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Histogram Equalisation

The algorithm Given an image as below, derive the intensity mapping that will as best

as possible equalise the image histogram. The image histogram is asshown on the right. There are 8 possible grey scale levels from 0 to 7.

44444

3454335553

3454344444

f(I)

I

005146000

76543210

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Step 1: calculate the cumulative

frequency distribution

005146000f(I)

252525206000Cuf

I 76543210

Idea is to derive an intensity mapping that will make the CuF

turn into

a straight ramp.

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Step 2: Compare with the CuF of

an equalised histogram

005146000f(I)

252525206000Cuf

Feq

I

33334333

76543210

In this case the Equalised (ideal) histogram needs to have 25/8

pels in each bin = 3.125 but only integer numbers of pels are

possible (its frequency after all). So Ive added in one to make

it ok.

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Step 2: Compare with the CuF of

an equalised histogram

252525206000Cuf

33334333Feq

2522191613963CuFeq

005146000f(I)

I 76543210

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Step 2: Design the mapping

252525206000Cuf

33334333Feq

2522191613963CuFeq

005146000f(I)

I 76543210

For each intensity in the original image, find an intensity in the

transformed image that has as close as possible, the same amount

of Cumulative frequency.

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Step 2: Design the mapping

0025206000Cuf

33334333Feq

2522191613963CuFeq

005146000f(I)

I 76543210

So for intensity 0, this has Cuf 0, and the closest in the xformed ima

Is a CuFreq of 3. So Intensity 0 in input maps to intensity 0 in outpu

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Step 2: Design the mapping

252525206000Cuf

33334333Feq

2522191613963CuFeq

005146000f(I)

I 76543210

Intensity 1 in input maps to Intensity 0 in output.

[Sometimes you can decide not to map into previously used

intensities]

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Step 2: Design the mapping

252525206000Cuf

33334333Feq

2522191613963CuFeq

005146000f(I)

I 76543210

For each intensity in the original image, find an intensity in the

transformed image that has as close as possible, the same amount

of Cumulative frequency.

13

02

01

00

Output IInput I

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Step 2: Design the mapping

252525206000Cuf

33334333Feq

2522191613963CuFeq

005146000f(I)

I 76543210

4 -> 5

5413

02

01

00

Output IInput I

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Step 2: Design the mapping

252525206000Cuf

33334333Feq

2522191613963CuFeq

005146000f(I)

I 76543210

5 ->7

75

5413

02

01

00

Output IInput I

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Step 2: Design the mapping

252525206000Cuf

33334333Feq

2522191613963CuFeq

005146000f(I)

I 76543210

And so on

7776

75

5413

02

01

00

Output IInput I

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The new histogram

501400060F(EQ)

25202066660Cu(EQ)

2522191613963CuFeq

252525206000Cuf

33334333Feq

005146000f(I)

I 76543210

7776

75

5413

02

01

00

Output IInput I

Output histogram only use4 bins!

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The result

44444

34543

35553

34543

44444

55555

15751

17771

15751

55555

77

76

75

54

1302

01

00Output IInput I

Original Equalised