Histogram Processing
Dr. Praveen Sankaran
Department of ECE
NIT Calicut
December 31, 2012
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 1 / 20
Outline
1 Review
2 Histogram ProcessingHistogram EqualizationHistogram Matching (Speci�cation)
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 2 / 20
Review
Contrast
De�ned as the di�erence in intensity between the highest and thelowest intensity levels in an image.
Also can be explained as - the di�erence in luminance and/or colorthat makes an object (or its representation in an image or display)distinguishable.1
1http://en.wikipedia.org/wiki/Contrast_%28vision%29Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 3 / 20
Review
Some Calculations - Contrast
Let g be an M×N digital image with l = 0,1, ...,L−1 possible graylevels.
Dispersion → Image pixel value variance.
‖g −g‖2 = 1
MN∑M−1m=0 ∑
N−1n=0
(g [m,n]−g)2
g = 1
MN∑M−1m=0 ∑
N−1n=0
g [m,n]
Units are squared here.
Contrast = ‖g −g‖=√‖g −g‖2 → standard deviation.
Note that it would take an order O (MN) algorithm to �nd this.
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 4 / 20
Review
Histogram
Let g be an M×N digital image with l = 0,1, ...,L−1 possible graylevels.
c [l ] =the number of pixels with gray level l .
De�ne relative frequency
p [l ] = c[l ]MN
, ∑L−1l=0
p [l ] = 1
→ digital image gray level distribution.
The probability that a randomly selected pixel has value l .
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 5 / 20
Review
Contrast Stretching
Idea → expand the range of intensity levels in an image so that it spans thefull intensity range of the recording medium or display device.
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 6 / 20
Histogram Processing Histogram Equalization
Outline
1 Review
2 Histogram ProcessingHistogram EqualizationHistogram Matching (Speci�cation)
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 7 / 20
Histogram Processing Histogram Equalization
Variables so far
1 g = M×N digital image with l = 0,1, ...,L−1 possible gray levels.
2 r = intensities of image g to be processed, [0 · · ·L−1].
3 s = output intensity level of g after transformation, T (r).
4 c [l ] =the number of pixels with gray level l .
5 p [l ] = c[l ]MN
, gives gray level distribution of image g .
6 And more importantly, ∑L−1l=0
p [l ] = 1.
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 8 / 20
Histogram Processing Histogram Equalization
Monotonic and Strictly Monotonic Functions
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 9 / 20
Histogram Processing Histogram Equalization
Cumulative Distribution Function
p [l ] = The probability that a randomly selected pixel has value l .
pr (r) =probability density function of r .
ps (s) =probability density function of s.
Figure: Slope of the r -s curve gives the transform function, ps (s) = pr (r)∣∣ drds
∣∣CDF of r → s = (L−1)
∫ r0 pr (w)dw = area under the probability
density function curve of r .→ What type of curve is this going to be?
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 10 / 20
Histogram Processing Histogram Equalization
What about ps (s)?
ds
dr=
dT (r)
dr= (L−1)
d
dr
[∫ r
0pr (w)dw
]=(L−1)pr (r)
ps (s) = pr (r)∣∣drds
∣∣ps (s) = pr (r)
∣∣∣ 1(L−1)pr (r)
∣∣∣= 1L−1 → Uniform distribution.
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 11 / 20
Histogram Processing Histogram Equalization
Discrete Equivalent
From earlier we have a general form, p [l ] = c[l ]MN
.
So modifying for current speci�cs, pr [lr ] =cr [lr ]MN
.
ls = T (lr ) = (L−1)∑lj=0 pr [jr ], l = 0,1, · · ·L−1.
Histogram equalization or histogram linearization.
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 12 / 20
Histogram Processing Histogram Equalization
Process Example
Objective → intensity values that cover the entire gray scale.Note → see the error creeping in?
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 13 / 20
Histogram Processing Histogram Equalization
Real Image Output Example
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 14 / 20
Histogram Processing Histogram Matching (Speci�cation)
Outline
1 Review
2 Histogram ProcessingHistogram EqualizationHistogram Matching (Speci�cation)
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 15 / 20
Histogram Processing Histogram Matching (Speci�cation)
Some More Continuous Function Theory!
s = T (r) = (L−1)∫ r0 pr (w)dw → Histogram equalization.
Let z be another random variable like r ,s, such that,
G (z) = (L−1)∫z
0pz (t)dt = s
G (z) = T (r)!→ z = G−1 (T (r)) = G−1 (s)
What does all this mean anyway?
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 16 / 20
Histogram Processing Histogram Matching (Speci�cation)
Some More Continuous Function Theory!
s = T (r) = (L−1)∫ r0 pr (w)dw → Histogram equalization.
Let z be another random variable like r ,s, such that,
G (z) = (L−1)∫z
0pz (t)dt = s
G (z) = T (r)!→ z = G−1 (T (r)) = G−1 (s)
What does all this mean anyway?
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 16 / 20
Histogram Processing Histogram Matching (Speci�cation)
Some More Continuous Function Theory!
s = T (r) = (L−1)∫ r0 pr (w)dw → Histogram equalization.
Let z be another random variable like r ,s, such that,
G (z) = (L−1)∫z
0pz (t)dt = s
G (z) = T (r)!→ z = G−1 (T (r)) = G−1 (s)
What does all this mean anyway?
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 16 / 20
Histogram Processing Histogram Matching (Speci�cation)
Explanation
To start with we have the image values we have to modify → r .
Idea → modify the image based on the histogram of some other image(with values z)!
Why? →because we like this second image for some reason!
1 Find CDF of Image 1, to perform histogram equalization (r → s).
2 Find CDF of Image 2.
3 Now assuming that values s is on the y-axis of image 2 CDF, �nd thecorresponding nearest z value.
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 17 / 20
Histogram Processing Histogram Matching (Speci�cation)
Illustration
2
2http://en.wikipedia.org/wiki/File:Histogram_matching.PNGDr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 18 / 20
Histogram Processing Histogram Matching (Speci�cation)
Summary
Histograms.
Monotonic/Strictly Monotonic.
Cumulative distribution/density function of an image.
Equalization.
Speci�cation/Matching.
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 19 / 20
Histogram Processing Histogram Matching (Speci�cation)
Questions
3.1, 3.2, 3.3, 3.4, 3.5
3.6, 3.7, 3.11
Dr. Praveen Sankaran (Department of ECE NIT Calicut )DIP Winter 2013 December 31, 2012 20 / 20
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