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Blossoms

Blossoms. CS 419 Advanced Topics in Computer Graphics John C. Hart Borrowed somewhat from Tom Sederberg’s notes. de Casteljau. p 1. p 12. p 2. p 012. 1- t. de Casteljau algorithm evaluates a point on a Bezier curve by scaffolding lerps

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Blossoms

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  1. Blossoms CS 419 Advanced Topics in Computer Graphics John C. Hart Borrowed somewhat from Tom Sederberg’s notes

  2. de Casteljau p1 p12 p2 p012 1-t • de Casteljau algorithmevaluates a point on aBezier curve byscaffolding lerps • Blossoming renames thecontrol and intermediatepoints, like p12, using apolar form, like p(0,t,1) p123 p0123 p01 t p23 p0 p3

  3. de Casteljau p(1,0,0) p12 p(1,1,0) p012 1-t • de Casteljau algorithmevaluates a point on aBezier curve byscaffolding lerps • Blossoming renames thecontrol and intermediatepoints, like p12, using apolar form, like p(0,t,1) p123 p0123 p01 t p23 p(0,0,0) p(1,1,1)

  4. de Casteljau p(1,0,0) p(1,t,0) p(1,1,0) p012 1-t • de Casteljau algorithmevaluates a point on aBezier curve byscaffolding lerps • Blossoming renames thecontrol and intermediatepoints, like p12, using apolar form, like p(0,t,1) p123 p0123 p(t,0,0) t p(1,1,t) p(0,0,0) p(1,1,1)

  5. de Casteljau p(1,0,0) p(1,t,0) p(1,1,0) p(t,t,0) 1-t • de Casteljau algorithmevaluates a point on aBezier curve byscaffolding lerps • Blossoming renames thecontrol and intermediatepoints, like p12, using apolar form, like p(0,t,1) p(1,t,t) p0123 p(t,0,0) t p(1,1,t) p(0,0,0) p(1,1,1)

  6. de Casteljau p(1,0,0) p(1,t,0) p(1,1,0) p(t,t,0) 1-t • de Casteljau algorithmevaluates a point on aBezier curve byscaffolding lerps • Blossoming renames thecontrol and intermediatepoints, like p12, using apolar form, like p(0,t,1) p(1,t,t) p(t,t,t) p(t,0,0) t p(1,1,t) p(0,0,0) p(1,1,1)

  7. Blossoming Rules p(1,0,0) p(1,t,0) p(1,1,0) p(t,t,0) 1-t • # of parameters = degree • Order doesn’t matter p(a,b,c) = p(b,a,c) • Linear in any parameter ap(a,b,c) = p(aa,b,c) = p(a,ab,c) p(a,b+c,d) = p(a,b,d) + p(a,c,d) • Lerping: (1-t)p(0,0,0) + tp(1,0,0)= p(0 (1-t),0,0) + p(1 t,0,0) = p(0 (1-t) + 1 t,0,0)= p(t,0,0) p(1,t,t) p(t,t,t) p(t,0,0) t p(1,1,t) p(0,0,0) p(1,1,1)

  8. Evaluation p(1,0,0) p(1,t,0) p(1,1,0) p(t,t,0) 1-t • Goal is to find p(t) bydiagonalization, bymanipulating blossomsinto p(t,t,t) • de Casteljau algorithm blossomsinto Bernstein polynomials p(1,t,t) p(t,t,t) p(t,0,0) t p(1,1,t) p(0,0,0) p(1,1,1) p(t) = p(t,t,t) = (1-t) p(t,t,0) + tp(t,t,1) = (1-t)[(1-t) p(t,0,0) + tp(t,0,1)] + t [(1-t) p(t,0,1) + tp(t,1,1)] = (1-t)2p(t,0,0) + 2 (1-t) tp(t,0,1) + t2p(t,1,1) = (1-t)2[(1-t)p(0,0,0)+tp(1,0,0)]+2 (1-t)t[(1-t)p(0,0,1)+tp(1,0,1)]+t2[(1-t)p(0,1,1)+tp(1,1,1)] = (1-t)3p(0,0,0) + 3 (1-t)2tp(0,0,1) + 3 (1-t) t2p(0,1,1) + t3p(1,1,1)

  9. B-Spline Blossoms • Three segments: [2,3],[3,4],[4,5] • Points within each segment influenced by four surrounding control points • Knots influenced by three surrounding control points • Need two extra knots ateach end of knot vector(in general, d-1 extraknots at each end) p(2,3,4) p(3,4,5) t=3 t=4 p(1,2,3) p(4,5,6) t=2 t=5 p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 4 5 6 7]

  10. Bohm Blossoms • Trick: Think of each segmentas a Bezier curve p(2,3,4) p(3,4,5) t=3 t=4 p(1,2,3) p(4,5,6) t=2 t=5 p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 4 5 6 7]

  11. Bohm Blossoms • Trick: Think of each segmentas a Bezier curve p(2,3,4) p(3,4,5) p(3,3,3) p(4,4,4) p(1,2,3) p(4,5,6) p(2,2,2) p(5,5,5) p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 4 5 6 7]

  12. Bohm Blossoms • Trick: Think of each segmentas a Bezier curve • Where should the othertwo control points gofor the [2,3] segment? p(2,3,4) p(3,4,5) p(3,3,3) p(1,2,3) p(4,5,6) p(2,2,2) p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 4 5 6 7]

  13. Bohm Blossoms • Trick: Think of each segmentas a Bezier curve • Where should the othertwo control points gofor the [2,3] segment? • Need to find: p(2,2,3) = 2/3 p(1,2,3) + 1/3 p(4,2,3) p(3,2,3) = 1/3 p(1,2,3) + 2/3 p(4,2,3) p(4,2,3) p(3,4,5) p(3,2,3) p(3,3,3) p(2,2,3) p(1,2,3) p(4,5,6) p(2,2,2) p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 4 5 6 7]

  14. Bohm Blossoms • Where are the endpoints located? • Need to find: p(2,1,2) = 1/3 p(0,1,2) + 2/3 p(3,1,2) p(3,4,3) = 2/3 p(3,4,2) + 1/3 p(3,4,5) p(2,2,2) = 1/2 p(2,1,2) + 1/2 p(2,3,2) p(3,3,3) = 1/2 p(3,2,3) + 1/2 p(3,4,3) • Reveals how to turn a B-splineinto a Bezier curve! p(3,4,2) p(3,4,3) p(3,4,5) p(3,2,3) p(3,3,3) p(2,3,2) p(3,1,2) p(2,2,2) p(2,1,2) p(0,1,2) knot vector: [0 1 2 3 4 5 6 7]

  15. Knot Insertion • Suppose we want to add aknot at t = 3.5 p(2,3,4) p(3,4,5) t=3 t=4 p(1,2,3) p(4,5,6) t=2 t=5 p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 4 5 6 7]

  16. Knot Insertion • Suppose we want to add aknot at t = 3.5 • Then we need new cp’sp(2,3,3.5), p(3,3.5,4)and p(3.5,4,5)and can get rid ofp(2,3,4) and p(3,4,5) p(3,3.5,4) p(2,3,4) p(3,4,5) p(3.5,4,5) p(2,3,3.5) t=3 t=4 p(1,2,3) p(4,5,6) t=2 t=5 p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 4 5 6 7]

  17. Knot Insertion • Suppose we want to add aknot at t = 3.5 • Then we need new cp’sp(2,3,3.5), p(3,3.5,4)and p(3.5,4,5)and can get rid ofp(2,3,4) and p(3,4,5) p(3,3.5,4) p(3.5,4,5) p(2,3,3.5) t=3.5 p(1,2,3) p(4,5,6) t=2 t=5 p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 3.5 4 5 6 7]

  18. de Boor Algorithm • What if we want toevaluate p(3.5)? • Then create a triple knotat t = 3.5 and figure outwhere to put the controlpoint p(3.5,3.5,3.5) p(3,3.5,4) p(3.5,4,5) p(2,3,3.5) t=3.5 p(1,2,3) p(4,5,6) t=2 t=5 p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 3.5 4 5 6 7]

  19. de Boor Algorithm • What if we want toevaluate p(3.5)? • Then create a triple knotat t = 3.5 and figure outwhere to put the controlpoint p(3.5,3.5,3.5) • Need p(3,3.5,3.5)and p(3.5,3.5,4) • Also subdivides B-spline into[0,1,2,3,3.5,3.5,3.5] and[3.5,3.5,3.5,4,5,6,7] p(3,3.5,4) p(3.5,3.5,4) p(3,3.5,3.5) p(3.5,4,5) p(2,3,3.5) p(3.5,3.5,3.5) p(1,2,3) p(4,5,6) p(0,1,2) p(5,6,7) knot vector: [0 1 2 3 3.5 3.5 3.5 4 5 6 7]

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