Histogram Processing


Histogram of a digital image with gray levels in
the range [0,L-1] is a discrete function
Where



h(rk) = nk
rk : the kth gray level
nk : the number of pixels in the image having gray
level rk
h(rk) : histogram of a digital image with gray levels rk
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Normalized Histogram

Dividing each of histogram at gray level rk by
the total number of pixels in the image, n
p(rk) = nk / n



For k = 0,1,…,L-1
p(rk) gives an estimate of the probability of
occurrence of gray level rk
The sum of all components of a normalized
histogram is equal to 1
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Histogram Processing



Basic for numerous spatial domain
processing techniques
Used effectively for image enhancement
Information inherent in histograms also
is useful in image compression and
segmentation
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h(rk) or p(rk)
Example
rk
Dark image
Components of
histogram are
concentrated on the
low side of the gray
scale.
Bright image
Components of
histogram are
concentrated on the
high side of the gray
scale.
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Example
Low-contrast image
histogram is narrow
and centered toward
the middle of the
gray scale
High-contrast image
histogram covers broad
range of the gray scale
and the distribution of
pixels is not too far from
uniform, with very few
vertical lines being much
higher than the others
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Histogram Equalization

an output image is obtained by mapping
each pixel with level rk in the input image
into a corresponding pixel with level sk in
the output image
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Example
before
after
Histogram
equalization
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Example
No. of pixels
6
2
3
3
2
4
2
4
3
4
3
2
3
5
3
4
2
2
4
2
4x4 image
Gray scale = [0,9]
5
1
Gray level
0 1 2 3 4 5 6 7 8 9
histogram
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Gray
Level(j)
No. of
pixels
0
1
2
3
4
5
6
7
8
9
0
0
6
5
4
1
0
0
0
0
0
0
6
11
15
16
16
16
16
16
k
n
j 0
j
k
nj
j 0
n
s
sx9
6
0
0
0
0
11
15 16 16 16 16 16
/
/
/
/
/
/
/
/
16 16 16 16 16 16 16 16
3.3
3
6.1
6
8.4
8
9
9
9
9
9
9
Example
No. of pixels
6
3
6
6
3
8
3
8
6
4
6
3
6
9
3
8
2
3
8
3
Output image
Gray scale = [0,9]
5
1
0 1 2 3 4 5 6 7 8 9
Gray level
Histogram equalization
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Histogram Equalization
Approach
Fig: a) Original Image (b) Original Histogram
(c) Equalized Histogram (d) Enhanced Image
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Image Enhancement by Histogram
Modification Technique
For gray levels that assume discrete values, we
deal with probabilities given by the relation
nk
p r (rk ) 
n
0 = rk= 1
k=0,1,…..L-1,
---------- (10)
Where L is number of levels, pr(rk) is the
probability of the kth gray level, nk is the total
number of times this level appears in the image ,
and n is called the total number of pixels in the
image.
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A plot of pr (rk) versus rk is usually called a
histogram, and the technique used for
obtaining a uniform histogram is known as
histogram
equalization
or
histogram
linearization.
The discrete form of equalization techniques
given by the relation
k
nj
j 0
n
s k  T (rk )  
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Example: suppose that a 64*64, 8-levels image has the graylevel distribution shown in table 1. The histogram of these
gray levels is shown in fig.3(a)
Table (1)
rk
r0 = 0
nk
790
1023
Pr(rk) = nk/n
0.19
0.25
850
0.21
656
0.16
329
0.08
245
0.06
122
0.03
81
0.2
r1=1/7
r2=2/7
r3=3/7
r4=4/7
r5=5/7
r6=6/7
r7=1
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Figure No.3 Illustrated of the histogram –equalization method
(a) Original Histogram (b) Transformation function
(c) Equalization Histogram.
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The transformation function is obtained by using
Eq. (8). For instance.
0
s o  T (ro )   p r (r j )
j 0
=pr(r0)
=0.19
Similarly,
1
s1  T (r1 )   pr (rj )
j 0
=pr(r0) + pr(r1)
=0.44
And
`
s2=0.65
s3=0.81
s4=0.89
s5=0.95
s6=0.98
s7=1.00
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The in formation function has form shown in fig.3 (b).
Since only eight equally spaced levels are allowed in this case, each of
the transformed values must be assigned to its closest valid level. Thus
we have
s0=1/7 s4=6/7
s1=3/7 s5=1
s2=5/7 s6=1
s3=6/7 s7=1
It is noted that there only five distinct histogram-equalization gray
levels. Redefining the notation to take this into account yields the levels
s0=1/7 s4=1
s1=3/7
s2=5/7
s3=6/7
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Since ro=0 was mapped to s o =1/7, there are 790 transformed
pixels with this new value . Also there are 1023 pixels with
value s1 =3/7and 850 pixels with value s2 = 5/7
How ever , since both levels r3and r4 were mapped to s3 = 6/7
, there are now 656+122=81=985 pixels with this new value.
Similarly, there are 245 + 122 +81 = 448 pixels with value
s4 =1 .Dividing these numbers by n= 4096 yield the histogram
shown in Fig.3(c) . Since a histogram is an approximation to a
probability density function, perfecting flat results are seldom
obtained when working with discrete levels.
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