1. Projective geometry of 2-space DLT alg HZ 4.1 Rectification HZ 2.7 Hierarchy of maps Invariants HZ 2.4 Projective transform HZ 2.3 Behaviour at infinity Primitives pt/line/conic HZ 2.2

2. Computing a homography given 2 images of a scene, we would like to compute the homography H that relates their pixels –excellent example of the system-of- equations approach to compute a matrix (the homography) as a null vector H is a 3x3 matrix with 8 dof (9 – 1) –Hx = x’ how will we solve for these 8 degrees of freedom? solve for H using 4 point correspondences (x 1,x 1 ’),..., (x 4,x 4 ’) –bootstrap from 4 matches to all matches how will we find 4 point correspondences? SIFT (later) can also rectify using these ideas (Fig 2.4) HZ87-88, HZ35, Figure 4.9a,b

3. Lessons of DLT using SIFT for feature detection and feature matching developing the linear system of equations appproach learning to precondition data learning how to solve least squares –we have learned how to solve for null space learning RANSAC

4. Cross product as matrix multiplication we know that cross product may be interpreted as a determinant it is even more useful to interpret it as a skew-symmetric matrix a x b = [0-a 2 a 1 ] [b 0 ] [a 2 0 -a 0 ] [b 1 ] [-a 1 a 0 0] [b 2 ] this is now a matrix multiplication and fits seamlessly into our algebraic schemes we abbreviate this matrix [a x ] so that a x b = [a x ] b cute: a is the null vector of [a x ] (there’s no avoiding null space!) mnemonic: lower triangle is a2,-a1,a0 later we will interpret cross product as a tensor! HZ581, Appendix A

5. Reducing H to a null space idea: use Hx i = x i ’ to solve for H Hx i = x i ’ (which is not true in projective space; think scale) becomes x i ’ x Hx i = 0 let x i ’ = (x i ’, y i ’, w i ’), and use the skew-symmetric representation: x i ’ x Hx i = [0- w i ’ y i ’] [h 1 t x i ] [w i ’ 0 -x i ’] [h 2 t x i ] [-y i ’ x i ’ 0] [h 3 t x i ]OR [0 -w i ’ y i ’] [x i t h 1 ] [w i ’ 0 -x i ’] [x i t h 2 ] = 0 [-y i ’ x i ’ 0] [x i t h 3 ]OR [0 -w i ’x i t y i ’x i t ] [h 1 ] [w i ’x i t 0 -x i ’x i t ] [h 2 ] = 0 [h 3 ]OR (having removed the 3 rd equation, since it is linearly dependent on the others) A i h = 0 –where A i is a 2x9 matrix and h is the row-major unrolling of H this interprets H as the null vector of a 2x9 matrix

6. Preamble to DLT algorithm Note: x i ’ * row1 + y i ’ * row2 = -w i ’ * row3 DLT (Direct Linear Transformation) algorithm uses this transformation of point correspondences into a solution for the homography H A i h = 0 encodes the restriction x i ’ x Hx i = 0 (x i ’ and Hx i are equivalent) –but now in the form of a linear system (no cross product, H isolated) set the projective coordinate w = 1 in all of the points collect 4 of these pt-correspondence matrices A i, yielding an 8x9 matrix A, with a one-dimensional null space solve Ah = 0 for null vector h using SVD (see above) notice how this reverses the point of view: change the typical “solve Hx i = x i ’ for x i “ into “solve Hx i = x i ’ for H” –translate the matrix into a vector by unrolling HZ88-89

7. DLT algorithm 1.Find 4 point correspondences between 2 images of the same scene. 2.Precondition (see below). 3.Build the matrix A i from each correspondence (x i, x i ’). 4.Combine A1,...,A4 into A. 5.Find A’s null vector [see above: last column of V in SVD of A]. 6.This is the homography matrix H (encoded in row-major unrolling) that maps between the two images.

8. Preconditioning essential to normalize the images before applying DLT algorithm let { (x i, x i ’): i = 1,...,n } be the point correspondences translate and scale both pointsets translate centroid of { x i } to origin scale new { x i } so that average {dist( x i, origin)} = sqrt(2) –motivation: typical point looks like (1,1) do likewise to {x i ’}, but entirely independently –translate centroid to origin and scale so avg dist from origin is \sqrt(2) this normalization ensures that changes in x, y and w have comparable effect –e.g., if (100,100,1) is a feature point, changes in w will have 100 times more effect than changes in x or y HZ107-109

9. DLT with more than 4 correspondences we can solve for the homography using only 4 point correspondences, but SIFT will typically find many more if n>4 point correspondences are known, we can collect them into a 2n-by-9 matrix A unless these correspondences are calculated exactly (highly unlikely!), Ah=0 will not be solvable instead, minimize ||Ah|| subject to the constraint ||h||=1 –the constraint is added to avoid the zero solution this optimal solution is the last column of V in the singular value decomposition A = UDV^t –serendipitously, same technique as original DLT algorithm the vector h is reconstructed into H (see above) that is, DLT algorithm remains exactly the same with 4 replaced by n HZ90-91, 593

10. Outliers and noise it is better to use all of the computed correspondences, because some of them will be unreliable –cannot guess which 4 are reliable –can use RANSAC to determine the highly unreliable correspondences (the outliers) –then use all of the remaining (still noisy) correspondences –when data is noisy, better to use lots of data Who wants to be a millionaire? studio audience poll