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Bezier Blossoms CS 418 Interactive Computer Graphics John C. Hart de Casteljau de Casteljau algorithm p 1 evaluates a point on a p 12 Bezier curve by p 2 p 012 1- t scaffolding lerps p 123 p 0123 Blossoming renames the p 01 control

  1. Bezier Blossoms CS 418 Interactive Computer Graphics John C. Hart

  2. de Casteljau • de Casteljau algorithm p 1 evaluates a point on a p 12 Bezier curve by p 2 p 012 1- t scaffolding lerps p 123 p 0123 • Blossoming renames the p 01 control and intermediate points, like p 12 , using a t p 23 polar form, like p (0, t ,1) p 0 p 3

  3. de Casteljau • de Casteljau algorithm p (1,0,0) evaluates a point on a p 12 p (1,1,0) Bezier curve by p 012 1- t scaffolding lerps p 123 p 0123 • Blossoming renames the p 01 control and intermediate points, like p 12 , using a t p 23 polar form, like p (0, t ,1) p (0,0,0) p (1,1,1)

  4. de Casteljau • de Casteljau algorithm p (1,0,0) p (1, t ,0) evaluates a point on a p (1,1,0) Bezier curve by p 012 1- t scaffolding lerps p 123 p 0123 • Blossoming renames the p ( t ,0,0) control and intermediate points, like p 12 , using a t p (1,1, t ) polar form, like p (0, t ,1) p (0,0,0) p (1,1,1)

  5. de Casteljau • de Casteljau algorithm p (1,0,0) p (1, t ,0) evaluates a point on a p (1,1,0) Bezier curve by p ( t , t ,0) 1- t scaffolding lerps p (1, t , t ) p 0123 • Blossoming renames the p ( t ,0,0) control and intermediate points, like p 12 , using a t p (1,1, t ) polar form, like p (0, t ,1) p (0,0,0) p (1,1,1)

  6. de Casteljau • de Casteljau algorithm p (1,0,0) p (1, t ,0) evaluates a point on a p (1,1,0) Bezier curve by p ( t , t ,0) 1- t scaffolding lerps p (1, t , t ) • Blossoming renames the p ( t , t , t ) p ( t ,0,0) control and intermediate points, like p 12 , using a t p (1,1, t ) polar form, like p (0, t ,1) p (0,0,0) p (1,1,1)

  7. Blossoming Rules 1. # of parameters = degree p (1,0,0) p (1, t ,0) Cubic: p (_,_,_) p (1,1,0) p ( t , t ,0) 1- t 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. Blossoming Rules 1. # of parameters = degree p (1,0,0) p (1, t ,0) Cubic: p (_,_,_) p (1,1,0) p ( t , t ,0) 1- t 2. Order doesn’t matter p (1, t , t ) p ( t , t , t ) p ( t ,0,0) p ( a , b , c ) = p ( b , a , c ) t p (1,1, t ) p (0,0,0) p (1,1,1)

  9. Blossoming Rules 1. # of parameters = degree p (1,0,0) p (1, t ,0) Cubic: p (_,_,_) p (1,1,0) p ( t , t ,0) 1- t 2. Order doesn’t matter p (1, t , t ) p ( t , t , t ) p ( t ,0,0) p ( a , b , c ) = p ( b , a , c ) t p (1,1, t ) 3. Setting up the board p (0,0,0) p (0,0,0), p (0,0,1), p (1,1,1) p (0,1,1), p (1,1,1)

  10. Blossoming Rules 1. # of parameters = degree p (1,0,0) p (1, t ,0) Cubic: p (_,_,_) p (1,1,0) p ( t , t ,0) 1- t 2. Order doesn’t matter p (1, t , t ) p ( t , t , t ) p ( t ,0,0) p ( a , b , c ) = p ( b , a , c ) t p (1,1, t ) 3. Setting up the board p (0,0,0) p (0,0,0), p (0,0,1), p (1,1,1) p (0,1,1), p (1,1,1) 4. Winning the game p ( t , t , t )

  11. Placing Blossoms

  12. p (1,0,0) Evaluation p (1, t ,0) p (1,1,0) p ( t , t ,0) 1- t p ( t ) = p ( t , t , t ) 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)

  13. p (1,0,0) Evaluation p (1, t ,0) p (1,1,0) p ( t , t ,0) 1- t p ( t ) = p ( t , t , t ) p (1, t , t ) p ( t , t , t ) = (1- t ) p ( t , t ,0) + t p ( t , t ,1) p ( t ,0,0) t p (1,1, t ) p (0,0,0) p (1,1,1)

  14. p (1,0,0) Evaluation p (1, t ,0) p (1,1,0) p ( t , t ,0) 1- t p ( t ) = p ( t , t , t ) p (1, t , t ) p ( t , t , t ) = (1- t ) p ( t , t ,0) + t p ( t , t ,1) p ( t ,0,0) t p (1,1, t ) = (1- t )[(1-t) p ( t ,0,0) + t p ( t ,0,1)] + t [(1- t ) p ( t ,0,1) + t p ( t ,1,1)] p (0,0,0) p (1,1,1)

  15. p (1,0,0) Evaluation p (1, t ,0) p (1,1,0) p ( t , t ,0) 1- t p ( t ) = p ( t , t , t ) p (1, t , t ) p ( t , t , t ) = (1- t ) p ( t , t ,0) + t p ( t , t ,1) p ( t ,0,0) t p (1,1, t ) = (1- t )[(1-t) p ( t ,0,0) + t p ( t ,0,1)] + t [(1- t ) p ( t ,0,1) + t p ( t ,1,1)] p (0,0,0) p (1,1,1) = (1- t ) 2 p ( t ,0,0) + 2 (1- t ) t p ( t ,0,1) + t 2 p ( t ,1,1)

  16. p (1,0,0) Evaluation p (1, t ,0) p (1,1,0) p ( t , t ,0) 1- t p ( t ) = p ( t , t , t ) p (1, t , t ) p ( t , t , t ) = (1- t ) p ( t , t ,0) + t p ( t , t ,1) p ( t ,0,0) t p (1,1, t ) = (1- t )[(1-t) p ( t ,0,0) + t p ( t ,0,1)] + t [(1- t ) p ( t ,0,1) + t p ( t ,1,1)] p (0,0,0) p (1,1,1) = (1- t ) 2 p ( t ,0,0) + 2 (1- t ) t p ( t ,0,1) + t 2 p ( t ,1,1) = (1- t ) 2 [(1- t ) p (0,0,0)+ t p (1,0,0)]+2(1- t ) t [(1- t ) p (0,0,1)+ t p (1,0,1)]+ t 2 [(1- t ) p (0,1,1)+ t p (1,1,1)]

  17. p (1,0,0) Evaluation p (1, t ,0) p (1,1,0) p ( t , t ,0) 1- t p ( t ) = p ( t , t , t ) p (1, t , t ) p ( t , t , t ) = (1- t ) p ( t , t ,0) + t p ( t , t ,1) p ( t ,0,0) t p (1,1, t ) = (1- t )[(1-t) p ( t ,0,0) + t p ( t ,0,1)] + t [(1- t ) p ( t ,0,1) + t p ( t ,1,1)] p (0,0,0) p (1,1,1) = (1- t ) 2 p ( t ,0,0) + 2 (1- t ) t p ( t ,0,1) + t 2 p ( t ,1,1) = (1- t ) 2 [(1- t ) p (0,0,0)+ t p (1,0,0)]+2(1- t ) t [(1- t ) p (0,0,1)+ t p (1,0,1)]+ t 2 [(1- t ) p (0,1,1)+ t p (1,1,1)] = (1- t ) 3 p (0,0,0) + 3 (1- t ) 2 t p (0,0,1) + 3 (1- t ) t 2 p (0,1,1) + t 3 p (1,1,1)

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