1 1
(8 + 1 )− 8
= 4 .0 1 5 6 2 5
1 + 2 81 1 = 2 . 0 0 7 8 1 2 5
= 4 .0 1 5 6 2 5
1 + 2 81 1 = 2 . 0 0 7 8 1 2 5 .
Problem 8.33
O n e a p p r o a c h is t o im p le m e n t E q . ( 8 .3 - 1 ) u s in g F o u r ie r t r a n s fo r m s . U s in g t h e
p r o p e r tie s o f th e F o u r ie r tr a n s fo r m :
f
=
f
(1 − α ) f + α w
=
w
F
=
=
w
w
=
=
=
S o , v is ib le w a te
a s c a le d (b y α )
1 − α ) o f a n im
s u m . T h is is o
s p a tia l d o m a in
(1 − α ) f + α w
(1 − α ) f
(1 − α ) =
r m a r k in g in th e tr a n s fo r
v e r s i o n o f a w a t e r m a r k ’s
a g e ’s F o u r i e r t r a n s f o r m
b v io u s ly m o r e c o m p u ta
a p p r o a c h [E q . ( 8 .3 - 1 ) ].
m d o
F o u r
a n d
tio n a
f
m a
ie r
ta k
lly
+ = [α w ]
+ α = [w ]
in
tr
in
d
c a n
a n s f
g th
e m a
b e a c c o m
o r m to a s
e in v e r s e
n d in g th a
p lis
c a le
tra n
n th
h e d b y a
d v e r s io
s fo r m o
e e q u iv
d d
n (
f t
a le
in g
b y
h e
n t
Problem 8.34
A v a r ie ty o f m e th o d s fo r in s e r tin g in v is ib le w a te r m a r k s in to th e D F T c o e ffi c ie n ts
o f a n im a g e h a v e b e e n r e p o r te d in th e lite r a tu r e . H e r e is a s im p lifi e d o u tlin e o f
o n e in w h ic h w a te r m a r k in s e r tio n is d o n e a s fo llo w s :
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 6 4
II
I
III
IV
F ig u r e P 8 .3 4
1 . C r e a te a w a te r m a r k b y g e n e r a tin g a P - e le m e n t p s e u d o - r a n d o m s e q u e n c e
o f n u m b e r s , ! 1 , ! 2 , ..., ! P , t a k e n fr o m a G a u s s ia n d is t r ib u t io n w it h z e r o
m e a n a n d u n it v a r ia n c e .
2 . C o m p u te th e D F T o f th e im a g e to b e w a te r m a r k e d . W e a s s u m e th a t th e
tr a n s fo r m h a s n o t b e e n c e n te r e d b y p r e - m u ltip ly in g th e im a g e b y (− 1 )x + y .
3 . C h o
m id
c ie n
(th e
o s e P2 c o e
d le fr e q u e
ts in th e o
lo w fr e q u
ffi c ie n
n c y ra
rd e r s h
e n c y c
ts fro
n g e .
o w n
o e ffi c
m
T
in
ie
e a
h is
F ig
n ts
c h o f th e
is e a s ily
. P 8 .3 4 a
) in e a c h
fo u r q u a d r a n ts o f th e D F T in th e
a c c o m p lis h e d b y c h o o s in g c o e ffi n d s k ip p in g th e fi r s t K c o e ffi c ie n ts
q u a d ra n t.
4 . In s e r t th e fi r s t h a lf o f th e w a te r m a r k in to th e c h o s e n D F T c o e ffi c ie n ts , c
f o r 1 ≤ i ≤ P2 , i n q u a d r a n t s I a n d I I I o f t h e D F T u s i n g
c
5 . In
o f
c e
a d
s e r t th e
q u a d ra
s s m a in
d itio n ,
s e c o n d
n ts II a n
ta in s th
c o n s ta n
h a lf o
d IV o
e s y m
t α d e
f th e
f th e
m e tr
te r m
i
0
= c
w a te
D F T
y o f t
in e s t
i
(1 + α ! i)
r m
in
h e
h e
a r k in to th e c
a s im ila r m a n
tra n s fo r m o f
s tre n g th o f th
h o s
n e r
a re
e in
e n D
. N o t
a l- v a
s e r te
F T c o e ffi c
e th a t th is
lu e d im a g
d w a te r m
ie n
p r
e .
a rk
ts
o In
.
6 . C o m p u te th e in v e r s e D F T w ith th e w a te r m a r k e d c o e ffi c ie n ts r e p la c in g th e
u n m a r k e d c o e ffi c ie n ts .
i
1 6 5
W a te r m a r k e x tr a c tio n is p e r fo r m e d a s fo llo w s :
1 . L o c a te th e D F T c o e ffi c ie n ts c o n ta in in g th e w a te r m a r k b y fo llo w in g th e in s e r tio n p r o c e s s in th e e m b e d d in g a lg o r ith m .
2 . C o m p u t e t h e w a t e r m a r k !^ 1 , !^ 2 , . . . , !^
!^
i
^c
=
u s in g
P
i
− c
i
3 . C o m p u t e t h e c o r r e l a t i o n b e t w e e n ! a n d !^ a n d c o m p a r e t o a p r e - d e t e r m i n e d
th r e s h o ld T to d e te r m in e if th e m a r k is p r e s e n t.
Problem 8.35
T h e r e a r e m a n y w a y s to w a te r m a r k a n im a g e u s in g D W T s . A s im p le e x a m p le
th a t h a s b e e n r e p o r te d in th e lite r a tu r e u s e s th e fo llo w in g w a te r m a r k in s e r tio n
te c h n iq u e :
1 . C r e a te a w a te r m a r k b y g e n e r a tin g a 1 0 0 0 e le m e n t p s e u d o - r a n d o m s e q u e n c e
o f n u m b e r s , ! 1 , ! 2 , ..., ! 1 0 0 0 , t a k e n fr o m a G a u s s ia n d is t r ib u t io n w it h z e r o
m e a n a n d u n it v a r ia n c e .
2 . C o m p u te th
L s o th a t th e
1 0 0 0 . R e c a ll
w h e r e J = lo
a lg o r ith m , w
e L + 1 - le v e
n u m b e r o f
fro m c h a p te
g 2 N a n d th
e w ill c a ll th
l D W T o f t
a p p r o x im a
r 7 th a t th e
e im a g e is
e s e le c te d
h e im a g
tio n c o e
s e c o e ffi
o f s iz e N
c o e ffi c ie
e to b e w
ffi c ie n ts
c ie n ts w e
× N . F o
n ts c i (i )
a te r m
a t le v e
re d e n
r th e p
fo r 1 ≤
a rk e d . C h o o
l J − L is a b o
o te d W ' ( J −
u rp o s e s o f th
i ≤ 1 0 0 0 .
s e
u t
L ,m ,n )
is
3 . C o m p u te th e a v e r a g e o f th e s e le c te d a p p r o x im a tio n c o e ffi c ie n ts c .
4 . E m b e d w a te r m a r k ! in to th e s e le c te d a p p r o x im a tio n c o e ffi c ie n ts u s in g
c
i
0
= c + [c
i
− c ] (1 + α ! i)
w h e re α d e te r m in e s th e “ in te n s ity ” o f th e w a te r m a r k .
5 . C o m p u te th e w a te r m a r k e d im a g e b y ta k in g th e in v e r s e D W T w ith th e m a r k e d
a p p r o x im a tio n c o e ffi c ie n ts r e p la c in g th e o r ig in a l u n m a r k e d c o e ffi c ie n ts .
W a te r m a r k d e te c tio n is a c c o m p lis h e d a s fo llo w s :
1 . C o m p u te th e L + 1 - le v e l D W T o f th e im a g e in q u e s tio n .
C H A P T E R 8 . P R O B L E M S O L U T IO N S
1 6 6
2 . C o m p u t e t h e a v e r a g e c ^¯ a n d v a r i a n c e σ ( c ^ ) o f t h e l e v e l J − L a p p r o x i m a t i o n
c o e f fi c i e n t s , ^c 1 , ^c 2 , . . . , ^c 1 0 0 0 .
3 . E x t r a c t w a t e r m a r k !^ u s i n g
^c
!^
i
=
i
−
σ (c )
σ ( c^ )
^c ¯
c
i
−
− (c
i
−
¯c )
¯c
4 . C o m p u t e t h e c o r r e l a t i o n b e t w e e n ! a n d !^ a n d c o m p a r e t o a p r e - d e t e r m i n e d
th r e s h o ld T to d e te r m in e if th e m a r k is p r e s e n t.
Chapter 9
Problem Solutions
Problem 9.1
( a ) C o n v e r tin g a r e c ta n g u la r to a h e x a g o n a l g r id b a s ic a lly r e q u ir e s th a t e v e n a n d
o d d lin e s b e d is p la c e d h o r iz o n ta lly w ith r e s p e c t to e a c h o th e r b y o n e - h a lf th e
h o r iz o n ta l d is ta n c e b e tw e e n a d ja c e n t p ix e ls (s e e th e fi g u r e in th e p r o b le m s ta te m e n t). B e c a u s e in a r e c ta n g u la r g r id th e r e a r e n o p ix e l v a lu e s d e fi n e d a t th e
n e w lo c a tio n s , a r u le m u s t b e s p e c ifi e d fo r th e ir c r e a tio n . A s im p le a p p r o a c h
is to d o u b le th e im a g e r e s o lu tio n in b o th d im e n s io n s b y in te r p o la tio n (s e e S e c t io n 2 .4 .4 ) . T h e n , t h e a p p r o p r ia t e 6 - c o n n e c t e d p o in t s a r e p ic k e d o u t o f t h e e x p a n d e d a r r a y . T h e r e s o lu tio n o f th e n e w im a g e w ill b e th e s a m e a s th e o r ig in a l
( b u t t h e fo r m e r w ill b e s lig h t ly b lu r r e d d u e t o in t e r p o la t io n ) . F ig u r e P 9 .1 ( a ) illu s tr a te s th is a p p r o a c h . T h e b la c k p o in ts a r e th e o r ig in a l p ix e ls a n d th e w h ite
p o in ts a r e th e n e w p o in ts c r e a te d b y in te r p o la tio n . T h e s q u a r e s a r e th e im a g e
p o in ts p ic k e d fo r th e h e x a g o n a l g r id a r r a n g e m e n t.
(b ) R o ta tio n s in a 6 - n e ig h b o r a r r a n g e m e n t a re in v a r ia n t in 6 0
◦
in c re m e n ts .
( c ) Y e s . A m b ig u itie s a r is e w h e n th e r e is m o r e th a n o n e p a th th a t c a n b e fo llo w e d
fr o m o n e 6 - c o n n e c t e d p ix e l t o a n o t h e r. F ig u r e P 9 .1 ( b ) s h o w s a n e x a m p le , in
w h ic h th e 6 - c o n n e c te d p o in ts o f in te r e s t a r e in b la c k .
Problem 9.2
(a ) W ith re fe re n c e to th e d is c u s s io n in S e c tio n
a v o id m u ltip le p a th s th a t a r e in h e r e n t in 8 - c o n n
c o n n e c te d b o u n d a r ie s , th e s e m u ltip le p a th s m a
s ic p a t t e r n s s h o w n in F ig . P 9 .2 ( a ) . T h e s o lu t io n
o r- m is s tr a n s fo r m to d e te c t th e p a tte r n s a n d th
0 , th u s e lim in a tin g th e m u ltip le p a th s . A b a s ic s
1 6 7
2 .5 .2
e c tiv
n ife s
to th
e n to
e q u e
, m
ity .
t th
e p
c h
n c e
- c o n n e c tiv ity is u s e d
In o n e - p ix e l- th ic k , fu
e m s e lv e s in th e fo u r b
r o b le m is to u s e th e h
a n g e th e c e n te r p ix e l
o f m o r p h o lo g ic a l s te
to
lly
a itto
p s
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 6 8
(a)
(b)
F ig u r e P 9 .1
to a c c o m p lis h th is is a s fo llo w s :
1
=
A
1
=
A \ X
2
=
Y
1
2
=
Y
1
3
=
Y
2
3
=
Y
2
4
=
Y
3
4
=
Y
3
X
Y
X
Y
X
Y
X
Y
1
B
c
1
2
B
c
\ X
2
3
B
c
\ X
3
4
B
\ X
c
4
w h e re A is th e in p u t im a g e c o n ta in in g th e b o u n d a r y .
( b ) O n ly o n e p a s s is r e q u ir e d . A p p lic a tio n o f th e h it- o r - m is s tr a n s fo r m u s in g a
g iv e n B i fi n d s a ll in s ta n c e s o f o c c u r r e n c e o f th e p a tte r n d e s c r ib e d b y th a t s tr u c tu r in g e le m e n t.
(c ) T h e o rd e r d o e s m a tte r.
P 9 .2 (b ), a n d a s s u m e th a t w
p o in t a w ill b e d e le te d a n
s tr u c tu r in g e le m e n ts . If, o
d e le te d a n d p o in t a w ill r e
c e p ta b le ) m - p a th s .
F o r e
e a re
d p o
n th e
m a in
x a
tr
in
o
m p
a v e
t b
th e
. T h u
le , c o n
lin g fr o
w ill r e
r h a n d
s , w e w
s id
m
m a
, B
o u
e r th e s e q u e n c e o f p o
le ft to r ig h t. If B 1 is a p
in a fte r a p p lic a tio n o
3 is a p p lie d fi r s t , p o in
ld e n d u p w ith d iffe r e
in ts in F
p lie d fi r
f a ll o th
t b w ill
n t (b u t a
ig .
s t,
e r
b e
c -
1 6 9
(b)
F ig u r e P 9 .2
Problem 9.3
S e e F ig . P 9 .3 . K e e p in m in d t h a t e r o s io n is t h e s e t fo r m e d fr o m t h e lo c a t io n s
o f th e o r ig in o f th e s tr u c tu r in g e le m e n t, s u c h th a t th e s tr u c tu r in g e le m e n t is
c o n ta in e d w ith in th e s e t b e in g e r o d e d .
Problem 9.4
( a ) E r o s io n is s e t in te r s e c tio n . T h e in te r s e c tio n o f tw o c o n v e x s e ts is c o n v e x a ls o .
(
b
o
o
b ) S e
la c k
u s s e
f d ila
e F ig . P 9 .4 ( a ) . K
d o ts . T h e lin e s a
ts w o u ld b e , th e
tio n in th is c a s e
e e p in m
re s h o w n
y a re n o t
is n o t c o
in d
fo r
p a r
n v e
th a t th e d
c o n v e n ie
t o f th e s e
x b e c a u s e
ig ita l s e ts
n c e in v is u
ts b e in g c o
th e c e n te
in q u e s tio n a re
a liz in g w h a t th
n s id e re d h e re .
r p o in t is n o t in
th
e c
T h
th
e la r g e r
o n tin u e r e s u lt
e s e t.
( c ) S e e F ig . P 9 .4 ( b ) . H e r e , w e s e e t h a t t h e lo w e r r ig h t p o in t is n o t c o n n e c t e d t o
th e o th e rs .
( d ) S e e F ig . 9 .4 ( c ) . T h e t w o in n e r p o in t s a r e n o t in t h e s e t .
Problem 9.5
R e fe r t o F ig . P 9 .5 . T h e c e n t e r o f e a c h s t r u c t u r in g e le m e n t is s h o w n a s a b la c k
d o t.
( a ) T h is s o lu tio n w a s o b ta in e d b y e r o d in g th e o r ig in a l s e t (s h o w n d a s h e d ) w ith
th e s tr u c tu r in g e le m e n t s h o w n (n o te th a t th e o r ig in is a t th e b o tto m , r ig h t).
( b ) T h is s o lu tio n w a s o b ta in e d b y e r o d in g th e o r ig in a l s e t w ith th e ta ll r e c ta n g u la r s tr u c tu r in g e le m e n t s h o w n .
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 7 0
F ig u r e P 9 .3
(a)
(b)
(c)
F ig u r e P 9 .4
( c ) T h is s o lu tio
v e r tic a l lin e s u s
is s lig h tly ta lle r
d ila te d w ith th e
n w
in g
th a
c ir
a s
th
n
c u
o b ta in e
e re c ta n
th e c e n t
la r s tr u c
d
g u
e r
tu
b y fi
la r s
s e c t
r in g
rs t
tr u
io n
e le
e
c t
o
m
r o d in g th e im a g e s h o w n d o w n to tw o
u r in g e le m e n t (n o te th a t th is e le m e n ts
f th e “ U ” fi g u r e ). T h is r e s u lt w a s th e n
e n t.
( d ) T h is s o lu tio n w a s o b ta in e d b y fi r s t d ila tin g th e o r ig in a l s e t w ith th e la r g e d is k
s h o w n . T h e d ila te d im a g e w a s e r o d e d w ith a d is k w h o s e d ia m e te r w a s e q u a l to
o n e - h a lf th e d ia m e te r o f th e d is k u s e d fo r d ila tio n .
Problem 9.6
T h e s o lu t io n s t o ( a ) t h r o u g h ( d ) a r e s h o w n fr o m t o p t o b o t t o m in F ig . P 9 .6 .
Problem 9.7
( a ) T h e d ila te d im a g e w ill g r o w w ith o u t b o u n d .
( b ) A o n e - e le m e n t s e t ( i.e ., a o n e - p ix e l im a g e ) .
1 7 1
(a)
(b)
(c)
(d)
F ig u r e P 9 .5
Problem 9.8
( a ) T h e im a g e w ill e r o d e to o n e e le m e n t.
( b ) T h e s m a lle s t s e t th a t c o n ta in s th e s tr u c tu r in g e le m e n t.
Problem 9.9
T h e p r o o f, w h ic h c o n s is ts o f s h o w in g th a t th e e x p re s s io n
3
x 2 Z
fo llo w
m e n ts
(B )x ⊆
A , o r x
2
jx + b 2 A , fo r e v e r y b 2 B
s d ir e c tly fr o m
o f th e fo r m x
A . C o n v e r s e ly
+ b 2 A fo r e v
th
+ b
, (B
e r y
e d e
fo r
)x ⊆
b 2
4
≡
3
x 2 Z
2
j(B )
x
⊆ A
4
fi n itio n o f tr a n s la tio n b e c a u s e th e s e t ( B )x h a s e le b 2 B . T h a t is , x + b 2 A fo r e v e r y b 2 B im p lie s th a t
A im p lie s th a t a ll e le m e n ts o f ( B )x a r e c o n ta in e d in
B .
Problem 9.10
(a ) L e t x 2 A
B . T h e n , fr o m th e d e fi n itio n o f e r o s io n g iv e n in th e p r o b le m
s ta te m e n t, fo r e v e r y b 2 B , x + b 2 A . B u t, x + b 2 A im p lie s th a t x 2 (A )− b . T h u s ,
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 7 2
F ig u r e P 9 .6
<
fo r e v e r y b 2 B , x 2 (A )− b , w h ic h im p lie s th a t x 2
(A )− b . S u p p o s e n o w th a t
b 2 B
<
x 2
(A )− b . T h e n , fo r e v e r y b 2 B , x 2 (A )− b . T h u s , fo r e v e r y b 2 B , x + b 2 A
b 2 B
w h ic h , fr o m th e d e fi n itio n o f e r o s io n , m e a n s th a t x 2 A
(b ) S u p p o s e th a t x 2 A
x + b 2
(B )x ⊆ A
a ll e le m
( a ) , x <+
(
x 2
b 2 B
A . B
, s o
e n ts
b 2
A )− b
u t, a
th a t
o f (
A im
.
s s h
x 2
B )x
p lie
o w n
A
B
a re c
s th a
B
in P
=
o n t
t x
=
<
b 2 B
(A )
− b
B .
. T h e n , fo r e v e r y b
r o b le m 9 .9 ,
x 2 Z 2 j(B )x
a in e d in A ,
2 (A )− b . T h
x + b 2 A fo r
⊆ A
. S im ila
o r x + b 2 A
u s , if fo r e v e r
e v e
r ly ,
fo r
y b
r y
(B
e v
2
2 B , x 2 (A )
b 2
)x ⊆
e r y b
B , x
B im
A im
2 B
2 (A
− b
, o r
p lie s th a t
p lie s th a t
o r, a s in
)− b , th e n
1 7 3
Problem 9.11
T h e a p p r o a c h is to p r o v e th a t
3
x 2 Z
T h e e le m e n t
im p lie s th a t
p re c e d in g e q
b 2 B , th e n x
2
s o
fo r
u a
−
( B ^ ) x \ A 6= ;
f ( B^ ) x a r
s o m e b
tio n th a
b = a o r
e o f
2 B
t x =
x −
4
≡
th e
, x
a
b 2
3
x 2 Z
2
jx = a + b
fo r m x −
− b 2 A , o
+ b ). C o n
A , w h ic h
b f
r x
v e r
im
o r b
− b
s e ly
p lie
fo r a 2 A a n d b 2 B
2 B
= a
, if x
s th a
. T
fo
=
t (
h e
r s o
a +
B^ ) x
c o
m
b
\
4
.
n d i t i o n ( B ^ ) x \ A 6= ;
e a 2 A (n o te in th e
fo r s o m e a 2 A a n d
A 6= ; .
Problem 9.12
( a ) S u p p o s e t h a t x 2 A ⊕ B=. T h e n , f o r s o m e a 2 A a n d b 2 B , x = a + b . T h=u s , x 2
(A )b . O n th e o th e r h a n d , s u p p o s e th a t x 2
(A )b .
(A )b a n d , th e re fo re , x 2
T h e n , fo r s o m e b 2 B ,
a n a 2 A s u c h th a t x =
p r o b le m s ta te m e n t, a 2
=
(b ) S u p p o s e th a t x 2
b 2 B
b 2 B
b 2 B
2 (A )b . H o w e v e r, x 2 (A )b im p lie s th a t th e r e e x is ts
a + b . B u t, fr o m th e d e fi n itio n o f d ila tio n g iv e n in th e
A , b 2 B , a n d x = a + b im p ly th a t x 2 A ⊕ B .
x
(A )b . T h e n , fo r s o m e b 2 B , x 2 (A )b . H o w e v e r, x 2 (A )
im p lie s th a t th e r e e x is ts a n a 2 A
a 2 A a n d b 2 B , th e n x − b = a o r
N o w , s u p p o s e t h a t x 2 ( B ^ ) x \ A =6
s o m e b 2 B , x − b 2 A o r x − b = a
fo r s o m e a 2 A a n d b 2 B , th e n x
s u c h th a t x = a + b . B u t, if x =
x − b 2 A , w h ic h im p lie s th a t x 2
; . T h e c o n d i t i o n ( B ^ ) x \ A 6= ;
( i.e ., x = a + b ) fo r s o m e a =
2 A .
(A
2 (A )b a n d , th e re fo re , x 2
b
a + b fo r s o m e
( B ^ ) x \ A =6 ; .
im p lie s th a t fo r
B u t, if x = a + b
)b .
b 2 B
Problem 9.13
F r o m t h e d e fi n it io n o f d ila tio n , E q . ( 9 .2 - 3 ) ,
(A ⊕ B )c =
T h e c o m p le m e n t o f th
( B^ ) z \ A = ; . I n t u r n , (
i s , ( B^ ) z ⊆ A c , a n d w e c
3
4c
z ( B ^ ) z \ A 6= ;
.
e s e t o f z ’ s t h a t s a t i s f y ( B ^ ) z \ A =6 ; i s t h e s e t o f z ’ s s u c h t h a t
B^ ) z \ A = ; i m p l i e s t h a t t h a t ( B^ ) z i s c o n t a i n e d i n A c . T h a t
a n w r ite
3
( A ⊕ B ) c = z ( B^ )
= A c  B^
z
⊆ A
c
4
w h e r e t h e s e c o n d s t e p fo llo w s fr o m t h e d e fi n it io n o f e r o s io n , E q . ( 9 .2 - 1 ) . T h is
c o m p le te s th e p r o o f.
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 7 4
Problem 9.14
S ta r tin g w ith th e d e fi n itio n o f c lo s in g ,
(A • B )
c
=
[(A ⊕ B )
B ]
c
( A ⊕ B ) ⊕ B^
(A c
B^ ) ⊕ B^
A c ◦ B^ .
=
=
=
c
T h e p r o o f o f th e o th e r d u a lity p r o p e r ty fo llo w s a s im ila r a p p r o a c h .
Problem 9.15
( a ) E r o s io n o f a s e t A b y B is d e fi n e d a s th e s e t o f a ll v a lu e s o f tr a n s la te s , z , o f B
s u c h th a t (B )z is c o n ta in e d in A . If th e o r ig in o f B is c o n ta in e d in B , th e n th e s e t
o f p o in ts d e s c r ib in g th e e r o s io n is s im p ly a ll th e p o s s ib le lo c a tio n s o f th e o r ig in
o f B s u c h th a t ( B )z is c o n ta in e d in A . T h e n it fo llo w s fr o m th is in te r p r e ta tio n
(a n d th e d e fi n itio n o f e r o s io n ) th a t e r o s io n o f A b y B is a s u b s e t o f A . S im ila r ly ,
d i l a t i o n o f a s e t C b y B i s t h e s e t o f a l l l o c a t i o n s o f t h e o r i g i n o f B^ s u c h t h a t t h e
i n t e r s e c t i o n o f C a n d ( B^ ) z i s n o t e m p t y . I f t h e o r i g i n o f B i s c o n t a i n e d i n B , t h i s
im p lie s t h a t C is a s u b s e t o f t h e d ila t io n o f C b y B . F r o m E q . ( 9 .3 - 1 ) , w e k n o w
th a t A ◦ B = (A
B ) ⊕ B . L e t C d e n o te th e e r o s io n o f A b y B . It w a s a lr e a d y
e s ta b lis h e d th a t C is a s u b s e t o f A . F r o m th e p r e c e d in g d is c u s s io n , w e k n o w a ls o
th a t C is a s u b s e t o f th e d ila tio n o f C b y B . B u t C is a s u b s e t o f A , s o th e o p e n in g
o f A b y B (th e e r o s io n o f A b y B fo llo w e d b y a d ila tio n o f th e r e s u lt) is a s u b s e t o f
A .
( b ) F r o m E q . ( 9 .3 - 3 ) ,
>
f ( B ) z j( B )
C ◦ B =
a n d
z
⊆ C g
>
f ( B ) z j( B )
D ◦ B =
z
⊆ D g .
T h e r e fo r e , if C ⊆ D , it fo llo w s th a t C ◦ B ⊆ D ◦ B .
(c ) F r o m (a ), (A ◦ B ) ◦ B ⊆ (A ◦ B ). F r o m th e d e fi n itio n o f o p e n in g ,
(A ◦ B ) ◦ B
f (A ◦ B )
=
=
f [(A
B ) ⊕ B ]
f (A
=
(A
⊇
⊇
B g ⊕ B
B ) • B g ⊕ B
B ) ⊕ B
A ◦ B .
B g ⊕ B
1 7 5
B u t
(A ◦
th e
P ro
, th
B )
fa c
b le
e o
◦ B
t th
m 9
n ly w
= (A
a t th
.1 6 ( a
a y th a t (A ◦ B ) ◦ B ⊆ (A ◦ B ) a n d (A ◦ B ) ◦ B ⊇ (A ◦ B ) c a n h o ld is if
◦ B ). T h e n e x t to la s t s te p in th e p r e c e d in g s e q u e n c e fo llo w s fr o m
e c lo s in g o f a s e t b y a n o th e r c o n ta in s th e o r ig in a l s e t [th is is fr o m
)].
Problem 9.16
( a ) F r o m P r o b l e m 9 . 1 4 , ( A • B ) c = A c ◦ B^ , a n d , f r o m P r o b l e m 9 . 1 5 ( a ) , i t f o l l o w s
th a t
( A • B ) c = A c ◦ B^ ⊆ A c .
T a k in g th e c o m p le m e n t o f b o th s id e s o f th is e q u a tio n r e v e r s e s th e in c lu s io n s ig n
a n d w e h a v e th a t A ⊆ (A • B ), a s d e s ire d .
(b ) F ro m P ro
a n d B^ i n s t e a
D c ◦ B^ . T h e
re v e rs e s th e
(c ) U s in g re s
b le m
d o f C
re fo re
in c lu s
u lts fr
9 .1 5 ( b ) , if D c ⊆ C c , t h e n D c ◦
, D , a n d B . F r o m P r o b le m 9 .1
, if D c ⊆ C c th e n (D • B )c ⊆
io n , s o w e h a v e th a t if C ⊆ D ,
o m P r o b le m s 9 .1 4 a n d 9 .1 5 ,
3
(A • B ) • B
=
(A • B
3
=
(A c ◦
3
=
(A c ◦
B^ ⊆ C c ◦ B^ w h e r e
5 , (C • B )c = C c ◦
(C • B )c . T a k in
th e n (C • B ) ⊆ (D
) c ◦ B^
B^ ) ◦ B^
4c
B^ )
(A • B )
=
=
c
w e u s e d D c
B^ a n d ( D • B
g c o m p le m
• B ), a s d e s
, C c ,
)c =
e n ts
ire d .
4c
4c
c
(A • B ) .
w h e r e t h e t h ir d s t e p fo llo w s fr o m P r o b le m 9 .1 5 ( c ) a n d t h e fo u r t h s t e p fo llo w s
fr o m P r o b le m 9 .1 4 .
Problem 9.17
F ig u r e P 9 .1 7 s h o w s th e s o lu tio n . A lth o u g h th e im a g e s s h o w n c o u ld b e s k e tc h e d
b y h a n d , th e y w e r e d o n e in M A T L A B fo r c la r ity o f p r e s e n ta tio n . T h e M A T L A B
c o d e fo llo w s :
>> f = imread(’FigProb0917.tif’);
> > se = strel(’dis’, 11, 0); % Structuring element.
> > fa = imerode(f, se);
> > fb = imdilate(fa, se);
> > fc = imdilate(fb, se);
> > fd = imerode(fc, se);
1 7 6
C H A P T E R 9 . P R O B L E M S O L U T IO N S
F ig u r e P 9 .1 7
T h e s iz e o f th e o r ig in a l im a g e is 6 4 8 × 6 2 4 p ix e ls . A d is k s tr u c tu r in g e le m e n t o f
r a d iu s 1 1 w a s u s e d . T h is s tr u c tu r in g e le m e n t w a s ju s t la r g e e n o u g h to e n c o m p a s s e a c h n o is e e le m e n t, a s g iv e n in th e p r o b le m s ta te m e n t. T h e im a g e s s h o w n
in F ig . P 9 .1 7 a r e : ( a ) e r o s io n o f t h e o r ig in a l, ( b ) d ila t io n o f t h e r e s u lt , ( c ) a n o t h e r
d ila tio n , a n d fi n a lly (d ) a n e r o s io n . T h e m a in p o in ts w e a r e lo o k in g fo r fr o m
t h e s t u d e n t ’s a n s w e r a r e : T h e fi r s t e r o s i o n s h o u l d t a k e o u t a l l n o i s e e l e m e n t s
th a t d o n o t to u c h th e r e c ta n g le , s h o u ld in c r e a s e th e s iz e o f th e n o is e e le m e n ts
c o m p le te ly c o n ta in e d w ith in th e r e c ta n g le , a n d s h o u ld d e c r e a s e th e s iz e o f th e
r e c ta n g le . If w o r k e d b y h a n d , th e s tu d e n t m a y o r m a y n o t r e a liz e th a t s o m e
” im p e r fe c tio n s ” a r e le ft a lo n g th e b o u n d a r y o f th e o b je c t. W e d o n o t c o n s id e r
th is a n im p o r ta n t is s u e b e c a u s e it is s c a le - d e p e n d e n t, a n d n o th in g is s a id in th e
p r o b le m s ta te m e n t a b o u t th is . T h e fi r s t d ila tio n s h o u ld s h r in k th e n o is e c o m p o n e n ts th a t w e r e in c r e a s e d in e r o s io n , s h o u ld in c r e a s e th e s iz e o f th e r e c ta n g le ,
1 7 7
a n d s h o u ld
n o is e c o m p
fi n a l e r o s io
ro u n d e d c o
o g n iz e d b y
ro u n d th e c o
o n e n ts c o m p
n (b o tto m r ig
r n e r s in th e fi
th e s tu d e n t.
r n e rs
le te ly
h t) s h
n a l a
. T h e
a n d fu
o u ld t
n s w e r
n e
r t
h e
a r
x t d ila tio
h e r in c re
n d e c re a
e a n im p
n s h o u ld e lim in a
a s e th e s iz e o f th e
s e th e s iz e o f th e
o r ta n t p o in t th a t
te
re
re
s h
th
c ta
c ta
o u
e in te
n g le .
n g le .
ld b e
r n a l
T h e
T h e
re c -
Problem 9.18
It w a s p o s s ib le to r e c
c a u s e th e y w e re n o t
s tr u c tu r in g e le m e n t w
b e e n tr u e if th e o b je c
c o m p le te r e c o n s tr u c t
e r o d e d b y a c ir c le , w o
o n s tr u c
c o m p le
a s th e s
ts a n d s
io n , fo r
u ld n o t
t th e th r e e la r g e s q u
te ly e r o d e d a n d th e
a m e ( i.e ., t h e y w e r e s
tr u c tu r in g e le m e n ts
in s ta n c e , b y d ila tin g
b e p o s s ib le .
a re s to th e ir
g e o m e tr y o f
q u a re s ). T h is
w e re re c ta n g
a r e c ta n g le t
o r ig in a l s iz e b e th e o b je c ts a n d
a ls o w o u ld h a v e
u la r. H o w e v e r, a
h a t w a s p a r tia lly
Problem 9.19
S e le c t
a n d le
a g e . T
w h e re
b e th e
a o n
t th e
h e r
th e
s a m
e - p ix e l b o
o r ig in b e
e s u lt o f a p
t w o T ’s w e
e a s th e o r
rd
lo
p
re
ig
e r a ro u n
c a te d a t
ly in g th e
in p e r fe
in o f th e
d th e s tr u c tu r in g e le m
th e h o r iz o n ta l/ v e r tic a l
h it- o r- m is s tr a n s fo r m
c t r e g is tr a tio n . T h e lo c
s tr u c tu r in g e le m e n t.
e n
m
w
a t
t (s u
id p o
o u ld
io n o
b im
in t
b e
f th
a g e o f t
o f th is s
a s in g le
e p o in t
h e T ),
u b im p o in t
w o u ld
Problem 9.20
T h e
m e r
tim e
a s fo
k e
fo
, a
llo
y d
r m
b a
w s
iffe re n c e b e tw e e n th e L a k e a n d th e o th e r tw o fe a tu re s is th a t th e fo rs a c lo s e d c o n to u r. A s s u m in g th a t th e s h a p e s a r e p r o c e s s e d o n e a t a
s ic tw o - s te p a p p r o a c h fo r d iffe re n tia tin g b e tw e e n th e th re e s h a p e s is
:
S te p 1 . A p p ly a n e n d - p o in t d e te c to r to th e o b je c t. If n o e n d p o in ts a r e fo u n d ,
th e o b je c t is a L a k e . O th e r w is e it is a B a y o r a L in e .
S te p 2 . T h e re a re n u m e r o u s w a y s to d iffe re n tia te b e tw e e n a B a y a n d
O n e o f th e s im p le s t is to d e te r m in e a lin e jo in in g th e tw o e n d p o in ts o f
je c t. If th e A N D o f th e o b je c t a n d th is lin e c o n ta in s o n ly tw o p o in ts ,
u r e is a B a y . O th e r w is e it is a L in e . T h e r e a r e p a th o lo g ic a l c a s e s in w h
te s t w ill fa il, a n d a d d itio n a l ” in te llig e n c e ” n e e d s to b e b u ilt in to th e p r o c
th e s e p a th o lo g ic a l c a s e s b e c o m e le s s p r o b a b le w ith in c r e a s in g r e s o lu tio
th in n e d fi g u re s .
a L
th e
th e
ic h
e s s ,
n o f
in e .
o b fi g th is
b u t
th e
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 7 8
F ig u r e P 9 .2 1
Problem 9.21
( a ) T h e e n t i r e i m a g e w o u l d b e fi l l e d w i t h 1 ’s .
( b ) T h e b a c k g r o u n d w o u l d b e fi l l e d w i t h 1 ’s .
( c ) S e e F ig . P 9 .2 1 .
Problem 9.22
(a ) W ith re fe re n c e to
s u lts fr o m u s in g th e s
c o n n e c te d p a th (le ftm
tu r in g e le m e n t in F ig .
(b
s e
(2
th
th
) U s in g a 3 × 3 s tr u c
g m e n ts c h a r a c te r iz
,2 ) in F ig . 9 .1 5 ( e ) w
e w a y to th e fi n a l r
a t w o u ld tu r n F ig . 9
th e
tr u
o s t
9 .1
tu r
e d
o u
e s u
.1 5
e x
c tu
fi g
3 (b
a m
r in
u re
) fo
in g e
b y d
ld b e
lt in
(i) in
p le
g e
), w
r m
s h o w n
le m e n t
h e re a s
s a 4 -c o
le m e n t o f
ia g o n a lly a 1 in s te a
F ig . 9 .1 5 (
to a m u c h
in
in
th e
n n
F ig . P 9 .2 2
F ig . 9 .1 5 (c
b o u n d a r y
e c te d p a th
, th
) g e
re s u
(r ig
a l l 1 ’s w o u l d i n t r o d
c o n n e c te d p ix e ls .
d o f a 0 . T h a t v a lu
i). T h e r e w o u ld b e
m o re d is to r te d o b
e b o u n d a r y th
n e r a lly fo r m s
ltin g fr o m th e
h tm o s t fi g u re )
u c e c o r n e r p ix
F o r e x a m p le ,
e o f 1 w o u ld c
o t h e r 1 ’s i n t r
je c t.
a t re a n 8 s tr u c .
e ls in to
s q u a re
a r r y a ll
o d u c e d
Problem 9.23
(a ) If
d e te r
b la c k
n e c te
n e c te
th
m
p
d
d
e s p
in in
o in t
to it
c o m
h e r e s a r e n o t a llo w e d to to u c h
g w h ic h p o in ts a re b a c k g r o u n
o n th e b o u n d a r y o f th e im a g
u s in g a c o n n e c te d c o m p o n e n
p o n e n ts a r e la b e ls w ith a v a lu
, th e s o lu
d (b la c k )
e a n d d e t
t a lg o r ith
e d iffe re n
tio n o f th e
p o in ts . T o
e r m in e a ll
m (S e c tio n
t fro m 1 o r
p r o b le m s ta r ts b
d o th is , w e p ic k
b la c k p o in ts c o n
9 .5 .3 ) . T h e s e c o n
0 . T h e re m a in in
y
a
g
-
1 7 9
F ig u r e P 9 .2 2
b la c k
in g th
b e e n
p o in t
fi llin g
p o in ts a re in te r io r to
e h o le fi llin g a lg o r ith
tu r n e d in to w h ite p o
s a r e a lr e a d y k n o w n ,
th e s p h e re s w ith o u t
s p h e
m in
in ts .
th e y
h a v in
re s .
S e c t
T h e
c a n
g to
W e c
io n 9
a le r t
a ll b e
d o re
a n
.5 .2
s tu
tu
g io
fi ll a ll s p h e r e
u n til a ll in te
d e n t w ill r e a
r n e d s im p ly
n fi llin g a s a
s w ith w h ite b y a p p ly
r io r b la c k p o in ts h a v
liz e th a t if th e in te r io
in to w h ite p o in ts th u
s e p a ra te p ro c e d u re .
e
r
s
( b ) If th e s p h e r e s a r e a llo w e d to to u c h in a r b itr a r y w a y s , a w a y m u s t b e fo u n d to
s e p a r a te th e m b e c a u s e th e y c o u ld c r e a te “ p o c k e ts ” o f b la c k p o in ts s u r r o u n d e d
b y a ll w h ite o r s u r r o u n d e d b y a ll w h ite a n d p a r t o f th e b o u n d a r y o f th e im a g e .
T h e s im p le s t a p p r o a c h is to s e p a r a te th e s p h e r e s b y p r e p r o c e s s in g . O n e w a y
to d o th is is to e r o d e th e w h ite c o m p o n e n ts o f th e im a g e b y o n e p a s s o f a 3 × 3
s tr u c tu r in g e le m e n t, e ffe c tiv e ly c r e a tin g a b la c k b o r d e r a r o u n d th e s p h e r e s , th u s
“ s e p a r a tin g ” th e m . T h is a p p r o a c h w o r k s in th is c a s e b e c a u s e th e o b je c ts a r e
s p h e r ic a l, w h ic h im p lie s th a t th e y h a v e s m a ll a r e a s o f c o n ta c t. T o h a n d le th e
c a s e o f s p h e r e s to u c h in g th e b o r d e r o f th e im a g e , w e s im p ly s e t a ll b o r d e r p o in t
to b la c k . W e th e n p r o c e e d to fi n d a ll b a c k g r o u n d p o in ts T o d o th is , w e p ic k a
p o in t o n th e b o u n d a r y o f th e im a g e (w h ic h w e k n o w is b la c k d u e to p r e p r o c e s s in g ) a n d fi n d a ll b la c k p o in ts c o n n e c te d to it u s in g a c o n n e c te d c o m p o n e n t
a lg o r it h m ( S e c t io n 9 .5 .3 ) . T h e s e c o n n e c t e d c o m p o n e n ts a r e la b e ls w it h a v a lu e
d iffe r e n t fr o m 1 o r 0 . T h e r e m a in in g b la c k p o in ts a r e in te r io r to th e s p h e r e s . W e
c a n fi ll a ll s p h e r e s w ith w h ite b y a p p ly in g th e h o le fi llin g a lg o r ith m in S e c tio n
9 .5 .2 u n t il a ll s u c h in t e r io r b la c k p o in t s h a v e b e e n t u r n e d in t o w h it e p o in t s . T h e
a le r t s tu d e n t w ill r e a liz e th a t if th e in te r io r p o in ts a r e a lr e a d y k n o w n , th e y c a n
a ll b e tu r n e d s im p ly in to w h ite p o in ts th u s fi llin g th e s p h e r e s w ith o u t h a v in g to
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 8 0
d o re g io n
m a k e s th e
s u c h a n a r
is s u e in tr o
to s o lv e y e
e a s a re s m
s tu d e n t s h
fi llin g a s a s e p a r a te p r o c e d u r e . N o te th a t th e e r o s io n o f w h
b la c k a r e a s in te r io r to th e s p h e r e s g r o w , s o th e p o s s ib ility e
e a n e a r th e b o r d e r o f a s p h e r e c o u ld g r o w in to th e b a c k g r o u
d u c e s fu r th e r c o m p lic a tio n s th a t th e s tu d e n t m a y n o t h a v e
t. W e re c o m m e n d m a k in g th e a s s u m p tio n th a t th e in te r io r
a ll a n d n e a r th e c e n te r. R e c o g n itio n o f th e p o te n tia l p r o b le
o u ld b e s u ffi c ie n t in th e c o n te x t o f th is p r o b le m .
ite a r
x is ts t
n d . T
th e to
b la c k
m b y
e a s
h a t
h is
o ls
a rth e
Problem 9.24
D e n o te th e o r ig in a l im a g e b y A . C re a te a n im a g e o f th e s a m e s iz e a s th e o r ig in a l , b u t c o n s i s t i n g o f a l l 0 ’s , c a l l i t B . C h o o s e a n a r b i t r a r y p o i n t l a b e l e d 1 i n A ,
c a ll it p 1 , a n d a p p ly th e c o n n e c te d c o m p o n e n t a lg o r ith m . W h e n th e a lg o r ith m
c o n v e r g e s , a c o n n e c te d c o m p o n e n t h a s b e e n d e te c te d . L a b e l a n d c o p y in to B
th e s e t o f a ll p o in ts in A b e lo n g in g to th e c o n n e c te d c o m p o n e n ts ju s t fo u n d , s e t
th o s e p o in ts to 0 in A a n d c a ll th e m o d ifi e d im a g e A 1 . C h o o s e a n a r b itr a r y p o in t
la b e le d 1 in A 1 , c a ll it p 2 , a n d r e p e a t th e p r o c e d u r e ju s t g iv e n . If th e r e a r e K c o n n e c te d c o m p o n e n ts in th e o r ig in a l im a g e , th is p r o c e d u r e w ill r e s u lt in a n im a g e
c o n s i s t i n g o f a l l 0 ’s a f t e r K a p p l i c a t i o n s o f t h e p r o c e d u r e j u s t g i v e n . I m a g e B w i l l
c o n ta in K la b e le d c o n n e c te d c o m p o n e n ts .
Problem 9.25
F r o m S e c t io n 9 .5 .9 , t h e fo llo w in g e x p r e s s io n is a n im a g e H in w h ic h a ll t h e h o le s
in im a g e f h a v e b e e n fi lle d :
$
%c
H = R Df c ( F ) .
T h e
h o le
ta in
ta in
o n
s h
in g
in g
ly d
a v e
o n
o n
iffe r
b e e
ly th
ly th
e n c e b e tw e e n th is e x p re s s io n a n d th e o r ig in a l im a g e f is th a t th e
n fi lle d . T h e r e fo r e , th e in te r s e c tio n o f f c a n d H is a n im a g e c o n e (fi lle d ) h o le s . T h e c o m p le m e n t o f th a t r e s u lt is a n im a g e c o n e h o le s .
f h o le s =
&
f
c
\
$
R
D
f
c
(F )
%c ' c
.
F ig u r e P 9 .2 5 s h o w s t h e h o le s e x t r a c t e d fr o m F ig 9 .3 1 ( a ) . K e e p in m in d t h a t t h e
h o l e s i n t h e o r i g i n a l i m a g e a r e b l a c k . T h a t ’s t h e r e a s o n f o r t h e c o m p l e m e n t i n
th e p re c e d in g e q u a tio n .
1 8 1
F ig u r e P 9 .2 5
Problem 9.26
( a ) If a ll t h e b o r d e r p o in t s a r e 1 , t h e n F in E q . ( 9 .5 - 2 8 ) w ill b e a ll 0 s . T h e d ila t io n
o f F b y B w ill a ls o b e a ll 0 s , a n d th e in te r s e c tio n o f th is r e s u lt w ith f c w ill b e a ll
0 s a ls o . T h e n H , w h ic h is th e c o m p le m e n t, w ill b e a ll 1 s .
(b ) If a
e n tire
im a g e
w ith in
ll th e p o
im a g e is
s h o u ld b
h o le s , s o
in ts in
a h o le
e fi lle d
th e re
th
, s
w
s u
e b
o H
ith
lt is
o rd
in
1 s .
a s
e r o f
(a ) w
T h is
e x p e c
f a re 1 , th a t m e a n s th a t th e in te r io r o f th e
ill b e th e c o r r e c t r e s u lt. T h a t is , th e e n tir e
a lg o r ith m is n o t c a p a b le o f d e te c tin g h o le s
te d .
Problem 9.27
E r
is
s o
B^
o s io n is th e s e t o f p o in ts z s u c h th a t B , tr a n s la te d b y z , is c o n ta in e d in A . If
a s in g le p o in t, th is d e fi n itio n w ill b e s a tis fi e d o n ly b y th e p o in ts c o m p r is in g
e r o s io n o f A b y B is s im p ly A . S im ila r ly , d ila tio n is th e s e t o f p o in ts z s u c h th
( B^ = B i n t h i s c a s e ) , t r a n s l a t e d b y z , o v e r l a p s A b y a t l e a s t o n e p o i n t . B e c a u
B is a s in g le p o in t, th e o n ly s e t o f p o in ts th a t s a tis fy th is d e fi n itio n is th e s e t
p o in ts c o m p r is in g A , s o th e d ila tio n o f A b y B is A .
B
A ,
a t
s e
o f
Problem 9.28
T h e d a r k im a g e s tr u c tu r e s th a t a r e c o m p le te ly fi lle d b y m o r p h o lo g ic a l c lo s in g
r e m a in c lo s e d a fte r th e r e c o n s tr u c tio n .
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 8 2
Problem 9.29
C o n s id e r fi r s t th e c a s e fo r n = 1 :
E
(1 )
G
(F )
&$
E
=
%c ' c
(F )
c
c
[(F  B ) [ G ]
=
c
(F  B )c \ G c
F c ⊕ B^ \ G
=
=
=
$
D
=
w h e re th e th
s te p fo llo w s
th e fi fth s te p
th e d e fi n itio
g e o d e s ic e r o
o b ta in it in t
tio n e d , c o m
c o m p le m e n
(1 )
G
c
F
⊕ B
(1 )
G c
c
(F
)
c
c
c
c
\ G
%c
i r d s t e p f o l l o w s f r o m D e M o r g a n ’s l a w , ( A [ B ) c = A c \ B c , t h e f o u r t h
fr o m t h e d u a lity p r o p e r t y o f e r o s io n a n d d ila tio n ( s e e S e c t io n 9 .2 .3 ) ,
fo llo w s fr o m th e s y m m e tr y o f th e S E , a n d th e la s t s te p fo llo w s fr o m
(2 )
n o f g e o d e s ic d ila tio n . T h e n e x t s te p , E G (F ), w o u ld in v o lv e th e
s io n o f th e a b o v e r e s u lt. B u t th a t r e s u lt is s im p ly a s e t, s o w e c o u ld
e r m s o f d ila tio n . T h a t is , w e w o u ld c o m p le m e n t th e r e s u lt ju s t m e n p le m e n t G , c o m p u te th e g e o d e s ic d ila tio n o f s iz e 1 o f th e tw o , a n d
t th e r e s u lt. C o n tin u in g in th is m a n n e r w e c o n c lu d e th a t
&
$
%c ! ' c
(n )
(1 )
(n − 1 )
E G
=
D G c
(F )
E G
#%c
$
"
(1 )
(n − 1 )
F c
=
D G c D G c
.
S im ila r ly ,
D
(1 )
G
(F )
&$
D
=
(1 )
G
(F )
%c ' c
[(F ⊕ B ) \ G ]
=
(F ⊕ B )c [ G c
F c  B^ [ G
=
=
=
$
E
=
F
c
 B
[ G
%c
) .
(1 )
G c
(F
$
D
(n − 1 )
G
c
c
c
c
c
c
c
c
A s b e fo re ,
D
(n )
G
=
=
&
E
$
E
(1 )
G c
(1 )
G c
"
E
(n − 1 )
G c
(F )
F
c
%c ! ' c
#%c
.
1 8 3
Problem 9.30
D
G
R
(F )
(k )
=
D G (F )
"
$
(1 )
E G c E
$
(k )
E G c F
=
=
=
E
G
R
(k − 1 )
G c
%c
c
c
c
F
c
#%c
c
F
w h e r e w e u s e d t h e r e s u lt fr o m P r o b le m 9 .2 9 . T h e o t h e r d u a lit y p r o p e r t y is p r o v e d
in a s im ila r m a n n e r.
Problem 9.31
(a ) C o n s id e r th e c a s e w h e n n = 2
[(F  2 B )]
c
=
c
[(F  B )  B ]
( F  B ) c ⊕ B^
=
=
=
F
c
⊕ B^
F
c
⊕ 2 B^
⊕ B^
w h e r e th e s e c o n d a n d th ir d lin e s fo llo w fr o m th e d u a lity p r o p e r ty in E q . (9 .2 - 5 ).
F o r a n a r b itr a r y n u m b e r o f e r o s io n s ,
[(F  n B )]
c
=
[(F  (n − 1 ) B )  B ]
=
[ ( F  ( n − 1 ) B ) ] c ⊕ B^
w h ic h , w h e n e x p a n d e d , w ill y ie ld [(F  n B )]c = F
c
⊕ n B^ .
( b ) P r o v e d in a s im ila r m a n n e r.
Problem 9.32
O
(n )
R
(F )
=
D
F
R
=
(F  n B )
E
F
=
R
F
=
R
$
C
=
R
[F  n B ]
c
F
E
c
E
F
c
(n )
R
c
F
c
⊕ n B^
c
F
c
c
c
⊕ n B
%c
c
c
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 8 4
w h e re
(P r o b le
s te p fo
tio n o f
s im ila r
th e s e c o n d
m 9 .3 0 ) , t h
llo w s fr o m
c lo s in g b y
m a n n e r.
s te
e th
th e
re c o
p fo
ird s
s y m
n s tr
llo w s fr
te p fo llo
m e tr y o
u c tio n .
o m
w s
f B
T h
th e
fro m
, a n d
e o th
d u a
th e
th e
e r d
lity o f r e c o n s tr u c
r e s u lt in P r o b le m
la s t s te p fo llo w s
u a lity p r o p e r ty c a
tio n
9 .3 1
fro m
n b e
b y d
, th e
th e
p ro v
ila
fo
d e
e d
tio n
u r th
fi n iin a
Problem 9.33
( a ) F r o m E q . ( 9 .6 - 1 ) ,
f  b
c
c
=
f (x + s ,y + t )
m in
(s ,t ) 2 b
=
c
− f (x + s ,y + t )
m a x
−
(s ,t ) 2 b
=
m a x
(s ,t ) 2 b
− f (x + s ,y + t )
− f ⊕ b^
f c ⊕ b^ .
=
=
T h e s e c o n d s te p fo llo w s fr o m th e d e fi n itio n o f th e c o m p le m e n t o f a g r a y - s c a le
fu n c tio n ; th a t is , th e
th e m a x im u m o f th e
th e d e fi n itio n o f th e c
g r a y - s c a le d ila tio n in
s te p fo llo w s fr o m th e
p r o p e r ty is p r o v e d in
m in im u m o f a s e t o f n u m b e
n e g a tiv e o f th o s e n u m b e r s .
o m p le m e n t. T h e fo u r th s te p
E q . ( 9 .6 - 2 ) , u s in g t h e fa c t t h a
d e fi n itio n o f th e c o m p le m e n
a s im ila r m a n n e r.
rs
T
fo
t b^
is e q u a l to th
h e th ird s te p
llo w s fr o m th e
(x ,y ) = b (− x
t, − f = f c . T h e
e n e g a tiv e o f
fo llo w s fr o m
d e fi n itio n o f
− y ). T h e la s t
o th e r d u a lity
( b ) W e p r o v e th e s e c o n d d u a lity p r o p e r ty :
f ◦ b
c
=
f  b
=
=
f  b
=
T h e s
s io n ,
o th e r
(c ) W
1 :
e c o n d a
a n d th e
p ro p e r
e p ro v e
n d th ird s
la s t s te p
ty in th e p
th e fi rs t d
f
⊕ b^
c
f
c
c
c
⊕ b
 b^
 b^
• b^ .
te p s fo llo w fr o m th e
fo llo w s fr o m th e d e fi
r o b le m s ta te m e n t is
u a lity p r o p e r ty . S ta r
d u a lity
n itio n
p ro v e d
t w ith t
p r
o f
in
h e
o p e r t
c lo s in
a s im
a g e o
y o f d ila tio n a n d e r o g in E q . (9 .6 - 8 ). T h e
ila r m a n n e r.
d e s ic d ila tio n o f s iz e
1 8 5
D
(1 )
g
&$
D
=
f
(1 )
g
%c ' c
f
=
( f ⊕ b ) ^ g
=
−
c
c
c
c
 b ) _ g c
%c
(1 )
.
f c
g c
c
( f ⊕ b )c _ g
=
=
( f
$
E
=
c
c
− ( f ⊕ b ) _ − g
=
c
− ( f ⊕ b ) _ − g
c
T h e s e c o n d s te p fo llo w s fr o m th e d e fi n itio n o f g e o d e s ic d ila tio n . T h e th ir d s te p
fo llo w s fr o m th e fa c t th a t th e p o in t- w is e m in im u m o f tw o s e ts o f n u m b e r s is th e
n e g a tiv e o f th e p o in t- w is e m a x im u m o f th e tw o n u m b e r s . T h e fo u r th a n d fi fth
s te p s fo llo w fr o m th e d e fi n itio n o f th e c o m p le m e n t. T h e s ix th s te p fo llo w s fr o m
t h e d u a l i t y o f d i l a t i o n a n d e r o s i o n ( w e u s e d t h e g i v e n f a c t t h a t b^ = b ) . T h e l a s t
s te p fo llo w s fr o m th e d e fi n itio n o f g e o d e s ic e r o s io n .
(2 )
T h e n e x t s te p in th e ite r a tio n , D g
f , w o u ld in v o lv e th e g e o d e s ic d ila tio n o f
s iz e 1 o f th e p r e c e d in g r e s u lt. B u t th a t r e s u lt is s im p ly a s e t, s o w e c o u ld o b ta in
it in te r m s o f e r o s io n . T h a t is , w e w o u ld c o m p le m e n t th e r e s u lt ju s t m e n tio n e d ,
c o m p le m e n t g , c o m p u te th e g e o d e s ic e r o s io n o f th e tw o , a n d c o m p le m e n t th e
r e s u lt. C o n tin u in g in th is m a n n e r w e c o n c lu d e th a t
#%c
$
"
(1 )
(n − 1 )
f c
D (g n ) ( f ) = E g c E g c
.
T h e o th e r p r o p e r ty is p r o v e d in a s im ila r w a y .
(d ) W e p ro v e th e fi rs t p ro p e r ty :
R
g
D
( f )
D (g k ) ( f )
$
"
(1 )
E g c E
$
(k )
E g c f
$
R gE c f
=
=
=
=
(k − 1 )
f
c
g
c
#%c
%c
c
%c
c
.
T h e o th e r p r o p e r ty is p r o v e d in a s im ila r m a n n e r.
(e ) W e p r o v e th e fi r s t p r o p e r ty . C o n s id e r th e c a s e w h e n n = 2
f  2 b
c
=
=
f  B
=
f  b
c
⊕ b^
c
⊕ 2 b^
f
=
f
c
 b
⊕ b^
⊕ b^
c
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 8 6
w h e r e th e s e c o n d a n d th ir d lin e s fo llo w fr o m th e d u a lity p r o p e r ty in E q . (9 .6 - 5 ).
F o r a n a r b itr a r y n u m b e r o f e r o s io n s ,
f  n b
c
=
f  (n − 1 )b
=
w h ic h , w h e n e x p a n d e d , w ill y ie ld
p r o v e d in a s im ila r m a n n e r.
(f ) W e p ro v e th e fi rs t p ro p e r ty :
O
(n )
R
( f )
=
=
=
=
=
w h e re
g iv e n
lo w s f
c lo s in
m a n n
th e s e c o n
in (d ) , th e
ro m th e s y
g b y re c o n s
e r.
d s te p
th ird
m m e tr
tr u c tio
fo llo w s fr o m
s te p fo llo w s
y o f b , a n d t
n . T h e o th e r
 b
c
f  (n − 1 )b
c
f  n b
R Df
$
R
$
R
$
R
$
C
=
f
c
c
⊕ b^
⊕ n b^ . T h e o t h e r p r o p e r t y i s
f  n b
E
f
c
E
f
c
E
f
c
c
f
(n )
R
f
%c
⊕ n b^
c
⊕ n b
%c
f c n
th e d u
fro m th
h e la s t
d u a lity
a lity
e re s
s te p
p ro p
%c
c
f  n b
%c
o f re c o n s tr u
u lt in (e ), th
fo llo w s fr o m
e r ty c a n b e p
c tio n
e fo u r
th e d
ro v e d
b y
th
e fi
in
d ila tio n
s te p fo ln itio n o f
a s im ila r
Problem 9.34
T h e m e th o d is p r e d ic a te d o n im a g e o p e n in g s a n d c lo s in g s , h a v in g n o th in g to d o
w ith s p a tia l a r r a n g e m e n t. B e c a u s e th e b lo b s d o n o t to u c h , th e r e is n o d iffe r e n c e
b e t w e e n t h e fu n d a m e n t a l a r r a n g e m e n t in F ig . 9 .4 3 a n d t h e fi g u r e in t h is p r o b le m . T h e s te p s to th e s o lu tio n w ill b e th e s a m e . T h e o n e th in g to w a tc h is th a t
th e S E u s e d to r e m o v e th e s m a ll b lo b s d o n o t r e m o v e o r s e v e r e ly a tte n u a te la r g e
b lo b s a n d th u s o p e n a p o te n tia l p a th to th e b o u n d a r y o f th e im a g e la r g e r th a n
th e d ia m e te r o f th e la r g e b lo b s . In th is c a s e , d is k s o f r a d iu s 3 0 a n d 6 0 (th e s a m e
t h o s e u s e d in F ig . 9 .4 3 ) d o t h e jo b p r o p e r ly . T h e s o lu t io n im a g e s a r e s h o w n in
F ig . P 9 .3 4 . T h e e x p la n a tio n is th e s a m e a s F ig . 9 .4 3 . T h e M A T L A B c o d e u s e d to
g e n e r a te d th e s o lu tio n fo llo w s .
>>
>>
>>
>>
>>
f = imread(’FigP0934(blobs_in_circular_arrangement).tif’);
figure, imshow(f)
% Remove small blobs.
fsm = imclose(f, strel(’disk’,30));
figure, imshow(fsm)
1 8 7
F ig u r e P 9 .3 4
> > % Remove large blobs.
> > flrg = imopen(fsm, strel(’disk’,60));
> > figure, imshow(flrg)
> > % Use a morphological gradient to obtain the boundary
> > % between the regions.
> > se = ones(3);
> > grad = imsubtract(imdilate(flrg, se), imerode(flrg, se));
> > figure, imshow(grad)
> > % Superimpose the boundary on the original image.
> > idx = find(grad > 0);
> > final = f;
> > final(idx) = 255;
> > figure, imshow(final)
C H A P T E R 9 . P R O B L E M S O L U T IO N S
1 8 8
F ig u r e P 9 .3 5
Problem 9.35
(a ) T h e n o is e s p ik e s a re o f th e g e n e r a l fo r m s h o
p o s s ib ilitie s in b e tw e e n . T h e a m p litu d e is ir r e le v
o f th e n o is e s p ik e s is o f in te re s t. T o re m o v e th e s e
w ith a c y lin d r ic a l s tr u c tu r in g e le m e n t o f r a d iu s
F ig . P 9 .3 5 (b ). N o te th a t th e s h a p e o f th e s tr u c tu
k n o w n s h a p e o f th e n o is e s p ik e s .
( b ) T h e b a s ic s o lu t
th e v a r io u s p o s s ib
in g e le m e n t lik e th
d o th e jo b . N o te in
th is a p p r o a c h . T h
b e n e e d e d a n d , c o
w n in
a n t in
s p ik e
g re a te
r in g e
F ig .
th is
s w e
r th a
le m e
P 9 .3 5 ( a ) , w it h
c a s e ; o n ly th e
p e r fo r m a n o p
n R max , a s s h o
n t is m a tc h e d
o th e r
s h a p e
e n in g
w n in
to th e
io n is th e s a m e a s in (a ), b u t n o w w e h a v e to ta k e in to a c c o u n t
le o v e r la p p in g g e o m e t r ie s s h o w n in F ig . P 9 .3 5 ( c ) . A s t r u c t u r e o n e u s e d in (a ) b u t w ith r a d iu s s lig h tly la r g e r th a n 4 R m a x w ill
(a ) a n d (b ) th a t o th e r p a r ts o f th e im a g e w o u ld b e a ffe c te d b y
e b ig g e r R m a x , th e b ig g e r th e s tr u c tu r in g e le m e n t th a t w o u ld
n s e q u e n tly , th e g r e a te r th e e ffe c t o n th e im a g e a s a w h o le .
Problem 9.36
( a ) C o lo r th e im a g e b o r d e r p ix e ls th e s a m e c o lo r a s th e p a r tic le s (w h ite ). C a ll
th e r e s u ltin g s e t o f b o r d e r p ix e ls B . A p p ly th e c o n n e c te d c o m p o n e n t a lg o r ith m
1 8 9
( S e c t io n 9 .5 .3 ) . A ll c o n n e c t e d c o m p o n e n t s t h a t c o n t a in e le m e n t s fr o m B
p a r tic le s th a t h a v e m e r g e d w ith th e b o r d e r o f th e im a g e .
( b ) It is g iv e n th a t a ll p a r tic le s a r e
p r o b le m ; m o r e g e n e r a l a n a ly s is r e
a r e a (n u m b e r o f p ix e ls ) o f a s in g le
th e im a g e th e p a r tic le s th a t w e r e
th e c o n n e c te d c o m p o n e n t a lg o r it
p o n e n t. A c o m p o n e n t is th e n d e
p ix e ls is le s s th a n o r e q u a l to A + "
fo r v a r ia tio n s in s iz e d u e to n o is e .
a re
o f th e s a m e s iz e (th is is d o n e to s im p lify th e
q u ir e s to o ls fr o m C h a p te r 1 1 ). D e te r m in e th e
p a r tic le ; d e n o te th e a r e a b y A . E lim in a te fr o m
m e r g e d w ith th e b o r d e r o f th e im a g e . A p p ly
h m . C o u n t th e n u m b e r o f p ix e ls in e a c h c o m s ig n a te d a s a s in g le p a r tic le if th e n u m b e r o f
, w h e r e " is a s m a ll q u a n tity a d d e d to a c c o u n t
( c ) S u b tr a c t fr o m th e im a g e s in g le p a r tic le s a n d th e p a r tic le s th a t h a v e m e r g e d
w ith th e b o r d e r, a n d th e r e m a in in g p a r tic le s a r e o v e r la p p in g p a r tic le s .
Problem 9.37
A s g iv e n in th e p r o b le m s ta te m e n t, in te r e s t lie s in d e v ia tio n s fr o m th e r o u n d in
th e in n e r a n d o u te r b o u n d a r ie s o f th e w a s h e r s . It is s ta te d a ls o th a t w e c a n ig n o re e r r o r s d u e to d ig itiz in g a n d p o s itio n in g . T h is m e a n s th a t th e im a g in g s y s te m h a s e n o u g h r e s o lu tio n s o th a t o b je c tio n a b le a r tifa c ts w ill n o t b e in tr o d u c e d
a s a r e s u lt o f d ig itiz a tio n . T h e m e c h a n ic a l a c c u r a c y s im ila r ly te lls u s th a t n o a p p r e c ia b le e r r o r s w ill b e in tr o d u c e d a s a r e s u lt o f p o s itio n in g . T h is is im p o r ta n t if
w e w a n t to d o m a tc h in g w ith o u t h a v in g to re g is te r th e im a g e s .
T h e fi r s t s te p in th e s o lu tio n is th e s p e c ifi c a tio n o f a n illu m in a tio n a p p r o a c h .
B e c a u s e w e a re in te re s te d in b o u n d a r y d e fe c ts , th e m e th o d o f c h o ic e is a b a c k lig h tin g s y s te m th a t w ill p r o d u c e a b in a r y im a g e . W e a r e a s s u r e d fr o m th e p r o b le m s ta te m e n t th a t th e illu m in a tio n s y s te m h a s e n o u g h r e s o lu tio n s o th a t w e
c a n ig n o re d e fe c ts d u e to d ig itiz in g .
T h e n e x t s te p is to s p e c ify a c o m p a r is o n s c h e m e . T h e s im p le s t w a y to m a tc h
b in a r y im a g e s is to A N D o n e im a g e w ith th e c o m p le m e n t o f th e o th e r. H e r e , w e
m a tc h th e in p u t b in a r y im a g e w ith th e c o m p le m e n t o f th e g o ld e n im a g e (th is is
m o r e e ffi c ie n t th a n c o m p u tin g th e c o m p le m e n t o f e a c h in p u t im a g e a n d c o m p a r in g it to th e g o ld e n im a g e ). If th e im a g e s a r e id e n tic a l (a n d p e r fe c tly r e g is te r e d ) th e r e s u lt o f th e A N D o p e r a tio n w ill b e a ll 0 s . O th e r w is e , th e r e w ill b e 1 s
in th e a re a s w h e re th e tw o im a g e s d o n o t m a tc h . N o te th a t th is re q u ire s th a t
th e im a g e s b e o f th e s a m e s iz e a n d b e re g is te re d , th u s th e a s s u m p tio n o f th e
m e c h a n ic a l a c c u r a c y g iv e n in th e p r o b le m s ta te m e n t.
A s n o te d , d iffe r e n c e s in th e im a g e s w ill a p p e a r a s r e g io n s o f 1 s in th e A N D
im a g e . T h e s e w e g r o u p in to r e g io n s (c o n n e c te d c o m p o n e n ts ) b y u s in g th e a lg o r ith m g iv e n in S e c t io n 9 .5 .3 . O n c e a ll c o n n e c t e d c o m p o n e n ts h a v e b e e n e x -
1 9 0
C H A P T E R 9 . P R O B L E M S O L U T IO N S
tr a c te d , w e c a n c o m p a re th e m a g a in s t s p e c ifi e d c r ite r ia fo r a c c e p ta n c e o r re
je c tio n o f a g iv e n w a s h e r. T h e s im p le s t c r ite r io n is to s e t a lim it o n th e n u m b e
a n d s iz e (n u m b e r o f p ix e ls ) o f c o n n e c te d c o m p o n e n ts . T h e m o s t s tr in g e n t c r ite
r io n is 0 c o n n e c te d c o m p o n e n ts . T h is m e a n s a p e r fe c t m a tc h . T h e n e x t le v e l fo
“ r e la x in g ” a c c e p ta n c e is o n e c o n n e c te d c o m p o n e n t o f s iz e 1 , a n d s o o n . M o r
s o p h is tic a te d c r ite r ia m ig h t in v o lv e m e a s u r e s lik e th e s h a p e o f c o n n e c te d c o m
p o n e n ts a n d th e r e la tiv e lo c a tio n s w ith r e s p e c t to e a c h o th e r. T h e s e ty p e s o
d e s c r ip to r s a re s tu d ie d in C h a p te r 1 1 .
r
r
e
f
Chapter 10
Problem Solutions
Problem 10.1
E x p a n d f (x + Δ x ) in to a T a y lo r s e r ie s a b o u t x :
f (x ) + Δ x f 0(x ) +
f (x + Δ x ) =
(Δ x )
f
2 !
00
(x ) + ···
T h e in c r e m e n t in t h e s p a t ia l v a r ia b le x is d e fi n e d in S e c t io n 2 .4 .2 t o b e 1 , s o b y
le ttin g Δ x = 1 a n d k e e p in g o n ly th e lin e a r te r m s w e o b ta in th e r e s u lt
f 0(x ) =
f (x + 1 ) −
f (x )
w h ic h a g r e e s w it h E q . ( 1 0 .2 - 1 ) .
Problem 10.2
T h
y ie
e n
a +
in
e m a s k s
ld a v a lu
te d in th
2 w h e n
th e d ire c
w o u ld h a v e
e o f 0 w h e n
e d ire c tio n f
a m a s k is c e
tio n fa v o re d
th e
c e n
a v o
n te
b y
c
te
re
re
th
o e ffi c ie n ts s h o
re d o n a p ix e l
d b y th a t m a s k
d o n a o n e - p ix
a t m a s k .
w n in
o f a n
. C o n
e l g a p
F ig .
u n b r
v e rs e
in a
P 1 0 .2 .
o k e n 3 ly , th e r
3 - p ix e l
E a c h m a s k
p ix e l s e g m e
e s p o n s e w o
s e g m e n t o r
w o u ld
n t o r iu ld b e
ie n te d
Problem 10.3
T h e k
a d e fi
h a v e
th e le
p o in t
e y to s o lv in
n itio n o f e n
b e e n fo u n d
n g th s o f th e
s o f e v e r y p
g th is p r o b le m is to fi n d a
d p o in t) o f lin e s e g m e n ts
, th e D 8 d is ta n c e b e tw e e n
v a r io u s g a p s . W e c h o o s e
a ir o f s e g m e n ts a n d a n y s
1 9 1
ll
in
a
th
u c
e n d p o in ts (s e e
th e im a g e . O n
ll p a ir s o f s u c h
e s m a lle s t d is ta
h d is ta n c e le s s
S e c t io n 9 .5 .8
c e a ll e n d p o
e n d p o in ts g
n c e b e tw e e n
th a n o r e q u a
fo r
in ts
iv e s
e n d
l to
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
1 9 2
F ig u r e P 1 0 .2
K s a tis fi e
n u m e ro u
F o r e x a m
to b e le s s
lik e th is a
s u c h te s t
th is u p in
s th e s ta te m e n t o f th e p r o b le m . T h is is
s e m b e llis h m e n ts c a n b e a d d e d to b u ild
p le , it is p o s s ib le fo r e n d p o in ts o f d iffe r e
th a n K p ix e ls a p a r t, a n d h e u r is tic te s ts t
r e q u ite u s e fu l. A lth o u g h th e p r o b le m s t
s , th e y n o r m a lly a r e n e e d e d in p r a c tic e
c la s s if th is p r o b le m is a s s ig n e d a s h o m e
a r u d im e n ta r y s o lu tio n , a n d
in te llig e n c e in to th e p r o c e s s .
n t, b u t c lo s e ly a d ja c e n t, lin e s
h a t a tte m p t to s o r t o u t th in g s
a te m e n t d o e s n o t c a ll fo r a n y
a n d it is w o r th w h ile to b r in g
w o rk .
Problem 10.4
( a ) T h e lin e s w e r e th ic k e r th a n th e w id th o f th e lin e d e te c to r m a s k s . T h u s , w h e n ,
fo r e x a m p le , a m a s k w a s c e n te r e d o n th e lin e it ” s a w ” a c o n s ta n t a r e a a n d g a v e
a re s p o n s e o f 0 .
( b ) V ia c o n n e c tiv ity a n a ly s is .
Problem 10.5
(a ) T
p ro fi
in g g
b o rd
a g e s
h e fi r s t r o w in F ig . P 1 0 .5 s h o w s a s t e p , r a m p , a n d
le s th r o u g h th e ir c e n te r s . S im ila r ly , th e s e c o n d
r a d ie n t im a g e s a n d h o r iz o n ta l p r o fi le s th r o u g h
e r s in th e im a g e s a r e in c lu d e d fo r c la r ity in d e fi
; th e y a re n o t p a r t o f th e im a g e d a ta .
e d g e
ro w s
th e ir
n in g
im a g
h o w s
c e n te
th e b
e , a n
th e
rs . T
o rd e
d h
c o r
h e
rs o
o r iz
re s p
th in
f th
o n ta l
o n d d a rk
e im -
( b ) T h e t h ir d r o w in F ig . P 1 0 .5 s h o w s t h e a n g le im a g e s a n d t h e ir p r o fi le s .
A ll im a g e s w
u la r, th e p r o fi le
itiv e v a lu e s . T h
a n d d a rk a re a s
T h e b la c k in a ll
(2 5 5 b e c a u s e th
e re
o f
e g
c o r
o th
e s e
s c a le d fo r d is p la y , s o o
th e a n g le im a g e o f th e
r a y in th e s c a le d a n g le
re s p o n d to p o s itiv e a n
e r im a g e s d e n o te s 0 a n
a re 8 - b it im a g e s ). G r a
n ly th e s h a p e s a r e o f in te r e s t.
r o o f e d g e h a s z e r o , n e g a tiv e ,
im a g e d e n o te s z e r o , a n d th e
d e q u a lly n e g a tiv e v a lu e s , r e s
d th e p u re w h ite is th e m a x im
y s a r e v a lu e s in b e tw e e n .
In p a r tic a n d p o s th e lig h t
p e c tiv e ly .
u m v a lu e
1 9 3
Edges and their profiles
Gradient images and their profiles
Angleimages and their profiles
F ig u r e P 1 0 .5
Problem 10.6
F ig u r e P 1 0 .6 s h o w s t h e s o lu t io n .
Problem 10.7
F ig u r e P 1 0 .7 s h o w s t h e s o lu t io n .
Problem 10.8
( a ) In s p e c tio n o f th e S o b e l m a s k s s h o w s th a t g x = 0 fo r e d g e s o r ie n te d v e r tic a lly
a n d g y = 0 fo r e d g e s o r ie n te d h o r iz o n ta lly . T h e r e fo r e , it fo llo w s in th is c a s e th a t,
?
g y2 = g y , a n d s i m i l a r l y f o r h o r i z o n t a l e d g e s .
fo r v e r tic a l e d g e s , r f =
( b ) T h e s a m e a r g u m e n t a p p lie s to th e P r e w itt m a s k s .
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
1 9 4
n
n
F ig u r e P 1 0 .6
Problem 10.9
C o n s id e r fi r s t t h e S o b e l m a s k s o f F ig s . 1 0 .1 4 a n d 1 0 .1 5 . A s im p le w a y t o p r o v e
th a t th e s e m a s k s g iv e is o tr o p ic r e s u lts fo r e d g e s e g m e n ts o r ie n te d a t m u ltip le s
o f 4 5 ◦ is to o b ta in th e m a s k re s p o n s e s fo r th e fo u r g e n e r a l e d g e s e g m e n ts s h o w n
in F ig . P 1 0 .9 , w h ic h a r e o r ie n t e d a t in c r e m e n t s o f 4 5 ◦ . T h e o b je c t iv e is t o s h o w
th a t th e r e s p o n s e s o f th e S o b e l m a s k s a r e in d is tin g u is h a b le fo r th e s e fo u r e d g e s .
T h a t t h is is t h e c a s e is e v id e n t fr o m T a b le P 1 0 .9 , w h ic h s h o w s t h e r e s p o n s e o f
e a c h S o b e l m a s k to th e fo u r g e n e r a l e d g e s e g m e n ts . W e s e e th a t in e a c h c a s e
th e re s p o n s e o f th e m a s k th a t m a tc h e s th e e d g e d ire c tio n is (4 a − 4 b ), a n d th e
re s p o n s e o f th e c o r re s p o n d in g o r th o g o n a l m a s k is 0 . T h e re s p o n s e o f th e re m a in in g tw o m a s k s is e ith e r (3 a − 3 b ) o r (3 b − 3 a ). T h e s ig n d iffe re n c e is n o t
s ig n ifi c a n t b e c a u s e th e g r a d ie n t is c o m p u te d b y e ith e r s q u a r in g o r ta k in g th e
a b s o lu te v a lu e o f th e m a s k r e s p o n s e s . T h e s a m e lin e o f r e a s o n in g a p p lie s to th e
P re w itt m a s k s .
E d g e
d ire c tio n
H o r iz o n ta l
V e r tic a l
+ 4 5 ◦
− 4 5 ◦
H o r iz o
S o b e l
4 a −
0
3 a −
3 b −
n ta l
(g x )
4 b
3 b
3 a
T a b le P 1 0 .9
V e r tic a l
S o b e l (g y )
0
4 a − 4 b
3 a − 3 b
3 a − 3 b
S o
3
3
4
+ 4
b e l
a −
a −
a −
0
5
◦
(g 45 )
3 b
3 b
4 b
− 4
S o b e l
3 b −
3 a −
5
(g
◦
− 4 5
)
3 a
3 b
0
4 a − 4 b
Problem 10.10
C o n
a n d
p o in
re s p
s id e r
a g e n
t h a s
o n s e
fi rs t
e ra l
v a lu
o f th
th
3
e
e
e 3 ×
× 3 s
e , a s
3 × 3
3 s m
u b im
F ig .
m a s k
o o th in
a g e a r
P 1 0 .1 0
w h e n
g m a s k
e a w ith
s h o w s
its c e n
m e n tio n e
in te n s itie
. R e c a ll th
te r is a t th
d in th e p r o
s a th ro u g h
a t v a lu e e is
a t lo c a tio n .
b le
i ,
re
Ig n
m s ta te m
w h o s e c e
p la c e d b y
o r in g th e
e n t,
n te r
th e
1 / 9
1 9 5
F ig u r e P 1 0 .7
s c a le fa c to r, th e r e s p o n s e o f t
c + d + e + f + g + h + i).
T h e id e a w ith th e o n e - d im
a p ix e l b y th e re s p o n s e o f th e
m in d , th e m a s k [1 1 1 ] w o u ld
p ix e ls w ith v a lu e s b , e , a n d h ,
N e x t, w e p a s s th e m a s k
h e m a s k w h e n c e n te r e d a t th a t lo c a tio n is (a + b +
e n
m a
y ie
re s
s io
s k
ld
p e
n a l m a s k
w h e n it is
th e fo llo w
c tiv e ly : (a
is
c
in
+
th e s a
e n te re
g re s p
b + c )
m e :
d o n
o n s e
, (d +
W e r e p la c e th
th a t p ix e l. W
s w h e n c e n te
e + f ), a n d (
e v a lu e
ith th is
re d a t t
g + h +
o f
in
h e
i ).
⎤
⎡
1
⎥
⎦
⎢
⎣ 1
1
th r o u g h th e s e r e s u lts
r e s p o n s e w ill b e [(a +
r e s u lt p r o d u c e d b y th
R e tu r n in g n o w to
th e p ix e l w ith v a lu e e
th e o n e - d im e n s io n a l
. W h e n th is m a s k
b + c ) + (d + e + f
e 3 × 3 s m o o th in g
p r o b le m a t h a n d ,
, its re s p o n s e is g x
d iffe re n c in g m a s k
⎡
is c e n te re
) + (g + h
m a s k .
w h e n th e
= (g + 2 h
− 1
⎢
⎣ 0
1
⎤
⎥
⎦
d a t th e p ix e l w ith v a lu e e , its
+ i )], w h ic h is th e s a m e a s th e
g x S o b e l m a s k is c e n te re d a t
+ i ) − (a + 2 b + c ). If w e p a s s
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
1 9 6
th r o u g h th e im a g e , its r e s p o n s e w h e n its c e n te r is a t th e p ix e ls w ith v a lu e s d ,
e , a n d f , r e s p e c t iv e ly , w o u ld b e :( g − a ) , (h − b ), a n d (i − c ). N e x t w e a p p ly t h e
s m o o th in g m a s k [1 2 1 ] to th e s e r e s u lts . W h e n th e m a s k is c e n te r e d a t th e p ix e l
w ith v a lu e e , its r e s p o n s e w o u ld b e [( g − a ) + 2 (h − b ) + (i − c )] w h ic h is [( g +
2 h + i ) − (a + 2 b + c )]. T h is is th e s a m e a s th e re s p o n s e o f th e 3 × 3 S o b e l m a s k fo r
g x . T h e p r o c e s s to s h o w e q u iv a le n c e fo r g y is b a s ic a lly th e s a m e . N o te , h o w e v e r,
th a t th e d ir e c tio n s o f th e o n e - d im e n s io n a l m a s k s w o u ld b e r e v e r s e d in th e s e n s e
th a t th e d iffe r e n c in g m a s k w o u ld b e a c o lu m n m a s k a n d th e s m o o th in g m a s k
w o u ld b e a r o w m a s k .
Problem 10.11
( a ) T h e o p e r a to r s a r e a s fo llo w s (n e g a tiv e n u m b e r s a r e s h o w n u n d e r lin e d ):
1 1 1
1 1 0
1 0 1
0 1 1
1 1 1
1 1 0
1 0 1
0 1 1
0 0 0
1 0 1
1 0 1
1 0 1
0 0 0
1 0 1
1 0 1
1 0 1
1 1 1
0 1 1
1 0 1
1 1 0
1 1 1
0 1 1
1 0 1
1 1 0
( b ) T h e s o lu tio n is a s fo llo w s :
C o m p a s s g r a d ie n t o p e r a to r s
E
N E
N
N W
W
S W
S
S E
G r a d ie n t d ir e c tio n
N
N W
W
S W
S
S E
E
N E
Problem 10.12
( a ) T h e s o lu t io n is s h o w n in F ig . P 1 0 .1 2 ( a ) . T h e n u m b e r s in b r a c k e t s a r e v a lu e s
o f [g x , g y ].
F ig u r e P 1 0 .9
1 9 7
F ig u r e P 1 0 .1 0
( b ) T h e s o lu t io n is s h o w n in F ig . P 1 0 .1 2 ( b ) . T h e a n g le w a s n o t c o m p u t e d fo r t h e
tr iv ia l c a s e s in w h ic h g x = g y = 0 . T h e h is to g r a m fo llo w s d ir e c tly fr o m th is ta b le .
( c ) T h e s o lu t io n is s h o w n in F ig . P 1 0 .1 2 ( c ) .
Problem 10.13
( a ) T h e lo c a l a v e r a g e a t a p o in t (x , y ) in a n im a g e is g iv e n b y
f¯ ( x , y ) =
1
n
z
2
z i2 S
i
x y
w h e re S x y is th e re g io n in th e im a g e e n c o m p a s s e d b y th e n × n a v e r a g in g m a s k
w h e n it is c e n te r e d a t (x , y ) a n d th e z i a r e th e in te n s itie s o f th e im a g e p ix e ls in
th a t re g io n . T h e p a r tia l
@ f¯ / @ x =
f¯ ( x + 1 , y ) −
f¯ ( x , y )
is th u s g iv e n b y
@ f¯ / @ x =
T h e fi
in th e
ro w p
w r ite
rs
s
ic
th
t s
e c
k e
e
u m m
o n d s
d u p
p re c e
a tio n o n th
u m m a tio n
b y th e m a
d in g e q u a t
1
z
n
z i2 S
1
−
i
2
z
i
2
n
x + 1 ,y
z i2 S
x y
e r ig h t c a n b e in te r p r e te d a s c o n s is tin g o f a ll th e p ix e ls
m in u s th e p ix e ls in th e fi r s t r o w o f th e m a s k , p lu s th e
s k a s it m o v e d fr o m (x ,y ) to (x + 1 ,y ). T h u s , w e c a n
io n a s
@ f¯ / @ x =
1
n
z
2
z i2 S
x + 1 ,y
i
−
1
n
z
2
z i2 S
x y
i
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
1 9 8
F ig u r e P 1 0 .1 2
,
=
1
n
z
n − 1
2
f (x +
1
n
+
x y
2
n
=
i
k = y −
n − 1
2
y +
n − 1
2
f (x +
2
k = y −
1
n
2
(s u m o f p ix e ls in n e w r o w )
(s u m o f p ix e ls in 1 s t r o w )
2
n
y +
1
2 S
i
1
−
=
z
2
n − 1
2
n + 1
,k ) −
2
n + 1
,k ) −
2
y +
1
n
1
−
n
z
2
z i2 S
n − 1
2
f (x −
2
k = y −
f (x −
i
x y
n − 1
2
n − 1
,k ) .
2
n − 1
,k )
2
1 9 9
T h i s e x p r e s s i o n g i v e s t h e v a lu e o f @ f¯ / @ x a t c o o r d i n a t e s ( x , y ) o f t h e s m o o t h e d
im a g e . S im ila r ly ,
@ f¯ / @ y
=
1
z
2
n
,
=
z i2 S
1
i
1
+
n − 1
2
k = x −
n − 1
2
x +
n − 1
2
(s u m o f p ix e ls in n e w c o l)
n − 1
2
1
−
x +
1
z
2
n
z i2 S
f (k ,y −
k = x −
f (k ,y −
i
x y
n − 1
2
2
n
n + 1
) −
2
f (k ,y +
2
k = x −
i
x y
n + 1
) −
2
f (k ,y +
1
2
n
x y
2
n
n
z i2 S
(s u m o f p ix e ls in 1 s t c o l)
2
n
x +
1
=
z
z i2 S
z
2
n
-
2
−
=
x ,y + 1
1
n
1
−
i
n − 1
2
n − 1
)
2
n − 1
) .
2
T h e e d g e m a g n i t u d e i m a g e c o r r e s p o n d i n g t o t h e s m o o t h e d i m a g e f¯ ( x , y ) i s t h e n
g iv e n b y
( @ f¯ / @ x ) 2 + ( @ f¯ / @ y ) 2 .
M¯ ( x , y ) =
( b ) F r o m t h e p r e c e d i n g e q u a t i o n f o r @ f¯ / @ x , t h e m a x i m u m v a lu e t h i s t e r m c a n
h a v e is w h e n e a c h d iffe re n c e in th e s u m m a tio n is m a x im u m w h ic h , fo r a n m - b it
im a g e , is L = 2 m − 1 ( e .g ., 2 5 5 fo r a n 8 - b it im a g e ) . T h u s ,
@ f¯ / @ x
m a x
=
1
n
2
(n L ) =
L
n
a n d s im ila r ly fo r o th e r d e r iv a tiv e te r m . T h e r e fo r e ,
)
L 2
L 2
L p
M¯ m a x ( x , y ) =
+
=
2 .
2
2
n
n
n
F o r th e o r ig in a l im a g e , th e m a x im u m v a lu e th a t @ f / @ x c a n h a v e is L , a n d s im ila r ly fo r @ f / @ y . T h e r e fo r e ,
p
M max (x ,y ) = L 2
a n d th e r a tio is :
M¯
M
m a x
m a x
(x ,y )
=
(x ,y )
1
n
w h ic h s h o w s th a t th e e d g e s tre n g th o f a s m o o th e d im a g e c o m p a re d to th e o r ig in a l im a g e is in v e r s e ly p r o p o r tio n a l to th e s iz e o f th e s m o o th in g m a s k .
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 0 0
Problem 10.14
( a ) W e p r o c e e e d a s fo llo w s
A v e ra g e
r 2G (x ,y )
1
1
− 1
1
− 1
1
=
=
− 1
4
p
4
σ
4
2
σ
p
4 π − 4 π
=
0
p
2
−
y 2
2 σ 2
d x d y
d y
1
d y
e
−
x 2
2 σ 2
d x
d x d y
2 π σ
p
2
2 π σ × σ
2 π σ
σ 2
2
=
x 2 + y 2
2 σ 2
− 1
4
y 2
2 σ 2
−
e
−
e
1
−
e
x 2 + y 2
2 σ 2
−
− 1
2 π σ × σ
+
4
2
− 1
1
σ
e
− 1
y
2
2
d x
1
σ
1
− 2 σ
1
x 2
2 σ 2
−
e
1
−
=
2
− 1
+
2
+ y
σ
x
σ
2
x
− 1
1
1
=
r 2G (x ,y )d x d y
2 π σ
2
th e fo u r th lin e fo llo w s fr o m th e fa c t th a t
v a r ia n c e (z ) = σ
a n d
p
( b ) C o n v o lv in g
a s m u ltip ly in g
d o m a in . F r o m
fre q u e n c y d o m
fre q u e n c y d o m
s o its F o u r ie r tr
F o u r ie r tr a n s fo
2 π σ
1
1
p
=
2 π σ
1
1
a n im a g e f (x ,y )
th e F o u r ie r tr a n
C h a p te r 4 , w e k
a in is p r o p o r tio n
a in . W e s h o w e d
a n s fo r m a t th e o
r m s o f f (x ,y ) a n
2
e
−
z 2
2 σ 2
z
2
e
−
z 2
2 σ 2
d z
− 1
d z = 1 .
− 1
w ith r 2 G (x ,y
s fo r m s o f th e
n o w th a t th e
a l to th e v a lu e
in (a ) th a t th e
r ig in m u s t b e
d r 2G (x ,y ) e
) in th e s p a tia l d o m a in
s e tw o fu n c tio n s in th e
a v e r a g e v a lu e o f a fu n c
o f its tr a n s fo r m a t th e o
a v e r a g e v a lu e o f r 2 G (x
z e ro . T h e re fo re , th e p ro
v a lu a te d a t th e o r ig in is
is th e s a m e
fre q u e n c y
tio n in th e
r ig in o f th e
, y ) is z e r o ,
d u c t o f th e
z e ro , fro m
2 0 1
w h ic h it fo llo w s th a t th e c o n v o lu tio n o f th e s e tw o fu n c tio n s h a s a z e r o a v e r a g e
v a lu e .
(c ) T h
th e L a
w ith a
s o if it
v o lv in
e a n s w e r is y e s . F
p la c ia n m a s k s in
n y im a g e g iv e s a
s a v e r a g e is z e r o ,
g th is r e s u lt w ith
ro
th
re
th
a n
m P r o b le m 3 .1 6 w e
e p r o b le m s ta te m e
s u lt w h o s e a v e r a g e
is m e a n s th a t a ll th
y im a g e a ls o g iv e s a
k n o w
n t) w h
v a lu e
e e le m
r e s u lt
th a t c o
o s e c o
is z e r o
e n ts s u
w h o s e
n v o lv in g a
e ffi c ie n ts s
. T h e re s u
m to z e ro .
a v e ra g e v a
m a s k (lik
u m to z e r
lt is d ig ita
T h u s , c o n
lu e is z e r o
e
o
l,
.
Problem 10.15
( a ) L e t g L (x , y ) d e n o t e t h e L o G im a g e , a s in F ig . 1 0 .2 2 ( b ) . T h is im a g e h a s b o t h
p o s itiv e a n d n e g a tiv e v a lu e s , a n d w e k n o w th a t th e z e r o c r o s s in g s a r e s u c h th a t
a t le a s t a p a ir o f o p p o s in g lo c a tio n s in a 3 × 3 n e ig h b o r h o o d d iffe r in s ig n . C o n s id e r a b in a r y im a g e g P (x , y ) fo r m e d b y le ttin g g P (x , y ) = 1 if g L (x , y ) > 0 a n d
g P (x , y ) = 0 o t h e r w is e . F ig u r e P 1 0 .1 5 ( a ) s h o w s g P (x , y ) fo r t h e L a p la c ia n im a g e in F ig . 1 0 .2 2 ( b ) . T h e im p o r t a n t t h in g a b o u t t h is im a g e is t h a t it c o n s is t s
o f c o n n e c te d c o m p o n e n ts . F u th e r m o re (fo r th e 3 × 3 m a s k u s e d to d e te c t z e ro
c r o s s in g s a n d a th r e s h o ld o f 0 ) e a c h p o in t o n th e b o u n d a r y o f th e s e c o n n e c te d
c o m p o n e n t s is e it h e r a p o in t s o f t r a n s it io n b e t w e e n p o s it iv e a n d n e g a t iv e ( i.e .,
it is th e c e n te r p o in t o f th e m a s k ) o r it is a d ja c e n t to th e p o in t o f tr a n s itio n . In
o th e r w o r d s , a ll z e r o - c r o s s in g p o in ts a r e a d ja c e n t to th e b o u n d a r is o f th e c o n n e c te d c o m p o n e n ts ju s t d e s c r ib e d . B u t, b o u n d a r ie s o f c o n n e c te d c o m p o n e n ts
fo r m a c lo s e d p a th (a p a th e x is ts b e tw e e n a n y tw o p o in ts o f a c o n n e c te d c o m p o n e n t, a n d th e p o in ts fo r m in g th e b o u n d a r y o f a c o n n e c te d c o m p o n e n ts a re p a r t
o f th e c o n n e c te d c o m p o n e n t, s o th e b o u n d a r y o f a c o n n e c te d a c o m p o n e n t is a
c lo s e d p a th ) . F ig u r e P 1 0 .1 5 ( b ) s h o w s t h e c lo s e d p a th s fo r th e p r o b le m in q u e s t io n . C o m p a r e t h e s e c lo s e d p a t h s a n d t h e b in a r y r e g io n s in F ig . P 1 0 .1 5 ( a ) .
(b ) T h e a n s w e r
th e z e r o c r o s s in
tio n a n d a th re s
a r e e x p la in e d in
P o g g io , IE E E T r a
1 4 7 - 1 6 3 . L o o k in
u n d e r p in n in g s o
s tu d e n ts .
is y e s fo r fu n c tio n s th a t m e e t c e r ta in m ild c o n d itio n s , a n d
g m e th o d is b a s e d o n r o ta tio n a l o p e r a to r s lik e th e L o G fu n
h o ld o f 0 . G e o m e tr ic a l p r o p e r tie s o f z e r o c r o s s in g s in g e n e r
s o m e d e t a i l i n t h e p a p e r ” O n E d g e D e t e c t i o n ,” b y V . T o r r e a n d
n s . P a t t e r n A n a ly s is a n d M a c h in e In t e ll., v o l. 8 , n o . 2 , 1 9 8 6 , p
g u p th is p a p e r a n d b e c o m in g fa m ilia r w ith th e m a th e m a tic
f e d g e d e te c tio n is a n e x c e lle n t r e a d in g a s s ig n m e n t fo r g r a d u a
if
c a l
T .
p .
a l
te
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 0 2
F ig u r e P 1 0 .1 5
Problem 10.16
G 0(r )
− r − r 22
e 2σ
σ 2
G 00( r )
1
r 2
−
4
σ
σ
r 2 − σ 2
σ 4
=
r 2G (x ,y )
=
=
=
L e ttin g r
2
= x
2
+ y
2
2
r 2
2 σ 2
−
e
r 2
2 σ 2
−
e
w e o b ta in
r 2G (x ,y ) =
C o m p a r in g th is d e r iv a
re a s o n fo r th e d iffe re n
w ith re s p e c t to x a n d y
w ith r is b a s ic a lly w o r k
a s w o r k in g w ith a 1 - D
it tu r n s o u t, th e d iffe re
a s ig n ifi c a n t d iffe re n c e
x
2
2
+ y
σ
2
− σ
e
4
−
x 2 + y 2
2 σ 2
tio n w ith th e d e r iv a tio n o f E q .
c e is in th e d e r iv a tiv e s o f r a s o
. A n o th e r w a y o f e x p la in in g th e
in g w ith a r a d ia l s lic e o f a 2 - D fu
G a u s s ia n . T h e 2 - D n a tu re o f th
n c e o f o n ly 2 in th e n u m e r a to r
in th e fi lte r e d r e s u lt.
Problem 10.17
( 1 0 .2 - 1 5 ) w e s e e t h a t t h
p p o s e d to th e d e r iv a tiv e
d iffe re n c e is th a t w o r k in
n c tio n , w h ic h is th e s a m
e p r o b le m is th u s lo s t. A
m u ltip ly in g s ig m a m a k e
( a ) F r o m E q . ( 1 0 .2 - 2 6 ) , t h e D o G fu n c t io n is z e r o w h e n
1
2 π σ
1
−
2
e
x 2 + y 2
2 σ 2
1
=
1
2 π σ
2
−
2
e
x 2 + y 2
2 σ 2
2
.
e
s
g
e
s
s
2 0 3
T a k in g th e n a tu r a l lo g o f b o th s id e s y ie ld s ,
1
ln
2 π σ
1
2
x
−
2
2
+ y
2 σ
1
1
= ln
2
2 π σ
2
2
x
−
2
2
+ y
2 σ
2
.
2
C o m b in in g te r m s ,
x
2
1
2
+ y
2 σ
1
1
−
2
2 σ
2
1
= ln
2
= ln
2 π σ
1
σ
1
σ
2
2
F in a lly , s o lv in g fo r σ
2
1
2
σ
2
1
−
2
σ
1
= ln
2
2 π σ
2
2
.
2
2
T h e L o G fu n c t io n [E q . ( 1 0 .2 - 2 3 ) ] is z e r o w h e n x
p re c e d in g e q u a tio n ,
σ
1
− ln
2
2
+ y
=
2
2 σ
. T h e n , fro m th e
1
σ
2
2
σ
.
2
,
2
σ
σ
=
σ
1
2
1
2
σ
2
2
− σ
2
ln
2
1
σ
2
2
σ
2
w h ic h a g r e e s w it h E q . ( 1 0 .2 - 2 7 ) .
( b ) T o o b ta in a n e x p r e s s io n in te r m s o f k , w e le t σ
e q u a tio n :
σ
2
2
k
=
=
k
2
k
2
σ
k
4
2
σ
2
2
− σ
2
− 1
2
2
σ
2
k
ln
2
=
k σ
2
in th e p re c e d in g
2
σ
σ
ln (k
2
2
1
2
2
2
)
w ith k > 1 .
Problem 10.18
( a ) E q u a t io n ( 1 0 .2 - 2 1 ) c a n b e w r it t e n in t h e fo llo w in g s e p a r a b le fo r m
G (x ,y )
=
−
x 2 + y 2
2 σ 2
−
x 2
2 σ 2
e
−
y 2
2 σ 2
=
e
=
G (x )G (y ).
e
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 0 4
F r o m E q . ( 3 .4 - 2 ) a n d t h e p r e c e d in g e q u a t io n , t h e c o n v o lu t io n o f G (x , y ) a n d
f (x ,y ) c a n b e w r itte n a s
a
a
f (x ,y )
G (x ,y )
=
G (s ,t ) f (x − s ,y − t )
s = − a t = − a
a
a
=
e
s = − a t = − a
a
=
−
e
s 2
2 σ 2
s 2
2 σ 2
−
⎡
⎣
a = (n − 1 )/ 2 a n d n is th e
0 .2 - 2 1 ) . T h e e x p r e s s io n in s
2
2
e n tia l te r m , e − t /2 σ , w ith t
2
2
c o n v o lu tio n o f e − s /2 σ w ith
G (x ,y )
(b )
e a c
1 -D
im a
m u
m u
D ire c
h lo c a
c o n v
g e , fo
ltip lic
ltip lic
t im p le
tio n o f
o lu tio n
r a to ta
a tio n s a
a tio n s .
m e n ta tio n
f (x ,y ), s o
re q u ire s n
l o f n × M
re re q u ire d
T h e c o m p u
f (x − s ,y − t )
⎤
a
t 2
2 σ 2
−
e
s = − a
w h e re
E q . (1
e x p o n
is th e
t 2
2 σ 2
−
e
f (x − s ,y − t )⎦
t = − a
s iz
id e
h e
th
n ×
c k e
f (x
n s o
e o f th e
th e b ra
ro w s o f
e c o lu m
f (x ,y ) = G (x )
n m a
ts is th
,y ). T h
f th e re
s k
e 1
e n
s u
o b
-D
th
lt.
n
2
M N
2 n M N
b y s a m p lin
lu tio n o f th
r s u m m a tio
a n o th e r w a
g
e
n
y ,
f (x ,y ) .
G (y )
o f 2 - D c o n v o lu tio n r e q u ir e s n
th e to ta l n u m b e r o f m u ltip lic a
m u ltip lic a tio n s a t e a c h lo c a tio
× N fo r th e p a s s a lo n g th e r o w
fo r th e p a s s a lo n g th e c o lu m n s
ta tio n a l a d v a n ta g e , A , is th e n
A =
ta in e d
c o n v o
e o u te
S ta te d
2
tio
n o
s .
, fo
m
n
f
T
r
u ltip lic a
s is n 2 ×
e v e r y ro
h e n , n ×
a to ta l o
tio n s
M ×
w in t
M ×
f 2 n M
a t
N .
h e
N
N
n
=
2
w h ic h is in d e p e n d e n t o f im a g e s iz e . F o r e x a m p le , if n = 2 5 , A = 1 2 .5 , s o it t a k e s
1 2 .5 m o r e m u lt ip lic a t io n s t o im p le m e n t 2 - D c o n v o lu t io n d ir e c t ly t h a n it d o e s t o
im p le m e n t th e p r o c e d u r e ju s t o u tlin e d th a t u s e s 1 - D c o n v o lu tio n s .
Problem 10.19
( a ) A s E q . ( 1 0 .2 - 2 5 ) s h o w s , t h e fi r s t t w o s t e p s o f t h e a lg o r it h m c a n b e s u m m a r iz e d in to o n e e q u a tio n :
g (x ,y ) = r
2
[G (x ,y )
f (x ,y )].
U s in g th e d e fi n itio n o f th e L a p la c ia n o p e r a to r w e c a n e x p r e s s th is e q u a tio n a s
g (x ,y )
2
@
=
=
@ x
@
@ x
2
G (x ,y )
2
G (x )
f (x ,y ) +
@
2
@ y
2
G (y )
f (x ,y ) +
G (x ,y )
2
@
@ y
f (x ,y )
2
2
G (x )
G (y )
f (x ,y )
2 0 5
w h e r e t h e s e c o n d s t e p fo llo w s fr o m P r o b le m 1 0 .1 8 , w it h G (x ) = e
−
−
x 2
2 σ 2
a n d G (y ) =
y 2
2 σ 2
e
. T h e te r m s in s id e th e tw o b r a c k e ts a r e th e s a m e , s o o n ly tw o c o n v o lu tio n s
a r e r e q u ir e d t o im p le m e n t t h e m . U s in g t h e d e fi n it io n s in S e c t io n 1 0 .2 .1 , t h e
p a r tia ls m a y b e w r itte n a s
2
@
f
a n d
@
2
f
2
@ y
T h e fi rs t te r m
c o e ffi c ie n ts c o
s a m e c o e ffi c ie
h a v e th e fi n a l
=
2
@ x
=
f (x + 1 ) +
f (x − 1 ) − 2 f (x )
f (y + 1 ) +
f (y − 1 ) − 2 f (y ).
c a n b e im p le m e n te d v ia c o n v o lu tio n w ith a 1 × 3 m a s k h a v in g
e ffi c ie n ts , [1 − 2 1 ], a n d th e s e c o n d w ith a 3 × 1 m a s k h a v in g th e
n t s . L e t t i n g r x2 a n d r 2y r e p r e s e n t t h e s e t w o o p e r a t o r m a s k s , w e
r e s u lt:
g (x ,y ) = r
2
x
G (x )
f (x ,y ) + r
G (y )
2
G (x )
y
G (y )
f (x ,y )
w h ic h r e q u ir e s a to ta l o f fo u r d iffe r e n t 1 - D c o n v o lu tio n o p e r a tio n s .
( b ) If w e u s e th e a lg o r ith m a s s ta te d in th e b o o k , c o n v o lv in g a n M × N
w ith a n n × n m a s k w ill r e q u ir e n 2 × M × N m u ltip lic a tio n s (s e e th e s o lu
P r o b le m 1 0 .1 8 ) . T h e n c o n v o lu t io n w it h a 3 × 3 L a p la c ia n m a s k w ill a d d a
9 × M × N m u ltip lic a tio n s fo r a to ta l o f (n 2 + 9 ) × M × N m u ltip lic a tio n
c o m p o s in g a 2 - D c o n v o lu tio n in to 1 - D p a s s e s r e q u ir e s 2 n M N m u ltip lic
a s in d ic a t e d in t h e s o lu t io n t o P r o b le m 1 0 .1 8 . T w o m o r e c o n v o lu t io n s o f
s u ltin g im a g e w ith th e 3 × 1 a n d 1 × 3 d e r iv a tiv e m a s k s a d d s 3 M N + 3 M N =
m u ltip lic a tio n s . T h e c o m p u ta tio n a l a d v a n ta g e is th e n
A =
(n
2
+ 9 )M N
=
2 n M N + 6 M N
im a g e
tio n to
n o th e r
s . D e a tio n s ,
th e re 6 M N
2
+ 9
2 n + 6
n
w h ic h is in d e p e n d e n t o f im a g e s iz e . F o r e x a m p le , fo r n = 2 5 , A = 1 1 .3 2 , s o it
ta k e s o n th e o r d e r o f 1 1 tim e s m o r e m u ltip lic a tio n s if d ir e c t 2 - D c o n v o lu tio n is
u s e d .
Problem 10.20
( a ) T h e s o lu t io n is b a s ic a lly t h e s a m e a s in P r o b le m 1 0 .1 9 , b u t u s in g t h e 1 - D
m a s k s in F ig . 1 0 .1 3 t o im p le m e n t E q s . ( 1 0 .2 - 1 2 ) a n d ( 1 0 .2 - 1 3 ) .
( b ) S m o o th in g w ith a G a u s s ia n fi lte r (s te p 1 o f th e a lg o r ith m ) r e q u ir e s c o n v o lv in g a n M × N im a g e w ith a n n × n fi lte r m a s k w ill r e q u ir e n 2 × M × N m u ltip lic a t io n s a s e x p la in e d in t h e s o lu t io n o f P r o b le m 1 0 .1 9 . T h e n t w o c o n v o lu t io n s
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 0 6
w ith a 3 × 3 h o r iz o n ta l a n d v e r tic a l e d g e d e te c to r re q u ire s 2 (9 × M × N ) = 1 8 M N
m u ltip lic a tio n s , fo r a to ta l o f (n 2 + 1 8 ) × M × N m u ltip lic a tio n s . T h e m a g n itu d e im a g e r e q u ir e s tw o m u ltip lic a tio n s , o n e to c o m p u te @ f / @ x tim e s its e lf
a n d @ f / @ y tim e s its e lf, fo r a to ta l o f 2 M N m u ltip lic a tio n s . T h e to ta l fo r th e 2 D a p p r o a c h is th e n (n 2 + 2 0 )M N m u ltip lic a tio n s . T h e 1 - D c o n v o lu tio n fo r s te p
1 r e q u ir e s 2 n M N m u lt ip lic a t io n s ( s e e t h e s o lu t io n t o P r o b le m , 1 0 .1 9 ) . T h e y
c o m p o n e n t o f th e g r a d ie n t c a n b e im p le m e n te d u s in g th e d iffe r e n c e @ f / @ y =
f (x , y + 1 ) − f (x , y ), w h ic h , a s s t a t e d in S e c t io n 1 0 .2 .5 , c a n b e im p le m e n t e d b y
c o n v o lv in g th e s m o o th e d im a g e w ith a 1 - D m a s k w ith c o e ffi c ie n ts [− 1 1 ]. T h is
w ill r e q u ir e 2 M N m u ltip lic a tio n s . S im ila r ly , th e o th e r p a r tia l d e r iv a tiv e c a n b e
im p le m e n te d w ith a v e r tic a l m a s k w ith th e s a m e c o e ffi c ie n ts , fo r a to ta l o f 4 N M
m u ltip lic a tio n s . T h e s q u a r e s a r e th e s a m e a s fo r th e 2 - D im p le m e n ta tio n : 2 M N
m u ltip lic a tio n s . T h e to ta l fo r th e 1 - D im p le m e n ta tio n is th e n (2 n + 6 )M N . T h e
r a tio o f 2 - D to 1 - D m u ltip lic a tio n s is
A =
(n 2 + 2 0 )M N
(2 n + 6 )M N
n
=
2
+ 2 0
2 n + 6
w h ic h is in d e p e n d e n t o f im a g e s iz e . W h e n , fo r e x a m p le , n = 2 5 , t h e n A = 1 1 .5 2 ,
s o it w o u ld ta k e a p p r o x im a te ly 1 1 tim e s m o r e m u ltip lic a tio n s to im p le m e n t th e
fi r s t p a r t o f th e C a n n y a lg o r ith m w ith 2 - D c o n v o lu tio n s .
Problem 10.21
P a r t s ( a ) t h r o u g h ( e ) a r e s h o w n in r o w s 2 t h r o u g h 6 o f F ig . P 1 0 .2 1 .
Problem 10.22
(a ) E x p re s
in g te r m s
b = ρ / s in
a n d b o f a
re p re s e n ta
s x c o s θ + y s in
w ith th e s lo p e θ . T h is g iv e s θ
g iv e n lin e , th e
tio n o f th a t lin e
(b ) θ = c o t
− 1
(2 ) = 2 6 .6
◦
θ = ρ in th
in te rc e p t fo
= c o t− 1 (a )
p a ra m e te rs
e fo
r m ,
a n d
θ a
r m
y
ρ
n d
=
y =
a x
= b s
ρ c o
− (c
+ b
in θ
m p
o t θ )x
, g iv e
. O n
le te ly
+ ρ / s in θ
s a = − (c
c e o b ta in e
s p e c ify th
. E q u a
o t θ ) a n
d fro m
e n o r m
td
a
a l
.
a n d ρ = (1 ) s in θ = 0 .4 5 .
Problem 10.23
( a ) P o in t 1 h a s c o o r d in a t e s x = 0 a n d y = 0 . S u b s t it u t in g in t o E q . ( 1 0 .2 - 3 8 ) y ie ld s
ρ = 0 , w h ic h , in a p lo t o f ρ v s . θ , is a s tr a ig h t lin e .
( b ) O n ly th e o r ig in (0 , 0 ) w o u ld y ie ld th is r e s u lt.
2 0 7
Edges and their profiles
Gradient images and their profiles
Laplacian images and their profiles
Images fromSteps 1 and 2 of theMarr-Hildreth algorithmand their profiles
Images fromSteps 1 and 2 of theCanny algorithmand their profiles
Canny angleimages and their profiles
F ig u r e P 1 0 .2 1
2 0 8
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
( c ) A t θ = + 9 0 ◦ , it fo llo w s fr o m E q . ( 1 0 .2 - 3 8 ) t h a t x · (0 ) + y · (1 ) = ρ , o r y = ρ . A t
θ = − 9 0 ◦ , x · (0 ) + y · (− 1 ) = ρ , o r − y = ρ . T h u s th e r e fl e c tiv e a d ja c e n c y .
Problem 10.24
S u b d iv id in g th e θ - a x is in to K in c r e m e n ts g iv e s , fo r e v e r y p o in t (x k , y k ), K v a lu e s
o f ρ c o r r e s p o n d in g to th e K p o s s ib le v a lu e s o f θ . W ith n im a g e p o in ts , th is
m e th o d in v o lv e s n K c o m p u ta tio n s . T h u s th e p r o c e d u r e is lin e a r in n .
Problem 10.25
T h is p r o b le m is a n a tu r a l fo r th e H o u g h tr a n s fo r m , w h ic h is s e t u p a s fo llo w s :
T h e θ a x is is d iv id e d in to s ix s u b d iv is io n s , c o r re s p o n d in g to th e s ix s p e c ifi e d d ir e c tio n s a n d th e ir e r r o r b a n d s . F o r e x a m p le (b e c a u s e th e a n g le d ir e c tio n s s p e c ifi e d in th e p r o b le m s ta te m e n t a r e w ith r e s p e c t to th e h o r iz o n ta l) th e fi r s t b a n d
fo r a n g le θ e x te n d s fr o m − 3 0 ◦ to − 2 0 ◦ , c o r r e s p o n d in g to th e − 2 5 ◦ d ir e c tio n a n d
p
p
D to ρ = +
D , w h e re D is th e
its ± 5 ◦ b a n d . T h e ρ a x is e x te n d s fr o m ρ = −
la r g e s t d is ta n c e b e tw e e n o p p o s ite c o r n e r s o f th e im a g e , p r o p e r ly c a lib r a te d to
fi t th e p a r tic u la r im a g in g s e t u p u s e d . T h e s u b d iv is io n s in th e ρ a x is a r e c h o s e n
fi n e ly e n o u g h to r e s o lv e th e m in im u m e x p e c te d d is ta n c e b e tw e e n tr a c k s th a t
m a y b e p a r a lle l, b u t h a v e d iffe r e n t o r ig in s , th u s s a tis fy in g th e la s t c o n d itio n o f
th e p r o b le m s ta te m e n t.
S e t u p in th is w a y , th e H o u g h tr a n s fo r m c a n b e u s e d a s a “ fi lte r ” to c a te g o r iz e
a ll p o in ts in a g iv e n im a g e in to g r o u p s o f p o in ts in th e s ix s p e c ifi e d d ir e c tio n s .
E a c h g r o u p is th e n p r o c e s s e d fu r th e r to d e te r m in e if its p o in ts s a tis fy th e c r ite r ia
fo r a v a lid tr a c k : (1 ) e a c h g r o u p m u s t h a v e a t le a s t 1 0 0 p o in ts ; a n d (2 ) it c a n n o t
h a v e m o r e th a n th r e e g a p s , e a c h o f w h ic h c a n n o t b e m o r e th a n 1 0 p ix e ls lo n g
( s e e P r o b le m 1 0 .3 r e g a r d in g t h e e s t im a t io n o f g a p s o f a g iv e n le n g t h ) .
Problem 10.26
T h e e s s e n c e o f th e a lg o r ith m is to c o m p u te a t e a
a ll p ix e ls w h o s e in te n s itie s a r e le s s th a n o r e q u a l
s im ila r ly , th e m e a n v a lu e , m 2 , o f a ll p ix e ls w ith v a
L e t p i = n i/ n d e n o te th e i th c o m p o n e n t o f th e
th e n u m b e r o f p ix e ls w ith in te n s ity i , a n d n is th
im a g e . V a lid v a lu e s o f i a r e in th e r a n g e 0 ≤ i ≤ L
in te n s itie s a n d i is a n in te g e r. T h e m e a n s c a n b e
c h s te p th e m e a n v a lu e , m 1 , o f
to th e p r e v io u s th r e s h o ld a n d ,
lu e s th a t e x c e e d th e th r e s h o ld .
im a g e h is to g r a m , w h e re n i is
e to ta l n u m b e r o f p ix e ls in th e
− 1 , w h e re L is th e n u m b e r o n
c o m p u te d a t a n y s te p k o f th e
2 0 9
a lg o r ith m :
I (k − 1 )
m
1
(k ) =
i p
/ P (k )
i
i = 0
w h e re
I (k − 1 )
P (k ) =
p
i
i = 0
a n d
L − 1
m
2
(k ) =
i p
i
/ [1 − P (k )] .
i = I (k − 1 )+ 1
T h e te r m I (k − 1 ) is th e s m a lle s t in te g e r le s s th a n o r e q u a l to T (k − 1 ), a n d T (0 )
is g iv e n . T h e n e x t v a lu e o f th e th r e s h o ld is th e n
T (k + 1 ) =
2
1
[m
1
(k ) + m
2
(k )] .
Problem 10.27
A s s t a t e d in S e c t io n 1 0 .3 .2 , w e a s s u m e t h a t t h e in it ia l t h r e s h o ld is c h o s e n b e tw e e n th e m in im u m a n d m a x im u m in te n s itie s in th e im a g e . T o b e g in , c o n s id e r
t h e h is t o g r a m in F ig . P 1 0 .2 7 . It s h o w s t h e t h r e s h o ld a t t h e k t h it e r a t iv e s t e p , a n d
th e fa c t th a t th e m e a n m 1 (k + 1 ) w ill b e c o m p u te d u s in g th e in te n s itie s g r e a te r
th a n T (k ) tim e s th e ir h is to g r a m v a lu e s . S im ila r ly , m 2 (k + 1 ) w ill b e c o m p u te d u s in g v a lu e s o f in te n s itie s le s s th a n o r e q u a l to T (k ) tim e s th e ir h is to g r a m v a lu e s .
T h e n , T (k + 1 ) = 0 .5 [m 1 (k + 1 ) + m 2 (k + 1 )]. T h e p r o o f c o n s is t s o f t w o p a r t s . F ir s t ,
w e p r o v e th a t th e th r e s h o ld is b o u n d e d b e tw e e n 0 a n d L − 1 . T h e n w e p r o v e th a t
th e a lg o r ith m c o n v e r g e s to a v a lu e b e tw e e n th e s e tw o lim its .
T o p r o v e t h a t t h e t h r e s h o ld is b o u n d e d , w e w r it e T (k + 1 ) = 0 .5 [m 1 (k + 1 ) +
m 2 (k + 1 )]. If m 2 (k + 1 ) = 0 , th e n m 1 (k + 1 ) w ill b e e q u a l to th e im a g e m e a n ,
M , a n d T (k + 1 ) w ill e q u a l M / 2 w h ic h is le s s th a n L − 1 . If m 2 (k + 1 ) is z e r o , th e
s a m e w ill b e tr u e . B o th m 1 a n d m 2 c a n n o t b e z e r o s im u lta n e o u s ly , s o T (k + 1 )
w ill a lw a y s b e g r e a te r th a n 0 a n d le s s th a n L − 1 .
T o p r o v e c o n v e r g e n c e , w e h a v e to c o n s id e r th r e e p o s s ib le c o n d itio n s :
1 . T (k + 1 ) = T (k ), in w h ic h c a s e th e a lg o r ith m h a s c o n v e r g e d .
2 . T (k + 1 ) < T (k ), in w h ic h c a s e th e th r e s h o ld m o v e s to th e le ft.
3 . T (k + 1 ) > T (k ), in w h ic h c a s e th e th r e s h o ld m o v e s to th e r ig h t.
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 1 0
Histogram
values
}
}
0
T (k )
Intensities
used to compute
m 2(k +1)
Intensities
used to compute
m 1(k +1)
Intensity
L - 1
F ig u r e P 1 0 .2 7
In c a s e (2 ), w h e n th e th r e s h o ld v a lu e m o v e s to th e le ft, m 2 w ill d e c r e a s e o r s ta y
th e s a m e a n d m 1 w ill a ls o d e c r e a s e o r s ta y th e s a m e (th e fa c t th a t m 1 d e c r e a s e s
o r s t a y s t h e s a m e i s n o t n e c e s s a r i l y o b v i o u s . I f y o u d o n ’t s e e i t , d r a w a s i m p l e h i s to g r a m a n d c o n v in c e y o u r s e lf th a t it d o e s ), d e p e n d in g o n h o w m u c h th e th r e s h o ld m o v e d a n d o n th e v a lu e s o f th e h is to g r a m . H o w e v e r, n e ith e r th r e s h o ld c a n
in c r e a s e . If n e ith e r m e a n c h a n g e s , th e n T (k + 2 ) w ill e q u a l T (k + 1 ) a n d th e a lg o r ith m w ill s to p . If e ith e r (o r b o th ) m e a n d e c r e a s e s , th e n T (k + 2 ) < T (k + 1 ), a n d
th e n e w th r e s h o ld m o v e s fu r th e r to th e le ft. T h is w ill c a u s e th e c o n d itio n s ju s t
s ta te d to h a p p e n a g a in , s o th e c o n c lu s io n is th a t if th e th r e s h o ld s s ta r ts m o v in g le ft, it w ill a lw a y s m o v e le ft, a n d th e a lg o r ith m w ill e v e n tu a lly s to p w ith a
v a lu e T > 0 , w h ic h w e k n o w is th e lo w e r b o u n d fo r T . B e c a u s e th e th r e s h o ld a lw a y s d e c r e a s e s o r s to p s c h a n g in g , n o o s c illa tio n s a r e p o s s ib le , s o th e a lg o r ith m
is g u a r a n te e d to c o n v e r g e .
C a s e (3 ) c a u s e s th e th r e s h o ld to m o v e th e r ig h t. A n a r g u m e n t s im ila r to th e
p r e c e d in g d is c u s s io n e s ta b lis h e s th a t if th e th r e s h o ld s ta r ts m o v in g to th e r ig h t it
w ill e ith e r c o n v e r g e o r c o n tin u e m o v in g to th e r ig h t a n d w ill s to p e v e n tu a lly w ith
a v a lu e le s s th a n L − 1 . B e c a u s e th e th r e s h o ld a lw a y s in c r e a s e s o r s to p s c h a n g in g ,
n o o s c illa tio n s a r e p o s s ib le , s o th e a lg o r ith m is g u a r a n te e d to c o n v e r g e .
Problem 10.28
C o n s id e r
th a n L / 2
is th e m a
th e ra n g e
a n im a
a n d le s
x im u m
[0 ,L −
g e w h o s e
s th a n L −
in te n s ity
1 ]. S u p p o
in te n
1 , w
le v e
s e th
s itie
h e re
l (re
a t w
s (o f b o th b a c k g ro u n d a n d o b
L is th e n u m b e r o f in te n s ity
c a ll th a t w e w o r k w ith th e in
e c h o o s e th e in itia l th r e s h o ld
je
le
te
a
c ts ) a
v e ls a
n s ity
s T =
re
n
s
0
g re a te r
d L − 1
c a le in
. T h e n
2 1 1
Histogram
values
I
min
}
}
0
L /2
Intensities
of background
M
Intensity
L - 1
Intensities
of objects
F ig u r e P 1 0 .2 9
a ll p ix e ls w ill b e a s s ig n e d to c la s s 1 a n d n o n e to c la s s 2 . S o m 1 w ill b e th e m e a n
v a lu e , M , o f th e im a g e a n d m 2 w ill b e z e r o b e c a u s e n o p ix e ls w ill b e a s s ig n e d to
c la s s 2 . T w ill b e th e s u m o f th e tw o m e a n s d iv id e d b y 2 , o r M / 2 . H o w e v e r, th e
im a g e m e a n , M , is b e tw e e n L / 2 a n d L − 1 b e c a u s e th e r e a r e n o p ix e ls w ith v a lu e s
b e lo w L / 2 . S o , th e n e w th r e s h o ld , T = M / 2 w ill b e b e lo w L / 2 . B e c a u s e th e r e a r e
n o p ix e ls w ith v a lu e s b e lo w L / 2 , m 2 w ill b e z e r o a g a in , a n d m 1 w ill b e th e m e a n
v a lu e o f th e in te n s itie s a n d T w ill a g a in b e e q u a l to M / 2 . S o , th e a lg o r ith m w ill
te r m in a te w ith th e w r o n g v a lu e o f th r e s h o ld . T h e s a m e c o n c e p t is r e s ta te d m o r e
fo r m a lly a n d u s e d a s a c o u n t e r - e x a m p le in t h e s o lu t io n o f P r o b le m 1 0 .2 9 .
Problem 10.29
T h e v a lu e o f th e th e th r e s h o ld a t c o n v e r g e n c e is in d e p e n d e n t o f th e in itia l v a lu e
if th e in itia l v a lu e o f th e th r e s h o ld is c h o s e n b e tw e e n th e m in im u m a n d m a x im u m in t e n s it y o f t h e im a g e ( w e k n o w fr o m P r o b le m 1 0 .2 7 t h a t t h e a lg o r it h m
c o n v e r g e s u n d e r th is c o n d itio n ). T h e fi n a l th r e s h o ld is n o t in d e p e n d e n t o f th e
in itia l v a lu e c h o s e n fo r T if th a t v a lu e d o e s n o t s a tis fy th is c o n d itio n . F o r e x a m p le , c o n s id e r a n im a g e w it h t h e h is t o g r a m in F ig . P 1 0 .2 9 . S u p p o s e t h a t w e
s e le c t th e in itia l th r e s h o ld T (1 ) = 0 . T h e n , a t th e n e x t ite r a tiv e s te p , m 2 (2 ) = 0 ,
m 1 (2 ) = M , a n d T (2 ) = M / 2 . B e c a u s e m 2 (2 ) = 0 , it fo llo w s th a t m 2 (3 ) = 0 ,
m 1 (3 ) = M , a n d T (3 ) = T (2 ) = M / 2 . A n y fo llo w in g ite r a tio n s w ill y ie ld th e s a m e
r e s u lt, s o th e a lg o r ith m c o n v e r g e s w ith th e w r o n g v a lu e o f th r e s h o ld . If w e h a d
s t a r t e d w i t h I m in < T ( 1 ) < I m a x , t h e a lg o r i t h m w o u ld h a v e c o n v e r g e d p r o p e r ly .
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 1 2
Problem 10.30
( a ) F o r a u n ifo r m h is to g r a m , w e c a n v ie w th e in te n s ity le v e ls a s p o in ts o f u n it
m a s s a lo n g th e in te n s ity a x is o f th e h is to g r a m . A n y v a lu e s m 1 (k ) a n d m 2 (k )
a r e th e m e a n s o f th e tw o g r o u p s o f in te n s ity v a lu e s G 1 a n d G 2 . B e c a u s e th e h is to g r a m is u n ifo r m , th e s e a re th e c e n te r s o f m a s s o f G 1 a n d G 2 . W e k n o w fr o m th e
s o lu t io n o f P r o b le m 1 0 .2 7 t h a t if T s t a r t s m o v in g t o t h e r ig h t , it w ill a lw a y s m o v e
in th a t d ir e c tio n , o r s to p . T h e s a m e h o ld s tr u e fo r m o v e m e n t to th e le ft. N o w ,
a s s u m e th a t T (k ) h a s a r r iv e d a t th e c e n te r o f m a s s (a v e r a g e in te n s ity ). B e c a u s e
a ll p o in ts h a v e e q u a l " w e ig h t" (r e m e m b e r th e h is to g r a m is u n ifo r m ), if T (k + 1 )
m o v e s to th e r ig h t G 2 w ill p ic k u p , s a y , Q n e w p o in ts . B u t G 1 w ill lo s e th e s a m e
n u m b e r o f p o in ts , s o th e s u m m 1 + m 2 w ill b e th e s a m e a n d th e a lg o r ith m w ill
s to p .
(b ) T h e
r ith m a
o ld to w
to lo s e
s to p .
p r o v e is s im ila r to (a ) b e c a u s e
r r iv e s a t th e p o in t b e tw e e n th e
a r d o n e o r th e o th e r m e a n w ill
th e s a m e " m a s s " . T h u s , th e ir s u
th
tw
c a
m
e m o d
o m e a
u s e o n
w ill b e
e s a re
n s , fu r
e g ro u
th e s a
id
th
p
m
e n tic a l.
e r m o tio
to p ic k u
e a n d th
W h e n th e a
n o f th e th r
p a n d th e o
e a lg o r ith m
lg
e s
th
w
o h e r
ill
Problem 10.31
(a ) A 1 = A 2 a n d σ 1 = σ 2 = σ , w h ic h m a k e s th e tw o m o d e s id e n tic a l [th is is th e
s a m e a s P r o b le m 1 0 .3 0 ( b ) ].
( b ) Y o u k n o w fr o m P r o b le m 1 0
tic a l, th e a lg o r ith m w o u ld c o n
In th e p r e s e n t p r o b le m , th e m o
T h e n , a ll w e c a n s a y is th a t th e a
tw e e n th e m e a n s (fr o m S e c tio n
s o m e p o in t b e tw e e n th e m in im
A 1 a n d A 2 a re g re a te r th a n 0 , w
a p o in t s o m e w h e re b e tw e e n m
(c ) σ
1
> > σ
2
.3 0 t h a t if t h e m o d e s w e r e s y m m e t r ic a n d id e n v e r g e to th e p o in t m id w a y b e w te e n th e m e a n s .
d e s a r e s y m m tr ic b u t th e y m a y n o t b e id e n tic a l.
lg o r ith m w ill c o n v e r g e to a p o in t s o m e w h e r e b e 1 0 .3 .2 w e k n o w t h a t t h e a lg o r it h m m u s t s t a r t a t
u m a n d m a x im u m im a g e in te n s itie s ). S o , if b o th
e a r e a s s u r e d th a t th e a lg o r ith m w ill c o n v e r g e to
1 a n d m 2 .
. T h is w ill ” p u ll” t h e t h r e s h o ld t o w a r d m
1
d u r in g ite r a tio n .
2 1 3
Problem 10.32
(a )
σ
2
=
P 1 (m
=
P 1 (m
=
P 1 [m
=
P 1 [P 2 m
=
P 1 P
=
(m
B
=
=
(m
1
− m
(m
1
)2 + P 2 P
2
2
2
) [P 1 P 2 (P
− m
1
w e u s e d th e fa c ts th a t m
p a r t o f E q . ( 1 0 .3 - 1 5 ) .
G
)
2
2
2
− (P 1 m
2
]2 + P 2 [m
2
(m
1
+ P 2 P
2
2
)]2 + P 2 [m
2
) − P 2 m
2
2
)
G
]2 + P 2 [P 1 m
2
)2 [P 1 P
2
P 1 P 2 (m
(1 − P
− m
1
− m
1
− m
2
+ P 2 m
1
− P 2 m
1
− m
1
)2 + P 2 (m
G
− (P 1 m
1
2
2
− m
1
− P 1 m
2
− m
1
)
2
− P 1 m
2
1
+ P 2 m
1
]
1
+ P 2 m
1
2
m
(k ) =
a n d P
2
+ P
1
= 1 . T h is p r o v e s th e fi r s t
2
− m (k )
.
1 − P 1 (k )
G
L − 1
P 2 (k )
i p
L − 1
i p
1 − P 1 (k )
⎡
⎣
1 − P 1 (k )
=
i
i = k + 1
1
m
i
i = k + 1
1
=
L − 1
⎤
k
i p
i
i p i⎦
−
i = 0
i = 0
− m (k )
.
1 − P 1 (k )
G
T h e n ,
σ
2
=
P 1 P 2 (m
=
P 1 P
=
P 1 (1 − P 1 )
B
=
2
]
1
1
=
(1 − P 1 )]
+ P 1 )]
2
= P 1 m
=
2
2
T h is w e d o a s fo llo w s :
(k )
2
2
m
2
)]
2
(b ) F ir s t, w e h a v e to s h o w th a t
m
− m
2
2
1
− m
m
−
P 1
m
2
)
2
− m
1 − P 1
G
m − P 1 m G
P 1 (1 − P 1 )
(m G P 1 − m )2
.
P 1 (1 − P 1 )
2
2
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 1 4
Problem 10.33
F r o m , E q . ( 1 0 .3 - 1 5 ) ,
σ
A s s ta te d
m 1 = m 2
s ta n t), in
c a s e , P 1 (k
o n p o s itiv
s o m e v a lu
in
w
w
)[
e
e
th e
h ic h
h ic h
1 − P
a n d
o f k
2
=
P 1 (k )P
=
P 1 (k )[1 − P 1 (k )][m
B
(k )[m
2
(k ) − m
1
1
2
(k )]
2
(k ) − m
(k )]
2
b o o k , t h e m e a s u r e σ 2B ( o r η ) t a k e s a m i n i m u
im p lie s th a t th e r e is o n ly o n e c la s s o f p ix e ls
c a s e P 1 (k ) = 1 o r P 1 (k ) = 0 fo r s o m e v a lu e
1 ( k ) ] > 0 o r , e q u iv a le n t ly , f o r 0 < P 1 ( k ) < 1 , s o
b o u n d e d v a lu e s , a n d a fi n ite m a x im u m is g u a
if th e im a g e is n o t c o n s ta n t.
2
m v a lu e o f 0 w h e n
(th e im a g e is c o n o f k . In a n y o th e r
th e m e a s u re ta k e s
r a n te e d to e x is t fo r
Problem 10.34
F r o m t h e d e fi n it io n in E q . ( 1 0 .3 - 1 2 ) ,
η =
w h e re
σ
2
B
σ
σ
2
B
2
G
= P 1 (m 1 − m G )2 + P 2 (m
= P 1 P 2 (m 1 − m 2 )2
− m
2
G
)
2
a n d
L − 1
σ
2
G
=
(i − m
G
)2 p
i
.
i = 0
A s i n t h e t e x t , w e h a v e o m i t t e d k i n σ 2B f o r t h e s a k e o f n o t a t i o n a l c l a r i t y , b u t t h e
a s s u m p tio n is th a t 0 ≤ k ≤ L − 1 . A s e x p la in e d in th e b o o k , th e m in im u m v a lu e o f
η is z e r o , a n d it o c c u r s w h e n th e im a g e is c o n s ta n t. It re m a in s to b e s h o w n th a t
th e m a x im u m v a lu e is 1 , a n d th a t it o c c u r s fo r tw o - v a lu e d im a g e s w ith v a lu e s 0
a n d L − 1 .
F r o m t h e s e c o n d l i n e o f t h e e x p r e s s i o n f o r σ 2B w e s e e t h a t t h e m a x i m u m o c c u r s w h e n th e q u a n tity (m 1 − m 2 )2 is m a x im u m b e c a u s e P 1 a n d P 2 a re p o s itiv e .
T h e in te n s ity s c a le e x te n d s fr o m 0 to L − 1 , s o th e m a x im u m d iffe r e n c e b e tw e e n
m e a n s o c c u r s w h e n m 1 = 0 a n d m 2 = L − 1 . B u t th e o n ly w a y th is c a n h a p p e n is
if th e v a r ia n c e o f th e tw o c la s s e s o f p ix e ls is z e r o , w h ic h im p lie s th a t th e im a g e
o n ly h a s th e s e tw o v a lu e s , th u s p r o v in g th e a s s e r tio n th a t th e m a x im u m o c c u r s
o n ly w h e n th e im a g e is tw o - v a lu e d w ith in te n s ity v a lu e s 0 a n d L − 1 . It r e m a in s
to b e s h o w n th a t th e m a x im u m p o s s ib le v a lu e o f η is 1 .
2 1 5
W h e n m
1
= 0 a n d m
= L − 1 ,
2
2
σ
=
P 1 P 2 (m
=
P 1 P 2 (L − 1 )2 .
B
− m
1
2
)
2
F o r a n im a g e w ith v a lu e s 0 a n d L − 1 ,
L − 1
2
G
σ
=
(i − m
G
)2 p
(i − m
G
)2 P
i
i = 0
L − 1
k
=
+
1
i = 0
=
(i − m
)2 P
G
2
i = k + 1
(0 − m
)2 P
G
+ (L − 1 − m
1
)2 P
G
2
F r o m E q . ( 1 0 .3 - 1 0 )
m
G
= P 1 m 1 + P 2 m
= P 2 (L − 1 ).
2
S o ,
σ
2
G
=
(0 − m
=
P
2
2
G
)2 P
2
(L − 1 ) P
2
=
(L − 1 ) (P
=
=
+ (L − 1 − m
1
2
2
P
1
2
(L − 1 ) P 2 P 1 (P
P 2 P 1 (L − 1 )
)2 P
2
+ (L − 1 ) (1 − P 2 )2 P
1
2
G
+ P
2
2
1
2
P 2 )
+ P 1 )
2
w h e r e w e u s e d t h e f a c t t h a t P 1 + P 2 = 1 . W e s e e f r o m t h e p r e c e d i n g r e s u l t s f o r σ 2B
a n d σ G2 t h a t σ 2B / σ G2 = 1 w h e n t h e i m a g e i s t w o - v a l u e d w i t h v a l u e s 0 a n d L − 1 .
T h is c o m p le te s th e p r r o f.
Problem 10.35
(a ) L e t R 1 a n d R 2
a n d le s s o r e q u a l
s o it g e ts m a p p e d
in R 1 a re m a p p e d
is th a t a ll v a lu e s
T 0. T h e s e n s e o f
in te n s itie s in th e
d e n o
to T , r
b y th
t o R 10
i n R 10
th e in
tw o re
te th e re g io n s w h o s e p ix e l in te n s itie s a re g re a te r th a n T
e s p e c tiv e ly . T h e th r e s h o ld T is s im p ly a n in te n s ity v a lu e ,
e tr a n s fo r m a tio n fu n c tio n to th e v a lu e T 0 = 1 − T . V a lu e s
a n d v a l u e s i n R 2 a r e m a p p e d t o R 20 . T h e i m p o r t a n t t h i n g
a r e b e l o w T 0 a n d a l l v a l u e s i n R 20 a r e e q u a l t o o r a b o v e
e q u a litie s h a s b e e n r e v e r s e d , b u t th e s e p a r a b ility o f th e
g io n s h a s b e e n p re s e r v e d .
( b ) T h e s o lu tio n in (a ) is a s p e c ia l c a s e o f a m o r e g e n e r a l p r o b le m . A th r e s h o ld is s im p ly a lo c a tio n in th e in te n s ity s c a le . A n y tr a n s fo r m a tio n fu n c tio n th a t
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 1 6
p (z )
0
60
170
255
z
F ig u r e P 1 0 3 6
p r
th
s e
c e
e s
re
r v
s s
e r v e s th e o rd e r o f
s h o ld . T h u s , a n y
e th is o rd e r. T h e v
e d w ith th e tr a n s f
in te n s itie
m o n o to n
a lu e o f th
o r m a tio n
s w ill p r e s e r v e th e s e p a r a b iliy e s ta b lis h e d b y th e
ic fu n c tio n (in c r e a s in g o r d e c r e a s in g ) w ill p r e e n e w th r e s h o ld is s im p ly th e o ld th r e s h o ld p r o fu n c tio n .
Problem 10.36
W ith re fe
s ta n d a rd
o f th e im
tio n 1 0 .3
9 0 % s p e c
re n c e
d e v ia
a g e is
.2 o r
ifi c a t
to F
tio n
w id
O ts u
io n .
ig . P 1 0 .3 6 , t h e k e y is t o n o t e t h a t t h e
s a p a r t, s o th e v a lle y b e tw e e n th e tw o
e a n d d e e p . T h e s im p le g lo b a l th r e s h
’s a l g o r i t h m w i l l g i v e a p e r f o r m a n c e
In fa c t, th e s o lu tio n w ill g iv e c lo s e to
m e a n
m o d
o ld in
th a t
1 0 0 %
s a re m o re
e s o f th e h is
g a lg o r ith m
e a s ily e x c e
a c c u ra c y .
th a n
to g r
o f S
e d s
1 0
a m
e c th e
Problem 10.37
( a ) T h e fi r s t c o lu m n w o u ld b e b la c
re a s o n : A p o in t in th e s e g m e n te d
th a t p o in t e x c e e d s b a t th a t p o in t.
g r e a te r th a n 0 w ill b e s e t to 1 a n d a
p o in ts in th e im a g e th a t d o n o t e x c
p o in ts in th e fi r s t c o lu m n .
k a n d
im a g e
B u t b
ll o th e
e e d 0
a ll o th e r c o lu m
is s e t to 1 if th
= 0 , s o a ll p o in
r p o in ts w o u ld
a re th e p o in ts t
n s w o
e v a lu
ts in t
b e s e t
h a t a r
u ld b e
e o f th
h e im a
to 0 . B
e 0 , w h
w h ite . T h e
e im a g e a t
g e th a t a re
u t th e o n ly
ic h a re th e
( b ) T h e r ig h tm o s t c o lu m n w o u ld b e 0 a n d a ll o th e r s w o u ld b e 1 , fo r th e r e a s o n
s ta te d in (a ).
2 1 7
( c ) A s in (a ), a ll th e p ix e ls in th e fi r s t c o lu m n o f
0 . C o n s id e r a ll o th e r p ix e ls in th e fi r s t r o w o f t
m in d th a t n = 2 a n d b = 1 . B e c a u s e th e r a m
in in te n s ity to th e r ig h t, th e v a lu e o f a p ix e l a t
g re a te r th a n th e (m o v in g ) a v e r a g e o f th a t p ix
p o in ts a lo n g th e fi r s t r o w , w ith th e e x c e p tio n
th e s c a n re a c h e s th e e n d o f th e fi rs t ro w a n d
s e c o n d r o w , th e r e v e r s e c o n d itio n w ill e x is t a n
o f th e s e g m e n te d im a g e w ill b e la b e le d 0 . T h
th e s a m e a s in th e fi r s t r o w , s o th e n e t e ffe c t
c o n s is t o f a ll 0 s in th e fi r s t c o lu m n , a n d a ll 0 s
th e s e c o n d . A ll o th e r p ix e ls w ill b e 1 .
th e s e g m e n te d im a g e a r e la b e le d
h e s e g m e n te d im a g e a n d k e e p in
p in th e o r ig in a l im a g e in c re a s e s
lo c a tio n k in th e im a g e is a lw a y s
e l a n d its p r e d e c e s s o r. T h u s , a ll
o f th e fi r s t a r e la b e le d 1 . W h e n
re v e r s e s d ire c tio n a s it s ta r ts th e
d a ll p ix e ls a lo n g th e s e c o n d r o w
e c o n d itio n s in th e th ird r o w a re
is th a t th e s e g m e n te d im a g e w ill
in a lte r n a tin g r o w s , s ta r tin g w ith
( d ) T h e fi r s t p ix e l in th e fi r s t r o w o f th e s e g m e n te d p ix e l w ill b e 0 a n d th e r e s t w ill
b e 1 s fo r th e re a s o n s ta te d in (a ). H o w e v e r, in s u b s e q u e n t r o w s th e c o n d itio n s
c h a n g e . C o n s id e r th e c h a n g e in d ire c tio n b e tw e e n th e e n d o f th e fi r s t r o w a n d
th e r e v e r s e s c a n o f th e s e c o n d r o w . B e c a u s e n is n o w m u c h g r e a te r th a n 2 , it w ill
ta k e “ a w h ile " b e fo r e th e e le m e n ts in th a t r o w o f th e o r ig in a l im a g e w ill b e c o m e
s m a lle r th a n th e n r u n n in g a v e r a g e , w h ic h r e a c h e d is h ig h e s t v a lu e a t th e e n d o f
th e fi r s t r o w . T h e r e s u lt w ill b e K w h ite p ix e ls b e fo r e th e v a lu e s in th e o r ig in a l
im a g e d r o p b e lo w th e r u n n in g a v e r a g e a n d th e p ix e ls in th e s e g m e n te d im a g e
b e c o m e b la c k . T h is c o n d itio n is r e v e r s e d w h e n th e tu r n is m a d e fr o m th e s e c o n d
to th e th ir d r o w . K b la c k p ix e ls w ill n o w b e p r e s e n t in th e fi r s t r o w b e fo r e th e
v a lu e s in th e o r ig in a l im a g e in c r e a s e p a s t th e r u n n in g a v e r a g e . O n c e th e y d o , th e
c o r r e s p o n d in g lo c a tio n s in th e s e g m e n te d r e g io n w ill b e la b e le d w h ite th r o u g h
th e e n d o f th a t r o w . T h e s e g m e n te d im a g e w ill th u s lo o k a s fo llo w s : T h e fi r s t p ix e l
in th e fi r s t r o w w ill b e b la c k a n d a ll o th e r s w ill b e w h ite . T h e n e x t r o w w ill h a v e
a K b la c k p ix e ls o n th e le ft a n d K w h ite p ix e ls o n th e r ig h t, w ith a ll o th e r p ix e ls
b e in g b la c k . T h e le ft a n d r ig h t o f th e n e x t r o w w ill b e th e s a m e , b u t th e p ix e ls in
b e tw e e n w ill b e w h ite . T h e s e tw o a lte r n a tin g r o w p a tte r n s a r e a p p lic a b le to th e
re s t o f th e r o w s in th e s e g m e n te d im a g e .
Problem 10.38
T h e m e a n s a re a
te n s ity le v e ls . A
ra n g e a b o u t 1 7 0
in te n s ity p o p u la
th e o b je c ts is q u
to a s e e d a n y p ix
t 6 0 a n d 1
ra n g e o f
g iv e s [1 4 0
tio n s . S o
ite a d e q u
e l th a t is 8
7 0 , a
± 3 σ
2 0 0 ]
c h o o
a te .
-c o n
n d th e s ta n d a rd
a b o u t 6 0 g iv e s th
s o s ig n ifi c a n t s e p
s in g 1 7 0 a s th e s
O n e a p p r o a c h is
n e c te d to a n y p ix
d e
e
a r
e e
to
e l
v ia tio n o f t
ra n g e [3 0 ,
a tio n e x is ts
d v a lu e fr o
g r o w re g io
p r e v io u s ly
h e n o is e is 1 0 in 9 0 ] a n d a s im ila r
b e tw e e n th e tw o
m w h ic h to g r o w
n s b y a p p e n d in g
a p p e n d e d to th a t
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 1 8
s e e d , a n d w h o s e in te n s ity is 1 7 0 ± 3 σ .
Problem 10.39
T h e r e g io n s p lit t in g is s h o w n in F ig . P 1 0 .3 9 ( a ) . T h e c o r r e s p o n d in g q u a d t r e e is
s h o w n in F ig . P 1 0 .3 9 ( b ) .
Problem 10.40
T o o b ta in th e s p a r s e o u te r r e g io n w e s im p ly fo r m th e A N D o f th e m a s k w ith th e
o u te r r e g io n . B e c a u s e th e m a s k c o m p le te ly s e p a r a te s th e in n e r r e g io n fr o m th e
b a c k g r o u n d , its c o m p le m e n t w ill p r o d u c e tw o d is jo in t r e g io n s . W e fi n d th e s e
tw o r e g io n s b y s im p ly b y u s in g a c o n n e c te d - c o m p o n e n ts a lg o r ith m (s e e S e c tio n
9 .5 .3 ) . W e t h e n id e n t ify t h e r e g io n c o n t a in in g t h e b a c k g r o u n d b y d e t e r m in in g
w h ic h o f th e tw o r e g io n s c o n ta in s a t le a s t o n e p o in t o n th e b o u n d a r y o f th e im a g e (s a y , a n y o n e o f th e fo u r c o r n e r p o in ts ). W e th e n A N D a n im a g e c o n ta in in g
o n ly th is r e g io n w ith th e o r ig in a l to o b ta in th e b a c k g r o u n d . A N D in g th e o th e r
r e g io n w ith th e im a g e is o la te s o n ly th e in n e r r e g io n .
Problem 10.41
( a ) T h e e le m e n ts o f T [n ] a r e th e c o o r d in a te s o f p o in ts in th e im a g e b e lo w th e
p la n e g (x , y ) = n , w h e r e n is a n in te g e r th a t r e p r e s e n ts a g iv e n s te p in th e e x e c u tio n o f th e a lg o r ith m . B e c a u s e n n e v e r d e c r e a s e s , th e s e t o f e le m e n ts in T [n − 1 ]
is a s u b s e t o f th e e le m e n ts in T [n ]. In a d d itio n , w e n o te th a t a ll th e p o in ts b e lo w
th e p la n e g (x , y ) = n − 1 a r e a ls o b e lo w th e p la n e g (x , y ) = n , s o th e e le m e n ts
o f T [n ] a r e n e v e r r e p la c e d . S im ila r ly , C n (M i ) is fo r m e d b y th e in te r s e c tio n o f
C (M i ) a n d T [n ], w h e r e C (M i ) (w h o s e e le m e n ts n e v e r c h a n g e ) is th e s e t o f c o o r d in a te s o f a ll p o in ts in th e c a tc h m e n t b a s in a s s o c ia te d w ith re g io n a l m in im u m
M i . B e c a u s e th e e le m e n ts o f C (M i ) n e v e r c h a n g e , a n d th e e le m e n ts o f T [n ] a r e
n e v e r r e p la c e d , it fo llo w s th a t th e e le m e n ts in C n (M i ) a r e n e v e r r e p la c e d e ith e r.
In a d d itio n , w e s e e th a t C n − 1 (M i ) ⊆ C n (M i ).
(b
c a
e v
th
th
) T h is p
u s e (1 )
e r r e p la
e n u m b
e s a m e .
a r t o f th e p
n a lw a y s in
c e d ; a n d (3
e r o f e le m e
r o b le
c re a s
) T [n
n ts o
m is a
e s ; (2 )
− 1 ] ⊆
f b o th
n s w e re d b y th e s a m e
th e e le m e n ts o f n e ith
T [n ] a n d C n − 1 (M i) ⊆
C n (M i ) a n d T [n ] e ith
a rg u m e n
e r C n (M
C n (M i)
e r in c re a
t a s in (a
i ) n o r T [
, it fo llo w
s e s o r re
). B e
n ] a r
s th a
m a in
e
t
s
2 1 9
1
2
11
12
13
14
21
22
23
32
24
41
321
322
411
412
323
324
413
414
42
31
33
34
43
44
4
3
4
1
11
12
13
41
14
411
2
21
22
412
42
413
43
44
414
3
23
24
31
321
32
322
33
323
34
324
F ig u r e P 1 0 .3 9
Problem 10.42
U s in g th e te r m in o lo g y o f th
tw e e n tw o c a tc h m e n t b a s in s
H o w e v e r, th e h e a r t o f th e a lg
te n s ity le v e l in th e im a g e a n
th e e n tire to p o g r a p h y is e n c
is a b re a k th a t c a u s e s w a te r
b o u n d a r ie s a r e p r e c is e ly th e
a lg o r ith m a lw a y s p r o d u c e s c
e w a te r s h e d a lg o r ith m ,
w o u ld c a u s e w a te r b e tw
o r ith m is to b u ild a d a m
y tim e a b re a k in s u c h b
lo s e d b y s u c h a d a m , d a
to m e rg e b e tw e e n tw o
to p s o f th e d a m s , s o it
lo s e d b o u n d a r ie s .
a b re a k in a b o u n d a r y b e e e n th e tw o b a s in s to m e r g e .
h ig h e r th a n th e h ig h e s t in o u n d a r ie s o c c u r s . B e c a u s e
m s a r e b u ilt a n y tim e th e r e
re g io n s , a n d s e g m e n ta tio n
fo llo w s th a t th e w a te r s h e d
2 2 0
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
F ig u r e P 1 0 .4 3
2 2 1
Problem 10.43
T h e fi r s t s te p in th e
b u ild a d a m o f h e ig
e n d s o f th e fu n c tio n
b u ild a b o x o f h e ig h
s e ttin g C [1 ] = T [1 ].
th e w a te r le v e l). T h
q 1 = f g (2 )g .
Q
b
c
d
e
b
a
a p p lic a tio n o f th
h t m a x + 1 to p re
, a s s h o w n in F ig .
t m a x + 1 a ro u n d
In th is c a s e , T [1 ]
e r e is o n ly o n e c
N e x t, w e le t n = 2 a n d , a s
[2 ] = f q 1 ;q 2 g , w h e r e , fo r c la r
y s e m ic o lo n s . W e s ta r t c o n
o m p o n e n t in Q [2 ]. W h e n q
itio n 2 is s a tis fi e d a n d , th e re
m p ty s e t) s o c o n d itio n 1 is s
e c o m e s C [2 ] = f g (2 ); g (1 4 )g
r e s e p a r a te d b y s e m ic o lo n s .
W h e n n
f 2 ,3 ;1 0 ,1 1 ;
P r o c e e d in g
in to th e n e
d itio n 1 a n
tio n 2 a n d
f 2 ,3 ;1 0 ,1 1 ;
=
1 3
a
w
d
C
1 3
3 [F ig . P 1 0 .4
,1 4 g w h e re ,
s a b o v e , q 1 \
s e t to y ie ld
C [3 ] = f 2 ,3
[3 ] = f 2 ,3 ;1
,1 4 g .
e w
v e n
P 1 0
its
=
o n n
a te r s h e d s e g m e n ta tio n
t th e r is in g w a te r fr o m
.4 3 ( b ) . F o r a n im a g e fu n
b o r d e r. T h e a lg o r ith m
g (2 ) , a s s h o w n in F ig .
e c te d c o m p o n e n t in th
s h o w n in F ig . P 1 0 .4 3 ( d ) ,
ity , d iffe re n t c o n n e c te d c o
s tr u c tio n o f C [2 ] b y c o n s
= q 1 , th e te r m q \ C [1 ] is
fo re , C [2 ] = f g (2 )g . W h e n
a tis fi e d a n d w e in c o r p o r a
w h e re , a s a b o v e , d iffe re n t
a lg o r ith m is to
r u n n in g o ff th e
c tio n w e w o u ld
is in itia liz e d b y
P 1 0 .4 3 ( c ) ( n o t e
is c a s e : Q [1 ] =
T [2 ] = f g (2 ), g (1 4 )g a n d
m p o n e n ts a re s e p a ra te d
id e r in g e a c h c o n n e c te d
e q u a l to f g (2 )g , s o c o n q = q 2 , q \ C [1 ] = ; (th e
te q in C [2 ], w h ic h th e n
c o n n e c te d c o m p o n e n ts
3 (e )], T [3 ] = f 2 ,3 ,1 0 ,1 1 ,1 3 ,1 4 g a n d Q [3 ] =
in o r d e r to s im p lify th e n o ta tio n w e le t k
C [2 ] = f 2 g s a tis fi e s c o n d itio n 2 , s o q 1 is
C [3 ] = f 2 , 3 ; 1 4 g . S im ila r ly , q 2 \ C [2 ] = ;
; 1 0 , 1 1 ; 1 4 g . F in a lly , q 3 \ C [2 ] = f 1 4 g s a
0 , 1 1 ; 1 3 , 1 4 g . It is e a s ily v e r ifi e d th a t C
fq 1 ;q 2 ;q 3 g =
d e n o te g (k ).
in c o r p o r a te d
s a tis fi e s c o n tis fi e s c o n d i[4 ] = C [3 ] =
W h e n n = 5 [F ig . P 1 0 .4 3 ( f ) ], w e h a v e , T [5 ] = f 2 , 3 , 5 , 6 , 1 0 , 1 1 , 1 2 , 1 3 , 1 4 g a n d
Q [5 ] = f q 1 ;q 2 ;q 3 g = f 2 ,3 ;5 ,6 ;1 0 ,1 1 ,1 2 , 1 3 ,1 4 g (n o te th e m e r g in g o f tw o p re v io u s ly d is tin c t c o n n e c te d c o m p o n e n ts ). Is is e a s ily v e r ifi e d th a t q 1 \ C [4 ] s a tis fi e s
c o n d itio n 2 a n d th a t q 2 \ C [4 ] s a tis fi e d c o n d itio n 1 . P r o c e e d in g w ith th e s e tw o
c o n n e c te d c o m p o n e n ts e x a c tly a s a b o v e y ie ld s C [5 ] = f 2 , 3 ; 5 , 6 ; 1 0 , 1 1 ; 1 3 , 1 4 g u p
to th is p o in t. T h in g s g e t m o re in te re s tin g w h e n w e c o n s id e r q 3 . N o w , q 3 \ C [4 ] =
f 1 0 ,1 1 ;1 3 ,1 4 g w h ic h , b e c u a s e it c o n ta in s tw o c o n n e c te d c o m p o n e n ts o f C [4 ],
s a tis fi e s c o n d itio n 3 . A s m e n tio n e d p r e v io u s ly , th is is a n in d ic a tio n th a t w a te r
fr o m tw o d iffe r e n t b a s in s h a s m e r g e d a n d a d a m m u s t b e b u ilt to p r e v e n t th is
c o n d itio n . D a m b u ild in g is n o th in g m o r e th a n s e p a r a tin g q 3 in to th e tw o o r ig in a l c o n n e c te d c o m p o n e n ts . In th is p a r tic u la r c a s e , th is is a c c o m p lis h e d b y th e
d a m s h o w n in F ig . P 1 0 .4 3 ( g ) , s o t h a t n o w q 3 = f q 3 1 ; q 3 2 g = f 1 0 , 1 1 ; 1 3 , 1 4 g . T h e n ,
q 3 1 \ C [4 ] a n d q 3 2 \ C [4 ] e a c h s a tis fy c o n d itio n 2 a n d w e h a v e th e fi n a l r e s u lt fo r
n = 5 , C [5 ] = f 2 ,3 ;5 ,6 ;1 0 ,1 1 ;1 3 ;1 4 g .
C o n tin u in g in th e m a n n e r ju s t e x p la in e d y ie ld s th e fi n a l s e g m e n ta tio n r e s u lt
s h o w n in F ig . P 1 0 .4 3 ( h ) , w h e r e t h e “ e d g e s ” a r e v is ib le ( fr o m t h e t o p ) ju s t a b o v e
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 2 2
F ig u r e P 1 0 .4 4
th e w a te r lin e . A fi n a l p o s t- p r o c e s s in g s te p w o u ld r e m o v e th e o u te r d a m w a lls
to y ie ld th e in n e r e d g e s o f in te r e s t.
Problem 10.44
W ith re fe re n c e
a p o s itiv e , r a th
th e p o s itiv e A D
T h e im a g e o n t
it h e re fo r c o n v
to
e r
I [
h e
e n
E q . ( 1 0 .6 - 4 ) , w e s e e
th a n a n e g a tiv e , th r
s e e E q . ( 1 0 .6 - 3 ) ]. T h
r ig h t is th e p o s itiv e
ie n c e in m a k in g th e
th a t c o m p a r in
e s h o ld w o u ld y
e r e s u lt is s h o w
A D I fr o m F ig . 1
c o m p a r is o n .
g
ie
n
0
th
ld
in
.5 9
e n
th
th
(b
e g a tiv e
e im a g e
e le ft o f
). W e h a
A D
n e
F ig
v e
I a g a
g a tiv
. P 1 0
in c lu
in s t
e o f
.4 4 .
d e d
Problem 10.45
( a ) T r u e , a s s u m in g th a t th e th r e s h o ld is n o t s e t la r g e r th a n a ll th e d iffe r e n c e s e n c o u n te r e d a s th e o b je c t m o v e s . T h e e a s ie s t w a y to s e e th is is to d r a w a s im p le
r e fe r e n c e im a g e , s u c h a s th e w h ite r e c ta n g le o n a b la c k b a c k g r o u n d . L e t th a t
r e c ta n g le b e th e o b je c t th a t m o v e s . B e c a u s e th e a b s o lu te A D I im a g e v a lu e a t
a n y lo c a tio n is th e a b s o lu te d iffe r e n c e b e tw e e n th e r e fe r e n c e a n d th e n e w im a g e , it is e a s y to s e e th a t a s th e o b je c t e n te r s a r e a s th a t a r e b a c k g r o u n d in th e
r e fe r e n c e im a g e , th e a b s o lu te d iffe r e n c e w ill c h a n g e fr o m z e r o to n o n z e r o a t th e
n e w a r e a o c c u p ie d b y th e m o v in g o b je c t. T h u s , a s lo n g a s th e o b je c t m o v e s , th e
d im e n s io n o f th e a b s o lu te A D I w ill g r o w .
(b ) T
in g o
n e v e
[a s s u
th a n
r u e . T
b je c t
r e x c e
m in g
th e o
h e
b e
e d
, a
b je
p o s itiv e
c a u s e th
th e th re
s in E q .
c t].
A D I is s ta tio n
e d iffe re n c e s b
s h o ld in a r e a s
( 1 0 .6 - 3 ) , t h a t
a r y
e tw
th a
th e
a n
e e
t a
b
d e q
n th
re b
a c k g
u a
e r
a c
ro
l to
e fe r
k g ro
u n d
th e
e n c
u n
h a
d im e n s
e a n d th
d in th e
s lo w e r
io
e
re
in
n s o f t
m o v in
fe re n c
te n s it
h e m o
g o b je
e im a
y v a lu
v c t
g e
e s
2 2 3
( c ) T r u e . F r o m E q . ( 1 0 .6 - 4 ) , w e s e e t h a t d iffe r e n c e s b e t w e e n t h e b a c k g r o u n d
a n d t h e o b je c t a lw a y s w ill b e n e g a t iv e [a s s u m in g , a s in E q . ( 1 0 .6 - 4 ) , t h a t t h e
in te n s ity le v e ls in th e o b je c t e x c e e d th e v a lu e o f th e b a c k g r o u n d ]. A s s u m in g a ls o
th a t th e d iffe r e n c e s a r e m o r e n e g a tiv e th a n th e th r e s h o ld , w e s e e fo r th e s a m e
r e a s o n a s in (a ) th a t a ll n e w b a c k g r o u n d a r e a s o c c u p ie d b y th e m o v in g o b je c t
w ill h a v e n o n z e r o c o u n ts , th u s in c r e a s in g th e d im e n s io n o f th e n o n z e r o e n tr ie s
in th e n e g a tiv e A D I (k e e p in m in d th a t th e v a lu e s in th is im a g e a r e c o u n ts ).
Problem 10.46
C o n s id e r fi r s t th e fa c t th a t m o tio n in th e x - d ir e c tio n is z e r o . W h e n a ll c o m p o n e n ts o f a n im a g e a re s ta tio n a r y , g x (t ,a 1 ) is a c o n s ta n t, a n d its F o u r ie r tr a n s fo r m y ie ld s a n im p u ls e a t t h e o r ig in . T h e r e fo r e , F ig . 1 0 .6 3 w o u ld n o w c o n s is t s
o f a s in g le im p u ls e a t th e o r ig in . T h e o th e r tw o p e a k s s h o w n in th e fi g u r e w o u ld
n o lo n g e r b e p r e s e n t. T o h a n d le th e m o tio n in th e p o s itiv e y - d ir e c tio n a n d its
c h a n g e in th e o p p o s ite d ir e c tio n , r e c a ll th a t th e F o u r ie r tr a n s fo r m is a lin e a r p r o c e s s , s o w e c a n u s e s u p e r p o s itio n to o b ta in a s o lu tio n . T h e fi r s t p a r t o f m o tio n
is in th e p o s itiv e y - d ir e c tio n a t 1 p ix e l/ fr a m e . T h is is th e s a m e a s in E x a m p le
1 0 .2 8 , s o t h e p e a k s c o r r e s p o n d in g t o t h is p a r t o f t h e m o t io n a r e t h e s a m e a s t h e
o n e s s h o w n in F ig . 1 0 .6 4 . T h e r e v e r s a l o f m o t io n is in s t a n t a n e o u s , s o t h e 3 3 r d
fr a m e w o u ld s h o w th e o b je c t tr a v e lin g in e x a c tly th e o p p o s ite d ir e c tio n . T o h a n d le t h is , w e s im p ly c h a n g e a 2 t o − a 2 in E q . ( 1 0 .6 - 7 ) . B a s e d o n t h e d is c u s s io n
in c o n n e c t io n w it h E q . ( 1 0 .6 - 5 ) , a ll t h is c h a n g e w o u ld d o is p r o d u c e p e a k s a t
fr e q u e n c ie s u = − a 2 V 2 a n d K + a 2 V 2 . F r o m E x a m p le 1 0 .2 8 w e k n o w t h a t t h e
v a lu e o f a 2 is 4 . F r o m th e p r o b le m s ta te m e n t, w e k n o w th a t V 2 = 1 a n d K = 3 2 .
T h u s , w e h a v e t w o n e w p e a k s a d d e d t o F ig . 1 0 .6 4 : o n e a t u = − 4 a n d t h e o t h e r
a t u = 3 6 . A s n o te d a b o v e , th e o r ig in a l p e a k s c o r re s p o n d to th e m o tio n in th e
p o s itiv e y - d ir e c tio n g iv e n in th e p r o b le m s ta te m e n t, w h ic h is th e s a m e a s in
E x a m p le 1 0 .2 8 . N o t e t h a t t h e fr a m e c o u n t w a s r e s t a r t e d fr o m 0 t o 3 1 w it h t h e
c h a n g e in d ire c tio n .
Problem 10.47
R e c
o n e
tim
d e r
a t a
o f s
a n o
a ll th a t v e lo c ity is a v e c to r, w h o s e m a g n itu d e is s p e e d . F u n c tio n g x is a
- d im e n s io n a l " r e c o r d " o f th e p o s itio n o f th e m o v in g o b je c t a s a fu n c tio n o f
e (fr a m e r a te ). T h e v a lu e o f v e lo c ity (s p e e d ) is d e te r m in e d b y ta k in g th e fi r s t
iv a tiv e o f th is fu n c tio n . T o d e te r m in e w h e th e r v e lo c ity is p o s itiv e o r n e g a tiv e
s p e c ifi c tim e , n , w e c o m p u te th e in s ta n ta n e o u s a c c e le r a tio n (r a te o f c h a n g e
p e e d ) a t th a t p o in t; th a t is w e c o m p u te th e s e c o n d d e r iv a te o f g x . V ie w e d
th e r w a y , w e d e te r m in e d ire c tio n b y c o m p u tin g th e d e r iv a tiv e o f th e d e r iv a -
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
2 2 4
tiv e
th e
o r z
im a
s ig n
o f
ta n
e ro
g in
, w
g
g
.
a
h
. B u t, t
e n t h a s a
B e c a u s e
r y to its r
ic h is w h
x
h e d e r iv a tiv e a t
p o s itiv e s lo p e ,
g x is a c o m p le x
e a l p a r t. T h is r a
a t w e s ta r te d o u
a p
th e
q u
tio
t to
o in t is s im p ly th e ta n g
v e lo c ity is p o s itiv e ; o th
a n tity , its ta n g e n t is g iv
is p o s itiv e w h e n S 1x a n
p ro v e .
e n t a t th
e r w is e it
e n b y th e
d S 2x h a v
a t
is
r
e
p o
n e
a tio
th e
in t. If
g a tiv e
o f its
s a m e
Problem 10.48
S e t
in g
c a p
e n c
im a
A (0
n e n
tim
u p a s m a ll ta r g e t o f k n o w n , u
a re a , a t c o o rd in a te s (x 0 ,y 0 ).
tu re d b y th e s y s te m a n d c a n
e to th e im a g e m o d e l d is c u s
g in g th e ta r g e t w h e n th e b u
)R a n d f a n d R w ill b e k n o w
tia l te r m is k n o w n , s o c a n c
e t w ill b e g iv e n b y
n ifo r m r
T h e s m a
th u s b e
s e d in S e
lb is n e w
n in th e
o m p u te
"
f (x ,y ) = i (x ,y )r (x ,y ) =
It is g iv e n
la m p is n e
n e n t o f th
c o rre c te d
in g th e k n
t h a t O t s u ’s a l g o
w (a n d t = 0 ) s o
e p re v io u s e q u a
im a g e w ill b e a s
o w n re fl e c tiv ity
(N o
f (x
a re
im a
0
te :
,y
R ,
g e
0
n
(x
0
,y 0 ) =
A (t ) − t 2 e
−
[
(x − M / 2 )2 + (y − N / 2 )
r ith m w o r k s w e ll w h e n f =
, if w e c a n fo rc e th e tim e a n d
tio n to h a v e th e v a lu e A (0 ),
d e s ire d . C o n s id e r th e p o r tio
ta r g e t a t a n y tim e t n
"
f
e fl e c tiv ity , R , n e a r a c o r n e r o f th e v ie w ll ta r g e t w ill b e c o m e p a r t o f th e im a g e s
m o n ito re d b y th e s o ftw a re . W ith re fe rc t io n 2 .3 .4 , y o u c a n d e t e r m in e A (0 ) b y
(t = 0 ) b e c a u s e th e n f (x 0 ,y 0 ) = i R =
re g io n o f th e im a g e d ta r g e t. T h e e x p o it o n c e a n d s to re it. T h e im a g e a t a n y
A (t
n
) − t n2 e
[
−
(x
A (0
s p
th e
n o
)r
a t
n
f t
− M / 2 )2 + (y 0 − N / 2 )
0
2
]
#
r (x ,y ).
(x ,y )
ia lly v
s e g m
h e im
., w h e n th
in g c o m p o
a tio n o f th
a t c o n ta in
e
e
-
#
R .
]
2
( i.e
a r y
e n t
a g e
in p r a c tic e w e w o u ld u s e th e a v e r a g e v a lu e o f f a r o u n d (x 0 , y 0 ) in s te a d o f
) to r e d u c e th e e ffe c ts o f n o is e .) T h e e x p o n e n tia l te r m is k n o w n , a n d s o
f n (x 0 , y 0 ), a n d t n . S o w e c a n s o lv e fo r A (t n ). T h e n w e c a n c o r r e c t th e in p u t
a s fo llo w s :
g (x ,y )
A (0 ) f (x ,y )
2
e − [ (x 0 − M / 2 )
=
=
A ( t n ) − t n2
⎡
A (t ) − t 2 e
A (0 ) ⎣
A ( t n ) − t n2 e
[
−
−
[
+ (y 0 − N / 2 )
2
]
(x − M / 2 )2 + (y − N / 2 )
(x
0
2
− M / 2 )2 + (y 0 − N / 2 )
⎤
]
2
]
⎦r (x ,y ).
A t t = t n , g ( x , y ) = A ( 0 ) r ( x , y ) , a n d w e k n o w t h a t O t s u ’s a l g o r i t h m w i l l p e r f o r m
s a t i s f a c t o r i l y . B e c a u s e w e d o n ’t k n o w h o w A ( t ) b e h a v e s a s a f u n c t i o n o f t i m e a n d
2 2 5
th e la m p s a r e e x p e r im e n ta l, th e s a fe s t c o u r s e o f a c tio n is to p e r fo r m th
in g c o r r e c tio n a s m a n y tim e s a s p o s s ib le d u r in g th e tim e th e la m p s
a tin g . T h is re q u ire s fre q u e n t c o m p u ta tio n o f A (t n ) u s in g in fo r m a tio n
fr o m th e re fl e c tiv e ta r g e t, a s w e d id a b o v e . If c o n tin u o u s c o r re c tio n
a c o m p u ta tio n a l b u r d e n th a t is to o g r e a t, o r to o e x p e n s iv e to im p le m
p e r fo r m in g e x p e r im e n ts w ith v a r io u s la m p s c a n h e lp in d e te r m in in g
a c c e p ta b le p e r io d b e tw e e n c o r r e c tio n s . It is lik e ly th a t lo n g e r p e r io d s
c o r r e c tio n s w ill b e a c c e p ta b le b e c a u s e th e c h a n g e in p h y s ic a l d e v ic e s
is m u c h s lo w e r th a n th e s p e e d o f im a g in g d e v ic e s .
e p re c e d a re o p e ro b ta in e d
p re s e n ts
e n t, th e n
a lo n g e r
b e tw e e n
g e n e r a lly
Problem 10.49
(a ) It is g iv e n th a t 1 0 % o f th e im a g e a re a in th e h o r iz o n ta l d ire c tio n is o c c u
b y a b u lle t th a t is 2 .5 c m lo n g . B e c a u s e th e im a g in g d e v ic e is s q u a r e (2 5 6 ×
e le m e n ts ) th e c a m e r a lo o k s a t a n a r e a th a t is 2 5 c m × 2 5 c m , a s s u m in g n o o p
d is t o r t io n s . T h u s , t h e d is t a n c e b e t w e e n p ix e ls is 2 5 / 2 5 6 = 0 .0 9 8 c m / p ix e l.
m a x im u m s p e e d o f t h e b u lle t is 1 0 0 0 m / s e c = 1 0 0 ,0 0 0 c m / s e c . A t t h is s p e e d
b u lle t w ill t r a v e l 1 0 0 , 0 0 0 / 0 .9 8 = 1 .0 2 × 1 0 6 p ix e ls / s e c . It is r e q u ir e d t h a t t h e b
n o t t r a v e l m o r e t h a n o n e p ix e l d u r in g e x p o s u r e . T h a t is , ( 1 .0 2 × 1 0 6 p ix e ls / s e
K s e c ≤ 1 p ix e l. S o , K ≤ 9 .8 × 1 0 − 7 s e c .
p ie d
2 5 6
tic a l
T h e
, th e
u lle t
c ) ×
( b ) T h e fr a m e r a te m u s t b e fa s t e n o u g h to c a p tu r e a t le a s t tw o im a g e s o f th e
b u lle t in s u c c e s s iv e fr a m e s s o th a t th e s p e e d c a n b e c o m p u te d . If th e fr a m e
r a te is s e t s o th a t th e b u lle t c a n n o t tr a v e l a d is ta n c e lo n g e r (b e tw e e n s u c c e s s iv e
fr a m e s ) th a n o n e h a lf th e w id th o f th e im a g e , th e n w e h a v e th e c a s e s s h o w n in
F ig . P 1 0 .4 9 . In c a s e s A a n d E w e g e t t w o s h o t s o f t h e e n t ir e b u lle t in fr a m e s t 2 a n d
t 3 a n d t 1 a n d t 2 , r e s p e c tiv e ly . In th e o th e r c a s e s w e g e t p a r tia l b u lle ts . A lth o u g h
th e s e c a s e s c o u ld b e h a n d le d w ith s o m e p r o c e s s in g ( e .g ., b y d e te r m in in g s iz e ,
le a d in g a n d tr a ilin g e d g e s , a n d s o fo r th ) it is p o s s ib le to g u a r a n te e th a t a t le a s t
tw o c o m p le te s h o ts o f e v e r y b u lle t w ill b e a v a ila b le b y s e ttin g th e fr a m e r a te s o
th a t a b u lle t c a n n o t tr a v e l m o r e th a n o n e h a lf th e w id th o f th e fr a m e , m in u s th e
le n g t h o f t h e b u lle t . T h e le n g t h o f t h e b u lle t in p ix e ls is ( 2 .5 c m ) / ( 0 .0 9 8 c m / p ix e l)
≈ 2 6 p ix e ls . O n e h a lf o f th e im a g e fr a m e is 1 2 8 p ix e ls , s o th e m a x im u m tr a v e l
d is ta n c e a llo w e d is 1 0 2 p ix e ls . B e c a u s e th e b u lle t tr a v e ls a t a m a x im u m s p e e d o f
1 .0 2 × 1 0 6 p ix e ls / s e c , t h e m in im u m fr a m e r a t e is 1 .0 2 × 1 0 6 / 1 0 2 = 1 0 4 fr a m e s / s e c .
( c ) In a fl a s h in g s itu a tio n w ith a r e fl e c tiv e o b je c t, th e im a g e s w ill te n d to b e d a r k ,
w it h t h e o b je c t s h in in g b r ig h t ly . T h e t e c h n iq u e s d is c u s s e d in S e c t io n 1 0 .6 .1
w o u ld th e n b e q u ite a d e q u a te .
( d ) F ir s t w e h a v e to d e te r m in e if a p a r tia l o r w h o le im a g e o f th e b u lle t h a s b e e n
o b ta in e d . A fte r th e p ix e ls c o r r e s p o n d in g to th e o b je c t h a v e b e e n id e n tifi e d u s -
2 2 6
C H A P T E R 1 0 . P R O B L E M S O L U T IO N S
F ig u r e P 1 0 .4 9
in g m o tio n s e g m e n ta tio n , w e d e te r m in e if th e o b je c t r u n s in to th e le ft b o u n d a r y
( s e e th e s o lu tio n t o P r o b le m 9 .3 6 r e g a r d in g a m e t h o d fo r d e t e r m in in g if a b in a r y
o b je c t r u n s in to th e b o u n d a r y o f a n im a g e ). If it d o e s , w e lo o k a t th e n e x t tw o
fr a m e s , w ith th e a s s u r a n c e th a t a c o m p le te im a g e o f th e b u lle t h a s b e e n o b ta in e d in e a c h b e c a u s e o f th e fr a m e r a te in (b ). If th e o b je c t d o e s n o t r u n in to
th e le ft b o u n d a r y , w e a r e s im ila r ly a s s u r e d o f tw o fu ll s h o ts in tw o o f th e th r e e
fr a m e s . W e th e n c o m p u te th e c e n tr o id o f th e o b je c t in e a c h im a g e a n d c o u n t th e
n u m b e r o f p ix e ls b e tw e e n th e c e n tr o id s . B e c a u s e th e d is ta n c e b e tw e e n p ix e ls
a n d th e tim e b e tw e e n fr a m e s a re k n o w n , c o m p u ta tio n o f th e s p e e d is a tr iv ia l
p r o b le m . T h e p r in c ip a l u n c e r ta in ty in th is a p p r o a c h is h o w w e ll th e o b je c t is
s e g m e n te d . H o w e v e r, b e c a u s e th e im a g e s a r e o f th e s a m e o b je c t in b a s ic a lly th e
s a m e g e o m e tr y , c o n s is te n c y o f s e g m e n ta tio n b e tw e e n fr a m e s c a n b e e x p e c te d .
Chapter 11
Problem Solutions
Problem 11.1
( a ) T h e k e y to th is p r o b le m is to r e c o g n iz e th a t th e v a lu e o f e v e r y e le
c h a in c o d e is r e la tiv e to th e v a lu e o f its p r e d e c e s s o r. T h e c o d e fo r a
t h a t is t r a c e d in a c o n s is t e n t m a n n e r ( e .g ., c lo c k w is e ) is a u n iq u e c ir c
n u m b e r s . S ta r tin g a t d iffe r e n t lo c a tio n s in th is s e t d o e s n o t c h a n g e
tu r e o f th e c ir c u la r s e q u e n c e . S e le c tin g th e s m a lle s t in te g e r a s th e s ta r
s im p ly id e n tifi e s th e s a m e p o in t in th e s e q u e n c e . E v e n if th e s ta r tin
n o t u n iq u e , th is m e th o d w o u ld s till g iv e a u n iq u e s e q u e n c e . F o r e x a
s e q u e n c e 1 0 1 0 1 0 h a s th r e e p o s s ib le s ta r tin g p o in ts , b u t th e y a ll y ie ld
s m a lle s t in te g e r 0 1 0 1 0 1 .
m e n t in a
b o u n d a r y
u la r s e t o f
th e s tr u c tin g p o in t
g p o in t is
m p le , th e
th e s a m e
( b ) C o d e : 1 1 0 7 6 7 6 5 5 4 3 3 2 2 . T h e s ta r tin g p o in t is 0 , y ie ld in g th e s e q u e n c e
0 7 6 7 6 5 5 4 3 3 2 2 1 1 .
Problem 11.2
(a ) T h e
ja c e n t e
d ire c tio
w h ile to
c h a n g e
fi rs t d
le m e n
n , th e
p o in t
th e c o
iffe r e n c e o n ly c
ts o f th e c o d e .
fi r s t d iffe re n c e
o u t to s tu d e n ts
d e its e lf ).
o u n ts th
B e c a u s e
is in d e p e
th a t th e
e n
th
n d
a s s
u m b e r o f d
e c o u n tin g
e n t o f b o u n
u m p tio n h e
ire
p r
d a
re
c t
o c
r y
is
io n s t
e s s is
ro ta t
th a t r
h a t s e p
in d e p e
io n . (It
o ta tio n
a ra t
n d e
is w
d o e
e a
n t
o r t
s n
d o f
h o t
(b ) C o d e : 0 1 0 1 0 3 0 3 0 3 3 2 3 2 3 2 2 1 2 1 1 1 . D iffe re n c e : 3 1 3 1 3 3 1 3 1 3 0 3 1 3 1 3 0 3 1 3 0 0 . T h e
c o d e w a s tr e a te d a s a c ir c u la r s e q u e n c e , s o th e fi r s t e le m e n t o f th e d iffe r e n c e is
th e tr a n s itio n b e tw e e n th e la s t a n d fi r s t e le m e n t o f th e c o d e , a s e x p la in e d in th e
te x t.
2 2 7
C H A P T E R 1 1 . P R O B L E M S O L U T IO N S
2 2 8
F ig u r e P 1 1 .3
Problem 11.3
(a ) T h
fl e c tio
s tr u c t
lin e s ,
u r a tio
(b )
s u c
th e
d iff
c e n
e r u b b e r
n o f th e
u re o f th e
th is p r o d
n .
If a
h th
s itu
e re n
te re
-b a n d a p p
c e ll w a ll.
in n e r a n d
u c e s th e m
ro a c h fo rc e s th e p o
T h a t is , th e lo c a tio
o u te r w a lls . B e c a u s
in im u m - p e r im e te r
ly g o n to h a v e v e
n s o f th e v e r tic e
e th e v e r tic e s a re
p o ly g o n fo r a n y
r tic e s
s a re
jo in e
g iv e n
a t e v e r y
fi x e d b y
d b y s tra
w a ll c o n
in
th
ig h
fi g
c o r n e r o f a c e ll is c e n te r e d a t a p ix e l o n th e b o u n d a r y , a n
a t th e r u b b e r b a n d is tig h te n e d o n th e o p p o s ite c o r n e r, w e
a t io n in F ig . P 1 1 .3 . A s s u m in g t h a t t h e c e ll is o f s iz e d × d , t h
p
2 d
c e b e tw e e n th e p ix e l a n d th e b o u n d a r y in th a t c e ll is
p
d o n p ix e ls , th e m a x im u m d iffe r e n c e is ( 2 d )/ 2 .
d th e c e
w o u ld h
e m a x im
. If c e lls
ll
a
u
a
e
t
-
is
v e
m
re
Problem 11.4
(a ) W h e n th e B v e r tic e s a re m ir r o re d , th e y c o in c id e w ith th e tw o w h ite v e r tic e s
in th e c o r n e r s , s o th e y b e c o m e c o llin e a r w ith th e c o r n e r v e r tic e s . T h e a lg o r ith m
ig n o r e s c o llin e a r v e r tic e s , s o th e s m a ll in d e n ta tio n w ill n o t b e d e te c te d .
(b
th
R e
b e
B e
a n
s g
s g
tia
) W h e n th e in d e n ta tio n is d e e p e r th a n o n e p ix e l (b u t s till 1 p ix e l w id e ) w e h a v e
e s it u a t io n s h o w n in F ig . P 1 1 .4 . N o t e t h a t t h e B v e r t ic e s c r o s s a ft e r m ir r o r in g .
fe r r in g to th e b o tto m fi g u r e , w h e n th e a lg o r ith m g e ts to v e r te x 2 , v e r te x 1 w ill
id e n tifi e d a s a v e r te x o f th e M P P, s o th e a lg o r ith m is in itia liz e d a t th a t s te p .
c a u s e o f in itia liz a tio n , v e r te x 2 is v is ite d a g a in . It w ill b e c o llin e a r w ith W C
d V L , s o B C w ill b e s e t a t th e lo c a tio n o f v e r te x 2 . W h e n v e r te x 3 is v is ite d ,
n (V L , W C , V 3 ) w ill b e 0 , s o B C w ill b e s e t a t v e r te x 3 . W h e n v e r te x 4 is v is ite d ,
n (1 , 3 , 4 ) w ill b e n e g a tiv e , s o V L w ill b e s e t to v e r te x 3 a n d th e a lg o r ith m is r e in iliz e d . B e c a u s e v e r te x 2 w ill n e v e r b e v is ite d a g a in , it w ill n e v e r b e c o m e a v e r -
2 2 9
1
2
Order and location of vertices
beforemirroring the B vertices
4
3
1
3
Order and location of vertices
after mirroring the B vertices
4
2
F ig u r e P 1 1 .4
te x o f th e M P P. T h
in d e n ta tio n s 2 p ix
th e s e q u e n c e 1 − 3
in g c a u s e d b y th e
T h is is a g e n e r a l re
e n e x t M P
e ls o r g r e a
− 4 in th e
m ir r o r in g
s u lt fo r 1 -
P v e r te
te r in d
s e c o n d
o f th e
p ix e l w
x t
e p
fi g
tw
id e
o b e d e te c te d w
th a n d 1 p ix e l w
u re . T h u s , th e a
o B v e r tic e s b y
, 2 p ix e l (o r g re a
ill b e
id e w
lg o r it
k e e p
te r) d
v e r
ill b
h m
in g
e e p
(c ) W h e n th e B v e r tic e s o f a 1 - p ix e l d e e p p r o tr u s io n a re m ir r o r
a lig n e d w ith th e tw o c o n v e x v e r tic e s o f th e p r o tr u s io n , w h ic h
v e r tic e s o f th e M P P (o th e r w is e th e y w o u ld b e th e c o r n e r s o f th
lin e p a s s in g fr o m th e fi r s t c o n v e x v e r te x to th e p r e v io u s M P P v e
th a t th e c o r r e s p o n d in g m ir r o r e d B v e r te x w o u ld b e o u ts id e th e
n o t b e a v e r te x o f th e M P P. A s im ila r c o n c lu s io n is r e a c h e d b y p a
th e s e c o n d c o r n e r to th e n e x t v e r te x o f th e M P P, s o th e s e c o n d
c o u ld n o t b e p a r t o f th e M P P. T h e c o n c lu s io n is th a t th e s m a ll p
b e m is s e d b y th e a lg o r ith m .
te x 4 . T h
e re p re s e
s o lv e s th
o n ly o n e
in tr u s io
e re fo
n te d
e c ro
v e r t
n s .
re ,
b y
s s e x .
e d , th e y b e c o m e
a re k n o w n to b e
e p r o tr u s io n ). A
r te x w o u ld s h o w
M P P, s o it c o u ld
s s in g a lin e fr o m
m ir r o re d v e r te x
r o tr u s io n w o u ld
( d ) A d r a w in g s im ila r to th e o n e u s e d in (b ) w o u ld s h o w th a t b o th v e r tic e s o f th e
p r o tr u s io n w o u ld b e d e te c te d n o r m a lly fo r a n y p r o tr u s io n e x te n d in g fo r m o r e
th a n o n e p ix e l.
Problem 11.5
( a ) T h e r e s u ltin g p o ly g o n w o u ld c o n ta in a ll th e b o u n d a r y p ix e ls .
( b ) A c tu a lly , in b o th c a s e s th e r e s u ltin g p o ly g o n w o u ld c o n ta in a ll th e b o u n d a r y
p ix e ls .
C H A P T E R 1 1 . P R O B L E M S O L U T IO N S
2 3 0
F ig u r e P 1 1 .6
Problem 11.6
( a ) T h e s o lu t io n is s h o w n in F ig . P 1 1 .6 ( b ) .
( b ) T h e s o lu t io n is s h o w n in F ig . P 1 1 .6 ( c ) .
Problem 11.7
( a ) F r o m F ig . P 1 1 .7 ( a ) , w e s e e t h a t t h e d is t a n c e fr o m t h e o r ig in t o t h e t r ia n g le is
g iv e n b y
r (θ )
D
=
0
0
=
=
=
=
=
c o s θ
D
c o s (1 2
D
c o s (1 8
D
c o s (2 4
D
c o s (3 0
D
c o s (3 6
◦
0
◦
− θ )
◦
− θ )
◦
− θ )
◦
− θ )
◦
− θ )
0
0
0
0
0
0
0
0
0
◦
≤ θ < 6 0
6 0
◦
≤ θ < 1 2 0
◦
1 2 0
◦
≤ θ < 1 8 0
◦
1 8 0
◦
≤ θ < 2 4 0
◦
2 4 0
◦
≤ θ < 3 0 0
◦
3 0 0
◦
≤ θ < 3 6 0
◦
2 3 1
w h e r e D 0 is th e p e r p e n d ic u la r d
tr ia n g le , a n d D = D 0 / c o s (6 0 ◦ )
th e tr ia n g le a r e g iv e n , d e te r m in
p r o b le m , a n d D 0 (w h ic h is th e
e le m e n ta r y g e o m e tr y .
is ta n c e
= 2 D 0 .
in g th e
s a m e f
fr
O
e q
o r
o m
n c e
u a t
th e
th e
th
io n
th
F ig u r e P 1 1 .7
o r
e c
o f
re e
ig in to
o o rd in
e a c h s
s tr a ig
o n
a te
tra
h t
e o f th e
s o f th e
ig h t lin e
lin e s ) fo
s id
v e
is
llo
e s o f th e
r tic e s o f
a s im p le
w s fro m
C H A P T E R 1 1 . P R O B L E M S O L U T IO N S
2 3 2
( b ) F r o m F ig . P 1 1 .7 ( c ) ,
r (θ )
=
=
=
=
=
=
=
=
B
2 c o s θ
◦
0
A
2 c o s (9 0
A
2 c o s (θ
B
2 c o s (1 8
B
2 c o s (θ
A
2 c o s (2 7
A
2 c o s (θ
B
2 c o s (3 6
◦
' ≤ θ < 9 0
− θ )
◦
− 1 8 0 ◦ )
◦
− θ )
− 2 7 0 )
◦
0
− θ )
◦
≤ θ < (1 8 0
◦
(1 8 0
− θ )
0
◦
9 0
− 9 0 ◦)
0
≤ θ < '
◦
− ' )
− ' ) ≤ θ < 1 8 0
1 8 0
◦
≤ θ < 1 8 0
◦
1 8 0
◦
+ ' ≤ θ < 2 7 0
2 7 0
◦
≤ θ < 2 7 0
2 7 0
◦
+ ' ≤ θ < 3 6 0 ◦ .
◦
◦
+ '
◦
+ '
w h e re ' = ta n − 1 (A / B ).
( c ) T h e e q u a t io n o f t h e e llip s e in F ig . P 1 1 .7 ( e ) is
2
x
a
2
2
y
+
= 1 .
2
b
W e a re in te re s te d in th e d is ta n c e fr o m th e o r ig in to a n a r b itr a r y p o in t (x ,y ) o n
th e e llip s e . In p o la r c o o r d in a te s ,
x = r c o s θ
a n d
y = r s in θ
w h e re r is th e d is ta n c e fr o m th e o r ig in to (x ,y ):
r =
2
x
2
+ y
.
S u b s titu tin g in to th e e q u a tio n o f th e e llip s e w e o b ta in
r
2
c o s 2 θ
+
a 2
2
r
s in
b 2
2
θ
= 1
fr o m w h ic h w e o b ta in th e d e s ir e d r e s u lt:
1
r (θ ) =
c o s θ
a
2
+
s in θ
b
2
1 / 2
.
W h e n b = a , w e h a v e th e fa m ilia r e q u a tio n o f a c ir c le , r (θ ) = a , o r x
2
+ y
2
= a
2
.
2 3 3
F ig u r e P 1 1 .8
F ig u r e P 1 1 .1 0
Problem 11.8
T h e s o lu t io n s a r e s h o w n in F ig . P 1 1 .8 .
Problem 11.9
(a ) In th e
( 1 1 .1 - 4 ) is
E q . ( 1 1 .1 a n d p 4 · p
is le ft u n c
le ft u n c h a
(b ) In th
p is le ft
th ird c a
S (p ) = 2
fi rs t c a s e , N (p ) =
s a tis fi e d a n d p is
4 ) is v io la te d a n d
6 · p 8 = 1 , s o c o n
h a n g e d . In th e fo
n g e d .
e fi
u n
s e
a n
rs t
c h a
(c 0)
d p
c a s e p
n g e d .
a n d (d
is le ft
5 , S (p ) = 1 , p 2 ·p 4 ·p 6 = 0 , a n d p 4 ·p 6 ·p
fl a g g e d fo r d e le tio n . In th e s e c o n d c a s e ,
p is le ft u n c h a n g e d . In th e th ir d c a s e p 2
d it io n s ( c ) a n d ( d ) o f E q . ( 1 1 .1 - 4 ) a r e v io
r th c a s e S (p ) = 2 , s o c o n d itio n (b ) is v io la
2 ·p 6 ·p 8
In th e s e
0) a r e v io
u n c h a n g
8 =
N (p
·p 4
la te
te d
0 , s o
) = 1
·p 6
d a n
a n d
E q .
, s o
= 1
d p
p is
= 1 s o c o n d it io n ( d 0) in E q . ( 1 1 .1 - 6 ) is v io la t e d a n d
c o n d c a s e N (p ) = 1 s o p is le ft u n c h a n g e d . In th e
la te d a n d p is le ft u n c h a n g e d . In th e fo u r th c a s e
e d .
Problem 11.10
( a ) T h e r e s u lt is s h o w n in F ig . P 1 1 .1 0 ( b ) .
( b ) T h e r e s u lt is s h o w n in F ig . P 1 1 .1 0 ( c ) .
C H A P T E R 1 1 . P R O B L E M S O L U T IO N S
2 3 4
Problem 11.11
( a ) T h e n u m b e r o f s y m b o ls in th e fi r s t d iffe r e n c e is e q u a l to th e n u m b e r o f s e g m e n t p r im itiv e s in th e b o u n d a r y , s o th e s h a p e o rd e r is 1 2 .
( b ) S ta r tin g a t th e to p le ft c o r n e r,
C h a in c o d e :
D iffe re n c e :
S h a p e n u m b e r :
0 0 0 3 3 2 1 2 3 2 1 1
3 0 0 3 0 3 3 1 1 3 3 0
0 0 3 0 3 3 1 1 3 3 0 3
Problem 11.12
W ith re fe re n
c o n ju g a te s y
g in h a v e th is
th a t th is c o n
b y p la c in g th
c e to C h
m m e tr ic
p ro p e r t
d itio n is
e o r ig in
a p te r
. O n ly
y . T h e
s a tis fi
a t th e
4 , th e D F T c a n b e
c o n to u rs th a t a re s
a x is s y s te m o f F ig .
e d fo r s y m m e tr ic fi
c e n te r o f g r a v ity o f
r e a l o n ly if th e
y m m e tr ic w ith
1 1 .1 9 w o u ld h a
g u re s . T h is c a n
th e c o n to u r.
d a ta s e q u e n c e
re s p e c t to th e o
v e to b e s e t u p
b e a c c o m p lis h
is
r is o
e d
Problem 11.13
S u p p o s e th a t w e h a
tic le w ill b e a t s o m e
s itu a tio n w ith w h ic
p o in t in th e c o m p le
th e m o v in g p a r tic le
d e s c r ib e s a p a th in
tio n o f a c u r v e b e c a
v e a p a r tic le m o v in g in th e x y - p la n e . A t e a c h tim
p o in t w h o s e c o o rd in a te s c a n b e w r itte n a s [x (y
h w e a r e d e a lin g is in th e c o m p le x p la n e . R e c a
x p la n e is w r itte n a t c = x + j y , w e c a n w r ite th e
in th e c o m p le x p la n e a s z (t ) = x (t ) + j y (t ). A s t
th e c o m p le x p la n e . T h is is c a lle d th e p a r a m e tr ic
u s e it d e p e n d s o n p a r a m e te r t .
e t th e
),y (t )].
llin g th
lo c a tio
v a r ie s ,
re p re se
p a rT h e
a t a
n o f
z (t )
n ta -
T h e " s ta n d a r d " e q u a tio n o f a c ir c le w ith c e n te r a t (b , c ) a n d r a d iu s r is g iv e n
b y (x − b )2 + (y − c )2 = r 2 . U s in g p o la r c o o r d in a te s , w e c a n w r ite
x (θ ) = b + r c o s θ
y (θ ) = c + r s in θ .
T h e n , th e c ir c le c a n b e r e p r e s e n te d in p a r a m e tr ic fo r m a s th e p a ir x (θ ), y (θ ) ,
w ith θ b e in g th e p a r a m e te r. T h e c ir c le c a n th u s b e r e p r e s e n te d in th e c o m p le x
p la n e , a s th e c u r v e
z (θ )
x (θ ) + j y (θ )
=
=
(b + c ) + r (c o s θ + j s in θ ).
2 3 5
U s in g o n ly t w o d e s c r ip t o r s in E q . ( 1 1 .2 - 5 ) g iv e s
s^ ( k )
=
=
2 π k
2 π k
1
+ j s in
a (0 ) + a (1 ) c o s
P
K
K
2 π k
a (1 )
2 π k
a (0 )
c o s
+
+ j s in
P
P
K
K
w h ic h w e s e e is in th e fo r m o f a c ir c le b y le ttin g (b + c ) = a (0 )/ P , r = a (1 )/ P , a n d
θ = 2 π k / K .
Problem 11.14
T h e m e a n is s u ffi c ie n t.
Problem 11.15
T w o e llip s e s w ith d iffe r e n t, s a y , m a jo r a x e s , h a v e s ig n a tu r e s w ith th e s a m e m e a n
a n d th ird s ta tis tic a l m o m e n t d e s c r ip to r s (b o th d u e to s y m m e tr y ) b u t d iffe re n t
s e c o n d m o m e n t (d u e to s p re a d ).
Problem 11.16
T h is p r o b le m c a n b e s o lv e d b y u s in g tw o d e s c r ip to r s : h o le s a n d th e c o n v e x d e fi c ie n c y ( s e e S e c t io n 9 .5 .4 r e g a r d in g t h e c o n v e x h u ll a n d c o n v e x d e fi c ie n c y o f a
s e t). T h e d e c is io n m a k in g p r o c e s s c a n b e s u m m a r iz e d in th e fo r m o f a s im p le
d e c is io n , a s fo llo w s : If th e c h a r a c te r h a s tw o h o le s , it is a n 8 . If it h a s o n e h o le
it is a 0 o r a 9 . O th e r w is e , it is a 1 o r a n X . T o d iffe re n tia te b e tw e e n 0 a n d 9 w e
c o m p u te th e c o n v e x d e fi c ie n c y . T h e p re s e n c e o f a ” s ig n ifi c a n t” d e fi c ie n c y (s a y ,
h a v in g a n a r e a g r e a te r th a n 2 0 % o f th e a r e a o f a r e c ta n g le th a t e n c lo s e s th e c h a r a c te r ) s ig n ifi e s a 9 ; o th e r w is e w e c la s s ify th e c h a r a c te r a s a 0 . W e fo llo w a s im ila r
p r o c e d u re to s e p a r a te a 1 fr o m a n X . T h e p re s e n c e o f a c o n v e x d e fi c ie n c y w ith
fo u r c o m p o n e n ts w h o s e c e n tr o id s a r e lo c a te d a p p r o x im a te ly in th e N o r th , E a s t,
W e s t, a n d E a s t q u a d r a n ts o f th e c h a r a c te r in d ic a te s th a t th e c h a r a c te r is a n X .
O th e r w is e w e s a y th a t th e c h a r a c te r is a 1 . T h is is th e b a s ic a p p r o a c h . Im p le m e n ta tio n o f th is te c h n iq u e in a re a l c h a r a c te r re c o g n itio n e n v ir o n m e n t h a s to
t a k e in t o a c c o u n t o t h e r fa c t o r s s u c h a s m u lt ip le ” s m a ll” c o m p o n e n t s in t h e c o n v e x d e fi c ie n c y d u e to n o is e , d iffe r e n c e s in o r ie n ta tio n , o p e n lo o p s , a n d th e lik e .
H o w e v e r, th e m a te r ia l in C h a p te r s 3 , 9 a n d 1 1 p r o v id e a s o lid b a s e fr o m w h ic h
to fo r m u la te s o lu tio n s .
C H A P T E R 1 1 . P R O B L E M S O L U T IO N S
2 3 6
Problem 11.17
(a ) C o - o c c u r re n c e m a tr ix :
1 9 6 0 0
2 0 0
2 0 0 0 0
0
( b ) N o r m a liz e th e m a tr ix b y d iv id in g e a c h c o m p o n e n t b y 1 9 6 0 0 +
= 3 9 8 0 0 :
0 .4 9 2 5
0 .0 0 5 0
0
0 .5 0 2 5
s o p
1 1
= 0 .4 9 2 5 , p
= 0 .0 0 5 , p
1 2
= 0 , a n d p
2 1
2 2
= 0 .5 0 2 5 .
(c ) D e s c r ip to r s :
( 1 ) M a x im u m p r o b a b ilit y : 0 .0 5 0 2 5 .
(2 ) C o r re la tio n : T h e m e a n s a re
2
m
2
=
i
r
p
i = 1
=
1 (p
=
i j
j = 1
+ p
1 1
) + 2 (p
1 2
2 1
+ p
2 2
)
1 (.4 9 2 5 + 0 .0 0 5 ) + 2 (0 + 0 .5 0 2 5 )
=
1 .5 0 2 5
a n d
2
m
2
=
j
c
p
i = 1
=
1 (p
=
i j
i = 1
+ p
1 1
) + 2 (p
2 1
1 2
+ p
2 2
)
1 .5 0 7 5 .
S im ila r ly , th e s ta n d a r d d e v ia tio n s a r e :
2
σ
2
2
=
(i − m
r
)
r
2
p
i = 1
=
(1 − m
=
i j
j = 1
r
)(p
1 1
+ p
1 2
p
i j
) + (2 − m
r
)(p
2 1
+ p
2 2
)
0 .5 0 0 0
a n d
2
σ
2
2
(j − m
c
=
j = 1
=
c
)
2
i = 1
=
(1 − m c )(p 1 1 + p 2 1 ) + (2 − m c )(p 1 2 + p 2 2 )
0 .4 9 9 9 .
2 0 0 +
2 0 0 0 0
2 3 7
T h e n , th e c o r r e la tio n m e a s u r e is c o m p u te d a s
2
2
(i − m
) j − m
r
σ
i = 1 j = 1
σ
r
p
c
=
σ
c
=
2
2
1
i j
σ
r
(i − m
c
) j − m
r
p
c
i j
i = 1 j = 1
0 .9 9 0 0 .
(3 ) C o n tra st:
2
2
2
i − j
p
=
(1 − 1 )2 p
=
0 .0 0 5 .
i j
+ (1 − 2 )2 p
1 1
+ (2 − 1 )2 p
1 2
2 1
+ (2 − 2 )2 p
2 2
i = 1 j = 1
(4 ) U n ifo r m ity :
2
2
2
i j
p
)2 + (p
=
(p
=
0 .4 9 5 1 .
1 1
1 2
)2 + (p
)2 + (p
2 2
p 2 1
+
2
p 2 2
1
2 1
)
2
i = 1 j = 1
(5 ) H o m o g e n e ity :
2
2
p
1 +
i = 1 i = 1
i j
i − j
p 1 2
+
2
p 1 1
+
1
=
=
0 .9 9 7 5 .
(6 ) E n tro p y :
2
2
p
−
i j
lo g
2
p
i j
=
−
p
1 1
lo g
p
1 1
+ p
+ p
2 2
lo g
2
1 2
lo g
p
2 2
2
p
1 2
+ p
2 1
lo g
2
p
2 1
i = 1 i = 1
=
w h e r e w e u s e d 0 lo g
2
2
1 .0 4 0 5
0 ≡ 0 .
Problem 11.18
W e c a n u s e t h e p o s i t i o n o p e r a t o r P : “ 2 m p i x e ls t o t h e r i g h t a n d 2 m p i x e ls b e lo w .”
C H A P T E R 1 1 . P R O B L E M S O L U T IO N S
2 3 8
Problem 11.19
(a ) T h e im a g e is
L
o
r
T
c
e t z 1 = 0 a n d z 2 = 1 . B e c a u
r d e r 2 × 2 . E le m e n t g 1 1 is th
ig h t o f a 0 . B y in s p e c tio n ,
h e to ta l n u m b e r o f p ix e ls
o - o c c u r re n c e m a tr ix is
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
s e th e re
e n u m b
g 11 = 0 .
s a tis fy in
a r e o n ly tw o in te n s ity le v e
e r o f p ix e ls v a lu e d 0 lo c a te d
S im ila r ly , g 1 2 = 1 0 , g 2 1 =
g th e p re d ic a te P is 2 0 , s o
0
1 / 2
G =
.
1 / 2
0
ls , m
o n e
1 0 , a
th e
a tr ix G
p ix e l t
n d g 22
n o r m a
is o f
o th e
= 0 .
liz e d
.
( b ) I n t h i s c a s e , g 1 1 i s t h e n u m b e r o f 0 ’s t w o p i x e l s t o t h e r i g h t o f a p i x e l v a l u e d
0 . B y in s p e c tio n , g 1 1 = 8 . S im ila r ly , g 1 2 = 0 , g 2 1 = 0 , a n d g 2 2 = 7 . T h e n u m b e r o f
p ix e ls s a tis fy in g P is 1 5 , s o th e n o r m a liz e d c o - o c c u r r e n c e m a tr ix is
8 / 1 5
0
G =
0
7 / 1 5
.
Problem 11.20
W h e n a s s ig n in g th is p r o b le m , th e In s tr u c to r m a y w is h to p o in t th e s tu d e n t to
th e re v ie w o f m a tr ic e s a n d v e c to r s in th e b o o k w e b s ite .
F r o m E q . ( 1 1 .4 - 6 ) ,
y = A (x − m x ) .
T h e n ,
m
E f y g = E f A (x − m x )g
=
y
A [E f x g − E f m x g ]
=
=
=
A (m
x
− m x )
0 .
T h is e s t a b lis h e s t h e v a lid it y o f E q . ( 1 1 .4 - 7 ) .
T o p r o v e t h e v a lid it y o f E q . ( 1 1 .4 - 8 ) , w e s t a r t w it h t h e d e fi n it io n o f t h e c o v a r ia n c e m a t r ix g iv e n in E q . ( 1 1 .4 - 3 ) :
C
y
= E f (y − m y )(y − m y )T g .
2 3 9
B e c a u s e m
y
= 0 , it fo llo w s th a t
C
A E f (x − m x )(x − m x )T g A
=
=
T
A C x A
v a lid ity o f E q . (1 1
r ia n c e m a tr ic e s a
h a t a re a l s y m m e
(w h ic h a r e e a s ily
r o w s o f m a tr ix A
C x A
g
E f [A (x − m x )][A (x − m x )]T g
=
S h o w in g th e
in g th a t c o v a
it is k n o w n t
e ig e n v e c to r s
c e d u re ). T h e
T
E f y y
=
y
T
.
.4 - 9 ) is a
re re a l a n
tr ic m a tr
o r th o n o
a re th e o
=
little m o r
d s y m m e
ix o f o rd e
r m a liz e d
r th o n o r m
e c o m p lic a te d . W e s ta r t b y n o t
tr ic . F r o m b a s ic m a tr ix a lg e b r a
r n h a s n lin e a r ly in d e p e n d e n
b y , s a y , th e G r a m - S c h m id t p r o
a l e ig e n v e c to r s o f C x . T h e n ,
C x [e 1 ,e 2 ,... e
=
[λ
=
A
e 1 ,λ
1
T
2
λ
⎢
⎢ 0
D = ⎢⎢ .
.
⎣ .
0
,
t
-
e 2 ,...,λ
n
]
e
n
n
]
D
w h e r e w e u s e d t h e d e fi n it io n o f a n e ig e n v e c t o r ( i.e ., C x e
a g o n a l m a tr ix c o m p o s e d o f th e e ig e n v a lu e s o f C x :
⎡
-
]
n
[C x e 1 ,C x e 2 ,... C x e
=
T
···
···
..
0
1
λ
2
..
.
..
.
λ
λ ie i ) a n d D is a d i-
⎥
⎥
⎥.
⎥
⎦
0
···
0
=
⎤
0
.
i
n
P r e m u ltip ly in g b o th s id e s o f th e p r e c e d in g e q u a tio n b y m a tr ix A g iv e s
A C x A
T
=
A A
=
w h e re w e u s e d th e fa c t th a
m a l v e c to rs . T h e re fo r, b e c a
m a tr ix th a t is p r o d u c e d b y
tr ix c o m p o s e d o f its e ig e n v
a s th e e ig e n v a lu e s o f C x . (R
d ia g o n a l te r m s ). T h e fa c t t
C y a re e q u a l to th e e ig e n v e
t A T A = A A T =
u s e C y = A C x A
d ia g o n a liz in g
e c to r s . T h e e ig
e c a ll th a t th e e
h a t C y e i = D e i
c to rs o f C x .
T
D
D
I b e c a u s e th e ro w s o f A a re o r th o n o r, w e h a v e s h o w n th a t C y is a d ia g o n a l
m a tr ix C x u s in g a tr a n s fo r m a tio n m a e n v a lu e s o f C y a r e s e e n to b e th e s a m e
ig e n v a lu e s o f a d ia g o n a l m a tr ix a r e its
= λ i e i s h o w s th a t th e e ig e n v e c to r s o f
T
C H A P T E R 1 1 . P R O B L E M S O L U T IO N S
2 4 0
Problem 11.21
T h e m e a
w h o s e c o
p a r tic u la
th e m e a n
n s q u
rre s p
r c a s e
s q u a
a re
o n d
, th
re e
e rr
in g
e fo
rro
o r, g iv e n b y E q . ( 1 1 .4 - 1 2 ) , is t h e s u m o f t h e e ig e n v a lu e s
e ig e n v e c to r s a re n o t u s e d in th e tr a n s fo r m a tio n . In th is
u r s m a lle s t e ig e n v a lu e s a r e a p p lic a b le ( s e e T a b le 1 1 .6 ) , s o
r is
6
e
λ
=
m s
= 1 7 2 9 .
j
j = 3
T
o
o
o
h e m a x im u
f a ll th e e ig
n ly th e tw o
f th e to ta l p
m e rro
e n v a lu
e ig e n v
o s s ib le
r o c c u r s w h e n K = 0 in E q . ( 1 1 .4 - 1 2 ) w h ic h t h e n is t h e s u m
e s , o r 1 5 0 3 9 in th is c a s e . T h u s , th e e r r o r in c u r re d b y u s in g
e c t o r s c o r r e s p o n d in g t o t h e la r g e s t e ig e n v a lu e s is ju s t 1 1 .5 %
e rro r.
Problem 11.22
T h is
4 0 9 6
m a tr
( 1 1 .4
p r o b le m is s im ila r to th
× 4 0 9 6 b e c a u s e th e im a g
ix is th e id e n tity m a tr ix ,
-1 2 ), th e m e a n s q u a re e r
e p re v io u s o n e . T h e c o v a r ia n c e m a tr ix is o f o rd e r
e s a re o f s iz e 6 4 × 6 4 . It is g iv e n th a t th e c o v a r ia n c e
s o a ll its 4 0 9 6 e ig e n v a lu e s a r e e q u a l to 1 . F r o m E q .
r o r is
4 0 9 6
e
2 0 4 8
λ
m s
=
j
λ
−
j = 1
=
i
i = 1
2 0 4 8 .
Problem 11.23
W h e n th e b o u n d a r y is s y m m e tr ic a b o u t th e b o th th e m a jo r a n d m in o r a x e s a n d
b o th a x e s in te r s e c t a t th e c e n tr o id o f th e b o u n d a r y .
Problem 11.24
A s o lu t i o n u s i n g t h e r e la t i o n s h i p ” c o n n e c t e d t o ,” i s s h o w n i n F i g . P 1 1 .2 4 .
Problem 11.25
W e c a n c o m p u te a m e a s u re o f te x tu re u s in g th e e x p re s s io n
R (x ,y ) = 1 −
1
1 + σ
2
(x ,y )
w h e re σ 2 (x ,y ) is th e in te n s ity v a r ia n c e c o m p u te d in a n e ig h b o r h o o d o f (x ,y ).
T h e s iz e o f th e n e ig h b o r h o o d m u s t b e s u ffi c ie n tly la r g e s o a s to c o n ta in e n o u g h
2 4 1
F ig u r e P 1 1 .2 4
s a m p le s to h a v e a s ta b le e s tim a te o f th e m e a n a n d v a r ia n c e . N e ig h b o r h o o d s o f
s iz e 7 × 7 o r 9 × 9 g e n e r a lly a r e a p p r o p r ia te fo r a lo w - n o is e c a s e s u c h a s th is .
B e c a u s e th e v a r ia n c e o f n o r m a l w a fe r s is k n o w n to b e 4 0 0 , w e c a n o b ta in a
n o r m a l v a lu e fo r R (x , y ) b y u s in g σ 2 = 4 0 0 in th e a b o v e e q u a tio n . A n a b n o r m a l r e g io n w ill h a v e a v a r ia n c e o f a b o u t (5 0 )2 = 2 , 5 0 0 o r h ig h e r, y ie ld in g a la r g e r
v a lu e o f R (x , y ). T h e p r o c e d u r e th e n is to c o m p u te R (x , y ) a t e v e r y p o in t (x , y )
a n d la b e l th a t p o in t a s 0 if it is n o r m a l a n d 1 if it is n o t. A t th e e n d o f th is p r o c e d u r e w e l o o k f o r c l u s t e r s o f 1 ’s u s i n g , f o r e x a m p l e , c o n n e c t e d c o m p o n e n t s
( s e e S e c t io n 9 .5 .3 r e g a r d in g c o m p u ta tio n o f c o n n e c t e d c o m p o n e n ts ) . If th e
a r e a (n u m b e r o f p ix e ls ) o f a n y c o n n e c te d c o m p o n e n t e x c e e d s 4 0 0 p ix e ls , th e n
w e c la s s ify th e s a m p le a s d e fe c tiv e .
Problem 11.26
T h is p r o b le m h a s fo u r m a jo r p a r ts . (1 ) D e te c tin g in d iv id u a l b o ttle s in a n im a g e ;
(2 ) fi n d in g th e to p e a c h b o ttle ; (3 ) fi n d in g th e n e c k a n d s h o u ld e r o f e a c h b o ttle ;
a n d (4 ) d e te r m in in g th e le v e l o f th e liq u id in th e r e g io n b e tw e e n th e n e c k a n d
th e s h o u ld e r.
(1 ) F in d in g in d iv id u a l b o ttle s . N o te th a t th e b a c k g r o u n d in th e s a m p le im a g e
is m u c h d a r k e r th a n th e b o ttle s . W e a s s u m e th a t th is is tr u e in a ll im a g e s . T h e n , a
s im p le w a y to fi n d in d iv id u a l b o ttle s is to fi n d v e r tic a l b la c k s tr ip e s in th e im a g e
h a v in g a w id th d e te r m in e d b y th e a v e r a g e s e p a r a tio n b e tw e e n b o ttle s , a n u m b e r
th a t is e a s ily c o m p u ta b le fr o m im a g e s r e p r e s e n ta tiv e o f th e a c tu a l s e tu p d u r in g
2 4 2
C H A P T E R 1 1 . P R O B L E M S O L U T IO N S
o p e r a tio n . W e c a n fi n d th e s e s tr ip e s in v a r io u s w a y s . O n e w a y is to s m o o th th e
im a g e to re d u c e th e e ffe c ts o f n o is e (w e a s s u m e th a t, s a y , a 3 × 3 o r 5 × 5 a v e r a g in g m a s k is s u ffi c ie n t). T h e n , w e r u n a h o r iz o n ta l s c a n lin e th r o u g h th e m id d le
o f th e im a g e . T h e lo w v a lu e s in th e s c a n lin e w ill c o r r e s p o n d to th e b la c k o r
n e a r ly b la c k b a c k g r o u n d . E a c h b o ttle w ill p r o d u c e a s ig n ifi c a n t r is e a n d fa ll o f
in te n s ity le v e l in th e s c a n lin e fo r th e w id th o f th e b o ttle . B o ttle s th a t a r e fu lly in
th e fi e ld o f v ie w o f th e c a m e r a w ill h a v e a p r e d e te r m in e d a v e r a g e w id th . B o ttle s
th a t a r e o n ly p a r tia lly in th e fi e ld o f v ie w w ill h a v e n a r r o w e r p r o fi le s , a n d c a n b e
e lim in a te d fr o m fu r th e r a n a ly s is (b u t w e n e e d to m a k e s u r e th a t th e tr a ilin g in c o m p le te b o ttle s a r e a n a ly z e d in th e n e x t im a g e ; p r e s u m a b ly , th e le a d in g p a r tia l
b o t t le w a s a lr e a d y p r o c e s s e d .) .
(2 ) F in d in g th e to p o f e a c h b o ttle . O n c e th e lo c a tio n o f e a c h (c o m p le te o r
n e a r ly c o m p le te ) b o ttle is d e te r m in e d , w e a g a in c a n u s e th e c o n tr a s t b e tw e e n
th e b o ttle s a n d th e b a c k g r o u n d to fi n d th e to p o f th e b o ttle . O n e p o s s ib le a p p r o a c h is to c o m p u te a g r a d ie n t im a g e (s e n s itiv e o n ly to h o r iz o n ta l e d g e s ) a n d
lo o k fo r a h o r iz o n ta l lin e n e a r th e to p o f th e g r a d ie n t im a g e . A n e a s ie r m e th o d
is to r u n a v e r tic a l s c a n lin e th r o u g h th e c e n te r o f th e lo c a tio n s fo u n d in th e p r e v io u s s te p . T h e fi r s t m a jo r tr a n s itio n in g r a y le v e l (fr o m th e to p o f th e im a g e ) in
th e s c a n lin e w ill g iv e a g o o d in d ic a tio n o f th e lo c a tio n o f th e to p o f a b o ttle .
(3 ) F in d in g th e n e c k a n d s h o u ld e r o f a b o ttle . In th e a b s e n c e o f o th e r in fo rm a tio n , w e a s s u m e th a t a ll b o ttle s a r e o f th e s a m e s iz e , a s s h o w n in th e s a m p le
im a g e . T h e n , o n c e w e n o w w h e r e th e to p o f a b o ttle is , th e lo c a tio n o f th e n e c k
a n d s h o u ld e r a r e k n o w n to b e a t a fi x e d d is ta n c e fr o m th e b o ttle to p .
(4 ) D e te r m in in g th e le v e l o f th e liq u id . T h e a re a d e fi n e d b y th e b o tto m o f th e
n e c k a n d th e to p o f th e s h o u ld e r is th e o n ly a r e a th a t n e e d s to b e e x a m in e d to d e te r m in e a c c e p ta b le v s . u n a c c e p ta b le fi ll le v e l in a g iv e n b o ttle . In fa c t, A s s h o w n
in th e s a m p le im a g e , a n a r e a o f a b o ttle th a t is v o id o f liq u id a p p e a r s q u ite b r ig h t
in a n im a g e , s o w e h a v e v a r io u s o p tio n s . W e c o u ld r u n a s in g le v e r tic a l s c a n lin e
a g a in , b u t n o te th a t th e b o ttle s h a v e a r e a s o f r e fl e c tio n th a t c o u ld c o n fu s e th is
a p p r o a c h . T h is c o m p u ta tio n is a t th e c o re o f w h a t th is s y s te m is d e s ig n e d to d o ,
s o a m o r e r e lia b le m e th o d s h o u ld b e u s e d . O n e a p p r o a c h is to th r e s h o ld th e a r e a
s p a n n in g a r e c ta n g le d e fi n e d b y th e b o tto m o f th e n e c k , th e s h o u ld e r, a n d s id e s
o f th e b o ttle . T h e n , w e c o u n t th e n u m b e r o f w h ite p ix e ls a b o v e th e m id p o in t o f
th is r e c ta n g le . If th is n u m b e r is g r e a te r th a n a p r e - e s ta b lis h e d v a lu e , w e k n o w
th a t e n o u g h liq u id is m is s in g a n d d e c la r e th e b o ttle im p r o p e r ly fi lle d . A s lig h tly
m o r e s o p h is tic a te d te c h n iq u e w o u ld b e to a c tu a lly fi n d th e le v e l o f th e liq u id .
T h is w o u ld c o n s is t o f lo o k in g fo r a h o r iz o n ta l e d g e in th e r e g io n w ith in th e b o ttle d e fi n e d b y th e s id e s o f th e b o ttle , th e b o tto m o f th e n e c k , a n d a lin e p a s s in g
m id w a y b e tw e e n th e s h o u ld e r a n d th e b o tto m o f th e n e c k . A g r a d ie n t/ e d g e lin k in g a p p r o a c h , a s d e s c r ib e d in C h a p te r 1 0 , w o u ld b e s u ita b le . N o te h o w e v e r,
2 4 3
th a t
c o m
ta tio
fi n d
if n o e
p le te ly
n to re
a n e d g
d g e is fo u n d , th e r e g io n is e ith e r fi lle d (d a r k v a lu e s in th e r e g io n ) o r
v o id o f liq u id (w h ite , o r n e a r w h ite v a lu e s in th e r e g io n ). A c o m p u s o lv e th e s e tw o p o s s ib le c o n d itio n s h a s to fo llo w if th e s y s te m fa ils to
e .
Problem 11.27
T h e k e y s p e c ifi c a tio n o f th e d e s ir e d s y s te m is th a t it b e a b le to d e te c t in d iv id u a l
b u b b le s . N o s p e c ifi c s iz e s a r e g iv e n . W e a s s u m e th a t b u b b le s a r e n e a r ly r o u n d ,
a s s h o w n in th e te s t im a g e . O n e s o lu tio n c o n s is ts o f (1 ) s e g m e n tin g th e im a g e ;
(2 ) p o s t- p r o c e s s in g th e r e s u lt; (3 ) fi n d in g th e b u b b le s a n d b u b b le c lu s te r s , a n d
d e te r m in in g b u b b le s th a t m e r g e d w ith th e b o u n d a r y o f th e im a g e ; (4 ) d e te c tin g
g r o u p s o f to u c h in g b u b b le s ; (5 ) c o u n tin g in d iv id u a l b u b b le s ; a n d (6 ) d e te r m in in g th e r a tio o f th e a r e a o c c u p ie d b y a ll b u b b le s to th e to ta l im a g e a r e a .
(1 ) S e g m e n tin g th e im a g e . W e a s s u m e th a t th e s a m p le im a g e is tr u ly r e p r e s e n ta tiv e o f th e c la s s o f im a g e s th a t th e s y s te m w ill e n c o u n te r. T h e im a g e
s h o w n in th e p r o b le m s ta te m e n t is ty p ic a l o f im a g e s th a t c a n b e s e g m e n te d b y
a g lo b a l t h r e s h o ld . A s s h o w n b y t h e h is t o g r a m in F ig . P 1 1 .2 7 , t h e in t e n s it y le v e ls
o f th e o b je c ts o f in te r e s t a r e h ig h o n th e g r a y s c a le . A s im p le a d a p tiv e th r e s h o ld m e th o d fo r d a ta th a t is th a t h ig h o n th e s c a le is to c h o o s e a th r e s h o ld e q u a l
to th e m e a n p lu s a m u ltip le o f th e s ta n d a r d d e v ia tio n . W e c h o s e a th r e s h o ld
e q u a l to m + 2 σ , w h ic h , fo r th e im a g e in th e p r o b le m s ta te m e n t, w a s 1 9 5 . T h e
s e g m e n t e d r e s u lt is s h o w n o n t h e r ig h t o f F ig . P 1 1 .2 7 . O b v io u s ly t h is is n o t t h e
o n ly a p p r o a c h w e c o u ld ta k e , b u t th is is a s im p le m e th o d th a t a d a p ts to o v e r a ll
c h a n g e s in in te n s ity .
( 2 ) P o s t - p r o c e s s in g . A s s h o w n in t h e s e g m e n t e d im a g e o f F ig . P 1 1 .2 7 , m a n y
o f th e b u b b le s a p p e a r a s b r o k e n d is k s , o r d is k s w ith in te r io r b la c k c o m p o n e n ts .
T h e s e a r e m o s tly d u e e ith e r to r e fl e c tio n o r a c tu a l v o id s w ith in a b u b b le . W e
c o u ld a tte m p t to b u ild a m o r p h o lo g ic a l p r o c e d u r e to r e p a ir a n d / o r fi ll th e b u b b le s . H o w e v e r, th is c a n tu r n in to a c o m p u ta tio n a lly e x p e n s iv e p r o c e s s th a t is
n o t w a r r a n te d u n le s s s tr in g e n t m e a s u r e m e n t s ta n d a r d s a r e r e q u ir e d , a fa c t n o t
m e n tio n e d in th e p r o b le m s ta te m e n t. A n a lte r n a tiv e is to c a lc u la te , o n a v e r a g e
(a s d e te r m in e d fr o m a s e t o f s a m p le im a g e s ), th e p e r c e n ta g e o f b u b b le a r e a s th a t
a r e fi lle d w ith b la c k o r h a v e b la c k ” b a y s ” w h ic h m a k e s th e ir b la c k a r e a s m e r g e
w ith th e b a c k g r o u n d . T h e n , o n c e th e d im e n s io n s o f e a c h b u b b le (o r b u b b le
c lu s te r ) h a v e b e e n e s ta b lis h e d , a c o r r e c tio n fa c to r b a s e d o n a r e a w o u ld b e a p p lie d .
( 3 ) F in d in g t h e b u b b le s . R e fe r to th e s o lu tio n t o P r o b le m 9 .3 6 . T h e s o lu tio n
is b a s e d o n c o n n e c te d c o m p o n e n ts , w h ic h a ls o y ie ld s a ll b u b b le s a n d b u b b le
c lu s te r s .
2 4 4
C H A P T E R 1 1 . P R O B L E M S O L U T IO N S
F ig u r e P 1 1 .2 7
(4 ) In o r d e r to d e te c t b u b b le c lu s te r s w e m a k e u s e o f s h a p e a n a ly s is . F o r e a c h
c o n n e c t e d c o m p o n e n t , w e fi n d t h e e ig e n a x e s ( s e e S e c t io n 1 1 .4 ) a n d t h e s t a n d a r d d e v ia tio n o f th e d a ta a lo n g th e s e a x e s (s q u a r e r o o t o f th e e ig e n v a lu e s o f th e
c o v a r ia n c e m a tr ix ). O n e s im p le s o lu tio n is to c o m p u te th e r a tio o f th e la r g e to
th e s m a ll v a r ia n c e o f e a c h c o n n e c te d c o m p o n e n t a lo n g th e e ig e n a x e s . A s in g le ,
u n ifo r m ly - fi lle d , p e r fe c tly r o u n d b u b b le w ill h a v e a r a tio o f 1 . D e v ia tio n s fr o m 1
in d ic a te e lo n g a tio n s a b o u t o n e o f th e a x e s . W e lo o k fo r e llip tic a l s h a p e s a s b e in g
fo r m e d b y c lu s te r s o f b u b b le s . A th r e s h o ld to c la s s ify b u b b le s a s s in g le v s . c lu s te r s h a s to b e d e te r m in e d e x p e r im e n ta lly . N o te th a t s in g le p ix e ls o r p ix e l s tr e a k s
o n e p ix e l w id e h a v e a s ta n d a rd d e v ia tio n o f z e r o , s o th e y m u s t b e p r o c e s s e d s e p a r a te ly . W e h a v e th e o p tio n o f c o n s id e r in g c o n n e c te d c o m p o n e n ts th a t c o n s is t
o f o n ly o n e p ix e l to b e e ith e r n o is e , o r th e s m a lle s t d e te c ta b le b u b b le . N o in fo r m a tio n is g iv e n in th e p r o b le m s ta te m e n t a b o u t th is . In th e o r y , it is p o s s ib le
fo r a c lu s te r to b e fo r m e d s u c h th a t its s h a p e w o u ld b e s y m m e tr ic a l a b o u t b o th
a x e s , in w h ic h c a s e th e s y s te m w o u ld c la s s ify th e c lu s te r a s a s in g le b u b b le . R e s o lu tio n o f c o n fl ic ts s u c h a s th is w o u ld r e q u ir e a d d itio n a l p r o c e s s in g . H o w e v e r,
th e r e is n o e v id e n c e in th e s a m p le im a g e to s u g g e s t th a t th is in fa c t is a p r o b le m .
B u b b le c lu s te r s te n d to a p p e a r a s e llip tic a l s h a p e s . In c a s e s w h e r e th e r a tio o f
th e s ta n d a r d d e v ia tio n s is c lo s e to th e th r e s h o ld v a lu e , w e c o u ld a d d a d d itio n a l
p r o c e s s in g to re d u c e th e c h a n c e s o f m a k in g a m is ta k e .
(5 ) C o u n tin g in d iv id u a l b u b b le s . A b u b b le th a t d o e s n o t m e r g e w ith th e b o r d e r o f th e im a g e o r is n o t a c lu s te r, is b y d e fi n itio n a s in g le b u b b le . T h u s , c o u n tin g th e s e b u b b le s is s im p ly c o u n tin g th e c o n n e c te d c o m p o n e n ts th a t h a v e n o t
b e e n ta g g e d a s c lu s te r s o r m e r g e d w ith th e b o u n d a r y o f th e im a g e .
(6 ) R a tio o f th e a r e a s . T h is r a tio is s im p ly th e n u m b e r o f p ix e ls in a ll th e c o n n e c te d c o m p o n e n ts p lu s th e c o r r e c tio n fa c to r s m e n tio n e d in (2 ), d iv id e d b y th e
to ta l n u m b e r o f p ix e ls in th e im a g e .
2 4 5
t
a
o
o
T h e p r
e c t. If, a s
b u b b le ,
n e p ix e l.
f o n e p ix
o b le m a ls o a
m e n tio n e d
th e n th e s m
F ro m th e p r
e l is 1 0 m m .
s k s fo r
in (4 ),
a lle s t b
o b le m
th e s iz e o f th e s m a lle s t
w e e le c t to c a ll a o n e - p
u b b le d im e n s io n d e te c
s ta te m e n t, 7 0 0 p ix e ls c o
b u b b le th e s y s te m c a n d e ix e l c o n n e c te d c o m p o n e n t
ta b le is th e p h y s ic a l s iz e o f
v e r 7 c m , s o th e d im e n s io n
Chapter 12
Problem Solutions
Problem 12.1
(a ) B y in s p
(1 .5 , 0 .3 )T ,
c o lo r, a n d
E q . ( 1 2 .2 - 5
e c tio n , th
m 2 = (4 .3
v ir g in ic a ,
). S u b s tit
e m e a
, 1 .3 )T
re s p e
u tin g
n v e c to rs o f th e
, a n d m 3 = (5 .5
c tiv e ly . T h e d e c
th e p re c e d in g v
d
1
(x )
=
x
T
d
2
(x )
=
x
T
d
3
(x )
=
x
T
1
−
m
2
−
m
3
−
1
m
2
1
2
1
2
m
T
m
T
m
T
1
3
2
th r e e c la s
, 2 .1 )T fo r
is io n fu n c
a lu e s o f m
s e
th
tio
e a
s a re
e c la
n s a
n v e
, a
s s
re
c t
p p
e s
o f
o rs
ro x
Ir is
th e
g iv
m
1
= 1 .5 x
1
+ 0 .3 x
2
− 1 .2
m
2
= 4 .3 x
1
+ 1 .3 x
2
− 1 0 .1
m
3
= 5 .5 x
1
+ 2 .1 x
2
− 1 7 .3 .
(b ) T h e d e c is io n b o u n d a r ie s a re g iv e n b y th e e q u a tio n s
d
1 2
(x )
=
d
1
(x ) − d
2
(x ) = − 2 .8 x
1
− 1 .0 x
2
+ 8 .9 = 0
d
1 3
(x )
=
d
1
(x ) − d
3
(x ) = − 4 .0 x
1
− 1 .8 x
2
+ 1 6 .1 = 0
d
2 3
(x )
=
d
2
(x ) − d
3
(x ) = − 1 .2 x
1
− 0 .8 x
2
+ 7 .2 = 0 .
F ig u r e P 1 2 .1 s h o w s a p lo t o f t h e s e b o u n d a r ie s .
Problem 12.2
F r o m th e d e fi n itio n o f th e E u c lid e a n d is ta n c e ,
D
j
@
@
( x ) = @x − m j @=
(x − m j )T (x − m j ) .1
2 4 7
/ 2
im
s e
fo
e s
a te ly , m 1 =
to s a , v e r s ir m g iv e n in
:
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
2 4 8
F ig u r e P 1 2 .1
B e c a u s e D j (x ) is n o n - n e g a tiv e , c h o o s in g th e s m a lle s t D
i n g t h e s m a l l e s t D 2j ( x ) , w h e r e
D
2
j
(x )
(x ) is th e s a m e a s c h o o s j
@
@
@x − m j @2 = ( x − m j ) T ( x − m j )
=
=
x
=
x
T
x − 2 x
T
x − 2
T
x
m
j
T
m
+ m
j
T
j
1
−
2
m
m
j
T
j
m
.
j
W e n o te th a t th e te r m x T x is in d e p e n d e n t o f j (th a t is , it is a c o n s ta n t w ith re s p e c t t o j i n D 2j ( x ) , j = 1 , 2 , . . . ) . T h u s , c h o o s i n g t h e m i n i m u m o f D 2j ( x ) i s e q u i v a "
#
l e n t t o c h o o s i n g t h e m a x i m u m o f x T m j − 12 m Tj m j .
Problem 12.3
T h e e q u a tio n o f th e d e c is io n b o u n d a r y b e tw e e n a p a ir o f m e a n v e c to r s is
d
i j
(x ) = x
T
(m
i
− m j ) −
1
2
(m
i
T
m
i
− m
T
j
m j ).
T h e m id p o in t b e t w e e n m i a n d m j is (m i + m j )/ 2 ( s e e F ig . P 1 2 .3 ) . F ir s t , w e s h o w
th a t th is p o in t is o n th e b o u n d a r y b y s u b s titu tin g it fo r x in th e a b o v e e q u a tio n
2 4 9
F ig u r e P 1 2 .3
a n d s h o w in g th a t th e r e s u lt is e q u a l to 0 :
1
2
(m
i
T
m
i
− m
T
j
m j ) −
2
1
(m
T
i
m
T
j
− m
i
m j )
1
=
2
1
−
=
N e x t, w e s h
b o u n d a r y . T
(m i − m j ) is
h y p e r p la n e
o w th a
h e re a r
in th e
w ith e q
t th e
e s e v
s a m e
u a tio
v e c
e ra l
d ir
n w
to r (m i
w a y s to
e c tio n a
1 x 1 + w
− m j ) is p e r
d o th is . P e r h
s th e u n it n o
2 x 2 + ...w n x n
u =
w
k w
T
(m
2
i
(m
m
i
− m
T
m
i
i
T
j
− m
m j )
T
j
m j )
0 .
p e n d ic u
a p s th e
r m a l to
+ w n + 1
la r to
e a s ie s
th e h y
= 0 , th
th e h y p e rp
t is to s h o w
p e r p la n e . F
e u n it n o r m
la n e
th a t
o r a
a l is
o
o
k
w h e r e w o = (w 1 , w 2 , ..., w n )T . C o m p a r in g t h e a b o v e e q u a t io n fo r d i j (x ) w it h t h e
g e n e r a l e q u a tio n o f a h y p e r p la n e ju s t g iv e n , w e s e e th a t w o = (m i − m j ) a n d
w n + 1 = − ( m Ti m i − m Tj m j ) / 2 . T h u s , t h e u n i t n o r m a l o f o u r d e c i s i o n b o u n d a r y i s
u =
(m
@
@m
− m j)
i
i
@
− m j@
w h ic h is in th e s a m e d ire c tio n a s th e v e c to r (m
i
− m j ). T h is c o n c lu d e s th e p r o o f.
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
2 5 0
F ig u r e P 1 2 .4
Problem 12.4
T h e s o l u t i o n i s s h o w n i n F i g . P 1 2 . 4 , w h e r e t h e x ’s a r e t r e a t e d a s v o l t a g e s a n d
t h e Y ’s d e n o t e i m p e d a n c e s . F r o m b a s i c c i r c u i t t h e o r y , t h e c u r r e n t s , I ’s , a r e t h e
p r o d u c ts o f th e v o lta g e s tim e s th e im p e d a n c e s .
Problem 12.5
A s s u m e th a t th e m a s k is o f s iz e J × K . F o r a n y v a lu e o f d is p la c e m e n t (s , t ), w e
c a n e x p r e s s th e a r e a o f th e im a g e u n d e r th e m a s k , a s w e ll a s th e m a s k w (x , y ),
in v e c to r fo r m b y le ttin g th e fi r s t r o w o f th e s u b im a g e u n d e r th e m a s k r e p r e s e n t
th e fi r s t K e le m e n ts o f a c o lu m n v e c to r a , th e e le m e n ts o f n e x t r o w th e n e x t K
e le m e n ts o f a , a n d s o o n . A t th e e n d o f th e p r o c e d u r e w e s u b tr a c t th e a v e r a g e
v a lu e o f th e in te n s ity le v e ls in th e s u b im a g e fr o m e v e r y e le m e n t o f a . T h e v e c to r
a is o f s iz e ( J × K ) × 1 . A s im ila r a p p r o a c h y ie ld s a v e c to r, b , o f th e s a m e s iz e , fo r
th e m a s k w (x ,y ) m in u s its a v e r a g e . T h is v e c to r d o e s n o t c h a n g e a s (s ,t ) v a r ie s
b e c a u s e th e c o e ffi c ie n ts o f th e m a s k a re fi x e d . W ith th is c o n s tr u c tio n in m in d ,
w e s e e t h a t t h e n u m e r a t o r o f E q . ( 1 2 .2 - 8 ) is s im p ly t h e v e c t o r in n e r - p r o d u c t
a T b . S im ila r ly , th e fi r s t te r m in th e d e n o m in a to r is th e n o r m s q u a r e d o f a , d e n o te d a T a = k a k 2 , w h ile th e s e c o n d te r m h a s a s im ila r in te r p r e ta tio n fo r b . T h e
c o r r e la tio n c o e ffi c ie n t th e n b e c o m e s
T
b
a )(b
T
a
γ (s ,t ) =
(a
T
b )
1 / 2
.
W h e n a = b (a p e r fe c t m a tc h ), γ (s ,t ) = k a k 2 / k a k k a k = 1 , w h ic h is th e m a x im u m
v a lu e o b ta in a b le b y th e a b o v e e x p r e s s io n . S im ila r ly , th e m in im u m v a lu e o c c u r s
2 5 1
w h e n a = − b , in w h ic h c a s e γ (s , t ) = − 1 . T h u s , a lth o u g h th e v e c to r a v a r ie s in
g e n e r a l fo r e v e r y v a lu e o f (s , t ), th e v a lu e s o f γ (s , t ) a r e a ll in th e r a n g e [− 1 , 1 ].
Problem 12.6
T h e s o lu tio n to th e fi r s t p a r t o f th is p r o b le m is b a s e d o n b e in g a b le to e x tr a c t
c o n n e c te d c o m p o n e n ts (s e e C h a p te r s 2 a n d 1 1 ) a n d th e n d e te r m in in g w h e th e r
a c o n n e c te d c o m p o n e n t is c o n v e x o r n o t (s e e C h a p te r 1 1 ). O n c e a ll c o n n e c te d
c o m p o n e n ts h a v e b e e n e x tr a c te d w e p e r fo r m a c o n v e x ity c h e c k o n e a c h a n d
r e je c t th e o n e s th a t a r e n o t c o n v e x . A ll th a t is le ft a fte r th is is to d e te r m in e if th e
r e m a in in g b lo b s a r e c o m p le te o r in c o m p le te . T o d o th is , th e r e g io n c o n s is tin g
o f t h e e x t r e m e r o w s a n d c o l u m n s o f t h e i m a g e i s d e c l a r e d a r e g i o n o f 1 ’s . T h e n
if th e p ix e l- b y - p ix e l A N D o f th is r e g io n w ith a p a r tic u la r b lo b y ie ld s a t le a s t o n e
r e s u lt th a t is a 1 , it fo llo w s th a t th e a c tu a l b o u n d a r y to u c h e s th a t b lo b , a n d th e
b lo b is c a lle d in c o m p le te . W h e n o n ly a s in g le p ix e l in a b lo b y ie ld s a n A N D
o f 1 w e h a v e a m a r g in a l r e s u lt in w h ic h o n ly o n e p ix e l in a b lo b to u c h e s th e
b o u n d a r y . W e c a n a r b itr a r ily d e c la r e th e b lo b in c o m p le te o r n o t. F r o m th e p o in t
o f v ie w o f im p le m e n ta tio n , it is m u c h s im p le r to h a v e a p r o c e d u r e th a t c a lls a
b lo b in c o m p le te w h e n e v e r th e A N D o p e r a tio n y ie ld s o n e o r m o r e r e s u lts v a lu e d
1 . A fte r th e b lo b s h a v e b e e n s c r e e n e d u s in g th e m e th o d ju s t d is c u s s e d , th e y
n e e d to b e c la s s ifi e d in to o n e o f th e th r e e c la s s e s g iv e n in th e p r o b le m s ta te m e n t. W e p e r fo r m th e c la s s ifi c a tio n p r o b le m b a s e d o n v e c to r s o f th e fo r m x =
(x 1 , x 2 )T , w h e r e x 1 a n d x 2 a r e , r e s p e c tiv e ly , th e le n g th s o f th e m a jo r a n d m in o r
a x is o f a n e llip tic a l b lo b , th e o n ly ty p e le ft a fte r s c r e e n in g . A lte r n a tiv e ly , w e c o u ld
u s e t h e e ig e n a x e s fo r t h e s a m e p u r p o s e . ( S e e S e c t io n 1 1 .2 .1 o n o b t a in in g t h e
m a jo r a x e s o r t h e e n d o f S e c tio n 1 1 .4 r e g a r d in g th e e ig e n a x e s .) T h e m e a n v e c to r
o f e a c h c la s s n e e d e d to im p le m e n t a m in im u m d is ta n c e c la s s ifi e r is g iv e n in th e
p r o b le m s ta te m e n t a s th e a v e r a g e le n g th o f e a c h o f th e tw o a x e s fo r e a c h c la s s
o f b lo b . If‘ th e y w e r e n o t g iv e n , th e y c o u ld b e o b ta in e d b y m e a s u r in g th e le n g th
o f th e a x e s fo r c o m p le te e llip s e s th a t h a v e b e e n c la s s ifi e d a p r io r i a s b e lo n g in g
to e a c h o f th e th r e e c la s s e s . T h e g iv e n s e t o f e llip s e s w o u ld th u s c o n s titu te a
tr a in in g s e t, a n d le a r n in g w o u ld c o n s is t o f c o m p u tin g th e p r in c ip a l a x e s fo r a ll
e llip s e s o f o n e c la s s a n d th e n o b ta in in g th e a v e r a g e . T h is w o u ld b e r e p e a te d fo r
e a c h c la s s . A b lo c k d ia g r a m o u tlin in g th e s o lu tio n to th is p r o b le m is s tr a ig h tfo r w a rd .
Problem 12.7
( a ) B e c a u s e it is g iv e n th a t th e p a tte r n c la s s e s a r e g o v e r n e d b y G a u s s ia n d e n s itie s , o n ly k n o w le d g e o f th e m e a n v e c to r a n d c o v a r ia n c e m a tr ix o f e a c h c la s s a r e
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
2 5 2
x2
6
+
d(x) =0
x1
6
F ig u r e P 1 2 .7
r e q u ir e d to s p e c ify th e B a y e s c la s s ifi e r. S u b s titu tin g th e g iv e n p a tte r n s in to E q s .
( 1 2 .2 - 2 2 ) a n d ( 1 2 .2 - 2 3 ) y ie ld s
1
m 1 =
1
m
5
=
2
C
5
=
1
1
0
0
1
− 1
= C
1
a n d
C
B e c a u s e C 1 = C 2 =
d is ta n c e c la s s ifi e r :
1
0
0
1
− 1
= C
2
.
I , th e d e c is io n fu n c tio n s a re th e s a m e a s fo r a m in im u m
d
1
T
(x ) = x
m
2
(x ) = x
T
m
2
1
−
1
a n d
d
=
2
−
1
2
T
m
2
m
1
2
T
m
m
2
1
= 1 .0 x
= 5 .0 x
1
1
+ 1 .0 x
+ 5 .0 x
2
2
− 1 .0
− 2 5 .0 .
T h e B a y e s d e c is io n b o u n d a r y is g iv e n b y th e e q u a tio n d (x ) = d
o r
d (x ) = − 4 .0 x 1 − 4 .0 x 2 + 2 4 .0 = 0 .
( b ) F ig u r e P 1 2 .7 s h o w s a p lo t o f t h e b o u n d a r y .
1
(x ) − d
2
(x ) = 0 ,
2 5 3
Problem 12.8
( a ) A s in P r o b le m 1 2 .7 ,
m
C
1
=
1
1
0
0
2
1
0
1
=
m
1
=
0
m
;
0
=
1
0
0
C
− 1
0
= 2
1
1
0
0
1
jC 1 j = 0 .2 5
;
a n d
C
= 2
2
1
0
0
;
1
− 1
C
2
=
1
2
1
0
0
1
jC 2 j = 4 .0 0 .
;
B e c a u s e t h e c o v a r ia n c e m a t r ic e s a r e n o t e q u a l, it fo llo w s fr o m E q . ( 1 2 .2 - 2 6 ) t h a t
A
1
1
2
0
T
x
x
d 1 (x ) =
−
ln (0 .2 5 ) −
0
2
2
2
=
1
−
2
2
ln (0 .2 5 ) − (x
2
+ x
1
)
2
a n d
d
2
(x )
=
1
−
=
ln (4 .0 0 ) −
2
−
1
2
A
1
ln (4 .0 0 ) −
x
2
1
4
(x
0 .5
0
T
2
1
2
+ x
2
0
0 .5
x
)
w h e r e th e te r m ln P (! j ) w a s n o t in c lu d e d b e c a u s e it is th e s a m e fo r b o th d e c is io n fu n c tio n s in th is c a s e . T h e e q u a tio n o f th e B a y e s d e c is io n b o u n d a r y is
d (x ) = d
1
(x ) − d
2
(x ) = 1 .3 9 −
4
3
(x
2
1
+ x
2
2
) = 0 .
( b ) F ig u r e P 1 2 .8 s h o w s a p lo t o f t h e b o u n d a r y .
Problem 12.9
T h e b a s ic m e c h a n ic s a r e t h e s a m e a s in P r o b le m 1 2 .6 , b u t w e h a v e t h e a d d itio n a l re q u ire m e n t o f c o m p u tin g c o v a r ia n c e m a tr ic e s fr o m th e tr a in in g p a tte r n s
o f e a c h c la s s .
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
2 5 4
x2
d(x) =0
+
x1
1.36
F ig u r e P 1 2 .8
Problem 12.10
F r o m b a s ic p r o b a b ility th e o r y ,
p (c / x )p (x ).
p (c ) =
x
F o r a n y p a tte r n b e lo n g in g to c la s s !
j
, p (c / x ) = p (!
p (!
p (c ) =
j
j
/ x ). T h e re fo re ,
/ x )p (x ).
x
S u b s titu tin g in to th is e q u a tio n th e fo r m u la p (!
p (x / !
p (c ) =
j
j
/ x ) = p (x / !
)p (!
j
j
)p (!
j
)/ p (x ) g iv e s
).
x
B e c
m iz
fo r
th e
th e
a u s e
in g p
j = 1
c la s s
p ro b
th
(x
,2
fr
a b
e a rg u m e n t o f
/ ! j )p (! j ) fo r
, ..., W , a n d u s
o m w h ic h x c a
ility o f e r r o r is
th e s u m m
e a c h j . T h
e th e la r g e
m e , th e n p
m in im iz e
io n is p o s itiv e , p (c
is , if fo r e a c h x w e
v a lu e e a c h tim e a
) w ill b e m a x im iz e
d b y th is p r o c e d u re .
a t
a t
s t
(c
) is m a x im iz
c o m p u te p (
s th e b a s is f
d . S in c e p (e
e d b y m a x
x / ! j )p (!
o r s e le c tin
) = 1 − p (c
i)
j
g
),
Problem 12.11
( a ) F o r c la s s ! 1 w e le t y (1 ) = (0 , 0 , 0 , 1 )T , y (2 ) = (1 , 0 , 0 , 1 )T , y (3 ) = (1 , 0 , 1 , 1 )T ,
y (4 ) = (1 , 1 , 0 , 1 )T . S im ila r ly , fo r c la s s ! 2 , y (5 ) = (0 , 0 , 1 , 1 )T , y (6 ) = (0 , 1 , 1 , 1 )T ,
y (7 ) = (0 ,1 ,0 ,1 )T , y (8 ) = (1 ,1 ,1 ,1 )T . T h e n , u s in g c = 1 a n d
w (1 ) = (− 1 ,− 2 ,− 2 ,0 )
T
2 5 5
it fo llo w s fr o m E q s . ( 1 2 .2 - 3 4 ) t h r o u g h ( 1 2 .2 - 3 6 ) t h a t :
w
w
w
w
w
w
w
w
(1
(2
(3
(4
(5
(6
(7
(8
)T y (
)T y (
)T y (
)T y (
)T y (
)T y (
)T y (
)T y (
1 ) =
2 ) =
3 ) =
4 ) =
5 ) =
6 ) =
7 ) =
8 ) =
0 ,
w
w
w
w
w
w
w
w
0 ,
0 ,
2 ,
2 ,
− 2 ,
0 ,
− 3 ,
(2 )
(3 )
(4 )
(5 )
(6 )
(7 )
(8 )
(9 )
= w
= w
= w
= w
= w
= w
= w
= w
(1 )
(2 )
(3 )
(4 )
(5 )
(6 )
(7 )
(8 )
+ y (
+ y (
+ y (
= (1
− y (
= (−
− y (
= (1
1 ) =
2 ) =
3 ) =
,− 2
5 ) =
1 ,−
7 ) =
,− 3
(− 1 ,− 2 ,−
(0 ,− 2 ,− 2
(1 ,− 2 ,− 1
,− 1 ,3 )T ;
(− 1 ,− 2 ,−
2 ,− 2 ,2 )T
(1 ,− 3 ,− 2
,− 2 ,1 )T .
2 ,1 )T ;
,2 )T ;
,3 )T ;
2 ,2 )T ;
;
,1 )T ;
A c o m p le te ite r a tio n th r o u g h a ll p a tte r n s w ith n o e r r o r s w a s n o t a c h ie v e d . T h e r e fo r e , th e p a tte r n s a r e r e c y c le d b y le ttin g y (9 ) = y (1 ), y (1 0 ) = y (2 ), a n d s o o n , w h ic h
g iv e s
w
w
w
w
w
w
w
w
(9
(1
(1
(1
(1
(1
(1
(1
)T y
0 )T
1 )T
2 )T
3 )T
4 )T
5 )T
6 )T
A g a in , a c o m p le
s o th e p a tte r n s a
g iv e s :
w (1 7
w (1 8
w (1 9
w (2 0
w (2 1
(9
y (
y (
y (
y (
y (
y (
y (
) =
1 0
1 1
1 2
1 3
1 4
1 5
1 6
)
)
)
)
)
)
)
1 ,
=
=
=
=
=
=
=
w
w
w
w
w
w
w
w
2 ,
0 ,
1 ,
1 ,
− 4 ,
− 2 ,
− 2 ,
(1
(1
(1
(1
(1
(1
(1
(1
0 )
1 )
2 )
3 )
4 )
5 )
6 )
7 )
= w
= w
= w
= w
= w
= w
= w
= w
(9
(1
(1
(1
(1
(1
(1
(1
) =
0 )
1 )
2 )
3 )
4 )
5 )
6 )
(1 ,−
(1 ,
y (1
(2 ,
y (1
(2 ,
(2 ,
(2 ,
=
+
=
−
=
=
=
3 ,−
− 3 ,
1 ) =
− 3 ,
3 ) =
− 3 ,
− 3 ,
− 3 ,
2 ,1
− 2 ,
(2 ,
− 1 ,
(2 ,
− 2 ,
− 2 ,
− 2 ,
)T ;
1 )T
− 3
2 )T
− 3
1 )T
1 )T
1 )T
;
,− 1 ,2 )T ;
;
,− 2 ,1 )T ;
;
;
.
te ite r a tio n o v e r a ll p a tte r n s w ith o u t a n e r r o r w a s n o t a c h ie v e d ,
r e r e c y c le d b y le ttin g y (1 7 ) = y (1 ), y (1 8 ) = y (2 ), a n d s o o n , w h ic h
)T y
)T y
)T y
)T y
)T y
(1 7
(1 8
(1 9
(2 0
(2 1
) = 1
) = 3
) = 1
) = 0
) = 0
,
,
,
,
,
w
w
w
w
w
(1 8
(1 9
(2 0
(2 1
(2 2
) = w
) = w
) = w
) = w
) = w
(1 7
(1 8
(1 9
(2 0
(2 1
) = (
) = (
) = (
) + y
) − y
2 ,−
2 ,−
2 ,−
(2 0
(2 1
3 ,−
3 ,−
3 ,−
) =
) =
2 ,1
2 ,1
2 ,1
(3 ,−
(3 ,−
)T ;
)T ;
)T ;
2 ,− 2 ,2 )T ;
2 ,− 3 ,1 )T .
It is e a s ily v e r ifi e d th a t n o m o r e c o r r e c tio n s ta k e p la c e a fte r th is s te p , s o w (2 2 ) =
(3 , − 2 , − 3 , 1 )T is a s o lu tio n w e ig h t v e c to r.
(b ) T h e d e c is io n s u r fa c e is g iv e n b y th e e q u a tio n
w
T
y = 3 y
1
− 2 y
2
− 3 y
3
+ 1 = 0 .
A s e c t io n o f t h is s u r fa c e is s h o w n s c h e m a t ic a lly in F ig . P 1 2 .1 1 . T h e p o s it iv e s id e
o f th e s u r fa c e fa c e s th e o r ig in .
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
2 5 6
F ig u r e P 1 2 .1 1
Problem 12.12
W e s ta r t b y ta k in g th e p a r tia l d e r iv a tiv e o f J w ith re s p e c t to w :
@ J
=
@ w
1
2
y s g n (w
T
y ) − y
w h e re , b y d e fi n itio n , s g n (w T y ) = 1 if w T y > 0 , a n d s g n (w T y ) = − 1 o th e r w is e .
S u b s titu tin g th e p a r tia l d e r iv a tiv e in to th e g e n e r a l e x p re s s io n g iv e n in th e p r o b le m s ta te m e n t g iv e s
4
c 3
y (k ) − y (k )s g n w (k )T y (k )
w (k + 1 ) = w (k ) +
2
w h e re y (k ) is th e tr a in in g p a tte r n b e in g c o n s id e re d a t th e k th ite r a tiv e s te p . S u b s titu tin g th e d e fi n itio n o f th e s g n fu n c tio n in to th is r e s u lt y ie ld s
w (k + 1 ) = w (k ) + c
0
y (k )
if w (k )T y (k )
o th e r w is e
w h e r e c > 0 a n d w (1 ) is a r b itr a r y . T h is e x p r e s s io n a g r e e s w ith th e fo r m u la tio n
g iv e n in th e p r o b le m s ta te m e n t.
Problem 12.13
L e t th e tr a in in g s e t o f p a tte r n s b e d e n o te d b y y 1 ,y 2 ,... ,y N . It is a s s u m e d th a t th e
tr a in in g p a tte r n s o f c la s s ! 2 h a v e b e e n m u ltip lie d b y − 1 . If th e c la s s e s a r e lin e a r ly s e p a r a b le , w e w a n t to p r o v e th a t th e p e r c e p tr o n tr a in in g a lg o r ith m y ie ld s
2 5 7
a s o lu tio n w e ig h t v e c to r, w ∗ , w ith th e p r o p e r ty
∗ T
w
w h e re T 0 is a n o n
r ith m (w ith c = 1 )
w (k ) + y i(k ) o th e r
S u p p o s e th a t w
(th e s e a r e th e o n ly
m a y w r ite
n
is
w
e
e g a t
e x p
is e .
re ta
in d ic
y
i
≥ T
0
iv e th r e s h o ld . W ith th is n o ta tio n , th e P e r c e p tr o n a lg o re s s e d a s w (k + 1 ) = w (k ) if w T (k )y i (k ) ≥ T 0 o r w (k + 1 ) =
in o n ly th e v a lu e s o f k fo r w h ic h a c o r r e c tio n ta k e s p la c e
e s o f in te re s t). T h e n , re - a d a p tin g th e in d e x n o ta tio n , w e
w (k + 1 ) = w (k ) + y i(k )
a n d
T
w
(k )y i (k ) ≤ T 0 .
W ith th e s e s im p lifi c a tio n s in m in d , th e p r o o f o f c o n v e r g e n c e is a s fo llo w s : F r o m
th e a b o v e e q u a tio n ,
w (k + 1 ) = w (1 ) + y i(1 ) + y i(2 ) + ··· + y i(k ).
T a k in g th e in n e r p r o d u c t o f th e s o lu tio n w e ig h t v e c to r w ith b o th s id e s o f th is
e q u a tio n g iv e s
T
w
E a c h te r m y
i
T
(k + 1 )w
∗
T
= w
(1 )w
∗
T
+ y
∗
(1 )w
i
T
+ y
∗
(2 )w
i
T
i
(k )w ∗ .
( j )w ∗ , j = 1 , 2 , ..., k , is le s s t h a n T 0 , s o
w
T
∗
(k + 1 )w
T
≥ w
(1 )w
2
U s in g th e C a u c h y - S c h w a r tz in e q u a lity , k a k
T
w
(k + 1 )w
∗
2
k b k
∗
+ k T 0 .
2
@
@2 @ @2
≤ @w T ( k + 1 ) @ @w ∗ @
@
@
@w T ( k + 1 ) @2 ≥
A n o th e r lin e o f r e a s o
F ro m a b o v e ,
@
@w ( j + 1
b )2 , r e s u lts in
T
≥ (a
o r
o r
+ ··· + y
w
T
(k + 1 )w
k w ∗k
2
∗
2
.
@
@2
n i n g l e a d s t o a c o n t r a d i c t i o n r e g a r d i n g @w T ( k + 1 ) @.
@ @2
@2
) @ = @w ( j ) @ + 2 w
@
@ @ @
@w ( j + 1 ) @2 − @w ( j ) @2 = 2 w
T
T
@
@2
( j ) y i ( j ) + @y i ( j ) @
@
@2
( j ) y i ( j ) + @y i ( j ) @ .
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
2 5 8
L e t Q = m a x jjy i ( j ) jj2 . T h e n , b e c a u s e w
T
(j )y i(j ) ≤
i
T 0 ,
@
@ @ @
@w ( j + 1 ) @2 − @w ( j ) @2 ≤ 2 T
+ Q .
0
A d d in g th e s e in e q u a litie s fo r j = 1 , 2 , . . . , k y ie ld s
@
@
@w ( j + 1 ) @2 ≤ k w ( 1 ) k
2
+ [2 T
@
T h i s i n e q u a l i t y e s t a b l i s h e s a b o u n d o n @w ( j + 1 )
la r g e k w ith th e b o u n d e s ta b lis h e d b y o u r e a r lie r
la r g e r th a n k m , w h ic h is a s o lu tio n to th e e q u a tio
w
T
∗
(k + 1 )w
+ k
T h is e q u a tio n s a y s th a t k m is
a lg o r ith m c o n v e r g e s in a fi n ite
th e p a tte r n s o f th e tr a in in g s e t
N o te : T h e s p e c ia l c a s e w ith
U n d e r th is c o n d itio n w e h a v e
T
T
@2
@t h a t c o n fl i c t s f o r s u f fi c i e n t l y
in e q u a lity . In fa c t, k c a n b e n o
n
2
0
2
= k w (1 )k
2
k w ∗k
w
m
+ Q ] k .
0
fi n ite ,
n u m b
a r e lin
T 0 =
th u s p r o v in
e r o f s te p s t
e a r ly s e p a r a
0 is p r o v e d
∗
(k + 1 )w
T
≥ w
+ [2 T
+ Q ] k
m
.
g th a t th e p e rc e p tr o n tr a in in g
o a s o lu tio n w e ig h t v e c to r w ∗ if
b le .
in a s lig h tly d iffe r e n t m a n n e r.
∗
(1 )w
0
+ k a
w h e re
a = m in
y
i
T
i
∗
(j )w
.
B e c a u s e , b y h y p o th e s is , w ∗ is a s o lu tio n w e ig h t v e c to r, w e k n o w th a t
0 . A ls o , b e c a u s e w T ( j )y i ( j ) ≤ (T = 0 ),
@
@ @ @
@w ( j + 1 ) @2 − @w ( j ) @2 ≤
y
i
T
(j )w
∗
≥
@
@
@y i ( j ) @2
≤
Q .
T h e re s t o f th e p r o o f re m a in s th e s a m e . T h e b o u n d o n th e n u m b e r o f s te p s is th e
v a lu e o f k m th a t s a tis fi e s th e fo llo w in g e q u a tio n :
w
T
(1 )w
∗
+ k
k w ∗k
2
m
a
2
= k w (1 )k
2
+ Q k
m
.
2 5 9
Problem 12.14
T h e s in g le d e c is io n fu n c tio n th a t im p le m e n ts a m in im u m d is ta n c e c la s s ifi e r fo r
tw o c la s s e s is o f th e fo r m
1
d ij (x ) = x T (m i − m j) −
( m Ti m i − m Tj m j ) .
2
T h u s , fo
a n d , w h
a re o n t
(m i − m
fu n c tio n
r a p a r tic u la r
e n d ij(x ) < 0
h e b o u n d a r y
j ) a n d w n + 1
in th e fo r m
p a tte r n v e c to
, x is a s s ig n e d
(h y p e r p la n e )
= − 12 ( m Ti m i −
r x , w h e n d ij (x ) >
to c la s s ! 2 . V a lu
s e p a r a tin g th e t
m Tj m j ) , w e c a n
T
d (x ) = w
x − w
n + 1
0 , x is
e s o f x
w o c la
e x p re s
a s s
fo r
s s e s
s th
ig
w
.
e
n e d
h ic
B y
a b o
to c la s s ! 1
h d ij (x ) = 0
le ttin g w =
v e d e c is io n
.
T h is is r e c o g n iz e d a s a lin e a r d e c is io n fu n c tio n in n d im e n s io n s , w h ic h is im p le m e n te d b y a s in g le la y e r n e u r a l n e tw o r k w ith c o e ffi c ie n ts
w
k
= (m
i k
− m
n + 1
= −
a n d
θ = w
)
j k
1
k = 1 ,2 ,... ,n
T
(m
2
m
i
i
T
j
− m
m j ).
Problem 12.15
T h e
T h e
th e n
fo r t
a p p r
id e a
e q u
w o p
d
o a c
is t
a te
a tte
i j
h to
o c o
c o e
r n c
(x )
s o lv in g th is p r
m b in e th e d e c
ffi c ie n ts . F o r e
la s s e s is o b ta in
=
d
−
(x ) = ln P (! i ) − ln P (!
(x ) − d
i
1
2
(m
i
o b le m is b a s ic a lly th e s a m e a s in P r o b le m 1 2 .1 4 .
is io n fu n c tio n s in th e fo r m o f a h y p e r p la n e a n d
q u a l c o v a r ia n c e m a tr ic e s , th e d e c is io n fu n c tio n
e d E q . ( 1 2 .2 - 2 7 ) :
j
− m j )T C
− 1
(m
j
) + x
T
C
− 1
(m
i
− m j )
− m j).
i
A s in P r o b le m 1 2 .1 4 , t h is is r e c o g n iz e d a s a lin e a r d e c is io n fu n c t io n o f t h e fo r m
T
d (x ) = w
x − w
n + 1
w h ic h is im p le m e n te d b y a s in g le la y e r p e r c e p tr o n w ith c o e ffi c ie n ts
w
k
= v
k = 1 ,2 ,... ,n
k
a n d
θ = w
w h e re th e v
k
n + 1
= ln P (! i ) − ln P (!
j
) + x
a r e e le m e n ts o f th e v e c to r
v = C
− 1
(m
i
− m j).
T
C
− 1
(m
i
− m j)
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
2 6 0
F ig u r e P 1 2 .1 7
Problem 12.16
(a ) W h e n P (! i) = P (!
(b ) N o . T h e m in im u
th e p e r p e n d ic u la r b
d e n s itie s a re k n o w n
m u m d e c is io n fu n c
d e lta r u le fo r tr a in in
it c a n n o t b e e x p e c te
j
m
is e
, t
tio
g a
d t
) a n d C = I.
d is ta
c to r
h e B
n in
n e u
o y ie
n c e c la s s ifi e r im p le m e n ts a d e c is io n fu n c tio n th a t is
o f th e lin e jo in in g th e tw o m e a n s . If th e p r o b a b ility
a y e s c la s s ifi e r is g u a r a n te e d to im p le m e n t a n o p tith e m in im u m a v e r a g e lo s s s e n s e . T h e g e n e r a liz e d
r a l n e tw o r k s a y s n o th in g a b o u t th e s e tw o c r ite r ia , s o
ld t h e d e c is io n fu n c t io n s in P r o b le m s 1 2 .1 4 o r 1 2 .1 5 .
Problem 12.17
T h e c la s s e s a n d b o u n d a r y n e e d e d t o s e p a r a t e t h e m a r e s h o w n in F ig . P 1 2 .1 7 ( a ) .
T h e b o u n d a r y o f m in im u m c o m p le x ity in th is c a s e is a tr ia n g le , b u t it w o u ld
b e s o tig h t in th is a r r a n g e m e n t th a t e v e n s m a ll p e r tu r b a tio n s in th e p o s itio n o f
th e p a tte r n s c o u ld r e s u lt in c la s s ifi c a tio n e r r o r s . T h u s , w e u s e a n e tw o r k w ith
th e c a p a b ility to im p le m e n t 4 s u r fa c e s (lin e s ) in 2 D . T h e n e tw o r k , s h o w n in F ig .
P 1 2 .1 7 ( b ) , is a n e x t e n s io n o f t h e c o n c e p t s d is c u s s e d in t h e t e x t in c o n n e c t io n
w it h F ig . 1 2 .2 2 . In t h is c a s e , t h e o u t p u t n o d e a c t s lik e a n A N D g a t e w it h 4 in p u t s .
T h e o u tp u t n o d e o u tp u ts a 1 (h ig h ) w h e n th e o u tp u ts o f th e p re c e d in g 4 n o d e s
a r e a ll h ig h s im u lta n e o u s ly . T h is c o r r e s p o n d s to a p a tte r n b e in g o n th e + s id e o f
a ll 4 lin e s a n d , th e r e fo r e , b e lo n g in g to c la s s ! 1 . A n y o th e r c o m b in a tio n y ie ld s a
0 (lo w ) o u tp u t, in d ic a tin g c la s s ! 2 1 .
Problem 12.18
A ll th
(x 1 ,x
n o r a
u s e d
(B e c a
a t
T
2 )
x is
to
u s
is n e e d e d
, w h e re x 1
o f th e b lo
tr a in a n e
e th e p a tte
is
is
b s
u r
r n
to
th
c o
a l
s a
g e n e
e le n
m p r
n e tw
re in
ra te fo r e
g th o f th e
is in g th e
o r k u s in g
2 D , it is u
a c h c la s s tr a in in g v e c to r s o f th
m a jo r a x is a n d x 2 is th e le n g th
tr a in in g s e t. T h e s e v e c to r s w o u
, fo r e x a m p le , th e g e n e r a liz e d
s e fu l to p o in t o u t to s tu d e n ts th
e fo r m x =
o f th e m ild th e n b e
d e lta r u le .
a t th e n e u -
2 6 1
r a l n e tw o r k c o u ld b e d e
b e p lo tte d , th e d e c is io n
its c o e ffi c ie n ts u s e d to s
a p a r t w ith re s p e c t to th
m e n tin g a lin e a r d e c is io
s ig n e d b y in
b o u n d a r y o
p e c ify th e n e
e ir s p re a d , s
n fu n c tio n c
s p e c tio n in th e s
f m in im u m c o m
u ra l n e tw o rk . In
o m o s t lik e ly a s
o u ld d o t h e jo b .)
e n s e th a
p le x ity o
th is c a s e
in g le la y
t th e c la s s e s c o u ld
b ta in e d , a n d th e n
th e c la s s e s a r e fa r
e r n e tw o r k im p le -
Problem 12.19
T h is p r o b le m , a lth o u g h it is a s im p le e x e r c is e in d iffe r e n tia tio n , is in te n d e d to
h e lp th e s tu d e n t fi x in m in d th e n o ta tio n u s e d in th e d e r iv a tio n o f th e g e n e r a liz e d d e lt a r u le . F r o m E q . ( 1 2 .2 - 5 0 ) , w it h θ 0 = 1 ,
h
j
(I j ) =
1 + e
1
$2
−
N K
k = 1
w
O
j k
%.
+ θ
k
j
B e c a u s e , fr o m E q . ( 1 2 .2 - 4 8 ) ,
N
I
K
=
j
w
O
j k
k
k = 1
it fo llo w s th a t
h
j
1
(I j ) =
[
−
1 + e
I j + θ
.
]
j
T a k in g th e p a r tia l d e r iv a tiv e o f th is e x p re s s io n w ith re s p e c t to I
h
0
j
@ h
(I j ) =
(I j )
j
@ I
j
=
[
−
e
$
1 + e
I j + θ
−
[
j
]
I j + θ
j
]
F r o m E q . ( 1 2 .2 - 4 9 )
O
j
= h
j
(I j ) =
1
[
−
1 + e
I j + θ
]
j
.
It is e a s ily s h o w n th a t
O j (1 − O j) =
e
$
1 + e
−
[
−
I j + θ
[
I j + θ
s o
h
T h is c o m p le te s th e p r o o f.
0
j
j
(I j ) = O j (1 − O j).
]
j
]
%2
%2 .
j
g iv e s
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
2 6 2
Problem 12.20
T h e fi rs
n o n -n e
fa c t th a
T o p
t p
g a
t k
ro
a r
tiv
is
v e
t o
e ,
in
th
f E q . (1 2
s o D (A ,
fi n ite w
e th ird p
.3 - 3
B ) =
h e n
a r t
) is p r o
1 / k ≥
(a n d o
w e u s e
v e d
0 .
n ly
th e
b y n o tin
S im ila r ly
w h e n ) th
d e fi n itio
g th a t t
, th e s e
e s h a p
n o f D
h e
c o
e s
to
d e
n d
a re
w r
g r e e o f s im ila r ity , k , is
p a r t fo llo w s fr o m th e
id e n tic a l.
ite
D (A ,C ) ≤ m a x [D (A , B ),D (B ,C )]
a s
1
k
≤ m a x
a c
k
1
,
a b
k
1
b c
o r, e q u iv a le n tly ,
k
a c
≥ m in [k
a b
,k
b c
]
w h e r e k i j is th e d e g r e e o f s im ila r ity b e tw e e n s h a p e i a n d s h a p e j . R e c a ll fr o m
th e d e fi n itio n th a t k is th e la r g e s t o r d e r fo r w h ic h th e s h a p e n u m b e r s o f s h a p e i
a n d s h a p e j s t ill c o in c id e . A s F ig . 1 2 .2 4 ( b ) illu s t r a t e s , t h is is t h e p o in t a t w h ic h
th e fi g u re s ” s e p a r a te ” a s w e m o v e fu r th e r d o w n th e tre e (n o te th a t k in c re a s e s
a s w e m o v e fu r th e r d o w n th e tre e ). W e p r o v e th a t k a c ≥ m in [k a b ,k b c ] b y c o n tr a d ic tio n . F o r k a c ≤ m in [k a b , k b c ] to h o ld , s h a p e A h a s to s e p a r a te fr o m s h a p e
C b e fo re (1 ) s h a p e A s e p a ra te s fro m s h a p e B , a n d (2 ) b e fo re s h a p e B s e p a ra te s
fr o m s h a p e C , o th e r w is e k a b ≤ k a c o r k b c ≤ k a c , w h ic h a u to m a tic a lly v io la t e s t h e c o n d it io n k a c < m in [k a b , k b c ]. B u t , if ( 1 ) h a s t o h o ld , t h e n F ig . P 1 2 .2 0
s h o w s th e o n ly w a y th a t A c a n s e p a r a te fr o m C b e fo r e s e p a r a tin g fr o m B . T h is ,
h o w e v e r, v io la te s (2 ), w h ic h m e a n s th a t th e c o n d itio n k a c < m in [k a b , k b c ] is v io la te d (w e c a n a ls o s e e th is in th e fi g u r e b y n o tin g th a t k a c = k b c w h ic h , s in c e
k b c < k a b , v io la te s th e c o n d itio n ). W e u s e a s im ila r a r g u m e n t to s h o w th a t if
(2 ) h o ld s th e n (1 ) is v io la te d . T h u s , w e c o n c lu d e th a t it is im p o s s ib le fo r th e
c o n d itio n k a c < m in [k a b , k b c ] to h o ld , th u s p r o v in g th a t k a c ≥ m in [k a b , k b c ] o r,
e q u iv a le n tly , th a t D (A ,C ) ≤ m a x [D (A , B ), D ( B ,C )].
Problem 12.21
Q = 0 im p lie s t h a t m a x ( jA j , jB j) = M . S u p p o s e t h a t jA j > jB j.
t h a t jA j = M a n d , t h e r e f o r e , t h a t M > jB j. B u t M is o b t a in e d
B , s o it m u s t b e b o u n d e d b y M ≤ m in (jA j , jB j). B e c a u s e w e
jA j > jB j, t h e c o n d it io n M ≤ m in ( jA j , jB j) im p lie s M ≤ jB j.
t h e a b o v e r e s u lt , s o t h e o n ly w a y f o r m a x ( jA j , jB j) = M t o
T h is , in tu r n , im p lie s th a t A a n d B m u s t b e id e n tic a l s tr in
jA j = jB j = M m e a n s t h a t a ll s y m b o ls o f A a n d B m a t c h . T h e
if A ≡ B th e n Q = 0 fo llo w s d ir e c tly fr o m th e d e fi n itio n o f Q .
T h e n , it m u s t
b y m a tc h in g
h a v e s tip u la te
B u t th is c o n tr
h o ld is if jA j
g s (A ≡ B ) b e
c o n v e rs e re s u
fo llo w
A a n d
d th a t
a d ic ts
= jB j.
c a u s e
lt th a t
2 6 3
F ig u r e P 1 2 .2 0
Problem 12.22
T h
s h
to
b y
v a
s o
e r e a r e v a r io u s p o s s ib le a p p r o a c h e s to th is p r o b le m , a n d o u r s tu d e n ts h a v e
o w n o v e r th e y e a r s a te n d e n c y to s u r p r is e u s w ith n e w a n d n o v e l a p p r o a c h e s
p r o b le m s o f th is ty p e . W e g iv e h e r e a s e t o f g u id e lin e s th a t s h o u ld b e s a tis fi e d
m o s t p r a c tic a l s o lu tio n s , a n d a ls o o ffe r s u g g e s tio n s fo r s p e c ifi c s o lu tio n s to
r io u s p a r ts o f th e p r o b le m . D e p e n d in g o n th e le v e l o f m a tu r ity o f th e c la s s ,
m e o f th e s e m a y b e o ffe r e d a s ” h in ts ” w h e n th e p r o b le m is a s s ig n e d .
B e c a u s e s p e e d a n d c o s t a r e e s s e n tia l s y s te m s p e c ifi c a tio n s , w e c o n c e p tu a liz e a b in a r y a p p r o a c h in w h ic h im a g e a c q u is itio n , p re p r o c e s s in g , a n d s e g m e n ta tio n a r e c o m b in e d in to o n e b a s ic o p e r a tio n . T h is a p p r o a c h le a d s u s to g lo b a l
th r e s h o ld in g a s th e m e th o d o f c h o ic e . In th is p a r tic u la r c a s e th is is p o s s ib le b e c a u s e w e c a n s o lv e th e in s p e c tio n p r o b le m b y c o n c e n tr a tin g o n th e w h ite p a r ts
o f th e fl a g (s ta r s a n d w h ite s tr ip e s ). A s d is c u s s e d in C h a p te r 1 0 , u n ifo r m illu m in a tio n is e s s e n tia l, e s p e c ia lly w h e n g lo b a l th r e s h o ld in g is u s e d fo r s e g m e n ta tio n . T h e s tu d e n t s h o u ld m e n tio n s o m e th in g a b o u t u n ifo r m illu m in a tio n , o r
c o m p e n s a tio n fo r n o n u n ifo r m illu m in a tio n . A d is c u s s io n b y th e s tu d e n t o f c o lo r
fi lte r in g to im p r o v e c o n tr a s t b e tw e e n w h ite a n d (r e d / b lu e / b a c k g r o u n d ) p a r ts o f
a n im a g e is a p lu s in th e d e s ig n .
T h e fi r s t s te p is to s p e c ify th
r e q u ir e d to d e te c t th e s m a lle s t c o
c a u s e th e im a g e s a re m o v in g a n d
is n e c e s s a r y to s p e c ify a fi e ld o f
c o n ta in a t le a s t o n e c o m p le te fl
e n o u g h s o th a t n o fl a g s a re m is s e
T h e fi e ld o f v ie w h a s to b e w id e
a c r o s s th a n tw o fl a g s p lu s th e m
w id th , W , o f th e v ie w in g a re a m u
e s iz e o f th e v ie w in g a r e a , a n d th e r e s o lu tio n
m p o n e n ts o f in te re s t, in th is c a s e th e s ta r s . B e th e e x a c t lo c a tio n o f e a c h fl a g is n o t k n o w n , it
v ie w th a t w ill g u a r a n te e th a t e v e r y im a g e w ill
a g . In a d d itio n , th e fr a m e r a te m u s t b e fa s t
d . T h e fi r s t p a r t o f th e p r o b le m is e a s y to s o lv e .
e n o u g h to e n c o m p a s s a n a r e a s lig h tly g r e a te r
a x im u m s e p a r a tio n b e tw e e n th e m . T h u s , th e
s t b e a t le a s t W = 2 (5 ) + 2 .0 5 = 1 2 .1 in . If w e u s e
2 6 4
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
a s t a n d a r d C C D c a m e r a o f r e s o lu t io n 6 4 0 × 4 8 0 e le m e n t s a n d v ie w a n a r e a 1 2 .8
in . w id e , th is w ill g iv e u s a s a m p lin g r a te o f a p p r o x im a te ly 5 0 p ix e ls / in c h , o r 2 5 0
p ix e ls a c r o s s a s in g le fl a g . V is u a l in s p e c tio n o f a ty p ic a l fl a g w ill s h o w th a t th e
b lu e p o r tio n o f a fl a g o c c u p ie s a b o u t 0 .4 tim e s th e le n g th o f th e fl a g , w h ic h in
th is c a s e g iv e s u s a b o u t 1 0 0 p ix e ls p e r lin e in th e b lu e a r e a . T h e r e is a m a x im u m
o f s ix s ta r s p e r lin e , a n d t h e b lu e s p a c e b e tw e e n th e m is a p p r o x im a te ly 1 .5 tim e s
t h e w id t h o f a s t a r, s o t h e n u m b e r o f p ix e ls a c r o s s a s t a r is 1 0 0 / ([1 + 1 .5 ] × 6 ) ' 6
p ix e ls / s ta r.
T h e n e x t tw o p r o b le m s a r e to d e te r m in e th e s h u tte r s p e e d a n d th e fr a m e r a te .
B e c a u s e th e n u m b e r o f p ix e ls a c r o s s e a c h o b je c t o f in te r e s t is o n ly 6 , w e fi x th e
b lu r a t le s s th a n o n e p ix e l. F o llo w in g th e a p p r o a c h u s e d in th e s o lu tio n o f P r o b le m 1 0 .4 9 , w e fi r s t d e t e r m in e t h e d is t a n c e b e t w e e n p ix e ls a s (1 2 .8 in ) / 6 4 0 p ix e ls =
0 .0 2 in / p ix e l. T h e m a x im u m s p e e d o f t h e fl a g s is 2 1 in / s e c . A t t h is s p e e d , t h e
fl a g s t r a v e l 2 1 / 0 .0 2 = 1 , 0 5 0 p ix e ls / s e c . W e a r e r e q u ir in g t h a t a fl a g n o t t r a v e l
m o r e th a n o n e p ix e l d u r in g e x p o s u r e ; th a t is (1 , 0 5 0 p ix e ls / s e c ) × T s e c ≤ 1 p ix e l.
S o , T ≤ 9 .5 2 × 1 0 − 4 s e c is t h e s h u t t e r s p e e d n e e d e d .
T h e fr a m e r a te m u s t b e fa s t e n o u g h to c a p tu re a n im a g e o f e v e r y fl a g th a t
p a s s e s t h e in s p e c t io n p o in t . It t a k e s a fl a g (2 1 in / s e c )/ (1 2 .8 in ) ' 0 .6 s e c t o c r o s s
th e e n tir e fi e ld o f v ie w , s o w e h a v e t o c a p tu r e a fr a m e e v e r y 0 .3 s e c in o r d e r to
g u a r a n te e th a t e v e r y im a g e w ill c o n ta in a w h o le fl a g , a n d th a t n o fl a g w ill b e
m is s e d . W e a s s u m e th a t th e c a m e r a is c o m p u te r c o n tr o lle d to fi r e fr o m a c lo c k
s ig n a l. W e a ls o m a k e th e s ta n d a r d a s s u m p tio n th a t it ta k e s 1 / 3 0 s e c ' 3 3 0 ×
1 0 − 4 s e c to re a d a c a p tu re d im a g e in to a fr a m e b u ffe r. T h e re fo re , th e to ta l tim e
n e e d e d t o a c q u ir e a n im a g e is (3 3 0 + 9 .5 ) × 1 0 − 4 ' 3 4 0 × 1 0 − 4 s e c . S u b t r a c t in g
t h is q u a n t it y fr o m t h e 0 .3 s e c fr a m e r a t e le a v e s u s w it h a b o u t 0 .2 7 s e c to d o a ll
th e p r o c e s s in g re q u ire d fo r in s p e c tio n , a n d to o u tp u t a n a p p r o p r ia te s ig n a l to
s o m e o th e r p a r t o f th e m a n u fa c tu r in g p r o c e s s .
B e c a u s e a g lo b a l th r e s h o ld in g fu n c tio n c a n b e in c o r p o r a te d in m o s t d ig itiz e r s a s p a r t o f th e d a ta a c q u is itio n p r o c e s s , n o a d d itio n a l tim e is n e e d e d to g e n e r a te a b in a r y im a g e . T h a t is , w e a s s u m e th a t th e d ig itiz e r o u tp u ts th e im a g e in
b in a r y fo r m . T h e n e x t s te p is to is o la te th e d a ta c o r r e s p o n d in g to a c o m p le te
fl a g . G iv e n th e im a g in g g e o m e tr y a n d fr a m e r a te d is c u s s e d a b o v e , fo u r b a s ic
b in a r y im a g e c o n fi g u r a tio n s a r e e x p e c te d : (1 ) p a r t o f a fl a g o n th e le ft o f th e im a g e , fo llo w e d b y a w h o le fl a g , fo llo w e d b y a n o th e r p a r tia l fl a g ; (2 ) o n e e n tir e fl a g
to u c h in g th e le ft b o r d e r, fo llo w e d b y a s e c o n d e n tir e fl a g , a n d th e n a g a p b e fo r e
th e r ig h t b o rd e r ; (3 ) th e o p p o s ite o f (2 ); a n d (4 ) tw o e n tire fl a g s , w ith n e ith e r fl a g
to u c h in g th e b o u n d a r y o f th e im a g e . C a s e s (2 ), (3 ), a n d (4 ) a r e n o t lik e ly to o c c u r
w ith a n y s ig n ifi c a n t fr e q u e n c y , b u t w e w ill c h e c k fo r e a c h o f th e s e c o n d itio n s . A s
w ill b e s e e n b e lo w , C a s e s (2 ) a n d (3 ) c a n b e h a n d le d th e s a m e a s C a s e (1 ), b u t,
g iv e n th e tig h t b o u n d s o n p r o c e s s in g tim e , th e o u tp u t e a c h tim e C a s e (4 ) o c c u r s
2 6 5
w ill b e to r e je c t b o th fl a g s .
T o h a n d le C a s e (1 ) w e h a v e to id e n tify a w h o le fl a g ly in g b e tw e e n tw o p a r tia l fl a g s . O n e o f th e q u ic k e s t w a y s to d o th is is to r u n a w in d o w a s lo n g a s th e
im a g e v e r tic a lly , b u t n a r r o w in th e h o r iz o n ta l d ir e c tio n , s a y , c o r r e s p o n d in g to
0 .3 5 in . ( b a s e d o n t h e w in d o w s iz e 1 / 2 o f [1 2 .8 − 1 2 .1 ]) , w h ic h is a p p r o x im a t e ly
(0 .3 5 )(6 4 0 )/ 1 2 .8 ' 1 7 p ix e ls w id e . T h is w in d o w is u s e d lo o k fo r a s ig n ifi c a n t g a p
b e t w e e n a h i g h c o u n t o f 1 ’s , a n d i t i s n a r r o w e n o u g h t o d e t e c t C a s e ( 4 ) . F o r
C a s e (1 ), th is a p p r o a c h w ill p r o d u c e h ig h c o u n ts s ta r tin g o n th e le ft o f th e im a g e , th e n d r o p to v e r y fe w c o u n ts (c o r re s p o n d in g to th e b a c k g r o u n d ) fo r a b o u t
tw o in c h e s , p ic k u p a g a in a s th e c e n te r (w h o le fl a g ) is e n c o u n te r e d , g o lik e th is
fo r a b o u t fi v e in c h e s , d r o p a g a in fo r a b o u t tw o in c h e s a s th e n e x t g a p is e n c o u n t e r e d , t h e n p i c k u p a g a i n u n t i l t h e r i g h t b o r d e r i s e n c o u n t e r e d . T h e 1 ’s b e t w e e n
th e tw o in n e r g a p s c o r r e s p o n d to a c o m p le te fl a g a n d a r e p r o c e s s e d fu r th e r b y
t h e m e t h o d s d i s c u s s e d b e l o w ; t h e o t h e r 1 ’s a r e i g n o r e d . ( A m o r e e l e g a n t a n d p o te n tia lly m o r e r u g g e d w a y is to d e te r m in e a ll c o n n e c te d c o m p o n e n ts fi r s t, a n d
th e n lo o k fo r v e r tic a l g a p s , b u t tim e a n d c o s t a r e fu n d a m e n ta l h e r e ). C a s e s (2 )
a n d (3 ) a r e h a n d le d in a s im ila r m a n n e r w ith s lig h tly d iffe r e n t lo g ic , b e in g c a r e fu l t o is o la t e t h e d a t a c o r r e s p o n d in g t o a n e n t ir e fl a g ( i.e ., t h e fl a g w it h a g a p o n
e a c h s id e ). C a s e (4 ) c o r re s p o n d s to a g a p - d a ta - g a p - d a ta - g a p s e q u e n c e , b u t, a s
m e n tio n e d a b o v e , it is lik e ly th a t tim e a n d c o s t c o n s tr a in ts w o u ld d ic ta te r e je c tin g b o th fl a g s a s a m o r e e c o n o m ic a l a p p r o a c h th a n in c r e a s in g th e c o m p le x ity
o f th e s y s te m to h a n d le th is s p e c ia l c a s e . N o te th a t th is a p p r o a c h to e x tr a c tin g
1 ’s i s b a s e d o n t h e a s s u m p t i o n t h a t t h e b a c k g r o u n d i s n o t e x c e s s i v e l y n o i s y . I n
o th e r w o r d s , th e im a g in g s y s te m m u s t b e s u c h th a t th e b a c k g r o u n d is r e lia b ly
s e g m e n te d a s b la c k , w ith a c c e p ta b le n o is e .
W ith r e fe r e n c e to F ig . 1 .2 3 , th e p r e c e d in g d is c u s s io n h a s c a r r ie d u s th r o u g h
th e s e g m e n ta tio n s ta g e . T h e a p p r o a c h fo llo w e d h e r e fo r d e s c r ip tio n , r e c o g n itio n , a n d th e u s e o f k n o w le d g e , is tw o fo ld . F o r th e s ta r s w e u s e c o n n e c te d c o m p o n e n t a n a ly s is . F o r th e s tr ip e s w e u s e s ig n a tu r e a n a ly s is . T h e s y s te m k n o w s
th e c o o r d in a te s o f tw o v e r tic a l lin e s w h ic h c o n ta in th e w h o le fl a g b e tw e e n th e m .
F ir s t, w e d o a c o n n e c te d c o m p o n e n ts a n a ly s is o n th e le ft h a lf o f th e r e g io n (to
s a v e tim e ) a n d fi lte r o u t a ll c o m p o n e n ts s m a lle r a n d la r g e r th a n th e e x p e c te d
s iz e o f s ta r s , s a y (to g iv e s o m e fl e x ib ility ), a ll c o m p o n e n ts le s s th a n 9 (3 × 3 ) p ix e ls a n d la r g e r th a n 6 4 (8 × 8 ) p ix e ls . T h e s im p le s t te s t a t th is p o in t is to c o u n t th e
n u m b e r o f re m a in in g c o n n e c te d c o m p o n e n ts (w h ic h w e a s s u m e to b e s ta r s ). If
th e n u m b e r is 5 0 w e c o n tin u e w ith th e n e x t te s t o n th e s tr ip e s . O th e r w is e , w e
r e je c t th e fl a g . O f c o u r s e , th e lo g ic c a n b e m a d e m u c h m o r e c o m p lic a te d th a n
th is . F o r in s ta n c e , it c o u ld in c lu d e a r e g u la r ity a n a ly s is in w h ic h th e r e la tiv e lo c a tio n s o f th e c o m p o n e n ts a r e a n a ly z e d . T h e r e a r e lik e ly to b e a s m a n y a n s w e r s
h e r e a s th e r e a r e s tu d e n ts in th e c la s s , b u t th e k e y o b je c tiv e s h o u ld b e to b a s e
2 6 6
C H A P T E R 1 2 . P R O B L E M S O L U T IO N S
th e a n a ly s is o n a r u g g e d m e th o d s u c h a s c o n n e c te d c o m p o n e n t a n a ly s is .
T o a n a ly z e th e s tr ip e s , w e a s s u m e th a t th e fl a g s a r e p r in te d o n w h ite s to c k
m a te r ia l. T h u s , “ d r o p p in g a s tr ip e ” m e a n s c r e a tin g a w h ite s tr ip e tw ic e a s w id e
a s n o r m a l. T h is is a s im p le d e fe c t d e te c ta b le b y r u n n in g a v e r tic a l s c a n lin e in
a n a r e a g u a r a n te e d to c o n ta in s tr ip e s , a n d th e n lo o k in g a t th e in te n s ity s ig n a tu r e fo r th e n u m b e r o f p u ls e s o f th e r ig h t h e ig h t a n d d u r a tio n . T h e fa c t th a t th e
d a ta is b in a r y h e lp s in th is r e g a r d , b u t th e s c a n lin e s h o u ld b e p r e p r o c e s s e d to
b r id g e s m a ll g a p s d u e to n o is e b e fo r e it is a n a ly z e d . In s p ite o f th e ± 1 5 ◦ v a r ia tio n in d ir e c tio n , a r e g io n , s a y , 1 in . to th e r ig h t o f th e b lu e r e g io n is in d e p e n d e n t
e n o u g h o f th e r o ta tio n a l v a r ia tio n in te r m s o f s h o w in g o n ly s tr ip e s a lo n g a s c a n
lin e r u n v e r tic a lly in th a t r e g io n .
It is im p o r ta n t th a t a n y a n s w e r to th is p r o b le m s h o w a w a r e n e s s o f th e lim its
in a v a ila b le c o m p u ta tio n tim e . B e c a u s e n o m e n tio n is m a d e in th e p r o b le m
s ta te m e n t a b o u t a v a ila b le p r o c e s s o r s , it is n o t p o s s ib le to e s ta b lis h w ith a b s o lu te
c e r ta in ty if a s o lu tio n w ill m e e t th e r e q u ir e m e n ts o r n o t. H o w e v e r, th e s tu d e n t
s h o u ld b e e x p e c te d to a d d r e s s th is is s u e . T h e g u id e lin e s g iv e n in th e p r e c e d in g
s o lu tio n a r e a m o n g th e fa s te s t w a y s to s o lv e th e p r o b le m . A s o lu tio n a lo n g th e s e
lin e s , a n d a m e n tio n th a t m u ltip le s y s te m s m a y b e r e q u ir e d if a s in g le s y s te m
c a n n o t m e e t th e s p e c ifi c a tio n s , is a n a c c e p ta b le s o lu tio n to th e p r o b le m .