Pointwise Image Operations
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1 Poinwise Image Operaions Binary Image Analysis Jana Kosecka hp://cs.gmu.edu/~kosecka/cs482.hml - Lookup able mach image inensiy o he displayed brighness values Manipulaion of he lookup able differen Visual effecs mapping is ofen non-linear image is no changed whie black Binary Images Go from gray level images o black whie. Indusrial applicaions Simple asks couning objecs, regions, conneced componens, hinning, hickening 1. Selec pixels as foreground pixels and background pixels Conras gamma Conras Brighness 1
2 Image segmenaion Image segmenaion by hresholding Ideally, objec pixels would be black (0 inensiy) and background pixels whie (imum inensiy) Bu his rarely happens pixels overlap regions from boh he objec and he background, yielding inensiies beween pure black and whie - edge blur cameras inroduce noise during imaging - measuremen noise poaoes have non-uniform hickness, giving variaions in brighness in X-ray - model noise Bu if he objecs and background occupy differen ranges of gray levels, we can mark he objec pixels by a process called hresholding: Le F(i,j) be he original, gray level image B(i,j) is a binary image (pixels are eiher 0 or 1) creaed by hresholding F(i,j) B(i,j) = 1 if F(i,j) < B(i,j) = 0 if F(i,j) >= We will assume ha he 1 s are he objec pixels and he 0 s are he background pixels Quanizaion Hisogram Hisogram frequency gray-level -> empirical disribuion of he inensiy values h[i] number of pixels of inensiy i Hisogram equalizaion making hisogram fla Passing he image hrough look-up able of he form of cumulaive disribuion funcion Image quanized o 5 differen gray-levels 2 levels 2
3 How do we choose he hreshold for segmenaion? Hisogram (h) - gray level frequency disribuion of he gray level image F. h F (g) = number of pixels in F whose gray level is g H F (g) = number of pixels in F whose gray level is <=g h(g) peak ideal h valley inensiy, g peak observed hisogram P-ile mehod in some applicaions we know approximaely wha percenage, p, of he pixels in he image come from objecs migh have one poao in he image, or one characer. H F can be used o find he gray level, g, such ha ~p% of he pixels have inensiy <= g Then, we can examine h F in he neighborhood of g o find a good hreshold (low valley poin) Triangle algorihm Peak and valley mehod Find he wo mos prominen peaks of h g is a peak if h F (g) > h F (g ± g), g = 1,..., k Le g 1 and g 2 be he wo highes peaks, wih g 1 < g 2 Find he deepes valley, g, beween g 1 and g 2 g is he valley if h F (g) <= h F (g ), g,g in [g 1, g 2 ] Use g as he hreshold A line is consruced beween he imum of he hisogram a brighness b and he lowes value b min = (p=0)% in he image. The disance d beween he line and he hisogram h[b] is compued for all values of b from b = b min o b = b. The brighness value b o where he disance beween h[b o ] and he line is imal is he hreshold value. This echnique is paricularly effecive when he objec pixels produce a weak peak in he hisogram. 3
4 Hand selecion selec a hreshold by hand a he beginning of he day use ha hreshold all day long! Many hreshold selecion mehods in he lieraure Probabilisic mehods make parameric assumpions abou objec and background inensiy disribuions and hen derive opimal hresholds Srucural mehods Evaluae a range of hresholds wr properies of resuling binary images one wih sraighes edges, mos easily recognized objecs, ec. Local hresholding apply hresholding mehods o image windows An advanced probabilisic hreshold selecion mehod - minimizing Kullback informaion disance Suppose he observed hisogram, f, is a mixure of he gray levels of he pixels from he objec(s) and he pixels from he background in an ideal world he hisogram would conain jus wo spikes (his depends of he class of images/objecs) bu measuremen noise model noise (e.g., variaions in ink densiy wihin a characer) edge blur (misalignmen of objec boundaries wih pixel boundaries and opical imperfecions of camera) spread hese spikes ou ino hills Kullback informaion disance Kullback informaion disance f(g) Make a parameric model of he shapes of he componen hisograms of he objecs(s) and background Parameric model - he componen hisograms are assumed o be o Gaussian p o and p b are he proporions of he image ha comprise he objecs and background µ o and µ b are he mean gray levels of he objecs and background σ o and σ b - are heir sandard deviaions µ o µ b σ o σ b fo(g) = f b( g) = g 1/2( g µo ) 2 po σo e 2πσo p 1/2( g u b b e σ b 2π σ b ) 2 Now, if we hypohesize a hreshold,, hen all of hese unknown parameers can be approximaed from he image hisogram. Le f(g) be he observed and normalized hisogram f(g) = percenage of pixels from image having gray level g po() = g= 0 f (g) µo() = f (g)g µb() = f (g)g g= 0 pb() = 1 p0() g= +1 4
5 Kullback informaion disance So, for any hypohesized, we can predic wha he oal normalized image hisogram should be if our model (mixure of wo Gaussians) is correc. P (g) = p o f o (g) + p b f b (g) The oal normalized image hisogram is observed o be f(g) So, he quesion reduces o: deermine a suiable way o measure he similariy of P and f hen search for he ha gives he highes similariy Kullback informaion disance A suiable similariy measure is he Kullback direced divergence, defined as K() = f (g)log[ f (g) g=0 P(g) ] If P maches f exacly, hen each erm of he sum is 0 and K() akes on is minimal value of 0 Gray levels where P and f disagree are penalized by he log erm, weighed by he imporance of ha gray level (f(g)) An alernaive - minimize probabiliy of error An alernaive - mimimize probabiliy of error Using he same mixure model, we can search for he ha minimizes he prediced probabiliy of error during hresholding Two ypes of errors background poins ha are marked as objec poins. These are poins from he background ha are darker han he hreshold objec poins ha are marked as background poins. These are poins from he objec ha are brigher han he hreshold For each reasonable hreshold compue he parameers of he wo Gaussians and he proporions compue he wo probabiliy of errors Find he hreshold ha gives minimal overall error mos equal errors f o eb() = pb eo() = po g =0 g= +1 fb(g) fo(g) f b 5
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