Intensity Histogram

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Intensity Histogram

This histogram is a graph showing the number of pixels in an image at

each different intensity value found in that image.

For an 8-bit greyscale image there are 256 different possible intensities,

and so the histogram will graphically display 256 numbers showing thedistribution of pixels amongst those greyscale values.

The histogram of a digital

image with gray levels in

the range [0,L-1] is a

discrete function h(rk) = n

k,

where rk

is the kth gray

level and nk

is the number

of pixels in the image

having gray level rk.


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s ogram rocess ng - owWorks

The operation is very simple. The image is scanned in a single pass

and a running count of the number of pixels found at each intensity

value is kept. This is then used to construct a suitable histogram.

The histogram of an 8-bit image, for example can be thought of as a

table with 256 entries, or bins, indexed from 0 to 255. In bin 0 we

record the number of times a gray level of 0 occurs; in bin 1 we

record the number of times a grey level of 1 occurs, and so on, up to

bin 255. See algorithm 6.3

2. For all pixels (x,y) of the image f, increment hf[f(x,y)] by 1 .

1. Assign zero values to all element of the

array hf ;


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Another way to get the histogram is to

use the C code, as following:

char image[rows][cols];

int histogram[256];

int row, col, i;

for (i = 0; i < 256; i++) histogram[j]=0;

for (row = 0; row < rows; row++)for(col = 0; col < cols; col++)

histogram[(int) image[row][col]++;


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Intensity Histogram - Guidelines for UseHistograms have many uses. One of the more common is to decide what

value of threshold to use when converting a greyscale image to a binary

one by thresholding. If the image is suitable for thresholding then thehistogram will be bi-modal--- i.e. the pixel intensities will be clustered

around two well separated values. A suitable threshold for separating

these two groups will be found somewhere in between the two peaks in

the histogram. If the distribution is not like this then it is unlikely that agood segmentation can be produced by thresholding.

The intensity histogram for the input image is
http://www.dai.ed.ac.uk/HIPR2/images/wdg2.gif


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Intensity Histogram - Guidelines for Use

The object being viewed is dark in color and it is placed on a light

background, and so the histogram exhibits a good bi-modal

distribution. One peak represents the object pixels, one represents the

background.
http://www.dai.ed.ac.uk/HIPR2/images/wdg2.gif


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The histogram is the the same but with the y-axisexpanded to show more detail. It is clear that a threshold

value of around 120 should segment the picture nicely as

can be seen in
http://www.dai.ed.ac.uk/HIPR2/images/wdg2thr2.gif


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Intensity Histogram - Guidelines for UseThe histogram of image is

This time there is a significant incident illumination gradient across the

image, and this blurs out the histogram. The bi-modal distribution has

been destroyed and it is no longer possible to select a single global

threshold that will neatly segment the object from its background.Two failed

thresholding

segmentations are

shown in
http://www.dai.ed.ac.uk/HIPR2/images/wdg3thr2.gifhttp://www.dai.ed.ac.uk/HIPR2/images/wdg3thr1.gifhttp://www.dai.ed.ac.uk/HIPR2/images/wdg3.gif


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The histogram is used and altered by many imageenhancement operators. Two operators which are

closely connected to the histogram are contrast

stretching and histogram equalization They are

based on the assumption that an image has to use thefull intensity range to display the maximum contrast.


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Intensity Histogram and Contrast stretching

Contrast stretching takes an image in which the

intensity values don't span the full intensity range and

stretches its values linearly.

The histogram shows that most of the pixels have rather high intensity

values.


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Contrast stretching the image yields

which has a clearly improved contrast.

The corresponding histogram is


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Multiplication of

gray levels by a

constant gain will

spread out the

histogram evenly if

a>1, increasing thespacing between

occupied bins,or

compress it if a


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Intensity Histogram and Contrast stretchingIf we expand they-axis, as was done in

We can see that now the pixel values are distributed over the

entire intensity range. Due to the discrete character of the pixelvalues, we can't increase the number of distinct intensity

values.

That is the reason why the stretched histogram shows the gaps

between the single values.


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The present image also has low contrast. However, if we look atits histogram, we see that the entire intensity range is used and

we therefore cannot apply contrast stretching. On the other hand,

the histogram also shows that most of the pixels values are

clustered in a rather small area, whereas the top half of theintensity values is used by only a few pixels.


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Intensity Histogram and histogram

equalizationThe idea ofhistogram equalization is that the pixels should

be distributed evenly over the whole intensity range, i.e. the

aim is to transform the image so that the output image has a

flathistogram. The image results from the histogramequalization


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Intensity Histogram - Conclusions

1. consider the image intensities as randomvariables with a probability density function

(pdf).2. we can estimate the pdf from the empirical

data given in the image itself .

3. record the frequency distribution of gray

levels in an image.

o for a b-bit image, you need an array of size2b

o loop through every pixel, recording the


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Intensity Histogram - Conclusions

4. normalize the histogram by dividing eachentry by the total number of pixels

- gives an estimate for the pdf

- each element of the array gives the probabilityof that gray level occurring at a randomlyselected pixel .

5. contains global information about the image

6. discards all spatial information

7. an image has only one histogram, but manyimages may have the same histogram


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Cumulative histogram -Conclusions

Cumulative histogram

- each array element gives the number ofpixels with a gray-level less than or equal tothe gray level corresponding to the arrayelement

- easily constructed from the histogram

=

=j

i

ij hc0

Cumulative frequencies, cj, are computed from histogram

counts, hi using,


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Cumulative histogram -Conclusions

- cumulative histogram has a steep slope

in densely populated parts of the

histogram

- cumulative histogram has a gradual

slope in sparsely populated parts of thehistogram


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Effect of gray-level mapping onhistogram - Conclusions

1. adding a bias shifts the histogram

2. gain > 1 stretches the histogram (increasingcontrast)

3. gain < 1 compresses histogram (reducingcontrast)

4. nonlinear mapping stretches some regionsand compresses others


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Histogram Equalization

Histogram modeling techniques (e.g. histogram equalization)provide a sophisticated method for modifying the dynamic range

and contrast of an image by altering that image such that its

intensity histogram has a desired shape. Unlike contrast

stretching , histogram modeling operators may employ non-linearand non-monotonic transfer functions to map between pixel

intensity values in the input and output images. Histogram

equalization employs a monotonic, non-linear mapping which

re-assigns the intensity values of pixels in the input image such

that the output image contains a uniform distribution of

intensities (i.e. a flat histogram). This technique is used in image

comparison processes (because it is effective in detail enhancement)

and in the correction of non-linear effects introduced by, say, a

digitizer or display system.


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Histogram Equalization

Histogram equalization involves finding a grey scale

transformation function that creates an output image with a

uniform histogram (or nearly so).


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How do we determine this grey scale transformation function?Assume our grey levels are continuous and have been normalized to

lie between 0 (black) and 1 (white).

We must find a transformation Tthat maps grey values rin the input ima

Fto grey valuess = T(r) in the transformed image .

It is assumed that

T is single valued and monotonically increasing, and

for

The inverse transformation froms to ris given by :

r= T-1(s).


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An example of such a transfer function is

illustrated in the Figure


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Histogram Equalization - Discrete FormulationWe first need to determine the probability distribution

of grey levels in the input image.

where nk

is the number of pixels having grey level k, andNis the

total number of pixels in the image.

The transformation now becomes

Note that

,the index k=0,1,2,255, andThe values ofs

kwill have to be scaled up by 255 and rounded to the

nearest integer so that the output values of this transformation will

range from 0 to 255. Thus the discretization and rounding ofsk

to the

nearest integer will mean that the transformed image will not have a


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Histogram Equalization - Discrete Formulation

The mapping function we need is obtained simply by

rescaling the cumulative histogram so that its values

lie in the range 0-255.

Thus, an image which is transformed using its

cumulative histogram yields an output histogram

which is flat .

See Algorithm 6.4 p 125 Effords book


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The original image and its histogram, and the equalized versions. Both

images are quantized to 64 grey levels.


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Histogram EqualizationGuidelines for Use

The histogram confirms what we can see by visual inspection: this imagehas poor dynamic range. (Note that we can view this histogram as a

description of pixel probability densities by simply scaling the vertical

axis by the total number of image pixels and normalizing the horizontal

axis using the number of intensity density levels (i.e. 256). However, the

shape of the distribution will be the same in either case.)


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Histogram EqualizationGuidelines for UseIn order to improve the contrast of this image, without affecting the

structure (i.e. geometry) of the information contained therein, wecan apply the histogram equalization operator.


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Histogram EqualizationGuidelines for UseNote that the histogram is not flat (as in the examples from the

continuous case) but that the dynamic range and contrast have been

enhanced. Note also that when equalizing images with narrowhistograms and relatively few gray levels, increasing the dynamic range

has the adverse effect of increasing visual grainyness. Compare this

result with that produced by the linearcontrast stretching operator

Aftercontrast stretchingAfterequalization operator


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Histogram Equalization - Example

Although the contrast on the building is acceptable, the sky region is

represented almost entirely by light pixels. This causes most

histogram pixels to be pushed into a narrow peak in the upper

graylevel region.


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Histogram Equalization - ExampleThe histogram equalization

operator defines a mapping

based on the cumulativehistogram
http://www.dai.ed.ac.uk/HIPR2/images/bld1heq1.gif


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While histogram equalization has enhanced the contrast of the sky

regions in the image, the picture now looks artificial because there isvery little variety in the middle graylevel range. This occurs because

the transfer function is based on the shallow slope of the cumulative

histogram in the middle graylevel regions (i.e. intensity density levels

100 - 230) and causes many pixels from this region in the original

image to be mapped to similar graylevels in the output image.

After

histogram

equalization


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We can improve on this if we define a mapping

based on a sub-section of the image which

contains a better distribution of intensity

densities from the low and middle range

graylevels. If we crop the image so as to isolate

a region which contains more building than

sky.

We can then define a histogram equalization

mapping for the whole image based on the

cumulative histogram of this smaller region.

Rather than saying that equalisation

flattens a histogram, it is more

accurate to say that it linearises the

cumulative frequency distribution.


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Since the cropped image contains a more even distribution of dark

and light pixels, the slope of the transfer function is steeper andsmoother, and the contrast of the resulting image is more natural. This

idea of defining mappings based upon particular sub-sections of the

image is taken up by another class of operators which performLocal

Enhancements


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Histogram Equalization - Conclusions

- use the cumulative histogram to generate a nonlinear

gray-level mapping- cumulative histogram has a steep slope in denselypopulated parts of the histogram

- cumulative histogram has a gradual slope in sparselypopulated parts of the histogram

- scale the entries based on bits per pixel and number ofpixels

- EqualizeOp.java

- convenient because no user input is required

- histogram equalization doesn't always get us the desired

results


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Histogram equalization is limited in that it is capable of producing only

one result: an image with a uniform intensity distribution. Sometimes it

is desirable to be able to control the shape of the output histogram inorder to highlight certain intensity levels in an image.This can be accomplished by the histogram specialization operator

which maps a given intensity distribution

into a desired distributionusing a histogram equalized image

The first step in histogram specialization, is to specify the desired

output density function and write a transformationg(c).

as an intermediate

stage

It is possible to combine these two transformations such that

the image need not be histogram equalized explicitly:

Then defines a mapping from the equalized

levels of the original image,

Histogram Specification


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Histogram Specification - Conclusions

1.We can specify the shape of the histogram we want our

image to have2. Specify (perhaps interactively) the histogram we

would like

3. compute the cumulative histogram from the desired

histogram

4. find the inverse of the desired cumulative histogram

(may not be single-valued)

5. two-step process

- perform histogram equalization on the image

- perform a gray-level mapping using the inverse of the

desired cumulative histogram


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Local enhancement

1. histogram equalization and histogram specification

techniques are based on gray-level distribution over the

entire image

2. gray-levels containing important information in a small

neighborhood (region of interest) may not be present insufficient quantities to affect the computation of a

mapping based on global information

3. at each pixel do the following

- compute the cumulative histogram based on a smallneighborhood around the pixel to be mapped

- apply histogram equalization using this cumulative

histogram


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Color processing

1.can apply histogram equalization to color images

2. don't want to apply it using the RGB color model

- equalizing R, G, and B bands independently

causes color shifts

3. must convert to a color model that separates

intensity information from color information (e.g.

HSI)

4. can then apply histogram equalization on theintensity band


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References

http://www.netnam.vn/unescocourse/computervision/22.htm

http://homepages.inf.ed.ac.uk/rbf/HIPR2/histeq.htm
http://www.netnam.vn/unescocourse/computervision/22.htmhttp://www.netnam.vn/unescocourse/computervision/22.htm

Intensity Histogram

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