The document discusses histograms and histogram equalization for digital image processing. It defines a histogram as estimating the probability distribution function of gray values in an image and providing insight into an image's contrast. Histogram equalization is introduced as a technique that transforms an image's gray values such that the transformed values are uniformly distributed, improving contrast by spreading out the most frequent intensities. The key steps of histogram equalization are outlined.

1. Histogram Processing
IT 472: Digital Image Processing, Lecture 7

2. Histogram
The i th histogram entry for a digital image is
M N
1
h(ri ) = χri (f [i, j]), 0 ≤ ri ≤ L − 1,
MN
i=1 j=1
where
χri (f [i, j]) = 1 if f [i, j] = ri
= 0 otherwise
Estimates the probability distribution function of gray values
in an image.
Gives a good idea of contrast in an image.
IT472: Lecture 7 2/17

3. Histogram
The i th histogram entry for a digital image is
M N
1
h(ri ) = χri (f [i, j]), 0 ≤ ri ≤ L − 1,
MN
i=1 j=1
where
χri (f [i, j]) = 1 if f [i, j] = ri
= 0 otherwise
Estimates the probability distribution function of gray values
in an image.
Gives a good idea of contrast in an image.
IT472: Lecture 7 2/17

4. Histogram
The i th histogram entry for a digital image is
M N
1
h(ri ) = χri (f [i, j]), 0 ≤ ri ≤ L − 1,
MN
i=1 j=1
where
χri (f [i, j]) = 1 if f [i, j] = ri
= 0 otherwise
Estimates the probability distribution function of gray values
in an image.
Gives a good idea of contrast in an image.
IT472: Lecture 7 2/17

5. Example of histograms
IT472: Lecture 7 3/17

6. Histogram Processing
Conclusion: A high contrast image will tend to have a wide
range of gray-values with a uniform distribution.
Given an input image with gray values treated as a random
variable r with pdf pR (r ), can we design a transformation T
such that the random variable
s = T (r )
is distributed uniformly, i.e. pS (s) = a, ∀s.
The answer is (almost) always YES and the algorithm to do
so is called Histogram Equalization.
IT472: Lecture 7 4/17

7. Histogram Processing
Conclusion: A high contrast image will tend to have a wide
range of gray-values with a uniform distribution.
Given an input image with gray values treated as a random
variable r with pdf pR (r ), can we design a transformation T
such that the random variable
s = T (r )
is distributed uniformly, i.e. pS (s) = a, ∀s.
The answer is (almost) always YES and the algorithm to do
so is called Histogram Equalization.
IT472: Lecture 7 4/17

8. Histogram Processing
Conclusion: A high contrast image will tend to have a wide
range of gray-values with a uniform distribution.
Given an input image with gray values treated as a random
variable r with pdf pR (r ), can we design a transformation T
such that the random variable
s = T (r )
is distributed uniformly, i.e. pS (s) = a, ∀s.
The answer is (almost) always YES and the algorithm to do
so is called Histogram Equalization.
IT472: Lecture 7 4/17

9. Histogram equalization
Assume continuous random variables 0 ≤ r , s ≤ 1. Let’s try to
search for a good T :
bijective,
monotonically increasing.
Crucial observation: Given that s0 = T (r0 ), how are
FS (s0 ) = P(s ≤ s0 ) and FR (r0 ) = P(r ≤ r0 ) related?
For all s = T (r ), FS (s) = FR (r ), since T is bijective and
monotonically increasing.
s0 r0
FS (s0 = T (r0 )) = pS (s) ds = FR (r ) = pR (r ) dr
0 0
What should pS (s) be?
s is uniformly distributed ⇒ pS (s) = 1, ∀s.
IT472: Lecture 7 5/17

10. Histogram equalization
Assume continuous random variables 0 ≤ r , s ≤ 1. Let’s try to
search for a good T :
bijective,
monotonically increasing.
Crucial observation: Given that s0 = T (r0 ), how are
FS (s0 ) = P(s ≤ s0 ) and FR (r0 ) = P(r ≤ r0 ) related?
For all s = T (r ), FS (s) = FR (r ), since T is bijective and
monotonically increasing.
s0 r0
FS (s0 = T (r0 )) = pS (s) ds = FR (r ) = pR (r ) dr
0 0
What should pS (s) be?
s is uniformly distributed ⇒ pS (s) = 1, ∀s.
IT472: Lecture 7 5/17

11. Histogram equalization
Assume continuous random variables 0 ≤ r , s ≤ 1. Let’s try to
search for a good T :
bijective,
monotonically increasing.
Crucial observation: Given that s0 = T (r0 ), how are
FS (s0 ) = P(s ≤ s0 ) and FR (r0 ) = P(r ≤ r0 ) related?
For all s = T (r ), FS (s) = FR (r ), since T is bijective and
monotonically increasing.
s0 r0
FS (s0 = T (r0 )) = pS (s) ds = FR (r ) = pR (r ) dr
0 0
What should pS (s) be?
s is uniformly distributed ⇒ pS (s) = 1, ∀s.
IT472: Lecture 7 5/17

12. Histogram equalization
Assume continuous random variables 0 ≤ r , s ≤ 1. Let’s try to
search for a good T :
bijective,
monotonically increasing.
Crucial observation: Given that s0 = T (r0 ), how are
FS (s0 ) = P(s ≤ s0 ) and FR (r0 ) = P(r ≤ r0 ) related?
For all s = T (r ), FS (s) = FR (r ), since T is bijective and
monotonically increasing.
s0 r0
FS (s0 = T (r0 )) = pS (s) ds = FR (r ) = pR (r ) dr
0 0
What should pS (s) be?
s is uniformly distributed ⇒ pS (s) = 1, ∀s.
IT472: Lecture 7 5/17

13. Histogram equalization
Assume continuous random variables 0 ≤ r , s ≤ 1. Let’s try to
search for a good T :
bijective,
monotonically increasing.
Crucial observation: Given that s0 = T (r0 ), how are
FS (s0 ) = P(s ≤ s0 ) and FR (r0 ) = P(r ≤ r0 ) related?
For all s = T (r ), FS (s) = FR (r ), since T is bijective and
monotonically increasing.
s0 r0
FS (s0 = T (r0 )) = pS (s) ds = FR (r ) = pR (r ) dr
0 0
What should pS (s) be?
s is uniformly distributed ⇒ pS (s) = 1, ∀s.
IT472: Lecture 7 5/17

14. Histogram equalization
Assume continuous random variables 0 ≤ r , s ≤ 1. Let’s try to
search for a good T :
bijective,
monotonically increasing.
Crucial observation: Given that s0 = T (r0 ), how are
FS (s0 ) = P(s ≤ s0 ) and FR (r0 ) = P(r ≤ r0 ) related?
For all s = T (r ), FS (s) = FR (r ), since T is bijective and
monotonically increasing.
s0 r0
FS (s0 = T (r0 )) = pS (s) ds = FR (r ) = pR (r ) dr
0 0
What should pS (s) be?
s is uniformly distributed ⇒ pS (s) = 1, ∀s.
IT472: Lecture 7 5/17

15. Histogram equalization
Assume continuous random variables 0 ≤ r , s ≤ 1. Let’s try to
search for a good T :
bijective,
monotonically increasing.
Crucial observation: Given that s0 = T (r0 ), how are
FS (s0 ) = P(s ≤ s0 ) and FR (r0 ) = P(r ≤ r0 ) related?
For all s = T (r ), FS (s) = FR (r ), since T is bijective and
monotonically increasing.
s0 r0
FS (s0 = T (r0 )) = pS (s) ds = FR (r ) = pR (r ) dr
0 0
What should pS (s) be?
s is uniformly distributed ⇒ pS (s) = 1, ∀s.
IT472: Lecture 7 5/17

16. Histogram equalization
Assume continuous random variables 0 ≤ r , s ≤ 1. Let’s try to
search for a good T :
bijective,
monotonically increasing.
Crucial observation: Given that s0 = T (r0 ), how are
FS (s0 ) = P(s ≤ s0 ) and FR (r0 ) = P(r ≤ r0 ) related?
For all s = T (r ), FS (s) = FR (r ), since T is bijective and
monotonically increasing.
s0 r0
FS (s0 = T (r0 )) = pS (s) ds = FR (r ) = pR (r ) dr
0 0
What should pS (s) be?
s is uniformly distributed ⇒ pS (s) = 1, ∀s.
IT472: Lecture 7 5/17

17. Histogram Equalization
s0 r0
∴ FS (s0 ) = 0 1 ds = s0 = FR (r0 ) = 0 pR (r ) dr
Histogram Equalization transformation
r0
s0 = FR (r0 ) = 0 pR (r ) dr
In general, if s = T (r ), where T is bijective, monotonically
dr
increasing and differentiable then pS (s) = pR (r ) ds
ds dT (r )
dr = dr
d r
= dr 0 pR (¯)r d¯
r
= pR (r ) Using Leibnitz rule .
This gives pS (s) = pR (r ) pR1 ) = 1 ⇒ s is uniformly
(r
distributed.
IT472: Lecture 7 6/17

18. Histogram Equalization
s0 r0
∴ FS (s0 ) = 0 1 ds = s0 = FR (r0 ) = 0 pR (r ) dr
Histogram Equalization transformation
r0
s0 = FR (r0 ) = 0 pR (r ) dr
In general, if s = T (r ), where T is bijective, monotonically
dr
increasing and differentiable then pS (s) = pR (r ) ds
ds dT (r )
dr = dr
d r
= dr 0 pR (¯)r d¯
r
= pR (r ) Using Leibnitz rule .
This gives pS (s) = pR (r ) pR1 ) = 1 ⇒ s is uniformly
(r
distributed.
IT472: Lecture 7 6/17

19. Histogram Equalization
s0 r0
∴ FS (s0 ) = 0 1 ds = s0 = FR (r0 ) = 0 pR (r ) dr
Histogram Equalization transformation
r0
s0 = FR (r0 ) = 0 pR (r ) dr
In general, if s = T (r ), where T is bijective, monotonically
dr
increasing and differentiable then pS (s) = pR (r ) ds
ds dT (r )
dr = dr
d r
= dr 0 pR (¯)r d¯
r
= pR (r ) Using Leibnitz rule .
This gives pS (s) = pR (r ) pR1 ) = 1 ⇒ s is uniformly
(r
distributed.
IT472: Lecture 7 6/17

20. Histogram Equalization
s0 r0
∴ FS (s0 ) = 0 1 ds = s0 = FR (r0 ) = 0 pR (r ) dr
Histogram Equalization transformation
r0
s0 = FR (r0 ) = 0 pR (r ) dr
In general, if s = T (r ), where T is bijective, monotonically
dr
increasing and differentiable then pS (s) = pR (r ) ds
ds dT (r )
dr = dr
d r
= dr 0 pR (¯)r d¯
r
= pR (r ) Using Leibnitz rule .
This gives pS (s) = pR (r ) pR1 ) = 1 ⇒ s is uniformly
(r
distributed.
IT472: Lecture 7 6/17

21. Histogram Equalization
s0 r0
∴ FS (s0 ) = 0 1 ds = s0 = FR (r0 ) = 0 pR (r ) dr
Histogram Equalization transformation
r0
s0 = FR (r0 ) = 0 pR (r ) dr
In general, if s = T (r ), where T is bijective, monotonically
dr
increasing and differentiable then pS (s) = pR (r ) ds
ds dT (r )
dr = dr
d r
= dr 0 pR (¯)r d¯
r
= pR (r ) Using Leibnitz rule .
This gives pS (s) = pR (r ) pR1 ) = 1 ⇒ s is uniformly
(r
distributed.
IT472: Lecture 7 6/17

22. Histogram equalization
nrk
For discrete gray values pR (rk ) = MN , 0 ≤ rk ≤ 255
k
Histogram equalization: sk = T (rk ) = (L − 1) j=0 pR (rj )
IT472: Lecture 7 7/17

23. Histogram equalization
nrk
For discrete gray values pR (rk ) = MN , 0 ≤ rk ≤ 255
k
Histogram equalization: sk = T (rk ) = (L − 1) j=0 pR (rj )
IT472: Lecture 7 7/17

24. Examples
IT472: Lecture 7 8/17

25. Examples
IT472: Lecture 7 9/17

26. Histogram Equalization
What happens if we apply histogram equalization twice on an
image?
Histogram equalization is idempotent!
IT472: Lecture 7 10/17

27. Histogram Equalization
What happens if we apply histogram equalization twice on an
image?
Histogram equalization is idempotent!
IT472: Lecture 7 10/17

28. Histogram Equalization issues
IT472: Lecture 7 11/17

29. Histogram Equalization issues
IT472: Lecture 7 12/17

30. Histogram Specification
IT472: Lecture 7 13/17

31. Histogram Specification
Can we build a transformation s = T (r ), where r follows the
density function pR (r ), such that s follows a particular
specified density function pS (s)?
Idea: Assume continuous random variables between 0 and 1.
Using Histogram equalization, we can compute z = T1 (r ),
¯
such that pZ (¯) = 1.
¯ z
Similarly, we can compute z = T2 (s), such that pZ (˜) = 1.
˜ ˜ z
What is the density function of the random variable
−1
z = T2 (T1 (r ))?
pZ (z) = pS (s)
Histogram Specification
−1
T = T2 · T1 achieves the specified density function.
IT472: Lecture 7 14/17

32. Histogram Specification
Can we build a transformation s = T (r ), where r follows the
density function pR (r ), such that s follows a particular
specified density function pS (s)?
Idea: Assume continuous random variables between 0 and 1.
Using Histogram equalization, we can compute z = T1 (r ),
¯
such that pZ (¯) = 1.
¯ z
Similarly, we can compute z = T2 (s), such that pZ (˜) = 1.
˜ ˜ z
What is the density function of the random variable
−1
z = T2 (T1 (r ))?
pZ (z) = pS (s)
Histogram Specification
−1
T = T2 · T1 achieves the specified density function.
IT472: Lecture 7 14/17

33. Histogram Specification
Can we build a transformation s = T (r ), where r follows the
density function pR (r ), such that s follows a particular
specified density function pS (s)?
Idea: Assume continuous random variables between 0 and 1.
Using Histogram equalization, we can compute z = T1 (r ),
¯
such that pZ (¯) = 1.
¯ z
Similarly, we can compute z = T2 (s), such that pZ (˜) = 1.
˜ ˜ z
What is the density function of the random variable
−1
z = T2 (T1 (r ))?
pZ (z) = pS (s)
Histogram Specification
−1
T = T2 · T1 achieves the specified density function.
IT472: Lecture 7 14/17

34. Histogram Specification
Can we build a transformation s = T (r ), where r follows the
density function pR (r ), such that s follows a particular
specified density function pS (s)?
Idea: Assume continuous random variables between 0 and 1.
Using Histogram equalization, we can compute z = T1 (r ),
¯
such that pZ (¯) = 1.
¯ z
Similarly, we can compute z = T2 (s), such that pZ (˜) = 1.
˜ ˜ z
What is the density function of the random variable
−1
z = T2 (T1 (r ))?
pZ (z) = pS (s)
Histogram Specification
−1
T = T2 · T1 achieves the specified density function.
IT472: Lecture 7 14/17

35. Histogram Specification
Can we build a transformation s = T (r ), where r follows the
density function pR (r ), such that s follows a particular
specified density function pS (s)?
Idea: Assume continuous random variables between 0 and 1.
Using Histogram equalization, we can compute z = T1 (r ),
¯
such that pZ (¯) = 1.
¯ z
Similarly, we can compute z = T2 (s), such that pZ (˜) = 1.
˜ ˜ z
What is the density function of the random variable
−1
z = T2 (T1 (r ))?
pZ (z) = pS (s)
Histogram Specification
−1
T = T2 · T1 achieves the specified density function.
IT472: Lecture 7 14/17

36. Histogram Specification
Can we build a transformation s = T (r ), where r follows the
density function pR (r ), such that s follows a particular
specified density function pS (s)?
Idea: Assume continuous random variables between 0 and 1.
Using Histogram equalization, we can compute z = T1 (r ),
¯
such that pZ (¯) = 1.
¯ z
Similarly, we can compute z = T2 (s), such that pZ (˜) = 1.
˜ ˜ z
What is the density function of the random variable
−1
z = T2 (T1 (r ))?
pZ (z) = pS (s)
Histogram Specification
−1
T = T2 · T1 achieves the specified density function.
IT472: Lecture 7 14/17

37. Histogram Specification
Can we build a transformation s = T (r ), where r follows the
density function pR (r ), such that s follows a particular
specified density function pS (s)?
Idea: Assume continuous random variables between 0 and 1.
Using Histogram equalization, we can compute z = T1 (r ),
¯
such that pZ (¯) = 1.
¯ z
Similarly, we can compute z = T2 (s), such that pZ (˜) = 1.
˜ ˜ z
What is the density function of the random variable
−1
z = T2 (T1 (r ))?
pZ (z) = pS (s)
Histogram Specification
−1
T = T2 · T1 achieves the specified density function.
IT472: Lecture 7 14/17

38. Histogram Specification: Digital Images
From Histogram equalization:
sk = T1 (rk ) = (L − 1) k pR (rj ).
j=0
Histogram equalization on the specified histogram:
T2 (zp ) = (L − 1) p pZ (zp )
i=0
You may need to round-off non-integer values.
−1
Computing zp = T2 · T1 (rk ), ∀rk may not be feasible when
working with digital images.
Find zp such that T2 (zp ) − sk is minimized for every sk .
There may not be a unique minimizer. In this case, use the
smallest sk .
IT472: Lecture 7 15/17

39. Histogram Specification: Digital Images
From Histogram equalization:
sk = T1 (rk ) = (L − 1) k pR (rj ).
j=0
Histogram equalization on the specified histogram:
T2 (zp ) = (L − 1) p pZ (zp )
i=0
You may need to round-off non-integer values.
−1
Computing zp = T2 · T1 (rk ), ∀rk may not be feasible when
working with digital images.
Find zp such that T2 (zp ) − sk is minimized for every sk .
There may not be a unique minimizer. In this case, use the
smallest sk .
IT472: Lecture 7 15/17

40. Histogram Specification: Digital Images
From Histogram equalization:
sk = T1 (rk ) = (L − 1) k pR (rj ).
j=0
Histogram equalization on the specified histogram:
T2 (zp ) = (L − 1) p pZ (zp )
i=0
You may need to round-off non-integer values.
−1
Computing zp = T2 · T1 (rk ), ∀rk may not be feasible when
working with digital images.
Find zp such that T2 (zp ) − sk is minimized for every sk .
There may not be a unique minimizer. In this case, use the
smallest sk .
IT472: Lecture 7 15/17

41. Histogram Specification: Digital Images
From Histogram equalization:
sk = T1 (rk ) = (L − 1) k pR (rj ).
j=0
Histogram equalization on the specified histogram:
T2 (zp ) = (L − 1) p pZ (zp )
i=0
You may need to round-off non-integer values.
−1
Computing zp = T2 · T1 (rk ), ∀rk may not be feasible when
working with digital images.
Find zp such that T2 (zp ) − sk is minimized for every sk .
There may not be a unique minimizer. In this case, use the
smallest sk .
IT472: Lecture 7 15/17

42. Histogram Specification: Digital Images
From Histogram equalization:
sk = T1 (rk ) = (L − 1) k pR (rj ).
j=0
Histogram equalization on the specified histogram:
T2 (zp ) = (L − 1) p pZ (zp )
i=0
You may need to round-off non-integer values.
−1
Computing zp = T2 · T1 (rk ), ∀rk may not be feasible when
working with digital images.
Find zp such that T2 (zp ) − sk is minimized for every sk .
There may not be a unique minimizer. In this case, use the
smallest sk .
IT472: Lecture 7 15/17

43. Histogram Specification: Digital Images
From Histogram equalization:
sk = T1 (rk ) = (L − 1) k pR (rj ).
j=0
Histogram equalization on the specified histogram:
T2 (zp ) = (L − 1) p pZ (zp )
i=0
You may need to round-off non-integer values.
−1
Computing zp = T2 · T1 (rk ), ∀rk may not be feasible when
working with digital images.
Find zp such that T2 (zp ) − sk is minimized for every sk .
There may not be a unique minimizer. In this case, use the
smallest sk .
IT472: Lecture 7 15/17

44. Histogram Specification example
IT472: Lecture 7 16/17

45. Histogram Specification example
IT472: Lecture 7 17/17