§ 位元平面
․三原色RGB可分解成R平面, G平面, B平面, 如右:
A
A
A
高灰階像素也可分解成八個位元平面,
假設256個灰階值表示成(g8g7g6g5g4g3g2g1)2 , 每一像素提供第i個位元, 即gi
以組成第i個位元平面(也就是第i張黑白影像), 如下:
B
B
B
B
Ex4: 給定一 4X4子影像:
A:
B
B
8
7
6
5
32
31
30
29
10
11
12
13
0
1
2
3
00001000
00000111
00000110
00000101
00100000
00011111
00011110
00011101
00001010
00001011
00001100
00001101
00000000
00000001
00000010
00000011
B
B
, 算出第三張位元平面?
⇒
0
1
1
1
0
1
1
1
0
0
1
1
0
0
0
0
․利用位元平面植入影像的缺點: 經過壓縮後, 所植入的影像容易遭到破壞;
解壓縮後所得影像, 常已破損; 即為數學上的One-way function。
§ Steganography and Watermark
․實際重疊高階四個位元平面(捨棄低階四個位元平面)所得影像, 肉眼幾乎分辨
不出差異; 故捨棄低階四個位元並不影響影像特徵太大(此乃因愈低階位元的
權重愈低, 所以影響影像特徵的機率愈小)。
例如, 某影像中有兩像素, 其灰階值為19310=(11000001)2 與 192=(11000000)2 ,
可把灰階值為37=(00100101)2 的第三個像素隱藏於前述影像中; 所得灰
階值為194=(11000010)2 與 197=(11000101)2 , 人眼幾乎察覺無異其影
像特徵。
․ 假設一個位元組可以隱藏一位元, 且影像術規則如下:
(1) 從浮水印讀出之位元為0, 則原影像對應位元組的最後兩位元由01/10, 改為
00/11。
(2) 從浮水印讀出之位元為1, 則原影像對應位元組的最後兩位元由00/11, 改為
01/10。
(3) 其餘情況則保持原狀
例如, 位元組11000000要隱藏位元1, 則改為11000001; 要隱藏0, 則位元組
11000000保持不變。
Ex5: 原影像為 24 710
21
42
8
66
34
10
12
, 想隱藏如右浮水印
A: 先改成二進制(如下), 再根據規則得
00011000
XX
XX
00101010
XX
XX
00100010
XX
XX
1
0
0
1
0
1
0
1
0
25
710
20
42
8
66
35
10
12
, 求出加入浮水印後
的十進位影像?
基本原理
․令B’為A隱藏在B後的結果, PSNR常用來評估B’和B的相似性;
2552
PSNR  10log
MSE
1
MSE  2
N
N 1 N 1
 B' ( x, y)  B( x, y
2
x 0 y 0
PSNR是不錯的失真表示法, 但無法充分反應紋理(texure)的失真情形; 所謂的
浮水印, 可把A看成標誌(logo)---而這標誌通常也是一種版權; 例如, NCKU
之於成大。Note: A的大小必須小於B ; 故必要時, 可把A先壓縮。
․設A為灰階影像且可被壓縮, 又為長條型矩陣; Rank(A)=m, 則 Singular Value
Decomposition of A 可表示成 A=U∑Vt , 其中 V and U is orthogonal.
 diag( ,
1
2
,..., n )
其中1, 2 ,..., n 為奇異值且滿足
 1   2  ...   m  0
 i  i
where
and
 m1   m 2  ...   n  0
i 為矩陣 AtA 的第 i 個 Eigenvalue
Ex6: Prove λi ≥ 0
 AX  ( AX ) AX  X A AX  X (X )  X X   X   
2
t
t
t
t
t
2
X
Ex7: Prove
A=U∑Vt
= (U1U2)  1
 0

先求正交矩陣V  (V1 ,V2 ),
2
AX
2
0
0 V1t 
 t   U1 1V1t
0 V2 
V1 為1 , 2 ,...,m 所算出的Eigenvectors, 即v1 , v2 ,...,vm
所構成, 也就是V1  (v1 , v2 ,...,vm );V2  (Vm1 ,Vm 2 ,...,Vn )是m1  0  m 2  ...  n
所求出之Eigenvectors所組成
 2 2
8 8
t
A

則
A
A

例如, 設
 2 2
8 8且特徵值1  16, 2  0;  1  1




 4,  2  0; eigenvectors : V1  (1,1) t , V2  (1,1) t
1 1 
1  1; u sin g


1
 V  (V1 ,V2 ) 
2
AV j   j u j  m eaning
sin ce
 1
 2

AtU  V 
t
1 
2 
t
SVD
Note : 欲解A  U  V t
AV  U 

AV1 1 2 2 
u1 


 1 4 2 2 

we
have
we
1  
2  
1  
2  
At u j  0; hence
 1
 2
t
of
A  U V  
 1
 2
(1)先解 (2)次解V
get
1 
2
1 
2 
u2 
1 

2   4 0 
 1  0 0 

2 
(3)末解U
1
2
1
2
1 
2
 1 
2 
․SVD被用於隱像術的原因: 乃因植入的影像A之奇異值, 可變得很小; 再把轉換
後的影像A’植入B, 則合成影像B’的SVD之奇異值, 仍以B的奇異值為主。Note:
前景取較大的奇異值; 即A’的奇異值接在B的後面, 如此A’就不易被察覺
形態學
․假設色調H為人臉特徵依據, 以訓練集(training images)測得皮膚色之色調範圍
可能顯得零碎; 吾人可利用形態學的opening與closing算子, 將太小且疏離的雜
訊刪除, 但將很靠近的區塊連接在一起; 加上頭髮的考慮, 進一步判定是否人臉
․closing算子會先進行dilation運算, 再作erosion運算; 效果是: 先擴張後, 區域旁
的小區域會被併在一起, 但離區域遠的小雜訊仍然處於孤立狀態。後經侵蝕運
算, 區域旁近距離的雜訊仍會存於新區域內, 但遠距離的雜訊則被侵蝕掉
․opening算子進行的順序恰相反, 有消除小塊雜點的功能; 能打斷以細線連接的
近距離兩區塊。原因是: 連接兩區域的細邊消失, 即使擴張兩區域也無法併合
․影像處理基本主題, 例如DCT、sampling theorem、aliasing等, 不在討論內
Digital Watermarking
․A digital watermark is a signal permanently embedded into
digital data (audio, images, and text) that can be detected or
extracted later by means of computing operations in order to
make assertions about the data. It has been developed to
protect the copyright of media signals.
․It is hidden in the host data in such a way that it is inseparable
from the data and so that it is resistant to many operations not
degrading the host document. Thus by means of watermarking,
the work is still accessible but permanently marked.
․It is derived from steganography, which means covered writing
Steganography is the science of communicating information
while hiding the existence of the communication.
․The goal of steganography is to hide an information message
inside harmless messages in such a way that it is not possible
even to detect that there is a secret message present. Watermarking is not like encryption in that the latter has the aim of
making messages unintelligible to any unauthorized persons
who might interpret them. Once encrypted data id decrypted,
the media is no longer protected.
Morphology
․Morphology means the form and structure of an object, or the
arrangements and interrelationships between the parts of an
object. Digital morphology is a way to describe or analyze the
shape of a digital (most often raster) object. The math behind it
is simply set theory.
․We can assume the existence of three color components (red,
green and blue) is an extension of a grey level, or each color
can be thought of as a separate domain containing new
information.
․Closing the red and blue images should brighten the green
images, and opening the green images should suppress the
green ones.
․Images consist of a set of picture elements (pixels) that collect
into groups having two-dimensional structure (shape).
Mathematical operations on the set of pixels can be used to
enhance specific aspects of the shapes so that they might be
(for example) counted or recognized.
․Erosion: Pixels matching a given pattern are deleted from the
image.
․Dilation: A small area about a pixel is set to a given pattern.
Binary Dilation: First marking all white pixels having at
least one black neighbor, and then
(simple)
setting all of the marked pixels to black.
(Dilation of the original by 1 pixel)
․In general the object is considered to be a mathematical set of
black pixels, written as A={(3,3),(3,4),(4,3),(4,4)} if the upper left
pixel has the index (0,0).
․Translation of the set A by the point x:  Ax  c c  a  x, a  A
For example, if x were at (1,2) then the first (upper left) pixel in
Ax would be (3,3)+(1,2)=(4,5); all of the pixels in A shift down by
one row and right by two columns.
ˆ  c c  a, a  A  Ac  c c  A
․Reflection: A
This is really a rotation of the object A by 180 degrees about the
origin, namely the complement of the set A.
․Intersection, union and difference (i.e.
the language of the set theory.

A  B c) correspond to

․Dilation: A  B  c c  a  b, a  A, b  B; the set B is called a
structuring element, and its composition defines the nature of
the specific dilation.
Ex1: Let B={(0,0),(0,1)}, A  B  C  ( A  {(0,0)})  ( A  {(0,1)})
(3,3)+(0,0)=(3,3), (3,3)+(0,1)=(3,4), … Some are duplicates.
B=
(0,0) added to A Adding (0,1) to A
A=
A=
After union
A=
Note: If the set B has a set pixel to the right of the origin, then a
dilation grows a layer of pixels on the right of the object.
To grow in all directions, we can use B having one pixel on
every side of the origin; i.e. a 3X3 square with the origin at the
center.
Ex2: Suppose A1={(1,1),(1,2),(2,2),(3,2),(3,3),(4,4)} and
B1={(0,-1),(0,1)}. The translation of A1 by (0,-1) yields
(A1)(0,-1)={(1,0),(1,1),(2,1),(3,1),(3,2),(4,3)} and
(A1)(0,1)={(1,2),(1,3),(2,3),(3,3),(3,4),(4,5)} as following.
B1=
(B1 not including the origin)
before
after
Note: (1) The original object pixels belonging to A1 are not
necessarily set in the result, (4,4) for example, due to
the effect of the origin not being a part of B1.
( A)b   ( B) a since dilation is
(2) In fact, A  B  b
B
aA
commutative. This gives a clue concerning a possible
implementation for the dilation operator. When the origin of B
aligns with a black pixel in the image, all of the image pixels that
correspond to black pixels in B are marked, and will later be
changed to black. After the entire image has been swept by B,
the dilation is complete. Normally the dilation is not computed in
place; that is, where the result is copied over the original image.
A third image, initially all white, is used to store the dilation while
it is being computed.
← Dilating →
(1st)
(Erosion)
(2nd)
(1st translation)
(Erosion)
⇒
(2nd)
⇒
(3rd)
(final)
Binary Erosion
• If dilation can be said to add pixels to an object, or to make it
bigger, then erosion will make an image smaller. Erosion can be
implemented by marking all black pixels having at least one
white neighbor, and then setting to white all of the marked
pixels. Only those that initially place the origin of B at one of the
members of A need to be considered. It is defined as
AB  c (B)c  A
Ex3: B={(0,0),(1,0)}, A={(3,3),(3,4),(4,3),(4,4)}
Four such
translations: B(3,3)={(3,3),(4,3)}
B(3,4)={(3,4),(4,4)}
B(4,3)={(4,3),(5,3)}
B(4,4)={(4,4),(5,4)}
Ex4: B2={(1,0)}, i.e. 0  B2 . The ones that result in a match are:
B(2,3)={(3,3)} B(2,4)={(3,4)} B(3,3)={(4,3)} B(3,4)={(4,4)}
Note: {(2,3),(2,4),(3,3),(4,4)} is not a subset of A, meaning the
eroded image is not always a subset of the original.
․Erosion and dilation are not inverse operations. Yet, erosion and
( AB)c  Ac  Bˆ
dilation are duals in the following sense:
․An issue of a “don’t care” state in B, which was not a concern
about dilation. When using a strictly binary structuring element
to perform an erosion, the member black pixels must
correspond to black pixels in the image in order to set the pixel
in the result, but the same is not true for a white pixel in B. We
don’t care what the corresponding pixel in the image might be
when the structuring element pixel is white.
Opening and Closing
․The application of an erosion immediately followed by a dilation
using the same B is referred to as an opening operation,
describing the operation tends to “open” small gaps or spaces
between touching objects in an image. After an opening using
simple the objects are better isolated, and might now be
counted or classified.
․Another using of opening: the removal of noise. A noisy greylevel image thresholded results in isolated pixels in random
locations. The erosion step in an opening will remove isolated
pixels as well as boundaries of objects, and the dilation step will
restore most of the boundary pixels without restoring the noise.
This process seems to be successful at removing spurious
black pixels, but does not remove the white ones.
․A closing is similar to an opening except that the dilation is
performed first, followed by an erosion using the same B, and
will fill the gaps or “close” them. It can remove much of the
white pixel noise, giving a fairly clean image. (A more complete
method for fixing the gaps may use 4 or 5 structuring elements,
and 2 or 3 other techniques outside of morphology.)
․Closing can also be used for smoothing the outline of objects in
an image, i.e. to fill the jagged appearances due to digitization
in order to determine how rough the outline is. However, more
than one B may be needed since the simple structuring element
is only useful for removing or smoothing single pixel irregularities. N dilation/erosion (named depth N) applications should
result in the smoothing of irregularities of N pixels in size.
․A fast erosion method is based on the distance map of each
object, where the numerical value of each pixel is replaced by
new value representing the distance of that pixel from the
nearest background pixel. Pixels on a boundary would have a
value of 1, being that they are one pixel width from a background pixel; a value of 2 meaning two widths from the background, and so on. The result has the appearance of a contour
map, where the contours represent the distance from the
boundary.
․The distance map contains enough information to perform an
erosion by any number of pixels in just one pass through the
image, and a simple thresholding operation will give any desired
erosion.
․There is another way to encode all possible openings as one
grey-level image, and all possible closings can be computed at
the same time. First, all pixels in the distance map that do NOT
have at least one neighbor nearer to the background and one
neighbor more distant are located and marked as nodal pixels.
If the distance map is thought of as a three-dimensional surface
where the distance from the background is represented as
height, then every pixel can be thought of as being peak of a
pyramid having a standardized slope. Those peaks that are not
included in any other pyramid are the nodal pixels.
․One way to locate nodal pixels is to scan the distance map,
looking at all object pixels; find the MIN and MAX value of all
neighbors of the target pixel, and compute MAX-MIN. If the
value is less than the MAX possible, then the pixel is nodal.
The “Hit and Miss” Transform
․It is a morphological operator designed to locate simple shapes
within an image. Though the erosion of A by S also includes
places where the background pixels in that region do not match
those of S, these locations would not normally be thought of as
a match.
․Matching the foreground pixels in S against those in A is “hit,”
and is accomplished with an erosion AS . The background
pixels in A are those found in Ac, and while we could use Sc as
the background for S in a more flexible approach is to specify
the background pixels explicitly in a new structuring element T.
A “hit” in the background is called a “miss,” and is found by
Ac T .
․What we need is an operation that matches both the foreground
and the background pixels of S in A, which are the pixels:
A  (S , T )  ( AS )  ( Ac T )
Ex5: To detect upper right corners. The figure (a) below shows
an image interpreted as being two overlapping squares.
(b) Foreground
structuring element
(a)
(e) Background S,
showing 3 pixels of
the corner
(c) Erosion of (a) by (b)
--the ‘hit’
(f) Erosion of (d)
by (e)--the ‘miss’
(d) Complement of (a)
(g) Intersection
of (c) and (f)--the result
Identifying Region Boundaries
․The pixels on the boundary of an object are those that have at
least one neighbor that belongs to the background. It can’t be
known in advance which neighbor to look for! A single structuring element can’t be constructed to detect the boundary. This is
in spite of the fact that an erosion removes exactly these pixels.
․The boundary can be stripped away using an erosion and the
eroded image can then be subtracted from the original, written
as: Boundary A  ( Asimple)
Ex6: Figure (h) results from the previous figure (a) after an erosion, and (i) shows (a)-(h): the boundary.
(a) of Ex5
(h)
(i)
Conditional Dilation
․There are occasions when it is desirable to dilate an object in
such a way that certain pixels remain immune. The forbidden
area of the image is specified as a second image in which the
forbidden pixels are black. The notation is A  (S , A) and is
computed in an iterative fashion: Ai  ( Ai 1  S )  A
A’: the set of forbidden pixels; Ai: the desired dilation
․One place where this is useful is in segmenting an image.
Ihigh: a very high threshold applying to an image-- a great many
will be missed.
Ilow: a very low threshold applying to the original image-- some
background will be marked.
R: a segmented version of the original-- a superior result than
using any single threshold in some cases, and obtained by:
R  I high  (simple, Ilow )
․Another application of conditional dilation is that of filling a
region with pixels, which is the inverse operation of boundary
extraction. It is to dilate until the inside region is all black, and
then combine with the boundary image to form the final result.
Fill  P  (Scross , Ac )
where P is an image containing only the seed pixel, known to be
inside the region to be filled, A is the boundary image and Scross
is the cross-shaped structuring element, (j) for example.
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)
Ex7: (i) boundary, (j) structuring element, (k) seed pixel iterated 0 of the process, (l) iteration 1, (m) iteration 2,
(n) iteration 3, (o) iteration 4, (p) iteration 5 and completed, (q) union of (i) with (p)-- the result
Counting Regions
․It is possible to count the number of regions in an binary image
using morphological operators, first discussed by Levialdi using
6 different structuring elements--4 for erosion naming L1~L4
and 2 for counting isolated “1” pixels (# operator). The initial
count of regions is the number of isolated pixels in the input
image, and the image of iteration 0 is A:
count0= #A, A0=A, countn= #An
The image of the next iteration is the union of the four erosions
of the current image: An1  ( AnL1 )  ( AnL2 )  ( AnL3 )  ( AnL4 )
The iteration stops when An becomes empty (all 0 pixels), and
the overall number of regions is the sum
of all of the values counti.
Ex8: Counting 8-connected,
(e), and (a)~(d): L1~L4
(a)
(b)
(c)
(d)
(e)
Grey-Level Morphology
․A pixel can now have any integer value, so the nice picture of
an image being a set disappears! The figures shows how the
dilated grey-level line (a) might appear to be (b), and was
computed as follows, A being the grey-level image to be dilated.
( A  S )[i, j ]  max{A[i  r , j  c] 
S[r , c][i  r , j  c]  A, [r , c]  S}
(a) Background is 0, and line
pixels have the value 20.
(b) Grey line after a dilation
․Process of the above computation:
(1) Position the origin of the structuring element over the first
pixel of the image being dilated.
(2) Compute the sum of each corresponding pair of pixel values
in the structuring element and the image.
(3) Find the maximum value of all of these sums, and set the
corresponding pixel in the output image to this value.
(4) Repeat this process for each pixel in the image being dilated.
․
Chromaticity Diagram
• Chromaticity is an objective specification of the quality of a
color regardless of its luminance, that is, as determined by its
colorfulness (or saturation, chroma, intensity, or excitation
purity) and hue.
• In color science, the white point of an illuminant or of a display
is a neutral reference characterized by a chromaticity; for
example, the white point of an sRGB display is an x,y chromaticity of [0.3127,0.3290]. All other chromaticities may be defined
in relation to this reference using polar coordinates. The hue is
the angular component, and the purity is the radial component,
normalized by the maximum radius for that hue.
• Purity is roughly equivalent to the term "saturation" in the HSV
color model. The property "hue" is as used in general color
theory and in specific color models such as HSV or HSL,
though it is more perceptually uniform in color models such as
Munsell, CIELAB or CIECAM02.
․Some color spaces separate the three dimensions of color into
one luminance dimension and a pair of chromaticity dimensions.
For example, the chromaticity coordinates are a and b in Lab
color space, u and v in Luv color space, x and y in xyY space,
etc. These pairs define chromaticity vectors in a rectangular 2space, unlike the polar coordinates of hue angle and saturation
that are used in HSV color space.
․On the other hand, some color spaces such as RGB and XYZ
do not separate out chromaticity; chromaticity coordinates such
as r and g or x and y can be calculated by an operation that
normalizes out intensity.
․The xyY space is a cross between the CIE XYZ color space and
its normalized chromaticity coordinates xyz, such that the
luminance Y is preserved and augmented with just the required
two chromaticity dimensions.
․The CIE (1931) diagram is a projection of a 3D color space,
called XYZ color space, to 2D. The light emitted by a device, or
light reflected from a surface consists of photons with different
wavelengths. The amount of photons with a certain wavelength,
λ, in a given light composition is represented by the function
C(λ). The CIE diagram comprises three funs μx(λ), μy(λ),
μz(λ), and is used for comparing colors produced by colorproducing devices, e.g. PC monitors, printers, and cameras.
The science of quantifying color is called colorimetry. The X, Y,
Z coordinates are found as follows:
X   C ( )  x d
Y   C ( )  y d
Z   C ( )  z d
x
X
X Y  Z
․The projection to the CIE diagram is obtained via
Y
Z
where x+y+z=1, it’so that only two of x, y,
y
z
X Y  Z
X Y  Z
z are independent, making the projection
a planar surface.
Texture
․A major characteristic is the repetition of a pattern or patterns
over a region. The pattern may be repeated exactly, or as a set
of small variations on the theme, possibly a function of position.
The goal of studying texture is to segment regions rather than
characterizing textures, determining which regions have texture
A and which have texture B. The result could be an image in
which texture has been replaced by a unique grey level or color.
․texton: the size, shape, color, and orientation of the elements of
the pattern.
Sometimes the difference between two textures is contained in
the degree of variation alone, or in the statistical distribution
found relating the textons. A region cannot display texture if it is
small compared with the size of a texton.
․The same texture at two different scales will be perceived as
two different textures, provided that the scales are different
enough. This leaves us with a problem of scale. As the scales
become closer together the textures are harder to distinguish,
and at some point they become the same.
Ex1: regions characterized by their textures– both are artificial.
(a)
․
(b)