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Published byEdward Ramsey Modified over 9 years ago
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Photoconsistency constraint C2 q C1 p l = 2 l = 3 Depth labels If this 3D point is visible in both cameras, pixels p and q should have similar intensities
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Photoconsistency neighborhood C1 p C2 q l = 2 l = 3 Depth labels Photoconsistency neighbors
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Data (photoconsistency) term Photoconsistency neighborhood N photo Arbitrary set of pairs of 3D points (same depth) Current implementation: if the projection of on C2 is nearest to q Our data penalty for configuration f is Note that f p = f q =l by definition of N photo
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Data term is regular
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Tsukuba images Our results, 4 interactions
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Comparison Our results, 10 interactionsBest results [SS ’02]
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Expectation-Maximization A powerful technique for many computer vision problems
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A simple problem – Line fitting Goal: To group a bunch of points into two “best-fit” line segments
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“Chicken-egg problem” If we knew which line each point belonged to, w could compute the best-fit lines.
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Chicken-egg problem If we knew what the two best-fit lines were, we could find out which line each point belonged to.
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Expectation-Maximization (EM) Initialize: Make random guess for lines Repeat: Find the line closest to each point and group into two sets. (Expectation Step) Find the best-fit lines to the two sets (Maximization Step) Iterate until convergence The algorithm is guaranteed to converge to some local optima
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Example:
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Converged!
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Multiway Cut for Stereo and Motion with Slanted Surfaces Stan Birchfield and Carlo Tomasi ICCV 1999
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Motivation Why does it look so bad? an image from a stereo pairdisparity map from graph cuts
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Solution Think of this as a segmentation Fit plane to each region to give more accurate results Once you have these planes, reassign pixels to get better fit an image from a stereo pairdisparity map from graph cuts
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Algorithm 1. Initialize a set of pixel labels Run graph cuts with integer disparities 2. Fit a plane to each region (connected component) They solve for an affine transformation that best aligns region in left image to corresponding region in right image 3. Assign labels (planes) to pixels Use graph cuts, of course! 4. Repeat Steps 2 & 3 until convergence This style of algorithm should look familiar...
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Stereo results
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Multimodal Stereo with Graph Cuts Kim, Kolmogorov and Zabih ICCV 2003
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Multimodal stereo Suppose the two cameras are different Internal parameters, or modalities There is some consistent mapping of intensities between them At the right disparity, I 1 (p) (I 2 (p+d))
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Just do it? Problem input has no assignment cost How can we tell how much p likes d ? D(p,d) = (I 1 (p) – µ(I 2 (p+d))) 2 Depends, obviously, on µ We could compute µ from right f Suggests an iterative (EM) approach Alternate between estimating the assignment costs D and the labeling f
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Joint intensity histogram Right intensity of corresponding pixel Left intensity
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EM-style approach When f is correct, the joint histogram will be highly “concentrated” And vice-versa For a given f, we can construct an assignment cost that tends to make the joint histogram more concentrated Iterative EM-style algorithm Find f given the assignments costs Find the assignment costs given f
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Right intensity of corresponding pixel Left intensity Good: Low cost Bad: High cost Assignment costs
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Formalizing this Depends on labeling Compute D n+1 from f n
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Properties Usable with other matching algorithms Can handle spatially varying µ Need decent D 0, and D n+1 D n For correspondence, easily true (why?) Related to Mutual Information With right formula for D
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Results with synthetic distortions
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Mutual information (MI) Very powerful method for multimodal registration (not correspondence) Find the affine transformation (warping) of I1 that makes it most similar to I2 A warping implies a joint histogram A disparity map is just a very complex warping MI measures joint histogram “concentration” Search via gradient descent Very successful in practice, and widely used
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Use MI for correspondence? No obvious way to apply it A disparity map is a warping with way too many parameters Turn MI into an assignment cost? MI( f ) depends on the joint histogram Each pixel doesn’t independently make the joint histogram concentrated
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Relationship with MI Suppose the joint histogram of f is similar to that of g Not equivalent to assuming f similar to g Take the first term of the Taylor expansion for MI( f ) centered at g This is a sum over pixels With the right choice of D, we can approximate MI Choice of D is fairly natural
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Summary Multimodal correspondence can be solved using energy minimization Assignment costs depend on labeling Approximation of MI EM-style use of graph cuts Experimental results look promising Not much in the way of guarantees!
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