(1)根据基本矩阵估计运动 1计算基本矩阵 00e0e02 E=le 12 01 20e21e2 e1 10 k+1 k+1 e 11 a k X=0 nk X=0 8对以上对应点求稳定解(实际经常使用 RANSAC算法)
(1) 根据基本矩阵估计运动 1. 计算基本矩阵 = 20 21 22 10 11 12 00 01 02 e e e e e e e e e E ( 1) 0 2 2 2 1 2 0 1 2 1 1 1 0 0 2 0 1 0 0 1 1 1 1 1 1 = + + + + + + e e e e e e e e e x x x y x y x y y y x y k k k k k k k k k k k k a X = 0 T k 0 , X = T An k 8对以上对应点求稳定解(实际经常使用 RANSAC算法)
(1)根据基本矩阵估计运动 1计算基本矩阵 00e0e02 E=le 12 01 20e21e2 e1 10 k+1 k+1 e 11 a k X=0 nk X=0 In reality, instead of solving AE=0 we seek e to ea minimize AE, least eigenvector ofAA
(1) 根据基本矩阵估计运动 1. 计算基本矩阵 = 20 21 22 10 11 12 00 01 02 e e e e e e e e e E ( 1) 0 2 2 2 1 2 0 1 2 1 1 1 0 0 2 0 1 0 0 1 1 1 1 1 1 = + + + + + + e e e e e e e e e x x x y x y x y y y x y k k k k k k k k k k k k a X = 0 T k 0 , X = T An k In reality, instead of solving , we seek E to minimize , least eigenvector of . AE = 0 AE A A
8-point algorithm To enforce that E is of rank 2, e is replaced by e that minimizes subject to E-E dete=0 It is achieved by SvD. Let e= U2V, where 0 00 ∑ 000 00 ,let∑ 0 00 0 00 then E=Ue'vlis the solution
8-point algorithm To enforce that E is of rank 2, E is replaced by E’ that minimizes subject to . E E − ' det ' 0 E = • It is achieved by SVD. Let , where , let then is the solution. E Σ = U V = 3 2 1 0 0 0 0 0 0 Σ = 0 0 0 0 0 0 0 Σ' 2 1 E ' Σ' = U V
8-point algorithm %o Build the constraint matrix A=x2(1,)*x1(1,)x2(1:).*x1(2,)x2(1,) x2(2,)*x1(1,)x2(2,)*x1(2,)x2(2,) (2,;) ones(npts, I)I U,DⅥ=svd(A) Extract fundamental matrix from the column ofv corresponding to the smallest singular value E=reshape(V( 9),, % Enforce rank2 constraint D,v= svd(e) E=U*diag(D(1,1)D(2,2)0])
8-point algorithm % Build the constraint matrix A = [x2(1,:)‘.*x1(1,:)' x2(1,:)'.*x1(2,:)' x2(1,:)' ... x2(2,:)'.*x1(1,:)' x2(2,:)'.*x1(2,:)' x2(2,:)' ... x1(1,:)' x1(2,:)' ones(npts,1) ]; [U,D,V] = svd(A); % Extract fundamental matrix from the column of V % corresponding to the smallest singular value. E = reshape(V(:,9),3,3)'; % Enforce rank2 constraint [U,D,V] = svd(E); E = U*diag([D(1,1) D(2,2) 0])*V';
Problem with 8-point algorithm 01 10 k+lk k+l.k +1 Vk+ k Vk+lk yA k+1 0 11 10000100101000-100010100-100 12 20 △ Orders of magnitude difference between column of data matrix >least-squares yields poor results 22
( 1) 0 2 2 2 1 2 0 1 2 1 1 1 0 0 2 0 1 0 0 1 1 1 1 1 1 = + + + + + + e e e e e e e e e x x x y x y x y y y x y k k k k k k k k k k k k Problem with 8-point algorithm ~10000 ~10000 ~100 ~10000 ~10000 ~100 ~100 ~100 1 ! Orders of magnitude difference between column of data matrix → least-squares yields poor results