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CS 395/495-26: Spring 2003

CS 395/495-26: Spring 2003

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CS 395/495-26: Spring 2003

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  1. CS 395/495-26: Spring 2003 IBMR: Week 4B Chapter 3: Estimation & Accuracy Jack Tumblin jet@cs.northwestern.edu

  2. Reminders • Project A due today; (before midnight)Homework submission instructions on website • Project B assigned today 4/24 (on website) Due 5/1 (1 week to complete it) • Take-Home Midterm starts 5/1 (Next Thurs.)Due Thurs 5/8 (1 week to complete it).

  3. Direct Linear Tranform (DLT) • For each point pair (xi,xi’) Solve Hx  x’ = 0 • Robust! Accepts any scale (x3=w1, x3’=w’1), finds anyH • ‘Vectorize’; make 2 rows of A per point pair • Solve Aih = 0 stacked up… [ 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] .................or written in book’s vector notation (pg 71, 3.1)............. [ 0T -w’i xiT y’I xiT ] [w’i xiT 0T -x’i xiT ] [ -y’i xiT -x’i xiT 0T ] h1 h2 h3

  4. Direct Linear Tranform (DLT) Solve Ah = 0 using Singular Value Decomp. • SVD(A) = USVT • Solution: h is V8, thelast column of V Assumes rank of A is 8; singular value s8 0 ‘Find Input & Output Axes, Linked by Scale’ b2 x2 x A = USVT v2 u3 v1 b1 v3 x1 b3 Ax=b u2 u1 Input (9 dim.) Output (8 dim.)

  5. Adding More Measurements If we use >4 point correspondences? Quick-and-sloppy: • Adds row pairs to our 8x9 matrix A: A·h = 0 • Use SVD to find Null space (Always gives an answer!) • Result: ‘Least squares’ solution • minimizes ‘algebraic distances’ Єi2between point pairs. || A h ||2 = Є2 =|| tall, near-zero vector ||2 = = iЄi2 = where,Єi2is error for i-th pt. correspondence: Єi2= || Hxi xi’ ||2 = || (2 rows of A)·h ||2 = ‘algebraic distance’ Є12 + Є22 + Є32 + …

  6. 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: (for P2) d(a,b) = dalgebraic(a,b) a3·b3

  7. Adding More Measurements Overall Strategy: • Overconstrain the answer H • Collect extra measurements (>4 point pairs, etc. …) • expect errors; use adjustable ‘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 ^ ^ ^ ^

  8. Error Measures & Corrections H ^ H xi IDEAL: x’i • One-Sided: • Perfect inputs xi, Flawed outputs xi’ • Find H to minimize i d(Hxi, xi’)2 • Two-sided (symmetric): • Flawed input xi, Flawed output xi’ • Interleave: find best H,find best H-1 to minimize i d(Hxi, xi’)2 + id(xi, H-1xi’)2 • Reprojection: • both input xi and output xi’ have errors • Unify ‘estimates’ of all: xi, xi’andH • DefineH for perfect matches: Hxi =xi’ • Minimize estimates .vs. real: i d(xi, xi)2 + d(x’i, xi’)2 H H, H-1 ^ ^ ^ ^ ^ ^ ^ ^ ^

  9. HOW? Reprojection in R4 • Rearrange Each point pair to make a 4-vector Xi: i-th point pair xi,xi’  Xi= (xi,xi’) = [x,y,x’,y’]T(assume w=w’=1) • 1) Define an R4 ‘measurement space’ for all Xi • (CAREFUL! this is NOT homogeneous P3: it’s true 4-D) • holds all possible point pairs, from perfect to horrid • 2) Find the “shape of perfection” for an H in R4: • WhatXi points in R4 are an error-free fit to H? • .e.g. find allXi points where H xixi’ = 0 • Choose an ‘estimate’Xi that is on this shape • 3) Find nearest estimateXi for measured point Xi ^ ^

  10. Find “Perfect Shape of H” in R4: x y x’ y’ Xi= • Rewrite one row of Ah=0 using Xi and get… one multilinear quadric in R4: • H sets 2 rows2 quadrics a ‘variety’ vH (‘variety’ = a collection of shapes in higher dimensions) • “Each H forms a variety in R4” XT QX + XTP + C = 0 P Q 00 a b00c d 0000 efgh XT X+ XT +C= 0 (assumes w=w’=1) (a,b,c,d,e,f,g,h,C are scalar constants)

  11. Find “Perfect Shape of H” in R4: • Recall a point pair Xi sets two rows of Ah=0: • Rewrite one row of Ah=0 using Xi= and get… -a (messy) quadric in R4, but -‘multilinear’; (only zeros are squared) -simpler if you include w’... [ 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] h1 h2 h3 x y x’ y’ P (assumes w=w’=1) Q 00 a b00c d 0000 efgh XT X+ XT +C= 0 a,b,c,d,e,f,g,h,C are scalar constants from H ?can you find them? XT QX + XTP + C = 0

  12. Find “Perfect Shape of H” in R4: ! NO ! It’s simpler if you include w’... • Recall a point pair Xi sets two rows of Ah=0: • Rewrite one row of Ah=0 using Xi= and get… -a (messy) quadric in R4, but -‘multilinear’; (only zeros are squared) -simpler if you include w’... [ 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] h1 h2 h3 x y x’ y’ w’ P (assumes w=w’=1) Q 00 a b00c d 0000 efgh XT X+ XT +C= 0 a,b,c,d,e,f,g,h,C are scalar constants from H ?can you find them? XT QX + XTP + C = 0

  13. Find “Perfect Shape of H” in R4: x y x’ y’ w’ • For the i-th point pair, DLT makes 2 rows Ai • In R4, point pair is Xi = [x y x’ y’]T(assume w=w’=1) • Augment Xi with w’; then R4+ quadrics are: XT Q X=0 or = 0 0T -w’i xiTy’I xiTw’i xiT 0T -x’i xiT -y’i xiT-x’i xiT 0T h1h2h3 0 0 0 Aih= 0 = xyx’y’w’  Q

  14. Find “Perfect Shape of H” in R4: • For the i-th point pair, DLT makes 2 rows Ai: • Write each row as an R4+ quadric: XT Qi X=0 0T -w’i xiTy’I xiTw’i xiT 0T -x’i xiT -y’i xiT-x’i xiT 0T h1h2h3 0 0 0 Row 1 Row 2 Row 3 Aih= Єi = Qi1for Row 1 Qi2for Row 2 Qi3 for Row 3 0 0 0 h31 -h210 0 0 h32 -h220 0 0 0 00 0 0 0wh220 0 0 0 wh33 0 0 -h31 0 h110 0 -h32 0 h120 0 0 0 -w h330 0 0 0 00 0 0 0’wh13 0 0 h21 -h11 00 0 h22 -h12 0 0 0 0 0 w h230 0 0 0 w h130 0 0 0 0

  15. R4 Quadrics for the Variety vH • For the i-th point pair, DLT makes 2 rows Ai: • H sets 2 rows2 quadrics a ‘variety’ vH • - “variety” = a collection of shapes in higher dimensions • - “Each H forms a variety in R4” means the perfect shape of H is the intersection of 2 (or 3) quadrics Variety of H is vH is two quadrics XT Qi X=0 Qi1for Row 1 Qi2for Row 2 Qi3 for Row 3 0 0 0 h31 -h210 0 0 h32 -h220 0 0 0 00 0 0 0wh220 0 0 0 wh33 0 0 -h31 0 h110 0 -h32 0 h120 0 0 0 -w h330 0 0 0 00 0 0 0w h13 0 0 h21 -h11 00 0 h22 -h12 0 0 0 0 0 w h230 0 0 0 w h130 0 0 0 0

  16. Reprojection: The Big Idea • Given H, find its R4‘variety’ vH(its 2 quadrics) • For each measured Xi, find nearest estimateXion vH by ‘projection’ • VERY nice result (robust!): • Minimizes input, output geometric error • Is also the ‘ML’ (maximum likelihood) estimate • Better than DLT: invariant to origin, scaling, similarity, BUT in R4, finding ‘nearest estimate’ d(Xi, vH) is a mess! Look at simplified case… ^

  17. Try R2 (not R4) first… Find d(Xi, vH) for R2 to an ‘x2 family’ curve: • Instead of (x,y)(x’,y’) correspondence, (x)(y) • Instead of Xi = (x,y,x’,y’), use Xi = (x,y) • Replace 4D ‘Variety’ vHwith 2D conic curve C • EstimateXi is on conic: XiT C Xi = 0 • Linefrom Xi to Xi is  to C tangent • Finding Xi from Xi, C isroot-finding! ^ ^ ^ ^ Xi ^ ^ Xi vH

  18. R4: Sampson Error Approx. CH(Xi)X CH(Xi)X ^ • Sampson error’s estimate: Xi = Xi + X • Re-name DLT result Aih = Єi asCH(X) = Єi = (Note: CH(X) is a stack of 2(or 3) rows of A ) • Use Taylor Series Approx: CH(Xi +X)  CH(X) + X + … • Find x for error 0:CH(Xi +X)  0 =Єi + X • But How? (2-,3- and 4-vectors here!) Єi1 Єi2 Єi3 CH(X) Єi X Xi x

  19. R4 Solutions: Sampson Error • CH(X) • X The CH(X) = ==error is a vector polynomial • Do Taylor Series at measured point Xi: • Follow -(error gradient) direction • distance*grad = error • Estimate Xi = Xi + x • Error vector x = -JT(JJT)-1Є (J = partial deriv. matrix...) Є1 Є2 =J x ^ Xi CH(X)

  20. R4 Quadrics for the Variety vH x y x’ y’ w’ • For the i-th point pair, DLT minimizes Єi: • In R4, point pair is Xi = [x y x’ y’]T(assume w=w’=1) • Augment Xi with w’; then R4 quadrics are: XiT Q Xi=Єi or = Єi 0T -w’i xiTy’I xiTw’i xiT 0T -x’i xiT -y’i xiT-x’i xiT 0T h1h2h3 Єi1 Єi2 Єi3 Aih= Єi = xyx’y’w’  Q

  21. R4 Quadrics for the Variety vH • For the i-th point pair, DLT minimizes Єi: • Write each row as an R4 quadric: XiT QiXi=Єi 0T -w’i xiTy’I xiTw’i xiT 0T -x’i xiT -y’i xiT-x’i xiT 0T h1h2h3 Єi1 Єi2 Єi3 Row 1 Row 2 Row 3 Aih= Єi = Qi1for Row 1 Qi2for Row 2 Qi3 for Row 3 0 0 0 h31 -w’h210 0 0 h32 -w’h220 0 0 0 00 0 0 0wh220 0 0 0 w’wh33 0 0 -h31 0 w’h110 0 -h32 0 w’h120 0 0 0 -w h330 0 0 0 00 0 0 0w’wh13 0 0 h21 -h11 00 0 h22 -h12 0 0 0 0 0 w h230 0 0 0 w h130 0 0 0 0

  22. R4 Quadrics for the Variety vH 0 0 a b c0 0 d e f0 0 0 0 g0 0 0 0 h0 0 0 0 j  Qiy  Qiy’  Qix’  Qix • Write each row as an R4 quadric: XT Qi X=0 • ALL have this form: Qi = • Quadric’s derivatives are easy: (Can also write as symmetric without changing meaning!) : : : : xyx’y’1 xyx’y’1 0 00000 00000 0 0 0 a0 0 0 0 b0 0 0 0 c 0 00000 00000 0 0 0 d0 0 0 0 e0 0 0 0 f 0 000 a0 000 d0 0 0 000 0 0 000 0 0 0 g 0 000 b0 000 e0 0 0 000 0 0 000 0 0 0 h x y x’ y’ 1 x y x’ y’ 1 x y x’ y’ 1 x y x’ y’ 1

  23. Sampson Error Approx. Єi1 Єi2 Єi3 CH(Xi)X xy x’ y’ CH2(Xi) CH2(Xi) CH2(Xi) CH2(Xi)x y x’ y’ CH1(Xi) CH1(Xi) CH1(Xi) CH1(Xi)x y x’ y’ ^ • Sampson error’s estimate: Xi = Xi + X • Re-name DLT result Aih = Єi asCH(X) = Єi = • Find x: solve Єi + X= 0 • Or as in Book: Єi + JX= 0 + = 0 Єi1 Єi2 Єi3 CH(X) Єi X Xi x (pg. 82) Find J terms from R4 quadrics, solve for X … CH2(Xi) CH2(Xi) CH2(Xi) CH2(Xi)x y x’ y’

  24. Sampson Error Approx. Єi1 Єi2 Єi3 CH(Xi)X xy x’ y’ CH2(Xi) CH2(Xi) CH2(Xi) CH2(Xi)x y x’ y’ CH1(Xi) CH1(Xi) CH1(Xi) CH1(Xi)x y x’ y’ ^ • Sampson error’s estimate: Xi = Xi + X • Re-name DLT result Aih = Єi asCH(X) = Єi = • Find x: solve Єi + x= 0 • Or as in Book: Єi + Jx= 0 + = 0 Єi1 Єi2 Єi3 CH(X) Єi X Xi x (pg. 82) !TROUBLE!rank=2, but 4 unknowns! CH2(Xi) CH2(Xi) CH2(Xi) CH2(Xi)x y x’ y’

  25. Sampson Error Approx. Єi1 Єi2 Єi3 xy x’ y’ CH1(Xi) CH1(Xi) CH1(Xi) CH1(Xi)x y x’ y’ CH2(Xi) CH2(Xi) CH2(Xi) CH2(Xi)x y x’ y’ ^ • Sampson error’s estimate: Xi = Xi + X • Re-name DLT result Aih = Єi asCH(X) = Єi = Or as in Book: Єi + Jx= 0 + = 0 • ANSWER: X = -JT(JJT)-1Є Єi1 Єi2 Єi3 CH(X) Єi X Xi x CH2(Xi) CH2(Xi) CH2(Xi) CH2(Xi)x y x’ y’ Answer: find shortestX: minimize ||X||2 by Lagrange multipliers…

  26. How to use it: • Initialize: use DLT to find H from point pairs • Revise: estimates by Sampson: Xi = Xi + X • Adjust: find new H from estimates (DLT) • Repeat until ||X||2 is small enough. Refinements: apply fraction of X on each iteration, so that H can change more... ^ ^

  27. END

  28. Aside: Lagrange Multipliers “Given only f(x) = 0, find shortest-length x” Semi-magical vector trick: • Length2 is scalar value: ||x||2 = xTx • Mix in an unknown vector (‘Lagrange multiplier’): xTx - 2 T f(x) = g(x,) =weird_length2

  29. Aside: Lagrange Multipliers “Given only f(x) = 0, find shortest-length x” Semi-magical vector trick: • Length2 is scalar value: ||x||2 = xTx • Mix in an unknown vector (‘Lagrange multiplier’): xTx - 2 T f(x) = g(x,) =weird_length2 Length2 (MUST be = 0)

  30. Aside: Lagrange Multipliers “Given only f(x) = 0, find shortest-length x” Semi-magical vector trick: • Length2 is scalar value: ||x||2 = xTx • Mix in an unknown vector (‘Lagrange multiplier’): xTx - 2 T f(x) = g(x,) =weird_length2 Find shortest ‘weird_length2’: • Find (g / x)= 0, Find (g / ) = 0, • Solve them for  (substitutions) • then use  in g(x,) to find x !Done! Length2 (MUST be = 0)

  31. More General Approach: • Define ‘Measurement space’ (e.g. R4) • Vector: a complete measurement set • Vector holds only known values, but • Can be ANYTHING: input,output, length, H, … • All possible measures (good or bad) • Define ‘Model’ • All possible error-free ‘perfect measurements’ such as the variety vH • Defined by a known function(e.g. CH(X) = 0 ) • A subset of measurement space

  32. More General Approach: • Gather points Xi in ‘measurement space’ • Replace Xi with ‘estimates’ Xi that: • Satisfy the model, and • minimize distance || Xi – Xi || • Book (pg 85) gives examples for: • Error in both images (what we just did) • Error in one image only • Maximum likelihood estimation: same as minimizing geometric error  same as minimizing reprojection error ^ ^

  33. DLT Non-Invariance !Accuracy depends on origin location! • ‘Non-optional’ Correction required: • Find centroid of measured points • Find average distance of points to centroid • Offset origin to centroid for all points • Scale about new origin to set average distance-to-origin to sqrt(2) • Reverse process after DLT. See book for proof. y x  x y (Outline on pg. 92)