Intensity Transformations (Histogram Processing)cnikou/Courses/Digital_Image... · C. Nikou –...

University of Ioannina - Department of Computer Science

Intensity Transformations(Histogram Processing)

Christophoros [email protected]

Digital Image Processing

2

C. Nikou – Digital Image Processing (E12)

Contents

Over the next few lectures we will look at image enhancement techniques working in the spatial domain:

– Histogram processing– Spatial filtering – Neighbourhood operations

3


Image Histograms

The histogram of an image shows us the distribution of grey levels in the imageMassively useful in image processing, especially in segmentation

Grey Levels

Freq

uenc

ies

4


Histogram ExamplesIm

ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

5


Histogram Examples (cont…)Im

ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

6



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

7



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

8



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

9



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

10



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

11



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

12



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

13



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

14


Histogram Examples (cont…)

• A selection of images and their histograms

• Notice the relationships between the images and their histograms

• Note that the high contrast image has the most evenly spaced histogram

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

15


Contrast Stretching

• We can fix images that have poor contrast by applying a pretty simple contrast specification

• The interesting part is how do we decide on this transformation function?

16


Histogram Equalisation

• Spreading out the frequencies in an image (or equalising the image) is a simple way to improve dark or washed out images.

• At first, the continuous case will be studied:– r is the intensity of the image in [0, L-1].– we focus on transformations s=T(r):• T(r) is strictly monotonically increasing.• T(r) must satisfy:

0 ( ) 1, 0 1T r L for r L≤ ≤ − ≤ ≤ −

17


Histogram Equalisation (cont...)

• The condition for T(r) to be monotonically increasing guarantees that ordering of the output intensity values will follow the ordering of the input intensity values (avoids reversal of intensities).

• If T(r) is strictly monotonically increasing then the mapping from s back to r will be 1-1.

• The second condition (T(r) in [0,1]) guarantees that the range of the output will be the same as the range of the input.

18



a) We cannot perform inverse mapping (from s to r).b) Inverse mapping is possible.

19



• We can view intensities r and s as random variables and their histograms as probability density functions (pdf) pr(r) and ps(s).

• Fundamental result from probability theory:– If pr(r) and T(r) are known and T(r) is

continuous and differentiable, then

1( ) ( ) ( )s r rdrp s p r p r

ds dsdr

= =

20



• The pdf of the output is determined by the pdf of the input and the transformation.

• This means that we can determine the histogram of the output image.

• A transformation of particular importance in image processing is the cumulative distribution function (CDF) of a random variable:

0

( ) ( 1) ( )r

rs T r L p w dw= = − ∫

21



• It satisfies the first condition as the area under the curve increases as r increases.

• It satisfies the second condition as for r=L-1we have s=L-1.

• To find ps(s) we have to compute

dsdr 0

( 1) ( )r

rdL p w dwdr

= − ∫ ( 1) ( )rL p r= −( )dT rdr

=

22



Substituting this result:

to

yields

( ) ( )s rdrp s p rds

=

( 1) ( )rds L p rdr

= −

1( ) ( )( 1) ( )s r

r

p s p rL p r

=−

1 , 0 11

s LL

= ≤ ≤ −−

Uniform pdf

23



The formula for histogram equalisation in the discrete case is given

where• rk: input intensity• sk: processed intensity• nj: the frequency of intensity j• MN: the number of image pixels.

0( ) ( 1) ( )

k

k k r jj

s T r L p r=

= = − ∑0

( 1) k

jj

L nMN =

−= ∑

24


Histogram Equalisation (cont...)Example

A 3-bit 64x64 image has the following intensities:

Applying histogram equalization: 0

0 0 00

( ) 7 ( ) 7 ( ) 1.33r j rj

s T r p r p r=

= = = =∑1

1 1 0 10

( ) 7 ( ) 7 ( ) 7 ( ) 3.08r j r rj

s T r p r p r p r=

= = = + =∑

0( ) ( 1) ( )

k

k k r jj

s T r L p r=

= = − ∑

25


Histogram Equalisation (cont...)Example

Rounding to the nearest integer:

0 1 2 3

4 5 6 7

1.33 1 3.08 3 4.55 5 5.67 66.23 6 6.65 7 6.86 7 7.00 7

s s s ss s s s

= → = → = → = →= → = → = → = →

26


Histogram Equalization (cont…)Example

Notice that due to discretization, the resulting histogram will rarely be perfectly flat. However, it will be extended.

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

27


Equalisation Transformation FunctionIm

ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

28


Equalisation ExamplesIm

ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

1

29


Equalisation Transformation Functions

The functions used to equalise the images in the previous example

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

30


Equalisation ExamplesIm

ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

2

31



The functions used to equalise the images in the previous example

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

32


Equalisation Examples (cont…)Im

ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002) 3

4

33



The functions used to equalise the images in the previous examples

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

34


Histogram Specification

• Histogram equalization does not always provide the desirable results.

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

• Image of Phobos (Mars moon) and its histogram.• Many values near zero in the initial histogram

35


Histogram Specification (cont...)Im

ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

Histogram equalization

36


Histogram specification (cont.)

• In these cases, it is more useful to specify the final histogram.

• Problem statement:– Given pr(r) from the image and the target

histogram pz(z), estimate the transformation z=T(r).

• The solution exploits histogram equalization.

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

37


Histogram specification (cont…)

•Equalize the initial histogram of the image:

•Equalize the target histogram:

•Obtain the inverse transform:In practice, for every value of r in the image:• get its equalized transformation s=T(r).• perform the inverse mapping z=G-1(s), where s=G(z) is the equalized target histogram.

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

0

( ) ( 1) ( )r

rs T r L p w dw= = − ∫

0

( ) ( 1) ( )r

zs G z L p w dw= = − ∫1( )z G s−= 1( ( ))G T r−=

( ) ( )G z T r=

38


Histogram specification (cont…)

The discrete case:

•Equalize the initial histogram of the image:

•Equalize the target histogram:

•Obtain the inverse transform:

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

0( ) ( 1) ( )

k

k k r jj

s T r L p r=

= = − ∑

1( )q kz G s−= 1( ( ))kG T r−=

( ) ( )G z T r=0

( 1) k

jj

L nMN =

−= ∑

0( ) ( 1) ( )

q

k q z ii

s G z L p r=

= = − ∑

39


Histogram Specification (cont...)Example

Consider again the 3-bit 64x64 image:

It is desired to transform this histogram to:

with

0 1 2 3

4 5 6 7

( ) 0.00 ( ) 0.00 ( ) 0.00 ( ) 0.15( ) 0.20 ( ) 0.30 ( ) 0.20 ( ) 0.15

z z z z

z z z z

p z p z p z p zp z p z p z p z

= = = == = = =

0 1 2 3 4 5 6 70, 1, 2, 3, 4, 5, 6, 7.z z z z z z z z= = = = = = = =

40



The first step is to equalize the input (as before):

The next step is to equalize the output:

Notice that G(z) is not strictly monotonic. We must resolve this ambiguity by choosing, e.g. the smallest value for the inverse mapping.

0 1 2 3 4 5 6 71, 3, 5, 6, 6, 7, 7, 7s s s s s s s s= = = = = = = =

0 1 2 3

4 5 6 7

( ) 0 ( ) 0 ( ) 0 ( ) 1( ) 2 ( ) 5 ( ) 6 ( ) 7

G z G z G z G zG z G z G z G z

= = = == = = =

41



Perform inverse mapping: find the smallest value of zq that is closest to sk:

0 0

1 1

2 2

3 3

4 4

5 5

6 6

7 7

( ) ( )1 ( ) 03 ( ) 05 ( ) 06 ( ) 16 ( ) 27 ( ) 57 ( ) 67 ( ) 7

k i qs T r G zs G zs G zs G zs G zs G zs G zs G zs G z

== == == == == == == == =

k qs z→

1 3→

3 4→

5 5→

6 6→

7 7→e.g. every pixel with value s0=1 in the histogram-equalized image would have a value of 3 (z3) in the histogram-specified image.

42



Notice that due to discretization, the resulting histogram will rarely be exactly the same as the desired histogram.

Imag

es ta

ken

from

Gon

zale

z &

Woo

ds, D

igita

l Im

age

Pro

cess

ing

(200

2)

43



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

Histogram equalizationOriginal image

44



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

Histogram equalization

45



ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

Specified histogram

Transformation function and its inverse

Resulting histogram

46


Local Histogram ProcessingIm

ages

take

n fro

m G

onza

lez

& W

oods

, Dig

ital I

mag

e P

roce

ssin

g (2

002)

• Image in (a) is slightly noisy but the noise is imperceptible.• HE enhances the noise in smooth regions (b).• Local HE reveals structures having values close to the values of the

squares and small sizes to influence HE (c).

47


Summary

We have looked at:– Different kinds of image enhancement– Histograms– Histogram equalisation– Histogram specification

Next time we will start to look at spatial filtering and neighbourhood operations

Intensity Transformations (Histogram Processing)cnikou/Courses/Digital_Image... · C. Nikou –...

Documents

Transcript of Intensity Transformations (Histogram Processing)cnikou/Courses/Digital_Image... · C. Nikou –...