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Image Segmentation Lecture 17 (page 1) 10.3 Thresholding 10.3.1 Foundation Segmentation problems requiring multiple thresholds are best solved using region growing methods Thresholding can be viewed as T = T [ x, y, p(x, y), f (x, y) ], where f (x, y) is gray-level at (x, y) and p(x, y) denotes some local property, for example average gray level in neighbourhood Image Segmentation Lecture 17 (page 2) A thresholded image g(x, y) is deﬁned as 1, f (x, y) > T g(x, y) = , 0, f (x, y) < T where 1 is object and 0 is background When T = T [ f (x, y) ], threshold is global When T = T [ p(x, y), f (x, y) ], threshold is local When T = T [ x, y, p(x, y), f (x, y) ], threshold is dynamic or adaptive 10.3.2 The Role of Illumination Again consider the model f (x, y) = i(x, y) r(x, y) In general, when only the reﬂectance component is present, the modes in the histogram can be more easily separated When the illumination component is present separation be- comes much more diﬃcult... Image Segmentation Lecture 17 (page 3) Explanation z(x, y) = ln f (x, y) = ln [ i(x, y) r(x, y) ] = ln i(x, y) + ln r(x, y) = i 0(x, y) + r 0(x, y) Image Segmentation Lecture 17 (page 4) Since i 0 is independent of r 0, we know from probability theory that hist (z) = hist (i 0) ∗ hist (r 0) Note that when i(x, y) ≡ constant, then i 0(x, y) = ln i(x, y) ≡ constant and hist (i 0) ≡ impulse so that hist (z) = hist (i 0) ∗ hist (r 0) ≈ hist (r 0) Therefore, when the illumination is uniform, the histogram is not substantially eﬀected With non-uniform illumination though, hist (i 0) ≡ pulse, which results in hist (z) being smoothed or smeared Image Segmentation Lecture 17 (page 5) When we have access to the source of illumination: Photograph any object with uniform reﬂectance g(x, y) = k i(x, y) and let f (x, y) r(x, y) h(x, y) = = g(x, y) k so that r(x, y) = k h(x, y) Thus, if r(x, y) can be segmented by using a single threshold T , then h(x, y) can be segmented by using a single threshold of value T /k 10.3.3 Basic Global Thresholding (1) Heuristic approach based on visual inspection of histogram (2) Algorithm used to obtain T automatically Image Segmentation Lecture 17 (page 6) Visual inspection of histogram Algorithm used to obtain T automatically (1) Select an initial estimate for T Group G1 (values > T ) (2) Segment image using T −→ Group G2 (values < T ) (3) Compute average gray level values for G1, G2 −→ µ1, µ2 (4) Compute a new threshold value T = 1 (µ1 + µ2) 2 (5) Repeat (2) through (4) until the diﬀerence in T in succes- sive iterations is smaller than T0 Image Segmentation Lecture 17 (page 7) Example 10.11: Segmentation using estimated global T Start with average gray level and T0 = 0 ˜ Algorithm results in T = 125.4 after 3 iterations, so let T = 125 Image Segmentation Lecture 17 (page 8) 10.3.4 Basic Adaptive Thresholding • Divide original image into subimages • Utilize a diﬀerent threshold to segment each subimage • Diﬃculties: Subdivision and subsequent threshold estimation Example 10.12: Image Segmentation Lecture 17 (page 9) • For subimages without boundaries, variance < 75, so when variance < 100, subimages treated as a single composite image • For subimages with boundaries, variance > 100, so when variance > 100, subimages treated separately • In both these cases T is obtained automatically with T0 midway between the minimum and maximum gray level Image Segmentation Lecture 17 (page 10) 10.3.5 Optimal Global and Adaptive Thresholding Assume two principal gray level regions... Mixture PDF describing overall gray level variation... p(z) = P1 p1(z) + P2 p2(z) P1: probability that pixel is object pixel P2: probability that pixel is background pixel P1 + P2 = 1 Select T that minimizes average error in making decision Probability in erroneously classifying background as object Z T E1(T ) = −∞ p2 (z) dz Probability in erroneously classifying object as background Z ∞ E2(T ) = T p1 (z) dz Image Segmentation Lecture 17 (page 11) Overall probability of error is E(T ) = P2 E1(T ) + P1 E2(T ) Threshold value for which the error is minimal P1 p1(T ) = P2 p2(T ) ... (∗∗) Approximate p1(z) and p2(z) with Gaussian densities... P1 2 2 P2 2 2 p(z) = √ e−(z−µ1) /2σ1 + √ e−(z−µ2) /2σ2 2π 2π Using this equation in (∗∗) results in AT 2 + BT + C = 0, where 2 2 A = σ1 − σ2 2 2 B = 2 (µ1 σ2 − µ2 σ1 ) 2 2 2 2 C = σ1 µ2 − σ2 µ2 + 2 σ1 σ2 ln (σ2 P1/σ1 P2) 2 1 Note that two threshold values are generally required 2 2 Also note that if σ 2 = σ1 = σ2 , one threshold is suﬃcient: µ1 + µ2 σ2 P 2 T = + ln 2 µ1 − µ2 P1 When P1 = P2 and/or σ = 0 the optimal threshold is the average of the means Image Segmentation Lecture 17 (page 12) Alternatively, a minimum mean square approach may be used to estimate a composite PDF from the image histogram 1 Xn ems = [ p(zi) − h(zi) ]2 n i=1 The reason for estimating the complete density is to determine the presence or absence of dominant modes in the PDF. Two dominant modes typically indicate the presence of edges in the image or subimage Example 10.13: Use of optimal thresholding for image seg- mentation Image Segmentation Lecture 17 (page 13) Example of cardioangiogram (1) Preprocessing (1.1) s = c log(1 + r) (1.2) Image subtraction (1.3) Image averaging (several images) −→ reduce random noise (2) Image subdivided into subimages and their histograms cal- culated (3) Unimodal histograms rejected and bimodal histograms ﬁt- ted by bimodal Gaussian density curves (4) At ﬁrst only those regions with bimodal histograms are assigned thresholds (5) Interpolate these thresholds to obtain thresholds for the other (unimodal) regions (6) Interpolate again so that every pixel has a threshold (7) Apply thresholds −→ binary image (8) Apply gradient operators −→ boundaries Image Segmentation Lecture 17 (page 14) Image Segmentation Lecture 17 (page 15) 10.3.6 Use of Boundary Characteristics for Histogram Improvement and Local Processing Consider only the edge pixels −→ “better” histograms Peaks of similar height Properties of “better” histograms: Symmetrical Deeper valleys −→ easier to obtain threshold 0 if |∇f | < T Let s(x, y) = ⊕ if |∇f | ≥ T and ∇2f ≥ 0 ª if |∇f | ≥ T and ∇2 f < 0 When we have a dark object on a light background: Pixels that are not edge pixels: s(x, y) = 0 Pixels on the dark side of the edge: s(x, y) = ⊕ Pixels on the light side of the edge: s(x, y) = ª A segmented binary image is then generated, where 1 is part of the object and 0 is part of the background... (· · ·)(ª, ⊕) (0 or ⊕) (⊕, ª)(· · ·) | {z } part of object Pixels in center bracket −→ 1, rest −→ 0 Image Segmentation Lecture 17 (page 16) Image Segmentation Lecture 17 (page 17) 10.4 Region-Based Segmentation Here we ﬁnd the regions directly 10.4.1 Basic formulation Let R represent the entire image region The segmentation process partitions R into n subregions, R1, R2, . . . , Rn, such that... (a) ∪n Ri = R i=1 (b) Ri is a connected region, i = 1, 2, . . . , n (c) Ri ∩ Rj = ∅ for all i and j, i 6= j (d) P (Ri) = TRUE for i = 1, 2, . . . , n (e) P (Ri ∪ Rj ) = FALSE for i 6= j Here P (Ri) is logical predicate deﬁned over all points in Ri (a) Every pixel must be in a region (b) All the points in a region must be “connected” (c) Regions must be disjoint (d) For example P (Ri) = TRUE if all the pixels in Ri have the same gray level (e) Regions Ri and Rj are diﬀerent in some sense Image Segmentation Lecture 17 (page 18) 10.4.2 Region growing • Start from a set of seed points and from these points grow the regions by appending to each seed those neighbouring pixels that have similar properties • The selection of the seed points depends on the problem. When a priory information is not available, clustering tech- niques can be used: compute the above mentioned properties at every pixel and use the centroids of clusters • The selection of similarity criteria depends on the problem under consideration and the type of image data that is available • Descriptors must be used in conjunction with connectivity (adjacency) information • Formulation of a “stopping rule”. Growing a region should stop when no more pixels satisfy the criteria for inclusion in that region. • When a model of the expected results is partially available, the consideration of additional criteria like the size of the region, the likeliness between a candidate pixel and the pixels grown so far, and the shape of the region can improve the performance of the algorithm. Image Segmentation Lecture 17 (page 19) Example 10.16: Application of region growing to weld in- spection Image Segmentation Lecture 17 (page 20) 10.4.3 Region splitting and merging Subdivide an image initially into a set of arbitrary, disjoint regions and then merge and/or split the regions in an attempt to satisfy the necessary conditions Let R represent entire image region and select a predicate P (1) Split into four disjoint quadrants any region Ri for which P (Ri) = FALSE (2) Merge any adjacent regions Rj and Rk for which P (Rj ∪ Rk ) = TRUE (3) Stop when no further merging or splitting is possible Several variations of this theme are possible Image Segmentation Lecture 17 (page 21) Example 10.17: Split and merge Deﬁne P (Ri) = TRUE if at least 80% of the pixels in Ri have the property |zi − mi| < 2σi If P (Ri) = TRUE, the value of all the pixels in Ri are set equal to mi Splitting and merging are done using the algorithm on the previous transparency Properties based on mean and standard deviation attempt to quantify the texture of a region Texture segmentation is based on using measures of texture for the predicates P (Ri)

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Lecture Notes in Computer Science, image segmentation, computer vision, International Conference, Image Processing, Pattern Recognition, Y. Chen, Lecture Notes, Remote Sensing, region growing

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posted: | 3/3/2011 |

language: | English |

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