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CS 395/495-26: Spring 2004

CS 395/495-26: Spring 2004. IBMR: Projective 2D DLT Jack Tumblin jet@cs.northwestern.edu. Recall: Projective Transform H. x. y. Apply the 3x3 matrix H x’ = Hx. 2D image (x,y)  Homog. coords [x,y, w ] T = x w=1. Homog. coords x’ = [x’,y’, w’ ] T w’=1  2D image (x’,y’).

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CS 395/495-26: Spring 2004

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  1. CS 395/495-26: Spring 2004 IBMR:Projective 2D DLT Jack Tumblin jet@cs.northwestern.edu

  2. Recall: Projective Transform H x y Apply the 3x3 matrix Hx’ = Hx 2D image (x,y)Homog. coords [x,y,w]T = x w=1 Homog. coordsx’ =[x’,y’,w’]Tw’=1 2D image (x’,y’) x x x2 (x/w,y/w) y x y y’ y x3 x x’ x’ y x1 (x’/w’,y’/w’) x

  3. Recall: Projective Transform: Hx = x’ h11 x + h12 y + h13 h31 x + h32 y + h33 h21 x + h22 y + h21 h31 x + h32 y + h33 x’2 x’3 x’1 x’3 H defines point pairs (x,x’) (correspondences) h11 h12 h13 h21 h22 h21 h31 h32 h33 x1 x2 x3 x’1 x’2 x’3 x1 x2 x3 x y w Input (or output) image is on central plane = = x = [x1, x2, x3] notation is confusing for multiple points! Instead, let’s use [x,y,w]: (so we can use subscripts to number the point pairs x,x’…) x’ = = y’ = = Write R2 expressions;(for each x,x’ pair) x’ (h31 x + h32 y + h33) = (h11 x + h12 y + h13) Rearrange, solve as a matrix problem… y’ (h31 x + h32 y + h33) = (h11 x + h12 y + h13)

  4. Recall: Projective Transform: Hx = x’ h11 x + h12 y + h13 h31 x + h32 y + h33 h21 x + h22 y + h21 h31 x + h32 y + h33 x’2 x’3 x’1 x’3 H defines point pairs (x,x’) (correspondences) h11 h12 h13 h21 h22 h21 h31 h32 h33 x1 x2 x3 x’1 x’2 x’3 x1 x2 x3 x y w Input (or output) image is on central plane = = let w=1 x’ = = y’ = = Write R2 expressions;(for each x,x’ pair) x’ (h31 x + h32 y + h33) = (h11 x + h12 y + h13) Rearrange, solve as a matrix problem… y’ (h31 x + h32 y + h33) = (h11 x + h12 y + h13)

  5. Recall: Projective Transform: Hx = x’ h11 x + h12 y + h13 h31 x + h32 y + h33 h21 x + h22 y + h21 h31 x + h32 y + h33 x’2 x’3 x’1 x’3 Finding H from point pairs (correspondences) h11 h12 h13 h21 h22 h21 h31 h32 h33 x1 x2 x3 x’1 x’2 x’3 x1 x2 x3 x y 1 Input (or output) image is on central plane = = x’ = = y’ = = Write R2 expressions;(for each x,x’ pair) x’ (h31 x + h32 y + h33) = (h11 x + h12 y + h13) Rearrange, solve as a matrix problem… y’ (h31 x + h32 y + h33) = (h11 x + h12 y + h13)

  6. Naïve Method: h11 h12 h13 h21 h22 h23 h31 h32 h33 4 point correspondence: Book shows: Rearrange: known vector (dot) unknown vector stack, solve for null space… But this assumes x1=x, x2=y, x3=w=1, There is a better way (DLT)… x’ (h31 x + h32 y + h33) = (h11 x + h12 y + h13) y’ (h31 x + h32 y + h33) = (h11 x + h12 y + h13) [x y 1 0 0 0 -x’x -x’y -x’] = 0 [0 0 0 x y 1 -y’x -y’y -y’]

  7. Direct Linear Tranform (DLT) • Naive method: Solve Hx – x’ = 0 • Requires constant scale (x3=1 always) • compare: • A BETTER WAY:DLT method: Solve Hx  x’ = 0 • Accepts any scale (x3=w1 OK…) h11 h12 h13 h21 h22 h23 h31 h32 h33 [ 0 0 0 x y 1 -y’x -y’y -y’ ] [ x y 1 0 0 0 -x’x -x’y -x’ ] [ 0 0 0 -w’x -w’y -w’w y’x y’y y’w] = 0 [w’x w’y w’w 0 0 0 -x’x -x’y -x’w]

  8. Direct Linear Tranform (DLT) • Naive method: Solve Hx – x’ = 0 • Requires constant scale (x3=1 always) • Compare to DLT: • DLT method: Solve Hx  x’ = 0 • Accepts any scale (x3=w1, x3’=w’1) h1 h2 h3 [ 0 0 0 x y 1 -y’x -y’y -y’ ] [ x y 1 0 0 0 -x’x -x’y -x’ ] [ 0 0 0 -w’x -w’y -w’w y’x y’y y’w] = 0 [w’x w’y w’w 0 0 0 -x’x -x’y -x’w] [ 0T -w’i xiT y’I xiT ] [w’i xiT 0T -x’i xiT ]

  9. Direct Linear Tranform (DLT) • DLT method: Solve H  x = 0 • Accepts any scale, any point (x3=w1 is OK) • ‘Pure’, Compatible –uses only P2terms. • Much better suited to error measurements. • Subtlety: • Won’t divide-by-zero if w=0 or h33=0(it happens!) • has a 3rd row; it is not degenerate if/when w’=0 • OK to use it…(Solve 8x12) 0T -w’i xiTy’I xiT w’i xiT0T -x’i xiT =0 y’i xiT -x’i xiT 0T h1 h2 h3

  10. STOP: Learn about SVDs …

  11. Matrix Math & SVDs • Matrix multiply is a change of coordinates • Rows of A = coordinate axis vectors • Ax = a dot product for each axis • input spaceoutput space • SVD ‘factors’ matrix A into 3 parts: • U columns: orthonormal axes for OUTPUT • V columns: orthonormal axes for INPUT • S: scale factors VU svd(A) = U S VT

  12. Actual ‘Robustness’ • Vectorizing (‘Flatten, Stack, Null Space’ method) works for almost ANY input! (Points, lines, planes, ||, , conics, quadrics, cross-ratios, vanishing points, twisted cubics… • Key to DLT formulation: (Hx  x’) = 0 • Rearrange as dot product: (known)·(unknown) = 0 • Be careful to have ENOUGH constraints tricky when you mix types: points, lines,… (pg 75) = 0 measured output measured inputtransformed 

  13. Deceptions: false ‘Robustness’… • Suppose we have 4 pt-correspondences: • Use DLT to write 8x9 (or 12x9) matrix A: Ah=0 • Solve for h null space. ALWAYS gives H matrix • But what if points are bad / fictional? • 3 collinear input pts, crooked out: IMPOSSIBLE! • Yet we get an H solution! Why? • A matrix rank is 7 or 6 rank 1 H result(s) • ‘Null Space’ of A may contain >1 H solution! • ‘Degenerate’ H solution of form aH1 + bH2… • Answer: SVD ranks A; reject bad point sets.

  14. Most Important SVD Ideas SVD solves Null-Space problems Ax = 0(‘flatten, stack, find null space’ to solve) • MANY problems can be written this way • Expandable! --more data points tall, ‘overconstrained’ A matrix --SVD finds optimal ‘least squares’ solution. Robustness: bane of all inverse methodsAccuracy vs. # of measurements (more isn’t always better! Outliers...) Data-rich images: quantity easier than qualitymore is easier than better) ?What links measurement errorsH errors? ?How can more measurements reduce error?

  15. How else can we use DLT? • I know this building consists of MANY parallel and perpendicular lines, corners, regularly spaced points. • How many different ways can we find to ‘rectify’ this image to get this one?

  16. Direct Linear Tranform (DLT) • Naïve method: Solves Hx – x’ = 0 • Uses constant scale (x3=w=1 in measurement) • Compare to DLT: • DLT method: Solve Hx  x’ = 0 • (a different identity: x’  x’ = 0) • Accepts any scale (x3=w1 OK, xx3) h11 h12 h13 h21 h22 h23 h31 h32 h33 [ 0 0 0 x y 1 -y’x -y’y -y’ ] [ x y 1 0 0 0 -x’x -x’y -x’ ] [ 0 0 0 -w’x -w’y -w’w y’x y’y y’w] = 0 [w’x w’y w’w 0 0 0 -x’x -x’y -x’w]

  17. Direct Linear Tranform (DLT) • Naïve method: Solve Hx – x’ = 0 • Requires constant scale (x3=1 always) • compare: • DLT method: Solve Hx  x’ = 0 • Accepts any scale (x3=w1, x3’=w’1) h1 h2 h3 [ 0 0 0 x y 1 -y’x -y’y -y’ ] [ x y 1 0 0 0 -x’x -x’y -x’ ] [ 0 0 0 -w’x -w’y -w’w y’x y’y y’w] = 0 [w’x w’y w’w 0 0 0 -x’x -x’y -x’w] [ 0T -w’i xiT y’I xiT ] [w’i xiT 0T -x’i xiT ]

  18. Direct Linear Tranform (DLT) • DLT method: Solve H  x = 0 • Accepts any scale, any point (x3=w1 is OK) • ‘Pure’ and Compatible -- P2terms only, no R2 • Much better suited to error measurements. • Subtlety: • Won’t divide-by-zero if w=0 or h33=0(it happens!) • has a 3rd row; it is not degenerate if w’=0 • OK to use it…(Solve 8x12) 0T -w’i xiTy’I xiT w’i xiT0T -x’i xiT =0 -y’i xiT -x’i xiT 0T h1 h2 h3

  19. Adding More Measurements How can we use >4 point correspondences? • Easy: • Add more rows to our 8x9 matrix A: A·h = 0 • Use SVD to find Null space (Always gives an answer!) • Result: ‘Least squares’ solution; error=A·h =  • minimizes || A h ||2 = i 2i where,iis error for i-th pt. correspondence: • i = || Hxi xi’ ||2 = || (2 rows of A)*h ||2 = ‘algebraic distance’ • ‘Algebraic Distance’ ? No geometric meaning!

  20. Adding More Measurements ( - )2+ ( - )2 d(a,b)2 = b2 b3 b1 b3 a1 a3 a2 a3 • 2D‘Algebraic Distance’ ?No geometric meaning! • 2D ‘Geometric Distance’ d(a,b)2is Better: measurable length in input or output space if a = (a1 a2 a3) and b = (b1 b2 b3), then define Turns out that: d(a,b)2 = dalgebraic(a,b) a3·b3 (Not very surprising)

  21. Adding More Measurements Iterative Strategy: • Overconstrain the answer H • Collect extra measurements (>4 point pairs, etc. …) • expect errors; call them ‘estimates’ x • Compute a 1st solution (probably by SVD) • Compute error d(Hx, x’)2, and use this to… • ‘Tweak’ answer Hand estimates x • Compute new answer • Stop when error < useful threshold ^ ^ ^ ^

  22. Using Estimates • Simplest: ‘one image’ transfer method: • Assume inputs are a perfect test pattern: only output pts are in error • Adjust output estimates x’ until d(Hx, x’)20 (note we re-compute H as x’ changes) • Better: ‘Symmetric’ transfer method: • Assume BOTH input x and output x’ has error. • Adjust BOTH input and output estimates x x’(note we also re-compute H as x, x’ change) • Stop when d(Hx,x’)2 + d(H-1x’, x)2is nearly 0 ^ ^ ^ ^ ^ ^ ^ ^

  23. END

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