1. Computer Aided Geometric Design Mika Hirsch
Blossoms
Computer Aided Geometric Design
Mika Hirsch

2. Blossoms are Polar forms
Definition of polar forms
de Casteljau algorithm
Subdivision
Degree raising
Derivation
Continuity
10 אוקטובר, 2019

3. Definition 1. – Polar form
Given F(t), a polynomial of degree at most n, the polar form of F(t), is the unique, symmetric, n-affine polynomial f (u1,…,un), satisfying the identity F(t) = f (t,…,t)
In contrast, F(t) itself is the diagonal form of f.
10 אוקטובר, 2019

4. Polar Form Properties
Symmetry f (u1,u2,...,un) = f (up(1),up(2) ,...,up(n)) for any permutation p of (1,2,…,n)
10 אוקטובר, 2019

5. Polar Form Properties
Multi-affine f(u1,u2,…,αuk+βwk,….,un) = αf(u1,u2,…,uk,…, un) + βf(u1,u2,…,wk,…, un) with α+β=1
Or more generally, f is multiaffine if
for all i=1,…,n, αj∊ℝ and
10 אוקטובר, 2019

6. Polar Form Properties
Diagonal F(t) = f(t,…,t)
10 אוקטובר, 2019

7. Example 1. – Polar forms of simple monomials
10 אוקטובר, 2019

8. Example 1. (Cont.) Affinity check… Symmetry and diagonal are trivial
10 אוקטובר, 2019

9. Example 2. Compute the polar form of G(t)= t3+3t2 -6t-8
The polar form is symmetric, triaffine polynomial g(t1,t2,t3) that satisfies g(t,t,t)=G(t)
For g to be triaffine, we must have the form
ci are constants in ℝ.
10 אוקטובר, 2019

10. G(t)= t3+3t2 -6t-8
Example 2.– (Cont.)
From the identity g(t,t,t)=G(t), we get c1 = 1, c2+c3+c4=3, c5+c6+c7=-6, c8=-8
Since g(t1,t2,t3) must be symmetric, we can add c2=c3=c4=1 and c5=c6=c7=-2.
We are left with the unique choice: g (t1,t2,t3) := t1t2t3+t1t2+t2t3+t1t3-2t1-2t2-2t3-8
10 אוקטובר, 2019

11. Theorem 1. – Existence and Uniqueness
Univariate polynomials F(t), of degree at most n are equivalent to symmetric, n-affine polynomials f (u1,u2,...,un) in the sense that, given a polynomial of either type, there exists a unique polynomial of the other type that satisfies the identity F(t) = f (t,t,…,t).
10 אוקטובר, 2019

12. Proof - Existence
In our given polynomial, we replace each ti by the expressions:
10 אוקטובר, 2019

13. Existence – general case
General polar form of a polynomial
Is given by:
Symmetry ensures uniqueness of construction.
10 אוקטובר, 2019

14. The de Casteljau Algorithm
Given points Pi, i = 0,...,3, our goal is to determine a curve G(t), for all values P1 P2
P0 = G(0)
P3 = G(1)
10 אוקטובר, 2019

15. The de Casteljau Algorithm (Cont.)
Let g be the polar form of G
We define P0 = g(0,0,0), P1=g(0,0,1),P2=g(0,1,1),P3=g(1,1,1)
P1=g(0,0,1)
P2=g(0,1,1)
P0 = g(0,0,0)
P3=g(1,1,1)
10 אוקטובר, 2019

16. The de Casteljau Algorithm (Cont.)
Since the polar form g is affine in its third argument, the line joining the points g(0,0,0) and g(0,0,1) must be the line g(0,0,t), tϵ[0,1]. The points g(0,t,1) and g(t,1,1) are constructed similarly.
g(0,0,1)
g(0,t,1)
g(0,1,1)
g(0,0,t)
g(t,1,1)
g(0,0,0)
g(1,1,1)
10 אוקטובר, 2019

17. The de Casteljau Algorithm (Cont.)
From symmetry g(0,0,t)=g(0,t,0) and g(0,t,1)=g(t,0,1) , so we can construct g(0,t,t) and g(t,t,1)
g(0,0,1)
g(0,t,1)=g(t,0,1)
g(0,1,1)
g(0,t,t)
g(t,t,1)
g(0,0,t)=g(0,t,0)
g(t,1,1)
g(0,0,0)
g(1,1,1)
10 אוקטובר, 2019

18. The de Casteljau Algorithm (Cont.)
In the last stage we calculate g(t,t,t)=G(t).
g(0,0,1)
g(0,t,1)=g(t,0,1)
g(0,1,1)
g(0,t,t)=g(t,t,0)
g(t,t,1)
g(t,t,t)=G(t)
g(0,0,t)=g(0,t,0)
g(t,1,1)
g(0,0,0)=G(0)
G(1)=g(1,1,1)
10 אוקטובר, 2019

19. Polar form to Bezier curve
Using the substitution t=(1-t) .0+t .1
and considering the affine invariance, we have
using symmetry:
10 אוקטובר, 2019

20. Polar form to Bezier curve (Cont.)
Continuing recursively, we get
10 אוקטובר, 2019

21. Polar form to Bezier curve (Cont.)
We have a constructed the polar form in terms of Bernstein blending functions!
10 אוקטובר, 2019

22. Corollary
The control points of a Bezier curve in the polar from, are given by the formula:
10 אוקטובר, 2019

23. Corollary - subdivide
A given Bezier curve γ(t) is subdivided at the point t into two Bezier curves γ1(t) and γ2(t) whose control points are given by the points on boundary of the de Casteljau table
10 אוקטובר, 2019

24. Subdivide (Cont.) In our example:
g(0,0,0), g(0,0,t),g(0,t,t),g(t,t,t) are the control points of γ1(t)
g(t,t,t)
(1-t) t
g(0,t,t)
g(t,t,1)
(1-t) t
g(0,0,t)
g(0,t,1)
g(t,1,1)
(1-t) t
g(0,0,0)
g(0,0,1)
g(0,1,1)
g(1,1,1)
10 אוקטובר, 2019

25. Subdivide (Cont.) In our example:
g(t,t,t), g(t,t,1), g(t,1,1), g(1,1,1) are the control points of γ2(t)
g(t,t,t)
(1-t) t
g(0,t,t)
g(t,t,1)
(1-t) t
g(0,0,t)
g(0,t,1)
g(t,1,1)
(1-t) t
g(0,0,0)
g(0,0,1)
g(0,1,1)
g(1,1,1)
10 אוקטובר, 2019

26. Degree raising - Polar form
Given a Bezier curve, G(t), of degree n, and its polar form g(t1,…,tn), we want to find a Bezier curve H(t) of degree n+1 such that H(t):=G(t)
10 אוקטובר, 2019

27. Lemma - Degree Raising using Polar form
If we build H(t) in the following manner:
H(t) = h(t1,…,tn+1)
Then H(t) is the degree raising of G(t) to degree n+1 i.e. H(t) is of degree n+1 and H(t):=G(t)
Leave out ti
10 אוקטובר, 2019

28. Degree Raising – Proof
10 אוקטובר, 2019

29. Degree Raising – General case
Given a Bezier curve, G(t), of degree n, and its polar form g(t1,…,tn), we can define a symmetric (n+d)-affine mapping h as
H(t), the diagonal of h(t1,…,tn+d), is the degree raised polynomial we want, i.e. H(t)=G(t) for any t, and is of degree (n+d).
10 אוקטובר, 2019

30. Degree Raising – General case proof.
h(t1,…,tn+d) is symmetric because every combination of n variables out of (n+d) variables are taken into account.
h(t1,…,tn+d) is affine since it is affine combination of
affine mappings.
10 אוקטובר, 2019

31. Example 3.
Degree Raise A Quadric To Quartic Bezier Curve Given a quadric Bezier curve, defined by 3 control points P0, P1 and P2. That is, the blossom has,
Define symmetric 4-affine mapping
10 אוקטובר, 2019

32. Example 3. (Cont.) The degree raised quartic curve is define as
And has the control points:
10 אוקטובר, 2019

33. Example 3. (Cont.)
10 אוקטובר, 2019

34. Definition –affine map of vectors
Consider the vector
We define recursively as
10 אוקטובר, 2019

35. Derivation The k-th derivative of F is gives as Note that
10 אוקטובר, 2019

36. Example
10 אוקטובר, 2019

37. Example (Cont.)
10 אוקטובר, 2019

38. Continuity Theorem: Ck-conditions
Let F(t) and G(t) be two polynomials of degree n, f and g their polar form respectively, and let u∊ℝ. Then the following two statements are equivalent:
F(t) and G(t) are Ck-conditions at u
f(u,…,u,u1,…,uk)=g(u,…,u,u1,…,uk)
for u1,…,uk∊ℝ
10 אוקטובר, 2019