Histogram Processing
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Histogram Processing
Histogram h( rk )  nk
rk is the k th intensity value
nk is the number of pixels in the image with intensity rk
nk
Normalized histogram p(rk ) 
MN
nk : the number of pixels in the image of
size M  N with intensity rk
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Histogram Equalization
The intensity levels in an image may be viewed as
random variables in the interval [0, L-1].
Let pr (r ) and ps ( s ) denote the probability density
function (PDF) of random variables r and s.
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Histogram Equalization
s  T (r )
0  r  L 1
a. T(r) is a strictly monotonically increasing function
in the interval 0  r  L -1;
b. 0  T (r )  L -1 for 0  r  L -1.
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Histogram Equalization
s  T (r )
0  r  L 1
a. T(r) is a strictly monotonically increasing function
in the interval 0  r  L -1;
b. 0  T (r )  L -1 for 0  r  L -1.
T (r ) is continuous and differentiable.
ps ( s)ds  pr (r )dr
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Histogram Equalization
r
s  T (r )  ( L  1)  pr ( w) dw
0
ds dT (r )
d  r


 ( L  1)
p
(
w
)
dw
r



0
dr
dr
dr 
 ( L  1) pr (r )
pr (r )dr pr (r )
1
pr (r )
ps ( s ) 



 ( L  1) pr (r )  L  1
 ds 
ds
 
 dr 
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Example
Suppose that the (continuous) intensity values
in an image have the PDF
 2r
,

2
pr (r )   ( L  1)
 0,

for 0  r  L-1
otherwise
Find the transformation function for equalizing
the image histogram.
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Example
r
s  T (r )  ( L  1)  pr ( w) dw
0
 ( L  1) 
r
0
2w
dw
2
( L  1)
2
r

L 1
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Histogram Equalization
Continuous case:
r
s  T (r )  ( L  1)  pr ( w)dw
0
Discrete values:
k
sk  T (rk )  ( L  1) pr (rj )
j 0
L 1 k
 ( L  1)

nj

MN j 0
j  0 MN
k
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nj
k=0,1,..., L-1
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Example: Histogram Equalization
Suppose that a 3-bit image (L=8) of size 64 × 64 pixels (MN = 4096)
has the intensity distribution shown in following table.
Get the histogram equalization transformation function and give the ps(sk)
for each sk.
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Example: Histogram Equalization
0
s0  T (r0 )  7 pr (rj )  7  0.19  1.33
1
s1  T (r1 )  7 pr (rj )  7  (0.19  0.25)  3.08
3
j 0
1
j 0
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s2  4.55  5
s3  5.67  6
s4  6.23  6
s5  6.65  7
s6  6.86  7
s7  7.00  7
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Example: Histogram Equalization
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