1. Computer Aided Geometric Design
Class Exercise #13
Advanced Topics - Blossoming 1

2. The Blossom of a Bezier Curve
Recall the de Casteljau algorithm, 𝑏 𝑖 ∈ ℝ 3 :
𝑏 0
1−𝑡
𝑏 0 1
1−𝑡 𝑡
𝑏 1
𝑏 0 2
1−𝑡 𝑡
𝑏 1 1
𝑏 0 3 =𝛾 𝑡
𝑏 2 ⋮ 𝑡
𝑏 1 2 ⋮
𝑏 2 1
𝑏 3 ⋮ 2

3. The Blossom of a Bezier Curve
Generalize by updating 𝑡 at each column:
𝑏 0
1− 𝑡 1
𝑏 𝑡 1
𝑡 1
𝑏 1
1− 𝑡 2
𝑏 𝑡 1 , 𝑡 2
𝑡 2
𝑏 𝑡 1
1− 𝑡 3
𝑏 𝑡 1 , 𝑡 2 , 𝑡 3
𝑏 2 ⋮
𝑡 3 ⋮
𝑏 𝑡 1 , 𝑡 2
𝑏 𝑡 1
𝑏 3 ⋮ 3

4. The Blossom of a Bezier Curve
We obtained a function:
𝑏 ≔ 𝑏 0 3 : 0,1 3 → ℝ 3 4

5. Q.1 Given the function 𝑏 𝑡 1 , 𝑡 2 , 𝑡 3 how can we find:
The corresponding Bezier curve?
The original control points?
The intermediate points of the de-Casteljau algorithm? 5

6. Solution For the Bezier curve choose 𝑡 1 = 𝑡 2 = 𝑡 3 =𝑡 and get:
𝛾 𝑡 =𝑏 𝑡,𝑡,𝑡 .
The curve is the blossom evaluated at the diagonal of the domain! 6

7. Solution For the control points:
First we know that a Bezier curve interpolates the first and last point, therefore:
𝑏 0 =𝛾 0 =𝑏 0,0,0 ,
𝑏 3 =𝛾 1 =𝑏 1,1,1 . 7

8. Solution What is 𝑏 0,0,1 ? 𝑏 0 𝑏 0 𝑏 1 𝑏 0 𝑏 1 𝑏 2 𝑏 1 𝑏 1 𝑏 2 𝑏 3 1 1
𝑏 1 1 1
𝑏 0
𝑏 1 1
𝑏 2
𝑏 1 1
𝑏 1 1
𝑏 2
𝑏 3 8

9. Solution Similarly: 𝑏 2 =𝑏 0,1,1 .
What about intermediate points of the de-Casteljau algorithm? 9

10. Solution 𝑏 0 =𝑏 0,0,0 𝑏 0 1 =𝑏 0,0,𝑡 𝑏 1 =𝑏 0,0,1 𝑏 0 2 =𝑏 0,𝑡,𝑡
𝑏 1 1 =𝑏 0,𝑡,1
𝑏 0 3 =𝑏 𝑡,𝑡,𝑡
𝑏 2 =𝑏 0,1,1
𝑏 1 2 =𝑏 𝑡,𝑡,1
𝑏 2 1 =𝑏 𝑡,1,1
𝑏 3 =𝑏 1,1,1 10

11. Solution The general fact is formulated as follows: 𝑏 𝑖 =𝑏 0 𝑛−𝑖 , 1 𝑖
𝑏 𝑖 =𝑏 0 𝑛−𝑖 , 1 𝑖
and the recursion rule is:
𝑏 0 𝑛−𝑟−𝑖 , 𝑡 𝑟 , 1 𝑖 = 1−𝑡 𝑏 0 𝑛−𝑟−𝑖+1 , 𝑡 𝑟−1 , 1 𝑖 +𝑡𝑏 0 𝑛−𝑟−𝑖 , 𝑡 𝑟−1 , 1 𝑖+1 11

12. Q.2 Prove that the blossom of a Bezier curve is: Symmetric
Multi-Affine. 12

13. Solution
We show the symmetry for two variables. The general case follows from the same arguments.
Consider:
𝑏 1
𝑎 𝑠 = 1−𝑠 𝑏 0 +𝑠 𝑏 1
𝑎 𝑡 = 1−𝑡 𝑏 0 +𝑡 𝑏 1
𝑏 𝑡
𝑎 𝑠
𝑏 𝑠
𝑎 𝑡
𝑏 2
𝑏 𝑠 = 1−𝑠 𝑏 1 +𝑠 𝑏 2
𝑏 0
𝑏 𝑡 = 1−𝑡 𝑏 1 +𝑡 𝑏 2 13

14. Solution
The geometry behind the proof is the following: the middle point can be interpolated first with 𝑡 and then with 𝑠, or vice versa.
𝑏 1
𝑏 𝑡
𝑎 𝑠
𝑏 𝑠
𝑎 𝑡
𝑏 2
𝑏 0 14

15. Solution Now: 𝑏 𝑡,𝑠 = 1−𝑠 𝑎 𝑡 +𝑠 𝑏 𝑡 𝑏 𝑠,𝑡 = 1−𝑡 𝑎 𝑠 +𝑡 𝑏 𝑠
𝑏 𝑡,𝑠 = 1−𝑠 𝑎 𝑡 +𝑠 𝑏 𝑡
𝑏 𝑠,𝑡 = 1−𝑡 𝑎 𝑠 +𝑡 𝑏 𝑠
From here compute and verify both expressions are the same (describing the same middle point!).
𝑎 𝑠 = 1−𝑠 𝑏 0 +𝑠 𝑏 1
𝑎 𝑡 = 1−𝑡 𝑏 0 +𝑡 𝑏 1
𝑏 𝑠 = 1−𝑠 𝑏 1 +𝑠 𝑏 2
𝑏 𝑡 = 1−𝑡 𝑏 1 +𝑡 𝑏 2 15

16. Solution
An Affine combination is a linear combination with coefficients that sum to 1.
(how is that different from a convex combination?) 16

17. Solution A function of one variable is called affine if:
𝑓 𝛼𝑥+ 1−𝛼 𝑦 =𝛼𝑓 𝑥 + 1−𝛼 𝑓 𝑦 .
A multivariate function is called multi affine if it is affine with respect to every variable. 17

18. Solution
From the symmetry – it suffices to show that the blossom is affine with respect to the first variable.
The proof is by induction using the deCasteljau recursion rule! 18

19. Blossoms as Polar Forms
Definition:
Given a polynomial 𝑓 𝑡 , the polar form of 𝑓 𝑡 is the polynomial in 𝑛 variables 𝐹 𝑢 1 ,…, 𝑢 𝑛 with the properties:
Symmetric
Multi-Affine
The diagonal: 𝐹 𝑡,…,𝑡 =𝑓 𝑡 19

20. Blossoms as Polar Forms
Theorem:
Every polynomial 𝑓 𝑡 indeed has a polar form 𝐹 and is unique. 20

21. Blossoms as Polar Forms
Idea behind the proof of existence:
𝑓 𝑡 = 𝑖=0 𝑛 𝑐 𝑖 𝑡 𝑖
Then:
𝐹 𝑡 1 ,…, 𝑡 𝑛 = 𝑖=0 𝑛 𝑐 𝑖 𝑛 𝑖 −1 𝑆⊂ 1,…,𝑛 , 𝑆 =𝑖 𝑗∈𝑆 𝑡 𝑗 21

22. Blossoms as Polar Forms
In practice, replace monomials in the following way:
𝑡↦ 𝑡 1 +…+ 𝑡 𝑛 𝑛
𝑡 2 ↦ 𝑡 1 𝑡 2 + 𝑡 2 𝑡 3 + 𝑡 1 𝑡 3 …+ 𝑡 𝑛−1 𝑡 𝑛 𝑛 2 22

23. Q.3
How can the blossom of a Bezier curve be used for subdivision at 𝑡? 23

24. Solution 𝑏 0 =𝑏 0,0,0 𝑏 0 1 =𝑏 0,0,𝑡 𝑏 1 =𝑏 0,0,1 𝑏 0 2 =𝑏 0,𝑡,𝑡
𝑏 1 1 =𝑏 0,𝑡,1
𝑏 0 3 =𝑏 𝑡,𝑡,𝑡
𝑏 2 =𝑏 0,1,1
𝑏 1 2 =𝑏 𝑡,𝑡,1
𝑏 2 1 =𝑏 𝑡,1,1
𝑏 3 =𝑏 1,1,1 24

25. Solution 𝑏 0 =𝑏 0,0,0 𝑏 0 1 =𝑏 0,0,𝑡 𝑏 1 =𝑏 0,0,1 𝑏 0 2 =𝑏 0,𝑡,𝑡
𝑏 1 1 =𝑏 0,𝑡,1
𝑏 0 3 =𝑏 𝑡,𝑡,𝑡
𝑏 2 =𝑏 0,1,1
𝑏 1 2 =𝑏 𝑡,𝑡,1
𝑏 2 1 =𝑏 𝑡,1,1
𝑏 3 =𝑏 1,1,1 25