3.3 Histogram Processing(page 142) h(r k )=n k. p(r k )=1

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1 Image enhancement in the spatial domain(3.3) SLIDE 1/18 Histogram 3.3 Histogram Processing(page 142) h(r k )=n k r k : kthgraylevel n k : numberofpixelsofgraylevelr k Normalization Discrete PDF MN: totalnumberofpixels p(r k )=n k /MN L 1 k=0 p(r k )=1 Histogram equalization: Histogram specification: 3.3.2(Development of method not discussed)

2 Image enhancement in the spatial domain(3.3) SLIDE 2/18 Examples

3 Image enhancement in the spatial domain(3.3) SLIDE 3/ Histogram Equalization First consider continuous functions and transformations of the form and assume that s=t(r), r [0,L 1] (a) T(r)monotonicallyincreasingforr [0,L 1] (Only requirement for histogram equilization) (b) T(r) [0,L 1]forr [0,L 1]

4 Image enhancement in the spatial domain(3.3) SLIDE 4/18 View gray levels as random variables p r (r): continuouspdfofr p s (s): continuouspdfofs IfT(r)continuousanddifferentiablethenp s (s)=p r (r) dr ds Considerthetransformationfunctions=T(r)=(L 1) r 0 p r (w)dw RHS is the cumulative distribution function (CDF) of r, and satisfies conditions(a) and(b). From Leibniz s rule... ds dr = dt(r) dr = (L 1) d dr = (L 1)p r (r) { r 0 } p r (w)dw p s (s) = p r (r) dr ds = p r (r) 1 (L 1)p r (r) 1 = L 1, s [0,L 1]

5 Image enhancement in the spatial domain(3.3) SLIDE 5/18 ForT(r)=(L 1) r 0 p r (w)dw,p s (s)isalwaysuniform,independentofp r (r)

6 Image enhancement in the spatial domain(3.3) SLIDE 6/18 p r (r)= Example 3.4 2r (L 1) 2, r [0,L 1] 0, otherwise s=t(r)=(l 1) r 0 p r (w)dw= 2 L 1 r o wdw= r2 L 1 Note: IfL=10,thenT(3)=3 2 /9=1. p s (s) = p r (r) dr ds 2r ( ) 1 ds = (L 1) 2 dr 2r ( d r 2 ) 1 = (L 1) 2 drl 1 2r = (L 1) (L 1) 2 2r = 1 L 1

7 Image enhancement in the spatial domain(3.3) SLIDE 7/18 Now consider discrete values... Recall p(r k )= n k MN, k=0,1,2,...,l 1 r Thediscreteversionofs=T(r)= p r (w)dw is 0 k s k = T(r k )=(L 1) p r (r j ) = (L 1) MN j=0 k n j, k=0,1,2,...,l 1 j=0 and is called histogram equalization NB:Thiswillnotproduceauniformhistogram,butwilltendtospreadout the histogram of the input image Advantages: Gray-level values cover entire scale(contrast enhancement) Fully automatic

8 Image enhancement in the spatial domain(3.3) SLIDE 8/18 Example 3.5 Consider3-bitimage(L=8)ofsize64 64pixels(MN =4096) 0 s 0 =T(r 0 )=7 p r (r j )=7p r (r 0 )=1.33 s 1 =T(r 1 )=7 j=0 1 p r (r j )=7p r (r 0 )+7p r (r 1 )=3.08 j=0

9 Image enhancement in the spatial domain(3.3) SLIDE 9/18 Rounding to nearest integer: s 0 = s 1 = s 2 = s 3 = s 4 = s 5 = s 6 = s 7 = p s (s 0 )=0; p s (s 1 )= ; p s(s 2 )=0; p s (s 3 )= ; p s(s 4 )=0; p s (s 5 )= ; p s(s 6 )= ; p s (s 7 )=

10 Image enhancement in the spatial domain(3.3) SLIDE 10/18 Example 3.6: Histogram equalization

11 Image enhancement in the spatial domain(3.3) SLIDE 11/18 Example 3.6: Transformation functions

12 Image enhancement in the spatial domain(3.3) SLIDE 12/ Histogram Matching(Specification) Some applications: hist. equalization not best approach So, generate processed image with specified histogram Development of the method: Not discussed Example 3.9: Histogram specification

13 Image enhancement in the spatial domain(3.3) SLIDE 13/18

14 Image enhancement in the spatial domain(3.3) SLIDE 14/18

15 Image enhancement in the spatial domain(3.3) SLIDE 15/ Local Enhancement(Previous methods(3.3.1 and 3.3.2) were global) Define square or rectangular neighbourhood(mask) and move the center from pixel to pixel For each neighbourhood... Calculate histogram of the points in the neighbourhood Obtain histogram equalization/specification function Map gray level of pixel centered in neighbourhood Canusenewpixelvaluesandprevioushisttocalculatenexthist Example 3.10: Enhancement using local histograms

16 Image enhancement in the spatial domain(3.3) SLIDE 16/ Use of Histogram Statistics for Image Enhancement With p(r i ) a normalized histogram, the nth moment of r (discrete) about itsmeanisdefinedas L 1 µ n (r)= (r i m) n p(r i ) wheremisthemeanvalueofr: m= i=0 L 1 i=0 r i p(r i ) Notethatµ 0 =1andµ 1 =0,andthatµ 2 isthevarianceσ 2 (r): µ 2 (r)= L 1 (r i m) 2 p(r i ) i=0 Mean: measure of average gray level Variance: measure of average contrast

17 Image enhancement in the spatial domain(3.3) SLIDE 17/18 Direct estimates from sample values sample mean and variance: m= 1 MN σ 2 = 1 MN M 1 x=0 M 1 x=0 N 1 y=0 N 1 y=0 f(x,y) [f(x,y) m] 2 Example 3.11: Shows that m and σ 2 obtained from histogram and sample values are the same Localmeanandvariance: Let(x,y)bethecoordinatesofapixelinanimage ands xy denoteasubimagecenteredat(x,y),withhistogramp Sxy,then σ 2 S xy = m Sxy = L 1 i=0 L 1 i=0 r i p Sxy (r i ) { ri m Sxy } 2pSxy (r i )

18 Image enhancement in the spatial domain(3.3) SLIDE 18/18 Example 3.12: Enhancement based on local statistics g(x,y)= { E f(x,y) ifmsxy [0,k 0 m G ]ANDσ Sxy [k 1 σ G,k 2 σ G ] f(x, y) otherwise m G : Globalmean;σ G : Globalstandarddeviation E=4.0; k 0 =0.4; k 1 =0.02; k 2 =0.4; (3 3)localregion