Digital Image Processing. Point Processing( 点处理 )

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1 Digital Image Processing Point Processing( 点处理 )

2 Point Processing of Images In a digital image, point = pixel. Point processing transforms a pixel s value as function of its value alone; it it does not depend on the values of the pixel s neighbors.

3 Point Processing of Images Brightness and contrast adjustment Gamma correction Histogram equalization Histogram matching Color correction.

4 Point Processing - gamma - brightness original + brightness + gamma histogram mod - contrast original + contrast histogram EQ

5 The Histogram of a Grayscale Image 灰度直方图 Let I be a 1-band (grayscale) image. I(r,c) is is an 8-bit integer between 0 and 255. Histogram, h I I, of I: I: a 256-element array, h II h I I (g), for g = 1, 1, 2, 2, 3, 3,, 256, is is an integer h I I (g) = number of of pixels in in I that have value g-1.

6 The Histogram of a Grayscale Image 灰度直方图 16-level (4-bit) image Resolution: 256*256 lower subscript: number of pixels with intensity g black regions mark pixels with intensity g

7 Sum( h(i) )=? The Histogram of a Grayscale Image Black marks pixels with intensity g Plot of histogram: number of pixels with intensity g

8

9 Histogram Examples

10 Histogram Examples

11 How is the histogram useful It is an easy method to statistically compare the contents of two images Pixelwise : Histogram : But

12 Very different images may have same histograms Histogram reflects the pixel intensity distribution, not the spatial distribution!

13 The Histogram of a Grayscale Image ( g +1) h I = the number of pixels in I with graylevel g. Luminosity

14 The Histogram of a Color Image 彩色直方图 If If I is is a 3-band( 波段 ) RGB image (truecolor, 24-bit). Each of of the 3 bands is is an integer between 0 and 255. Then I has 3 histograms, for R,G,B components respectively: h R (g+1) = # of of pixels in in I(:,:,1) with intensity value g h G (g+1) = # of of pixels in in I(:,:,2) with intensity value g h B (g+1) = # of of pixels in in I(:,:,3) with intensity value g

15 The Histogram of a Color Image There is is one one histogram per per color band R, R, G, G, & B. B. Luminosity histogram is is from 1 band = (R+G+B)/3 Luminosity h I L hi R h I G h I B Luminosity: 光度

16 The Histogram of a Color Image

17 Value or Luminance Histograms 光度直方图和亮度直方图 The value histogram( 光度直方图 ) of a 3-band (truecolor) image, I, is the histogram of the value image, 1 V, 3 ( r, c) = [ R( r, c) + G( r, c) + B( r c) ] Where R, G, and B are the red, green, and blue bands of I. The luminance histogram( 亮度直方图 ) of I is the histogram of the luminance image, L ( r, c) = R( r, c) G( r, c) B( r, c)

18 Value Histogram vs. Avg. of R,G,B Histograms R,G,B,&V histograms V & avg. avg. of of R,G,B R,G,B histograms Value histogram : average followed by statistics; Avg. of R,G,B histograms: statistics followed by average.

19 Multi-Band Histogram Calculator in Matlab % Multi-band histogram calculator function h=histogram(i) [R C B]=size(I); % Row/Column/Band size % allocate the histogram h=zeros(256,1,b); % range through the intensity values for g=0:255 h(g+1,1,:) = sum(sum(i==g)); % accumulate end return;

20 Point Processing( 点处理 ) s=t(r) r: gray-level at (x,y) in original image f(x,y) s: gray-level at (x,y) in processed image g(x,y) T is called gray-level transformation( 变换 ) or mapping( 映射 )

21 Point Operations via Functional Mappings Image: I Φ, Φ, point point operator operator J J = Φ[ I ] Pixel: Input I(r,c) function, function, ff Output J(r,c) If I and then ( r, c) = g f ( g) = k J ( r, c) = k.

22 Point Operations via Functional Mappings One-band Image J ( r, c) = f ( I ( r, c) ) for all pixels locations, ( r, c). Three-band Image J ( r, c, b ) = f I ( r, c, b ) J ( ), ( r, c, b) = f ( I( r, c, b) ) for b = 1,2,3 b and all ( r, c)., or

23 Point Operations using Look-up Tables( 对照表 ) A look-up table (LUT) implements a functional mapping. If k = f ( g) for g = 0, K,255, andif k values in, takes on { 0, K,255}, K To remap a 1-band image, I, to J : then the LUT that implements f is a 256x1 array whose (g +1) th value is k = f (g). J = LUT I ( ) +1

24 Point Operations = Look-up Table Ops output value input value E.g.: index input value output

25 Point Operations using Look-up Tables If If IIis is 3-band, then a) a) each band is is mapped separately using the the same LUT for for each band or or b) b) each band is is mapped using different LUTs one for for each band. J = LUT( I + 1), (:,:, b) = LUT ( I (:,:, b) + 1) for b = 1,2,3. a) J or b) b

26 Look-Up Tables( 对照表 ) input a pixel pixel with with this thisvalue cell index contents output is is mapped to to this thisvalue

27 How to Generate a Look-Up Table For example: Let Let σ a = x 2. { 0, K,255} e ( x; a) = a( x 127)/ 32 Or in Matlab: a = 2; x = 0:255; LUT = 255./ (1+exp(-a*(x-127)/32)); This is is just one example. Digital inverse: s=l-1-r Power-law transformation: s=cr γ Log transformation: s=clog(1+r)

28 Image Negative L-1 s Convert the color to its complement O L-1 r s=(l-1)-r What is the relationship between the histogram of the original and the negative image?

29 Power-law transformations( 指数型变换 ) Power-law transformation: s=cr γ

30 Log transformations( 对数型变换 ) Log transformation: s=clog(1+r)

31 Luminosity Point Processes: h I L Increase Brightness J k ( r, c) ( r, c) g, if I k ( r, c) + g < 256, if I k ( r, c) + g > 255 { 1, 2 } is the bandindex. I k + = 255 g 0 and k, g transform mapping

32 Point Processes: Decrease Brightness J k ( r, c) g 0 and k, 3 if I k ( r, c) g 0 ( r, c) g, if I k ( r, c) { 1, 2 } is the bandindex. 0, < = I k transform mapping 255-g

33 Point Processes: Increased Gamma J ( r, c) ( r, c) 1/ γ 255 I = for γ > transform mapping

34 Point Processes: Decreased Gamma J ( r, c) ( r, c) 1/ γ 255 I = for γ < m M transform mapping

35 Gamma Correction: Effect on Histogram

36 Point Processes: Increase Contrast Let J k Tk ( r, c) = a [ Ik ( r, c) 127] + 127, where a > , if Tk ( r, c) < 0, ( r, c) = Tk ( r, c), if 0 Tk ( r, c) 255, 255, if T ( r, c) > 255. k { 1, 2, 3} k transform mapping

37 Point Processes: Decrease Contrast T k ( r, c) where = a 0 [ I ( r, c) 127] k a < 1.0 and + 127, k { 1,2,3 } transform mapping

38 Point Processes: Contrast Stretch Let Then, J m m J I = min = min ( r, c) = ( M m ) J [ I ( r, c) ], M I = max[ I ( r, c) ], [ J ( r, c) ], M = max[ J ( r, c) ]. J ( r, c) I M I J m m I I + m J. M J m J m I 127 M I 255 transform mapping The watershed value m: The gray levels below m are darkened and the levels above m are brightened.

39 Contrast stretch Contrast Stretching: to get an image with higher contrast than the original image The gray levels below m are darkened and the levels above m are brightened m 0 m 255

40 Contrast stretch Limiting case: produces a binary image (two level) from the input image

41 How to realize the S-curve for constrast stretch Strategy 1: sine function y = sin(x) Strategy 2: inverse tangent function y = arctan(x) Strategy 3: inverse function of y=x 3 1 Strategy 4: sigmoid function y = a( x c) 1+ e y = a( x 127)/32 1+ e 255 x=0:0.1:10; y=sigmf(x,[2 4]); plot(x,y) xlabel('sigmf, P=[2 4]')

42 How to realize Contrast stretch Can you design a polynomial S-curve as the contrast stretch mapping function? e.g. f(x)=a n x n +a n-1 x n-1 + +a 2 x 2 +a 1 x+a 0 Requirement: f(0)=0 f(1)=1 f (0)=0 f (1)=0; 0 1 m 0 m 1

43 Let The Probability Density Function(pdf) A = 255 g = 0 R rows by C of an Image( 概率密度函数 ) h I g Note that since h I k k ( + 1). ( g +1) (the k th color band of image I ) with value g, A is the number of pixels in I. Then, I k columns then 1 pi g h k I g k A ( + 1) = ( + 1) is the number A = That is if R C. of I pixels in is the graylevel probability density function of I. is This is the probability that an arbitrary pixel from I k has value g. k pdf [lower case]

44 The Probability Density Function(pdf) of an Image p band band (g+1) is is the the fraction of of pixels in in (a (a specific band of) of) an an image that have intensity value g. g. p band band (g+1) is is the the probability that a pixel randomly selected from the the given band has intensity value g. g. the the sum of of the the histogram h band band (g+1) over all all g from 0 to to 255 is is equal to to the the number of of pixels in in the the image, the the sum of of p band band (g+1) over all all g is is p band band is is the the normalized histogram( 归一化直方图 )) of of the the band.

45 The Cumulative Distribution Function(cdf) of an Image( 累积概率分布函数 ) Let q = [q 1 q 2 q 3 ] = I(r,c) be the value of a randomly selected pixel from I. Let g be a specific graylevel. The probability that q k g is given by cdf [lower case] ( γ + 1) g g Ik 1 γ = 0 PI ( g + 1) = p ( γ + 1) = ( γ + 1) = k I h k Ik 255 γ= 0 A γ= 0 hi γ + k γ = 0 g h ( 1), where h Ik (γ +1) is the histogram of the kth band of I. This is is the probability that any given pixel from II kk has value less than or or equal to to g. g.

46 The Cumulative Distribution Function(cdf) of an Image P band band (g+1) is is the the fraction of of pixels in in (a (a specific band of) of) an an image that have intensity values less than or or equal to to g. g. P band band (g+1) is is the the probability that a pixel randomly selected from the the given band has an an intensity value less than or or equal to to g. g. P band band (g+1) is is the the cumulative (or (or running) sum of of p band band (g+1) from 0 through g inclusive. P band band (1) (1) = p band band (1) (1) and P band band (256) = 1; 1; P band band (g+1) is is non-decreasing( 非减函数 ). ). a.k.a. a.k.a. Probability Distribution Function(PDF). Note: the Probability Distribution Function (PDF, capital letters) and the Cumulative Distribution Function (CDF) are exactly the same things. Both PDF and CDF will refer to it. However, pdf (small letters) is the density function. a.k.a. : also known as

47 Point Processes: Histogram Equalization( 直方图均衡化 ) The normalized histogram is a probability density function (pdf) From the pdf, build the cumulative distribution function (cdf) pdf cdf

48 Histogram Equalization Histogram equalization tries to match the pdf of the result image to the uniform pdf. It is easier to implement by working on the cdf index value input output Any other Solutions?

49 Point Processes: Histogram Equalization Task: remap image I so that its histogram is as close to constant as possible ( γ ) Let +1 P I be the cumulative (probability) distribution function of I. Then J has, as closely as possible, the correct histogram if J ( r c) = 255 P I ( r, c), [ + 1]. The CDF itself is used as the LUT. I all all bands processed similarly

50 LUT

51

52

53 Point Processes: Histogram Equalization Luminosity before J ( r, c) = 255 P ( g + 1) I after

54 Histogram Equalization pdf CDF

55 Histogram Equalization pdf The The CDF CDF (cummulative distribution) is is the the LUT LUT for for remapping. LUT J ( r, c) = 255 P ( g + 1) I

56 Histogram Equalization pdf The The CDF CDF (cummulative distribution) is is the the LUT LUT for for remapping. LUT J ( r, c) = 255 P ( g + 1) I

57 Histogram Equalization

58 Challenge: Histogram Equalization on specified interval Task: remap image I with min = m I and max = M I so that its histogram is as close to constant as possible and has min = m J and max = M J. ( γ ) Let P I +1 be the cumulative (probability) distribution function of I. Then J has, as closely as possible, the correct histogram if Using intensity extrema ( r c) = ( M m ) [ I ( r, c) + 1] PI ( mi + 1) 1 P ( m + 1) P J + I, J J J I I m.

59 Point Processes: Histogram Matching( 直方图匹配 ) Task: remap image I so that it has, as closely as possible, the same histogram as image J. Because the images are digital it is not, in general, possible to make h I h J. Therefore, p I p J. Q: How, then, can the matching be done? A: By matching percentiles( 百分位数 ).

60

61

62 Matching Percentiles( 匹配百分位数 ) assuming a 1-band 1-band image image or or a single single band band of of a color color image. image. Recall: The CDF of image I is such that 0 P I ( g I ) 1. P I ( g I +1) = c means that c is the fraction of pixels in I that have a value less than or equal to g I. 100c is the percentile of pixels in I that are less than or equal to g I. To To match percentiles, replace all all occurrences of of value g II in in image I I with the the value, g J J,, from image J whose percentile in in J most closely matches the the percentile of of g II in in image I. I.

63 Matching Percentiles So, to create an image, K, from image I such that K has nearly the same CDF as image J do the following: If I(r,c) = g I then let K(r,c) = g J where g J is such that P I (g I ) > P J (g J -1) AND P I (g I ) P J (g J ). assuming assuming a a 1-band 1-band image image or or a a single single band band of of a a color color image. image. Example: Example: I(r,c) I(r,c) = 55 P I I (5) (5) = P J J (9) (9) = P J J (10) (10) = K(r,c) K(r,c) = P I ( g I ) P J ( g J ) g I g J

64 Example: Histogram Matching Image pdf Image with with intensity values g

65 Example: Histogram Matching Image CDF CDF I (g) * g *a.k.a Cumulative Distribution Function, CDF I.

66 Example: Histogram Matching Target pdf Target with with intensity values g

67 Example: Histogram Matching Target CDF CDF I (g) * g *a.k.a Cumulative Distribution Function, CDF J.

68 LUT Creation Image CDF Target CDF LUT 10

69 Input & Target CDFs, LUT and Resultant CDF P I ( g ) Input Input P J ( g ) Target Target g g LUT ( g ) LUT LUT P K ( g ) Result Result g g

70 Histogram Matching Algorithm assuming assuming a a 1-band 1-band image image or or a a single single band band of of a a color color image. image. [R,C] = size(i); K = zeros(r,c); g J = m J; for g I = m I to M I while g J < 255 AND PI ( gi + 1) P ( g + 1) < P ( g + 1) g J g = J J J +1; end K = K + g I = = end [ ( )] J g I I I < 1 AND This This directly matches image IIto to image J. J. ( I + ) ( + ) P g 1 : CDF of I I P g 1 : CDF of J. J J m M m M J J I I = min J, = max J, = min I, = max I. Better to to use use a LUT. See See next slide

71 Histogram Matching with a Lookup Table The algorithm on the previous slide matches one image to another directly. Often it is faster or more versatile to use a lookup table (LUT). Rather than remapping each pixel in the image separately, one can create a table that indicates to which target value each input value should be mapped. Then K = LUT[I+1] In Matlab if the LUT is a matrix with values from 0 to 255 and if image I is of type uint8, it can be remapped with the following code: K = uint8(lut(double(i)+1));

72 Look Up Table for Histogram Matching LUT = zeros(256,1); g J = 0; for g I = 0 to 255 while P ( g ) < P ( g + 1) end LUT end J g J g = J J + I I J 1 AND g < ; ( g + ) = g ; I 1 J This This creates a look-up table which can can then be be used to to remap the the image. ( I + ) ( J + ) ( g + ) P g 1 : CDF of I, I P g 1 : CDF of J, J LUT I 1 : Look- Up Table

73 Is there any relation between the LUT and the source/target image CDF? Example: Histogram Matching original target remapped

74 Break Get confused by more and more terms, abbreviations and acronyms? histogram, channel, band, luminance, binear, bicubic, percentiles; Interpolation, decimation, matching, equalization, normalization; LUT, cdf, pdf, vs., a.k.a.;

75 Homework III 1. Convert the following image to a visually meaningful one by point processing. Submit your code and the result image. 2. Realize your own Histogram equalization or matching algorithm. Submit your code and demo images. Compare your result with matlab function histeq().

76 Probability Density Functions of a Color Image Atlas-Mercury Atlas-Mercury red pdf green pdf blue pdf luminosity pdf

77 Cumulative Distribution Functions (CDF) red CDF green CDF blue CDF luminosity CDF

78 Probability Density Functions of a Color Image TechnoTrousers TechnoTrousers red pdf green pdf blue pdf luminosity pdf

79 Cumulative Distribution Functions (CDF) red CDF green CDF blue CDF luminosity CDF

80 Remap an Image to have the Luminance CDF of Another original target luminosity remapped

81 CDFs and the LUT Atlas-Mercury Luminosity CDF TechnoTrousers Luminosity CDF LUT (Luminosity) Atlas-Mercury to TechnoTrousers Atlas-Mercury Remapped Luminosity CDF

82 Effects of Luminance Remapping on CDFs Before After

83 Remap an Image to have the R/G/B CDFs of Another original target R, G, & B remapped

84 CDFs and the LUTs Atlas-Mercury Red PDF Atlas-Mercury Green PDF Atlas-Mercury Blue PDF TechnoTrousers Red PDF TechnoTrousers Green PDF TechnoTrousers Blue PDF LUT (Red) Atlas-Mercury to TechnoTrousers LUT (Green) Atlas-Mercury to TechnoTrousers LUT (Blue) Atlas-Mercury to TechnoTrousers Atlas-Mercury RGB Remapped Red PDF Atlas-Mercury RGB Remapped Green PDF Atlas-Mercury RGB Remapped Blue PDF

85 Effects of RGB Remapping on CDFs Before After

86 Remap an Image: To Have Two of its Color pdfs Match the Third original G & B R B & R G R & G B

87 Challenge: Display of HDRI 报告人 : 何小伟

88 什么是 HDR Dynamic Range( 动态范围 ): 是指图像中所包含的从 最亮 至 最暗 的比值, 也就是图像从 最亮 到 最暗 之间灰度划分的等级数 ; 动态范围越大, 所能表示的层次越丰富, 所包含的色彩空间也越广 高动态范围 (High Dynamic Range) 顾名思义就是从 最亮 到 最暗 可以达到非常高的比值

89 什么是 HDR 8bit 的动态范围 100:1 人眼所能感知的动态范围 :1 用 32bit float 来表示 理论上动态范围可到 10^76.8:1

90 什么是 HDR 亮的地方可以非常亮 HDR 暗的地方可以非常暗 亮暗部的细节都很明显

91 HDRI OpenEXR Industrial Light & Magic 扩展名为 (.exr) FP ^ -5 到 ^ 4 Radiance RGBE 扩展名为 (.hdr) 4 个通道, 每个通道 8bit Float TIFF 扩展名为 (.tif) 3 个通道每个通道为 FP32

92 Radiance RGBE rgb 为颜色通道,e 为亮度级别, 每个通道 8bit R = r * 2^(e 128-8) G = g * 2^(e 128-8) B = b * 2^(e 128-8)

93 什么是 HDR

94 Tone mapping 动态范围的压缩, 压缩后具有较好的全局对比度 ; 压缩后的图像具有真实感, 避免光晕 (halo) 等痕迹, 较好地保持局部细节

95 Tone mapping 全局映射 局部映射 分段线性映射 双边滤波

96 Tone mapping 存在信息丢失 关键 : 保留主要特征区域 如边缘信息, 而去除噪声

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