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a)
4 1 or 2 2 1 1 or 3 3 2 1 2 or 3 2 or 3 2 1 or 2 2 or 3 1 1 or 3 1 1 3 3 1 or 3 1 2 3 4 [0] -> [X 0 ] -> [-1 X 0] -> [0 -1 X 0] -> [-1 -1 X 0] -> [0] -> [1 0] -> [X 1 0] -> [0 X 1 0] [X]-> [-1 X] -> [-1 -1 X] -> [0 -1 -1 X] -> [-1 -1 -1 X]
b)
function [c, i] = extend_match (x, y, c1, c2, c3, oc) [c, i] = min([c1 + oc, c2 + oc, sqrt((x-y)^2) + c3]) ; Results) extend_match(1,1,3.7,3.9,3.5,0.01) ans = 3. extend_match(1,0,3.7,3.9,3.5,0.01) ans = 3.
c)
function d = stereo_1d(v1, v2, oc) d1 = length(v1) ; d2 = length(v2) ; d = [] ; CM = zeros(d1+1, d2+1) ; % For containing the matching cost MM = zeros(d1, d2) ; % For reconstructing the optimum path
for i = 0 : d for j = 0 : d if i == 0 || j == 0 CM(i+1,j+1) = oc * max([i j]) ; else [CM(i+1,j+1) MM(i, j)] = ... extend_match(v1(i), v2(j), CM(i, j+1), CM(i+1, j), CM(i, j), oc) ; end end end i = d1 ; j = d2 ; disparity = 0 ; while ( i ~= 0 && j ~= 0 ) switch MM(i,j) case 1 , disparity = disparity - 1 ; d = [' X ' d] ; i = i - 1 ; case 2 , disparity = disparity + 1 ; j = j - 1 ; case 3 , if disparity < 0 d = [' -' char('0' + abs(disparity)) ' ' d] ; else d = [' ' char('0' + disparity) ' ' d] ; end i = i -1 ; j = j -1 ; otherwise, sprintf('\nError occured during computation.\n') ; end end for j = 1 : i d = [ d 'X'] ; end Results) v1 = [1 0 1 1 0 1 1 0 0 1 1 0 1 1 1] ; v2 = [1 0 1 0 1 0 1 1 0 0 1 1 1 1 1] ; stereo_1d(v1, v2, 0.01) ans = 0 0 0 -1 -1 -1 -1 -1 -1 -1 -1 X 0 0 0
else [CM(i+1,j+1) MM(i, j)] = ... extend_match(v1(i), v2(j), CM(i, j+1), CM(i+1, j), CM(i, j), oc) ; end end end i = d1 ; j = d2 ; disparity = 0 ; sprintf('Done part 1') ; while ( i ~= 0 && j ~= 0 ) switch MM(i,j) case 1 , disparity = disparity - 1 ; d = [DL+10 d] ; i = i - 1 ; case 2 , disparity = disparity + 1 ; j = j - 1 ; case 3 , if disparity > DL d = [DL+10 d] ; elseif disparity < -DL d = [-DL-10 d] ; else d = [disparity d] ; end i = i -1 ; j = j -1 ; otherwise, sprintf('\nError occured during computation.\n') ; end end for j = 1 : i d = [DL+10 d] ; end function D = stereo_2DF(V1, V2, oc) if size(V1) ~= size(V2) error('DIM NEQ', 'The dimension of the images are not compatible', size(V1), size(V2)) ; end D = [];
[n_rows n_cols] = size(V1) ; for i = 1 : n_rows DM = stereo_1DF(V1(i,:) , V2(i,:), oc, 15) ; D = [D ; DM ] ; end D = D - min(min(D)) ; D = D / max(max(D)) * 255 ; D = uint8(D) ; Results)
Distance from P to line l : (v-1) / sqrt(2) Distance from O to Q : (v+1) / sqrt(2) Distance from O to R : 1 / sqrt(2) By the rule of similar triangle, Distance from R to S = u = ((v-1) / sqrt(2) ) / (v+1) = (v-1) / (v + 1) * 1/sqrt(2) [Another solution] First, think about the coordinate system of the case that the image plane along the line z=1 (the second image). We can have this coordinate system as a reference coordinate system. Then any point (x,z) in that plane would be mapped to v = x/z. If we observe that point (x,z) at the coordinate system of second image, it looks like the point is rotated around the y-axis – origin is the same- and the degree of the rotation is cosθ = 1/sqrt(2) , sinθ = 1/sqrt(2) , θ = π/4. (Think about the rotation matrix you’ve learned from the class) Therefore if we transform the (x,z) to the second image coordinate system, we will have x’ = (x-z) / sqrt(2) and z’ = (x+z) / sqrt(2). Then the projection of that point on x + z = 1 is u = x ‘/z ‘ * ( 1/sqrt(2) ) = (x-z) / (x+z) * (1/ sqrt(2)). v = x / z, x = zv u = (x-z) / (x+z) * (1/ sqrt(2)) = (v-1) / (v+1) * (1/ sqrt(2)) a) when v = 0, the corresponding u is (– 1 / sqrt(2)) b) (v-1) / (v+1) * (1/ sqrt(2)) Dist(O,Q) = (v+1) / sqrt(2) S O
R
Q
Dist(P,l) = (v-1) / sqrt(2) P = (v,1) Line l : (x – z) / sqrt(2) = 0
