Image Enhancement: Histogram Processingmanj/ece178-Fall2009/e178-L4.ppt.pdfHistogram Processing 7 Histogram Equalization ( i ) T(r) is single valued valued and monotonically increasing

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Image Enhancement: Histogram Processing

Reading: Chapter 3 (Spatial domain)

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Histogram Processing

•  Histogram Equalization •  Histogram Specification/Matching

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Histogram

p(rk)= nk/n

rk

nk : # pixels with gray level rk n : # pixels in the image

€

rk ∈ {0,1,2,…,L −1}

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Histogram

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Figure 3.15: histograms

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Histogram Modification

r : Input gray level [0, 1] s : Transformed gray level [0, 1]

s = T (r) T : Transformation function

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Histogram Equalization ( i ) T (r) is single valued valued and monotonically increasing in 0 ≤ r ≤ 1 ( ii ) 0 ≤ T (r) ≤ 1 for 0 ≤ r ≤ 1

[0, 1] T→ [0, 1] Inverse transformation : T −1(s)= r 0 ≤ s ≤ 1T −1(s) also satisfies ( i ) and ( ii )The gray levels in the image can be viewed as random variables takingvalues in the range [0,1].Let pr (r) : p.d.f. of input level r and let ps (s) : p.d.f. of s

s = T (r) ; ∴ ps (s) = pr (r) dr ds r=T −1 (s ) (from ECE 140)

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Equalization (contd.)

We are interested in obtaining a transformation function T( ) which transforms an arbitrary p.d.f. to an uniform distribution

pr(r) ps(s)

s

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Consider s = T (r) = pr (w) dw 0 ≤ r ≤ 10

r

∫

(Cumulative distribution function of r)ps (s) = pr (r)dr ds r=T −1 (s ) ;

dsdr

=ddr

pr (w)dw0

r

∫

= pr (r)

∴ ps (s) = pr (r)1

pr (r) r=T −1 (s )≡ 1 0 ≤ s ≤ 1

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Equalization: Example

2

1 r

pr(r)

pr (r) =−2r + 2 0 ≤ r ≤ 10 Else

s = T (r) = (2 − 2w) dw0

r

∫ = (2w − w2 ) 0r= 2r − r2

∴r2 − 2r + s = 0

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Equalization (example: contd.)

r =+2 ± 4 − 4s

2= 1± 1− s

r = T −1(s) = 1− 1− s as r ∈[0,1]

ps (s) = pr (r)drds

= (−2r + 2) dds

(1− 1− s)r=1− 1− s

= (−2r + 2) −12

(1− s)−1/2 (−1)

= (−2 + 2 1− s + 2) +12 1− s

= 1 0 ≤ r ≤ 1

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Equalized Histograms

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Fig 3.18: Transformation curves

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Equalization: Discrete Case

0

Number of levels

p r nn

r k L

L

s T r p r n n

r kk

k

k k r jj

j

k

j

k

( ) ; , ,...,

( ) ( )

= ≤ ≤ = −

→

= = ===∑∑

1 0 1 1

00

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Discrete Case: Example

64x64 image; 8 gray levels.

Notice that equalized histogram is not perfectly flat!

k r n n n Snn p s

ssssssss

k kk

kj

j

k

s k=

→ →

→ →

→ →

→ →

→ →

→ →

→ →

→ →

=∑0

17 0

17

37 1

27

57 2

37

67 3

47

67 3

57 4

67 4

77 4

0 0 790 019 019 0191 1023 0 25 0 0 252 850 0 21 0 65 0 213 656 016 0 81 0 244 329 0 08 0 89 0 245 245 0 06 0 95 1 0116 122 0 03 0 98 1 0117 81 0 02 10 1 011

( )

. . .. .44 .. . .. . .. . .. . .. . .. . .

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Equalization: Image Examples

Original, Equalized (64) Equalized (256)

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..and their histograms

Original, equalized (64 levels) Equalized (256 levels)

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Histogram specification

Suppose s = T (r) = pr (w) dw0

r

∫pr (r)→ Original histogram ; pz (z)→ Desired histogram

Let v = G(z) = pz (w)0

z

∫ dw and z = G−1(v)

But s and v are identical p.d.f.∴ z = G−1(v) = G−1(s) = G−1(T (r))

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Question

•  What happens when you apply equalization to an already equalized histogram –  In the continuous case? –  In the discrete case?

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Fig 3.19: Matching

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Matching: Summary

Steps: (1) Equalize the levels of original image (2) Specify the desired pz(z) and obtain G(z) (3) Apply z=G-1(s) to the levels s obtained in

step 1

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Matching: an example

z p z v G r n p zzzzz sz sz sz sz

k z k k k k z k 0 0 0

0 0 0 0 0 0 0.15 = 790 0.19 0.35 = 1023 0.25 0.65 = 850 0.21 0.85 = 985 0.24

17

37

57

67

( ) ( ) ( )

.

.

.

..

=

=

=

=

= ↔

= ↔

= ↔

= ↔

=

0

11

7

22

7

33

7 0

44

7 1

55

7 2

66

7 3

7

0 000

0 150 20 30 2

1 015 4 1.0 = 1 448 0.11 ↔ s

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Histogram Matching: example

Original image (Jenolan caves, blue mountain, Sydney, Ausralia)

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Color to grayscale

I=imread(‘sydney1.jpg); I1=rgb2gray(I); I1=imresize(I1,0.5); Imhist(I1);

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Desired & modified histograms

I2=histeq(I1,imhist(J)); Imhist(I2);

Imhist(J) J=some image

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Histogram modified image

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Fig 3.20: Another example

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Fig 3.21

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Fig 3.22