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CS 395/495-26: Spring 2004

CS 395/495-26: Spring 2004. IBMR: 2-D Conics: Introduction Jack Tumblin jet@cs.northwestern.edu. Intuition for Conics: 2D Ellipses. Still a circle/ellipse after P 2 projective transform (Better!) every transformed conic is another conic!. H 1. H 2. H 3.

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CS 395/495-26: Spring 2004

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

  2. Intuition for Conics: 2D Ellipses Still a circle/ellipse after P2 projective transform (Better!) every transformed conic is another conic! H1 H2 H3

  3. Intuition for Conics: 2D Ellipses 1/b 1/a • Cartesian axis Circle/Ellipse: a x2+b y2– r2= 0 • P2 axis Circle/Ellipse: a x12 + b x22 – r2 x32 = 0 • 5 parameters (deg. of freedom): • center point (cx, cy) • major, minor axes (1/a, 1/b) • orientation  

  4. Conics: Key Ideas • ‘Conics’== intersection of cone & plane: • Many possible shapes: circles, ellipses, parabola, hyperbola, degenerate lines & points In P2, Conics have Dual Forms: • ‘Point Conics’ defined by points on the curve • ‘Line Conics’ defined by lines tangent to curve

  5. Conics written as a Matrix • All points on any conic solve a 2D quadratic: ax2 + bxy + cy2 +dx +ey +f = 0 • In homogeneous coordinates: (mpy. all by x32) ax12 + bx1x2 + cx22 +dx1x3 +ex2x3 + fx32 = 0 xTCx = 0 • C is symmetric, 5DOF (not 6; ignores x3 scaling) • Can find any C from 5 homog. points... How?(SVD solves Null-space problems easily…) x1 x2 x3 a b/2 d/2 b/2 c e/2 d/2 e/2 f x1 x2 x3 = 0 (and how could conics help you find H?...)

  6. Familiar Conics: Circles, Ellipses... x1 x2 x3 a b/2 d/2 b/2 c e/2 d/2 e/2 f xT C x = 0 or • General form of point conic: ax2 + bxy + cy2 +dx +ey +f = 0 (or) ax12 + bx1x2 + cx22 +dx1x3 +ex2x3 + fx32 = 0 • Circle at d with radius r: 1x12 + bx1x2 + 1x22 +dx1x3 +ex2x3–r2x32 = 0; Cr = • ‘Degenerate’ Cases (C rank<3): lines, parabolas, hyperbolae... Example: ax12 + bx1x2 + 0x22 +dx1 +ex2+ fx32 = 0; x1 x2 x3 = 0 1 0 0 0 1 0 0 0 -r2 a 0 0 0 0 0 0 0 f (Try it—plot points for x3 = 1…)

  7. Point Conics: C • Matrix C makes conic curves from points x: xTCx = 0 C is a ‘point conic’ • The tangent line L for point x is: L = C x • P2 point conic transformed by H: C’ = H-T C H-1 (messy! requires an invertible H !) x L Prove it! Use the two equations above…

  8. ‘Dual’ or Line Conics: C • Matrix C* makes conic curves from lines L: LT C* L = 0 C is a ‘line conic’ • The tangent point x for line L is: x = C* L • Projective transform by H is: C*’ = H C* HT (better -- no H inversion required) • If C is non-singular (rank 3), then C* = C-1 x L Prove it! Use two equations above…

  9. Conic Summary • Matrix C identifies all the points on a conic: xTCx = 0 C is a ‘point conic’ • Given a point x on a conic curve, the homog. tangent line l is given by l = C x • Matrix C* identifies all lines tangent to a conic: lTC*l = 0C* is a ‘Dual Conic’ • Can you prove this? Try it! C* = C-1 (hint: use the boxed equations)

  10. Degenerate Conics 1 i 0 1 -i 0 1 0 0 0 1 0 0 0 0 x1 x2 x3 a b/2 d/2 b/2 c e/2 d/2 e/2 f xT C x = 0 or • General form of point conic: ax2 + bxy + cy2 +dx +ey +f = 0 (or) ax12 + bx1x2 + cx22 +dx1x3 +ex2x3 + fx32 = 0 • Very special ‘degenerate’ case: ‘infinite radius circle’ as x30, aka C 1x12 + bx1x2 + 1x22 +dx1x3 +ex2x3 +0x32 = 0 C = only solutions to xT Cx =0 are 2complexx vectors: and x1 x2 x3 = 0

  11. Conics: Key Ideas 1 0 0 0 1 0 0 0 0 L m • Transformed conics tell you how H will change angles: • one strange, degenerate conic C* can measure angle • Angles between line pairs yield transformed C*    C*= C*’=      0 L’ m’ H = HS HA HP 2D image space 2D world space

  12. Conic Weirdness I 1 i 0 1 -i 0 1 0 0 0 1 0 0 0 0 ‘Circular Points’ I= , J= ; and C = • ‘Ideal points’; 2 points at infinity / horizon, but… • Complex : real part on x1 axis, imag. on x2 • “Where infinite circle hits horizon line L [0,0,1]” • Where? imaginary! (infinite x,y axes, maybe?) ?WHY BOTHER? • To measure angles in projective space • To use angle correspondences to find HA,HP

  13. Conic Weirdness 2 1 i 0 1 -i 0 1 0 0 0 1 0 0 0 0 ‘Circular Points’ I= , J= ; and C = • C is a point conic, affected byHAHP only. • C* is line conic, but is the samematrix(!?!) • Is same as this outer product: IJT + JIT • Intersects L, the infinity line(proof: LC*L=0) • Affected byHP andHA only, ( HSdoes nothing) ALL transforms of C* have only 4DOF (pg 33 )

  14. Conics: Angle Measuring 1 0 0 0 1 0 0 0 0 • Matrix H transforms C* to another space: C*’=H C* HT Angles: • define ‘world space’ C* as • (for any two world-space homogeneous lines) L and m are perpendicular iff LT C* m = 0 • Angle  between lines L and m given by (Remember, it is simpler to transform a line conic) (LTC* m) (LT C* L)(mT C* m) cos() =

  15. Conics: Angle Measuring • Matrix H transforms C* to another space: C*’=H C* HT Angles: • ALSO true for transformedL, m and C*, so… • If you FIND C*’in image space, you can measure 2 image-space lines L’, m’ and find the angle  between them in world space! Use the same expression: (Remember, it is simpler to transform a line conic) (L’T C*’ m’) (L’T C*’ L’) (mT C*’ m’) cos() =

  16. ? How can we find C*’ without H ? One Answer: measure perpendicular lines… • We know 2 world-space lines L and m are perpendicular iff LT C* m = 0 • Also true for transformedL, m and C*: (e.g. in image space)L’T C*’ m’ = 0 Hmmm. L’ and m’ are known, C*’ unknown. Isn’t this a null-space problem? Once we have C*’ then what? See any other uses for conics in P2? 1 0 0 0 1 0 0 0 0

  17. Polar Lines and Pole Points • Line Conic C’s tangent line Lt at point xt by: C xt = Lt(given xt is on the conic:xtT C xt=0) (You can skip this) xt C Lt

  18. Polar Lines and Pole Points • Line Conic C’s tangent line Lt at point xt by: C xt = Lt(given xt is on the conic:xtT C xt=0) • But if x is NOT on the conic? try Cxp = Lp (You can skip this) xp Lp C

  19. Polar Lines and Pole Points • Line Conic C’s tangent line Lt at point xt by: C xt = Lt(given xt is on the conic:xtT C xt=0) • But if x is NOT on the conic? try Cxp = Lp • ‘Polar line’ Lp = conic at p1, p2(find them?ugly!) (You can skip this) p1 xp Lp C p2

  20. Polar Lines and Pole Points • Line Conic C’s tangent line Lt at point xt by: C xt = Lt(given xt is on the conic:xtT C xt=0) • But if x is NOT on the conic? try Cxp = Lp • ‘Polar line’ Lp = conic at p1, p2(to find them?ugly!) • p1, p2 tangent lines meet at ‘Pole point’ xp (You can skip this) p1 xp Lp C p2

  21. Polar Lines and Pole Points • Line Conic C’s tangent line Lt at point xt by: C xt = Lt(given xt is on the conic:xtT C xt=0) • But if x is NOT on the conic? try Cxp = Lp • ‘Polar line’ Lp = conic at p1, p2(to find them?ugly!) • p1, p2 tangent lines meet at ‘Pole point’ xp p1 Interesting, But why does book show it? xp Lp C p2

  22. Question: Can you apply DLT? • Choose pairs of lines that are known to be perpendicular in world space; measure those lines in image space. • Solve for C*’ using L’TC*’ m’ = 0 • We also know that HC*HT=C*’ Can you solve for H?Zisserman tried, but book method is confused, contains errors, and may be incorrect…

  23. END. STOP HERE

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

  25. Recall: Projective Transform H x y Goal: Find H for ‘Image Rectification’ Apply the 3x3 matrix Hx’ = Hx 2D image (x,y)Homog. coords [x,y,1]T = x Homog. coordsx’ =[x’,y’,1]T 2D image (x’,y’) x x x2 (x,y) y x y y’ y x3 x x’ x’ y (x’,y’) x1 x ‘World Plane’ ‘View Plane’

  26. 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

  27. Recall: The bits and pieces of H H has 8 independent variables (DOF) • Computer Vision Jargon (2D projective): Isometry --3DOF(2D translate tx,ty; 2D rotate z; ) Similarity --4DOF (add uniform scale s;) Affine --6DOF (add orientable scale s,/s, s/s) Projective--8DOF (changes x3; 3D-rotation-like) Rectification “up to a Similarity” (finds HP, HA)

  28. Conics: Key Ideas 1 0 0 0 1 0 0 0 0 L M • Transformed conics tell you how H changes angles • one strange, degenerate conic C* measures angle • Angles between line pairs yield transformed C* • We can compute HA and HP from transformed C*    C*= C*’=       L m HS HA HP 2D image space 2D world space

  29. Undoing H: Metric Rect. 1 L M • C*’ =  • First, ignore projective part: set v=0. • Choose two pairs of perpendicular lines L,m • Second, Use LT’ C*’ m’ = 0 to solve for K KKT  Kv    vTK  s KKT 0    0  s aka ‘Weak Perspective’ L m HS HA HP 2D image space 2D world space

  30. Undoing H: Metric Rectification 1 0 0 0 1 0 0 0 0 L M • C*’has only 4DOF (before,after ANY H) • All C*’ DOF are contained in (HA HP) • SVD can convert C*’ to (HA HP) matrix! • Task is then “find C*’, then use it to find H”    C*= C*’=      0 L m HS HA HP 2D image space 2D world space

  31. Undoing H: Metric Rect. 1 • H = HS HA HP = • Book shows how to write transformed C* as (simple, tedious): C*’ = where K is 2x2 symmetric (affine part: 2DOF) v is 2x1 vector, (projective part: 2DOF)    sR t 0T 1 K 0 0T 1 I 0 vT v            KKT  Kv    vTK  0

  32. Undoing H: Metric Rect. 1 • H = HS HA HP = • Book shows how to write transformed C* as (simple, tedious): C*’ = where K is 2x2 symmetric (affine part: 2DOF) v is 2x1 vector, (projective part: 2DOF)    sR t 0T 1 K 0 0T 1 I 0 vT v            !!!BUT DERIVATION & RESULT IS WRONG!!! SKIP ‘Metric Rectification I’ pg 35,36! KKT  Kv    vTK  0

  33. ‘Let SVDs Explain it All for You’ 1 0 0 0 1 0 0 0 0 SVD finds H for you (?) (pg 35) Book’s Method for finding HAHP from C*’ : • C*’= H C* HT is symmetric; • Assume SVD(C*’) = USVT = (U S ) I ( S VT ) = ( H ) C* ( HT) • BUT THIS DOESN’T WORK! • ( Recall C* = ) A = USVT ‘Find Input & Output Axes, Linked by Scale’

  34. Eigen -values,-vectors, Fixed pt & line • Formalizes ‘invariant’ notion: • if x is ‘fixed’ for H, then Hx only scales xH x = x( is a constant scale factor) • x is an ‘eigenvector’,  is its ‘eigenvalue’ • again, SVD helps you find them. • Elaborate topic (but not hard). Skip for now. (You can skip this)

  35. Book Errata Website: http://www.robots.ox.ac.uk/~az/HZbook/HZerrata.html pg. 35, eqn. 1.22 is dubious / erroneous; C*’ = HC*HT = (HSHAHP) C* (HSHAHP)T is the transformed line conic, but book uses C*’ = (HPHAHS) C* (HPHAHS)T instead, AND has error: = But I get: C*’ = KKT  Kv    vTK  ? (book: ?=0; errata website: ?=vTKKTv) KKT  KKTv    vTKKT  vTKKTv

  36. Dubious Book Aggravations: KKT  Kv    vTK  0* • Similarity transform HS leaves C* unchanged:C* =HS C* HST • H transforms C* to image space C*’ by:C*’=H C* HT = (HS HA HP) C* (HS HA HP)T • Book claims to derive C*’ = (HP HA) C* (HP HA)T (!?!?!) Then errs in simplifying to: 1 0 0 0 1 0 0 0 0

  37. SVDs and Conics • Conics (both C and C*) are symmetric; • SVD of any symmetric A is also symmetric:SVD(A) = USUT !Not Always! • Find conics’ singular values’ sign: si =0,1, or –1, to classify conic type: (pg 40) Si values | Equation | Type . imaginary-only circle/ellipse single real point (0,0,1)2 lines: x+/- y2 co-located lines: x=0 x2 + y2 + w2 = 0x2 + y2 - w2 = 0 x2 + y2 = 0x2 - y2 = 0x2 = 0 (1, 1, 1) (1, 1,-1) (1, 1, 0) (1,-1, 0) (1, 0, 0)

  38. Undoing H: Metric Rectification • (recall)H = HS HA HP = • Tedious algebra*(pg35) shows symmetry: C*’ =H C* HT = where K is 2x2 symmetric (affine part: 2DOF) v is 2x1 vector, (projective part: 2DOF) • ?But what do K and V really control? sR  t    0T 1 K  0    0T 1 I 0  vT v KKT  Kv    vTK  0* *==error?!?

  39. Undoing H: Metric Rectification • (recall)H = HS HA HP = • Tedious algebra (pg35) shows symmetry: C*’ =H C* HT = where K is 2x2 symmetric (affine part: 2DOF) v is 2x1 vector, (projective part: 2DOF) • ?But what do K and V really control? sR  t    0T 1 K  0    0T 1 I 0  vT v KKT  Kv    vTK  ? (book: ?=0; errata website: ?=vTKKTv)

  40. Undoing H: Metric Rectification • OK, then how do we find K and v? • (book: ?=0; errata website: ?=vTKKTv) • Choose known-perpendicular line pairs(Li, mi),then compute by: • Method 1a: (pg 36) Assume v=0, solve for K • Method 1b: (NOT in book)Assume K=I, solve for v • Method 2: Rearrange, solve for full C*’ then get HAHP using SVD. KKT Kv    vTK  ? C*’ =H C* HT =

  41. Undoing H: Metric Rect. 1a m1 m2 m3 s1 s2 s3 l1 l2 l3 Method 1a: (pg 36) Assume v=0, solve for K • Choose 2  line pairs (La,ma) and (Lb,mb) • Both pairs satisfy LT C*’ m = 0 or: • (note x3 term is ignored!) • ‘flatten’ to one equation: KKT  0    0  0 KKT  Kv    vTK  0 C*’ =H C* HT =  s1 s2 0 s2 s3 0 0 00 =0 l1m1 l2m1+l1m2 l2m2 = 0

  42. Undoing H: Metric Rect. 1a s1 s2 s3 Method 1a: (pg 36) Assume v=0, solve for K • ‘Stack’ to combine both line pairs; • Solve for s by SVD: find ‘input null space’ (Ax=0) • From s make C*’ = KKT, then extract HAHPusing SVD: recall C*’ =H C* HT , it is symmetric… KKT  Kv    vTK  0 s1 s2 0 s2 s3 0 0 00 C*’ =H C* HT =  la1ma1 la2ma1+la1ma2 la2ma2 = 0 lb1mb1 lb2mb1+lb1mb2 lb2mb2

  43. Undoing H: Metric Rect. 1b m1 m2 m3 l1 l2 l3 Method 1b: (not in book) Assume K=I, solve for v • Choose 2  line pairs (La,ma) and (Lb,mb) • These satisfy LT C*’ m = 0 or: • ‘flatten’ to one equation (messy), stack, solve for v1,v2 • Extract Hp from C*’ using SVD I  v    vT  0 KKT  Kv    vTK  0 C*’ =H C* HT =  1 0 v1 0 1 v2 v1 v2? =0

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