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CS 395/495-26: Spring 2004

Learn about the concept of 2D Homogeneous Coordinates and their importance in projective geometry. Discover their properties, including defining points at infinity and lines at infinity, as well as their role in view interpolation.

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CS 395/495-26: Spring 2004

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

  2. Recall: Scene & Image Light + 3D Scene: Illumination, shape, movement, surface BRDF,… 2D Image: Collection of rays through a point Image Plane I(x,y) Position(x,y) Angle(,) GOAL: a Reversible Mapping Scene angles image positions

  3. View Interpolation: How? • LATER: 3D Projection(Hartley,Zisserman Chapter 2) Find new views of that scene From a 3D scene view1 view2

  4. View Interpolation: How? • BUT FIRST, the simpler caseHartley, Zisserman Chapter 1: 2D Projection Find new views of that image From a flat 2D image, view1 view2

  5. Answer: 2D Homogeneous Coords Chapter 1: 2D Projection From a flat 2D image, Find new views of that image view1 view2 2D Homogeneous (x1 x2 x3) coordinates in P2 Cartesian (x,y) coordinates in R2

  6. Overview • Chapter 1: any 2D plane, as seen byany 3D camera view (texture mapping generalized) • Homogeneous Coordinates are Wonderful • Everything is a Matrix: • Conics (you can skip for now) • Aside: interpolation of pixels and points… • Transform your Point of View: H matrix • Parts of H: useful kinds of Transforms

  7. 2-D Homogeneous Coordinates y x • WHY? makes MUCH cleaner math! • Unifies lines and points • Puts perspective projection into matrix form • No divide-by-zero; defines horizon, infinities … in R2, write point x as But in P2, write same point x as where: x1 x2 x3 x y x = x1 / x3, y = x2 / x3, x3 = anything non-zero!(but usually defaults to 1) (x,y)

  8. Useful 3D Graphics Ideas • Every Homog. point (x1,x2,x3) describes a 3D ray • ‘Phantom’ dimension x3 is the ‘z-buffer’ value • 3D2D ‘hard-wired’: (Fig 1.1, pg 8) • “Homogeneous coords allows translation matrix:” (true. But there is more!) ‘Image Plane’ (x,y,-1) (xmax, ymax, -1) (0,0,-1) An ‘Ideal Point’ at (0,0,0) Center of Projection 1 0 3 0 1 5 0 0 1 x y = 1 x+3 y+5 1

  9. 2-D Homogeneous Coordinates WHAT?!Why x3? Why ‘default’ value of 1? • Look at lines in R2 : • ‘line’ == all (x,y) points where • scale by ‘k’  no change: • Using ‘x3’ for points UNIFIES notation: • line is a 3-vector named l • now point (x,y) is a 3-vector too, named x ax + by + c = 0 kax + kby + kc = 0 ax + by + c = 0 a b c x1 x2 x3 = 0 xT.l = 0

  10. 2D Homogeneous Coordinates Important Properties 1(see book for details) • 3 coordinates, but only 2 degrees of freedom(only 2 ratios (x1 / x3), (x2 / x3) can change) • DUALITY: points, lines are interchangeable • Line Intersections = point: (a 3D cross-product) • Point ‘Intersections’ = line: • Projective theorem for lines  theorem for points! l1 l2 = x x1 x2 = l

  11. Homogeneous Coordinates Important Properties 2(see book for details) • Neatly Sidesteps ‘divide-by-zero’: • in (x,y) space (R2) • Remember, we store (x1 ,x2, x3)T, then compute (x1 / x3), (x2 / x3)only if OK. • “OK, yeah, sure, but so what?” ...

  12. Homogeneous Coordinates Why is deferring divide-by-zero so important? • Cleanly Defines all ‘Points at Infinity’ or ‘Ideal Points’ (x,y,0)T • ==outermost pixels (+/-, +/-) of an infinite image plane ((x,y), or R2) • (x,y,0) is an entire plane of points in the (semi-bogus) 3D space (x1 ,x2, x3)T space • but it is a ‘circle around infinity’ in the (legitimate) P2 projective space (x1 /x3, x2 /x3)T • Note! ‘Center of Projection’ is an ideal point. (Do outer limits of image plane ‘wrap around’ to that point?!?!)

  13. Homogeneous Coordinates Why is deferring divide-by-zero so important? Cleanly defines one ‘Line at infinity’ l (0,0,1)T, or “0x +0y +1=0” • All ideal points are on l: proof? l•(x1,x2,0)T = 0 • Any line l intersects with l line at an ideal point • Two parallel lines l and l’ always meet at an ideal point Let l = (a,b,c)T and l’ = (a,b,c’)T …(page 7)

  14. Homogeneous Coordinates x1 x2 x3 x’1 x’2 x’3 Important Properties 3(see book for details) • View Interpolation (‘Projective Transform’) • ‘Central’ image plane (x,y,-1)T • Choose a known point x’ • Apply 3x3 Matrix H to (x,y,-1)Tto make point on OTHER plane • Ray through known point x’ pierces unknown point x 3D world plane x’ x H x = x’ ‘Image Plane’ (x,y,-1) (0,0,-1) h11 h12 h13 h21 h22 h21 h31 h32 h33 = (0,0,0)

  15. END

  16. Useful 3D Graphics Ideas • Every xy point describes a 3D ray in (x1,x2,x3) • ‘Phantom’ dimension x3 is the (lost) distance-to-camera • Homogeneous coords allows matrix for 2-D translation: • (BUT a 3x3 transform in R3Euclidean Space (x,y,z) can’t translate—only rotate,scale, skew!) ‘Central Image Plane’ (x,y,1) 1 0 3 0 1 5 0 0 1 x y = 1 x+3 y+5 1 x2 An ‘Ideal Point’ (0,0,0) at Center of Projection Ray -x3 y x1 x 1 0 3 0 1 5 0 0 1 x y = z x+z y+z z

  17. 2D Homogeneous Coordinates ‘Central Image Plane’ (x,y,1) x2 An ‘Ideal Point’ (0,0,0) at Center of Projection Ray -x3 y x1 x • 3 coords, but only 2 degrees of freedom (2DOF)(x1 / x3), (x2 / x3) • Ideal points == (x,y,0) (every line hits the horizon at an ideal point) • Line Intersections = point: (a 3D cross-product) • Point ‘Intersections’ = line: • Duality: P2 theorem for lines  P2 theorem for points! l1 l2 = x x1 x2 = l

  18. 2D Homogeneous Coordinates ‘Central Image Plane’ (x,y,1) x2 An ‘Ideal Point’ (0,0,0) at Center of Projection Ray -x3 y x1 x Left-Handed: use –x3 or (x,y,-1) image ctr, or... • 3 coords, but only 2 degrees of freedom (2DOF)(x1 / x3), (x2 / x3) • Ideal points == (x,y,0) • Line Intersections = point: (a 3D cross-product) • Point ‘Intersections’ = line: • Duality: P2 theorem for lines  P2 theorem for points! l1 l2 = x x1 x2 = l

  19. 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

  20. 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’

  21. Projective Transform H H matrix: Plane-to-Plane mapping What if H is unknown, but we have four or more pairs of x,x’ points? (see pg 15, Zisserman) x1, x2, x3, x4, x’1, x’2, x’3, x’4 … x1 x2 x3 h11 h12 h13 h21 h22 h21 h31 h32 h33 x’1 x’2 x’3 H x’ = x = x’ x ‘Image Plane’ (x,y,-1) (0,0,-1) (0,0,0)

  22. Projective Transform: H • Matrix H transforms Points: • Given a point x: (x1,x2,x3) (or in R2 : (x1/x3, x2/x3)) x’ = Hx • And matrix H-T transform Lines: • Given a line l: (a,b,c) in R2 : (a·x+ b·y + c = 0) in P2 : (a·x1 + b·x2 + c·x3 = 0) l’ = H-T l Recall point x is on line l iff xTl =0 Jargon: points transform ‘covariantly’; lines transform ‘contravariantly’

  23. The bits and pieces of H H has 8 independent variables (DOF) • Cartesian (‘3D’) decomposition: 2D Translation(tx, ty)3D Scale (sx,sy,sz)3D Rotation (x,y,z) • Computer Vision (Projective 2D) decomp. 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)

  24. The bits and pieces of H H has 8 independent variables (DOF) • Computer Graphics method (3x3 matrix): 2D Translation(tx, ty)3D Scale (sx,sy,sz)3D Rotation (x,y,z) • Computer Vision method(2D projective): Isometry --3DOF (2D translate tx,ty; 2Drotate z; ) Similarity --4DOF (add uniform scale s;) Affine --6DOF (add orientable scale s,/s, s/s) Projective--8DOF (changes x3; 3D-rotation-like) Affects only x1,x2

  25. The bits and pieces of H H has 8 independent variables (DOF), not 9! • Decomposable into 3 useful parts: (pg 22) H = HS HA HP = • Similarity HS: • 2D translate, rotate, uniform scale only (4DOF) • Affine HA: • non-uniform scale (2DOF) • Projective HP: • Projective coupling for x3 (2DOF)    sR t 0T 1 K 0 0T 1 I 0 vT v            

  26. The bits and pieces of H H has 8 independent variables (DOF), not 9! • Decomposable into 3 useful parts: (pg 22) H = HS HA HP = • Similarity HS: • 2D translate, rotate, uniform scale only (4DOF) s=scale, =translate details… • Affine HA: • non-uniform scale (2DOF) • Projective HP: • Projective coupling for x3 (2DOF) s  cos  -s sin tx s sin  s  cos ty 0 0 1 1 0 0 0 1 0 v1 v2vS 1 k2 0 k3 1 0 0 01

  27. Compare HA and HP sR  t    0T 1 K  0    0T 1 I 0  vT v L M (recall)H = HS HA HP = L (parallelogram) HA m (K matrix, 2DOF) 2D image space (square) L m 2D world space HP (quadrilateral) ( v vector) 2D image space

  28. 1-D Projective Geometry (?Why? We’ll use it soon…) • A ‘side’ view of 2D projective geometry • Convert R1 scalar b to a 2-vector inP1 • As with P2, we can transform points: • (use various H2x2 s to change cyan lines below) b 1 H2b H2a

  29. 1-D Projective Geometry a1 a2 b1 b2 What is invariant here? all the positions, lengths and ratios change with each transform… Answer: ‘Cross Ratio’: cross(a,b,c,d) • let a= , b= … cross = |a-b||c-d| |a-c||b-d| cross(a,’b,’c,d’) = cross(a’,b’,c’,d’) a’ b’ c’ d’ a b c d

  30. Image Rectifying: Undo parts of H x’ = H x where H = HS HA HP GOAL: Put world plane x’ into view plane • Affine Rect.; (find only HP(2DOF)); • Similarity Rect.; (find only HA HP (6DOF)); • Full Rect.; (find all of H (8DOF)); METHODS:(mix & match?) • Full: 4-point correspondence • ‘Vanishing Point’, Infinity line methods • Other feature points, conics, …

  31. Image Rectifying: Undo parts of H x’ = H x where H = HS HA HP GOAL: Put world plane x’ into view plane • Affine Rect.; (find only HP(2DOF)); • Similarity Rect.; (find only HA HP (6DOF)); • Full Rect.; (find all of H (8DOF)); METHODS:(mix & match?) • Full: 4-point correspondence • ‘Vanishing Point’, Infinity line methods • Conics and circular points Other DOF? make assumptions, or ignore

  32. Vanishing-Point Methods: 1 • In 2D image, find two pairs of lines that are parallel in ‘world’ space. H world space 2D image

  33. Vanishing-Point Methods: 1 • Find intersection (vanishing points) p1, p2 (compute line intersections with 3D cross-products (see last lecture) • Horizon line Ih connects p1, p2. Ih p1 p2 H world space 2D image

  34. Vanishing-Point Methods: 1 • H-T transforms world infinity linel (0,0,1)Tto Ih • Answer (see old book)find, copy the lh coordinates! Hp= 1 0 0 0 1 0 lh1 lh2 lh3 Ih p1 p2 H world space 2D image

  35. Vanishing-Point Methods: 1 Limitations: • Accurate point, line-finding can be tricky • Can have high error sensitivity: ‘twitchy’ • Only rectifies Hp-- but what if HA, HS needed? Ih p1 p2 H world space 2D image

  36. Vanishing-Point Methods: 2 No parallel lines? • find 2D image line with a known distance ratio • Use Cross-Ratio (in P1along that line) to • Find vanishing point distance at point d’ Ih p1 p2 d’ c’ H b’ a b c d(at infinity) a’ 2D image world space

  37. 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)

  38. 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 = = —!STOP! — Trucco Book Does This… Zisserman Book calls it “naïve”, gives better method: ‘DLT’: direct linear xfm.) 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)

  39. Projective Transform H Finding H from point pairs (correspondences) • We know that Hx’ = x, and • we know at least 4 point pairs x’ and x that satisfy it: • ATTEMPT 1:‘plug & chug’: make a matrix of x and x’ values… H x’ = x x1 x2 x3 h11 h12 h13 h21 h22 h21 h31 h32 h33 x’1 x’2 x’3 = H x’1 x1 x’2 x’3 x’4 x2 x3 x4 =

  40. END --Please download the starter code, compile and run. --Then you can get started on Project 1 right away.

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