digital image processingcnikou/courses/digital_image_processing/... · 2011. 3. 8. · 2 3 image...

I t it T f ti

Digital Image Processing

Intensity Transformations(Histogram Processing)

Christophoros Niko

University of Ioannina - Department of Computer Science

Christophoros Nikoucnikou@cs.uoi.gr

2 Contents

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

– Histogram processing– Spatial filtering – Neighbourhood operations

C. Nikou – Digital Image Processing (E12)

3 Image Histograms

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

Grey Levels

4 Histogram Examples

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5 Histogram Examples (cont…)g

6 Histogram Examples (cont…)

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• A selection of images and their histogramsg

g• Notice the relationships

between the images and their histograms

• Note that the high contrast

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image has the most evenly spaced histogram

15 Contrast Stretching

• We can fix images that have poor contrast by applying a pretty simple contrast y pp y g p y pspecification

• 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) 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

t t i t it l ill f ll th d ioutput 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.

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

• We can view intensities r and s as random variables and their histograms as probability g p ydensity 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

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

• 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:

( ) ( 1) ( )r

rs T r L p w dw= = − ∫

• 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

Substituting this result:ds

( ) ( )s rdrp s p rds

( 1) ( )rds L p rdr

Uniform pdf

yields

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

p s p rL p r

1 , 0 11

= ≤ ≤ −−

The formula for histogram equalisation in the discrete case is given

where• rk: input intensity• sk: processed intensity

0( ) ( 1) ( )

k k r jj

s T r L p r=

= = − ∑0

( 1) k

L nMN =

−= ∑

k p y• nj: the frequency of intensity j• MN: the number of image pixels.

24Histogram Equalisation (cont...)

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

Applying histogram equalization:

0( ) ( 1) ( )

k k r jj

s T r L p r=

= = − ∑

y g g0

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=

= = = + =∑

25Histogram Equalisation (cont...)

ExampleRounding to the nearest integer:

0 1 2 31.33 1 3.08 3 4.55 5 5.67 6s s s s= → = → = → = →0 1 2 3

4 5 6 76.23 6 6.65 7 6.86 7 7.00 7s s s s= → = → = → = →

26Histogram Equalization (cont…)

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

will be extended.

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27 Equalisation Transformation Functiong

28 Equalisation Examples

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29 Equalisation Transformation Functions

The functions used to equalise the images in the previous exampleg

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30 Equalisation Examples

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The functions used to equalise the images in the previous exampleg

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32 Equalisation Examples (cont…)

002) 3

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The functions used to equalise the images in the previous examplesg

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34 Histogram Specification

• Histogram equalization does not always provide the desirable results.

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

35 Histogram Specification (cont...)g

Histogram equalization

36 Histogram specification (cont.)

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

g• Problem statement:

– Given pr(r) from the image and the target histogram pz(z), estimate the transformation z=T(r).

Th l ti l it hi t li ti& W

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• The solution exploits histogram equalization.

37 Histogram specification (cont…)

•Equalize the initial histogram of the image:

( ) ( 1) ( )r

s T r L p w dw= = − ∫•Equalize the target histogram:

•Obtain the inverse transform:I ti f l f i th i&

( ) ( 1) ( )rs T r L p w dw∫

( ) ( 1) ( )r

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

( ) ( )G z T r=

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.

38 Histogram specification (cont…)

The discrete case:

•Equalize the initial histogram of the image:

•Equalize the target histogram:

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0( ) ( 1) ( )

k k r jj

s T r L p r=

= = − ∑( ) ( )G z T r=

( 1) k

L nMN =

−= ∑

( ) ( 1) ( )q

G L ∑

•Obtain the inverse transform:

1( )q kz G s−= 1( ( ))kG T r−=

( ) ( 1) ( )k q z ii

s G z L p r=

= = − ∑

39Histogram Specification (cont...)

ExampleConsider again the 3-bit 64x64 image:

It is desired to transform this histogram to:

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

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= = = = = = = =

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

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

The next step is to equalize the output:

0 1 2 3 4 5 6 7

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

= = = == = = =

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

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

( ) ( )1 ( ) 03 ( ) 05 ( ) 06 ( ) 16 ( ) 2

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

== == == == =

k qs z→

1 3→

3 4→

5 5→

6 ( ) 27 ( ) 57 ( ) 67 ( ) 7

s G zs G zs G zs G z

= == == == =

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.

ExampleNotice that due to discretization, the resulting histogram will rarely be exactly the same as the

desired histogram.

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Histogram equalizationOriginal image

44 Histogram Specification (cont...)

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Histogram equalization

Specified histogram

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Transformation function and its inverse

Resulting histogram

46 Local Histogram Processing

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• 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 enhancementDifferent kinds of image enhancement– Histograms– Histogram equalisation– Histogram specification

Next time we will start to look at spatial

filtering and neighbourhood operations

digital image processingcnikou/courses/digital_image_processing/... · 2011. 3. 8. · 2 3 image...

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