03-2 - histogram processing

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Histogram processingg p g

Spring 2008 ELEN 4304/5365 DIP 1

by Gleb V. Tcheslavski: [email protected]

Preliminaries

The histogram of a digital image with intensity levels in the range [0, L‐1] is a discrete function

( )k kh r n=

kth intensity value

Number of pixels in the image with intensity rk

Histograms are frequently normalized by the total number of pixels in the image. Assuming a M x N image, a normalized histogram


( ) , 0,1,... 1kk

np r k LMN

= = −

is related to probability of occurrence of rk in the image

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Preliminaries

Components are concentrated on the low (dark) side

Components are concentrated on the high (light) side

Narrow histogram typically concentrated around the middle

Components cover a wide range of intensities – high contrast


Histogram equalizationAssuming that r represents the intensity of an input image in the range [0, L-1] (black to white), we consider the intensity mapping

( ) 1T L 0 ≤ ≤( ) 1s T r r L= 0 ≤ ≤ −such that a) T(r) is monotonically increasing in [0, L-1] – to prevent reversal intensity artifacts;

b) T(r) being in [0, L-1] for r in [0, L-1] – to ensure the range of output intensity being the same as for input intensity.

F l


For example:

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Histogram equalizationThe image intensity levels can be viewed as random variables in [0, L-1]. Let pr(r) and ps(s) represent pdfs of rand s We need a transformation that would produce theand s. We need a transformation that would produce the output image with uniform ps(s) for an input image with an arbitrary pr(r).Note, histograms are approximations of pdfs!The desired transformation is

( ) ( 1) ( )r

s T r L p w dw= = − ∫


0

( ) ( 1) ( )rs T r L p w dw∫It can be shown that this transformation leads to the rv suniformly distributed in [0,L-1].

Histogram equalization

For instance: ( )22 0 1

1( )r

r r LLp r

⎧ ≤ ≤ −⎪ −= ⎨⎪ 0 otherwise⎪⎩

Therefore:2

0 0

2( ) ( 1) ( )1 1

r r

rrs T r L p w dw w dw

L L= = − = =

− −∫ ∫For discrete values, we work with normalized histogram

l d ti i t d f i t ti


values and summation instead of integration.

( ) , 0,1,... 1kr k

np r k LMN

≈ = −

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The discrete form of the equalizing transformation is

0 0

1( ) ( 1) ( ) 0,1,... 1k k

k k r j jj j

Ls T r L p r n k LMN= =

−= = − = = −∑ ∑

This mapping is called a histogram equalization or histogram linearization.

It’ t t i


It’s not necessary one-to-one mapping.

Histogram equalization: Ex

A simple 3-bit (L=8) 64 by 64 image has the intensity distributionIntensity levels are integers in [0, 7].y g [ ]

histogram transformation Equalized histogram


Values of the histogram equalization transform function are0

0 0 00

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

s T r p r p r=

= = = =∑

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Histogram equalization: ExSimilarly:

1

1 1 0 1( ) 7 ( ) 7 ( ) 7 ( ) 3.08r j r rs T r p r p r p r= = = + =∑0j=

and s2 = 4.55, s3 = 5.67, s4 = 6.23, s5 = 6.65, s6 = 6.86, s7 = 7.00. we notice that the transformation function has fractional values that need to be rounded to obtain a quantized histogram:

0 1 2 3

4 5 6 7

1.33 1; 3.08 3; 4.55 5; 5.67 6;6.23 6; 6.65 7; 6.86 7; 7.00 7.

s s s ss s s s

= → = → = → = →= → = → = → = →


4 5 6 76.23 6; 6.65 7; 6.86 7; 7.00 7.s s s s→ → → →Dividing these values by MN, we obtain a new histogram, No new intensity levels are allowed. Therefore, perfectly uniform results cannot be achieved for digital images.


Previously seen images

Histogram-equalized images

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images

And their histograms

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Histogram matching (specification)

Histogram equalization automatically determines a transformation function seeking to produce an output image with a uniform histogram.Another method is to generate an image having a specified histogram – histogram matching:1. Find the histogram pr(r) of the input image and determine its equalization transformation: r

Spring 2008 ELEN 4304/5365 DIP 11

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( ) ( 1) ( )r

rs T r L p w dw= = − ∫2. Use the specified pdf pz(r) of the output image to obtain the transformation function:


( ) ( 1) ( )z

zG z L p t dt s= − =∫0∫

3. Find the inverse transformation z = G-1(s) – the mapping from s to z:

[ ]1 1( ) ( )z G T r G s− −= =

4. Obtain the output image by equalizing the input image first; th f h i l i th li d i f th i

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then for each pixel in the equalized image, perform the inverse mapping to obtain the corresponding pixel of the output image.

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Spring 2008 ELEN 4304/5365 DIP 13

Original image of Phobos and its histogram

A histogram-equalized image and its histogram

Histogram matching

A large concentration of pixel values near zero in the originalvalues near zero in the original image leads to unsatisfactory results of histogram equalization (“washed out appearance”). Histogram matching may lead to much better results…

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However, this technique is a “trial-and-error” process. There are no step-by-step recipes…

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Local histogram processingSometimes, it is desired to apply histogram processing to a portion (portions) of am image…

Spring 2008 ELEN 4304/5365 DIP 15

Original image Global histogram equalization

Local histogram equalization using neighborhood of 3 x 3

Histogram statistics in image enhancement

Let the intensity in an image is represented by a discrete rv rin [0, L-1] and let p(ri) is the normalized histogram – estimate of pdf for the intensity.The nth statistical moment is

( )1

0

( ) ( )L

nn i i

i

r r m p rμ−

=

= −∑Mean value

For image intensities, a sample mean:

Spring 2008 ELEN 4304/5365 DIP 16

1 1

0 0

1 ( , )M N

x y

m f x yMN

− −

= =

= ∑∑

[ ]1 1

22

0 0

1 ( , )M N

x yf x y m

MNσ

− −

= =

= −∑∑and sample variance:

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Histogram statistics in image enhancement

As previously, we may specify global mean and variance (for the entire image) and local mean and variance for a specifiedthe entire image) and local mean and variance for a specified sub-image (subset of pixels).

Spring 2008 ELEN 4304/5365 DIP 17

Original imageResult of global histogram equalization

Result of local histogram equalization

Conclusion

Out of the methods discussed above, only theOut of the methods discussed above, only the global histogram equalization can be done completely automatically. The rest is … an art.

Spring 2008 ELEN 4304/5365 DIP 18

03-2 - histogram processing

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