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CS 395/495-26: Spring 2003

CS 395/495-26: Spring 2003. IBMR: Week 4A from Chapter 3: Point Correspondence and Direct Linear Transform (DLT) Jack Tumblin jet@cs.northwestern.edu. Proj_A Workaround. Programming ‘Pick & Drag’ is messy! KeyDown, KeyUp is easy (see ProjA_06) We have only a few points to drag;

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CS 395/495-26: Spring 2003

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  1. CS 395/495-26: Spring 2003 IBMR: Week 4A from Chapter 3: Point Correspondence and Direct Linear Transform (DLT) Jack Tumblin jet@cs.northwestern.edu

  2. Proj_A Workaround • Programming ‘Pick & Drag’ is messy! • KeyDown, KeyUp is easy (see ProjA_06) We have only a few points to drag; • Use number keys to select points: • ‘1’ for 1st point, ‘2’ for 2nd point, etc. • ‘SHIFT 1’ for 1st line, ‘SHIFT 2’ for 2nd line... • Use arrow keys to change point / line coords: • left/right for x1 value, • up/down for x2 value, • SHIFT up/down for x3 value, etc.

  3. Most Important 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?

  4. Recall: Rectification Undo parts of H x’ = H x where H = HS HA HP GOAL: Put world plane x’ into view plane x • Affine Rect.; (find only HP(2DOF)); • Metric Rect.; (find HA and HP (6DOF)); • Full Rect.; (find all: HSHAHP (8DOF)); METHODS: • Affine: ‘Vanishing Point’, Horizon line methods • Metric: Conics & Circular Points (!?!?) • Full: 4-point correspondence

  5. Finding H from Image Space L M • Affine Rect; (find only HP(2DOF)); 2 parallel line pairs2 points on horizon line • Metric Rect. (find HA and HP (4DOF)); 5 perpendicular line pairs C*’ • Full Rect.; (find all: HSHAHP (8DOF)); 4 (or more point pairs)  all H coeffs L m HS HA HP 2D image space 2D world space

  6. 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

  7. 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)

  8. Naïve Method: (pg 13-14) 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’]

  9. Direct Linear Tranform (DLT) • Naïve method: Solves Hx – x’ = 0 • Requires constant scale (x3=1 in measurement) • compare: • include the w, w’ terms: • look what happens when w’0! 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]

  10. Direct Linear Tranform (DLT) • Naïve method: Solves Hx – x’ = 0 • Requires constant scale (x3=1 in measurement) • compare: • 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]

  11. 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 ]

  12. 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

  13. Deceptive ‘Robustness’ • Suppose we have 4 pt-correspondences: • Use DLT to write 8x9 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. Actual ‘Robustness’ • Vectorizing (‘Flatten, Stack, Null Space’ method) works for almost ANY input! (Points, lines, planes, ||, , conics, quadrics, cross-ratios, vanishing points, twisted cubics… • Use 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 

  15. 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!

  16. 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)

  17. Adding More Measurements Overall 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 ^ ^ ^ ^

  18. 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 inputs and outputs have error. • Adjust BOTH input and output ests x x’(note we re-compute H as x, x’ change) • Stop when d(Hx,x’)2 + d(H-1x’, x)20 ^ ^ ^ ^ ^ ^ ^ ^

  19. END

  20. DLT: Four-Point Correspondence • 4 point correspondence: • Book shows: • Rearrange: known vector (dot) unknown vector • stack, solve for null space… 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’] h11 h12 h13 h21 h22 h23 h31 h32 h33

  21. Direct Linear Tranform (DLT) • Project 2 method: Solve Hx – x’ = 0 • Requires constant scale (x3=1 in measurement) • compare: • DLT method: Solve Hx  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]

  22. Direct Linear Tranform (DLT) • Project 2 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 ]

  23. Direct Linear Tranform (DLT) • DLT method: Solve H  x = 0 • Accepts any scale, any point (x3=w1 is OK) • ‘Pure’, Compatible -- P2terms only • 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

  24. Deceptive ‘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.

  25. Actual ‘Robustness’ • Vectorizing (‘Flatten, Stack, Null Space’ method) works for almost ANY input! (Points, lines, planes, ||, , conics, quadrics, cross-ratios, vanishing points, twisted cubics… • Use 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 

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