Blossoms Bezier Curve Construction Reminder Successive linear - - PowerPoint PPT Presentation

Blossoms

Bezier Curve Construction Reminder

• Successive linear

interpolations at point t

• Each point can be found

as a function of the parameter, t, and the control points,

• bt=∑

i=0 n

i ,nbi=∑

i=0 n



n i t

i1−t n−1bi

bi

Changing parameters

• We can interpolate

using a vector of values

• btb[t1,t 2...tn]

b0 b2 b1 b0

1[t]

b1

1[t]

b0

2[t ,t]

t=0.5

Changing parameters

• We can interpolate

using a vector of values

• This is the polar form of

function.

b0 b2 b1 b0

1[t1]

b1

1[t1]

b0

2[t1,t 2]

b0

2[t1,t3]

t1≈0.5 t 2≈0.25 t3≈0.75

btb[t1,t 2...tn]

Affect of changing Parameters

• Red:

; Green:

• Can also fix one or

more parameters

b[t ,t] b[t ,t

2]

Polar Representation

• For each parametric function univariate function,

, there is a comparable multivariate function

• Instead of one parameter, we expand the function to

have one parameter for each degree of freedom.

pt:ℝℝ

n

P[t0,t1 ,t 2...tn]= pt

Examples

• The polar form of a standard cubic polynomial is
• A bivariate quadratic function, in parametric and

polar form, is

pt=a0a1ta2t

2a3t 3

P[t1,t 2,t3]=a0 a1 3 t1t 2t3 a2 3 t1t2t 2t3t3t1a3t1t2t3 x= x , y , f  x=x

23xy2y 2

F [ x0, x1]=x0 x1 3 2 x0 y1 3 2 x1 y02 y0 y1

Blossoms

• A multivariate function

represents the polar form of a polynomial function.

• This polar form function is known as a blossom

b[t1,t 2,...t n]

Properties

• If

, then the function diverges to a standard parametric curve

– Example, using the standard cubic –

• This is the definition of the polar form of a function.

t 0=t1=...=tn P[t ,t ,t ]=a0 a1 3 ttt a2 3 t

2t 2t 2a3t 3

a0a1ta2t

2a3t 3= pt

Properties (Cont.)

• Symmetric

– Proof is Menalaos' Theorem

• Affine
• Both these properties can be extended to n degrees
• f freedom

b[s ,t]=b[t , s] b[ rs ,t]= b[r ,t]b[s ,t ], =1

Menelaos' Theorem

• We can define the

following points

• We will see shortly why the parameter

switching in the s terms is allowed

b[0,t]=1−tb0t b1 b[s ,0]=1−sb0sb1 b[1,t]=1−tb1t b2 b[s ,1]=1−sb1sb2

Menelaos' Theorem (cont.)

• We further define 2

more points

• Menalaos' theorem

states:

b[s ,t]=1−tb[s ,0]t b[s ,1] b[t , s]=1−sb[0,t]sb[1,t] b[s ,t]=b[t , s]

Menalaos' Theorem (Proof)

• Can switch the parameters (Symmetric)
• The extension to n degrees of freedom is trivial

(leave some parameters untouched

b[s ,t]=1−tb[s ,0]tb[s ,1] 1−t1−sb0[1−tst1−s]b1stb2 1−sb[0,t]sb[1,t]=b[t , s]

Multi-Affinity

• Blossoms are affine invariant with respect to the 1st

parameter.

• We then apply the property of symmetry to extend

to other parameters.

n-Degree Properties

• Symmetric

–

denotes a permutation of the n arguments

b[t1,t 2,...tn]=b[t1,t2,...t n] t1,t 2,...tn

n-Degree Properties (Cont

• Multiaffinity

– Example

b[ rs,*]= b[r ,*][s ,*], =1 b[t1,t 2,t3]=1−t1b[0,t 2,t3]t1b[1,t2,t3]

Properties (Cont)

• Diagonalization

–

• One last property: Vector notation

– – Example,

t1=t 2....=tn=t , b[t ,t ,... ,t]=b[t

〈n〉]

b[  b−a,*]=b[b ,*]−b[a ,*] a=0,b=1,b−a= 1 b[  1, r , s]=b[1,r , s]−b[0,r , s]

Leibniz Formula

• When given a blossom argument of form

we get the following formula:

 rs

〈n〉

b[ rs

〈n〉]=∑ i=0 n



n i 

i  n−i b[r 〈i〉 , s 〈n−i〉]

Leibniz Formula, Proof

• We prove via induction
• The initial case is trivial

n=1: b[ rs]= b[r]b[s]

Leibniz Formula, Cont.

• Inductive step:
• We transform the indexes, and

 r s

〈n1〉=∑ i=0 n



n i 

i  n−i b[ r s ,r 〈i〉 , s 〈n−i 〉]=...

∑

i=0 n



n i 

i1 n−i b[r 〈i1〉 , s 〈n−i 〉]∑ i=0 n



n i 

i  n1−i b[r 〈i〉 , s 〈n1−i〉]

 rs

〈n1〉=...

∑

i=0 n1



n i−1

i  n1−i b[r 〈i 〉, s 〈n1−i〉]∑ i=0 n1



n i 

i  n1−i b[r 〈i〉 , s 〈n1−i〉]

Leibniz Formula, Conclusion

• We last use the recursion
• We combine the 2 summations, and conclude
• 

n1 i = n i−1 n i  b[ rs

〈n1〉]=∑ i=0 n1



n1 i 

i  n1−i b[r 〈i 〉, s 〈n1−i 〉]

Uses of Blossoms

• Remember the de Casteljau Algorithm.
• Given a control polygon

and , this interpolation results in:

b0, b1,...bn t∈[0,1] b0, b1, b2, b3 b0

1, b1 1, b2 1

b0

2, b1 2

b0

3

• Add new parameter for each interpolation

b0, b1, b2, b3 b0

1[t1] ,

b1

1[t1] ,

b2

1[t1]

b0

2[t1,t 2] ,

b1

2[t1,t2]

b0

3[t1,t 2,t3]

Evaluation

• To evaluate, we must first check special arguments
• Can repeat for all control points:

[t1,t2,t3]=[0,0,0] bi=b[0

〈n−i〉 ,1 〈i〉]

b0, b1, b2, b3 b0, b1, b2 b0, b1 b0 b0=b[0,0,0]

Evaluation Cont

• We next evaluate at
• We continue in the same manner to get

[t1,t2,t3]=[0,0,t] b0, b1, b2, b3 b0, b1, b2 b0, b1 b0

1

b0

1=b[0,0,t]

b[0,0,0] , b[0,0,1] , b[0,1,1], b[1,1,1] b[0,0,t] , b[0,t ,1] , b[t ,1,1] b[0,t ,t] , b[t ,t ,1] b[t ,t ,t]

Bezier Curves with Blossoms

• We can now rewrite the recursion algorithm as
• Using the blossom formulation, and the substitution

We can now rewrite the Leibniz Formula

b[0

〈n−r−i〉 ,t 〈 r〉 ,1 〈i〉]=...

1−tb[0

〈n−r−i1〉 ,t 〈r−1〉 ,1 〈i〉]t b[0 〈n−r−i〉 ,t 〈r−1〉 ,1 〈i1〉]

t=1−t⋅0t⋅1 b[t

〈n〉]=∑ i=0 n



n i t

i1−t i−i b[0 〈n−i〉 ,1 〈i〉]=∑ i=0 n

i ,n⋅bi

Derivative of the Bezier Blossom Curve

• Tangent line is line connecting the last 2

interpolation points

b' t=n⋅b[t

〈n−1〉 ,

1]=n⋅b[t

〈n−1〉 ,1]−b[0,t 〈n−1〉]

Derivative, Proof

• We start with the following:

– t is a point on the real line is a 1D vector

 v d b' dt =lim

 v

1 ∣ v∣[b[t v

〈n〉]−b[t 〈n〉]]=...

lim

 v 

1 ∣ v∣[∑

i=0 n



n i b[t

〈n−i 〉 ,

v

〈i 〉]−b[t 〈n〉]]

b[t v

〈n〉]=∑ i=0 n

b[t

〈n−i〉 ,

v

〈i〉]

Proof, Cont.

• We cancel, rearrange, and factor, and:
• We observe that,

and that the limit annihilates the remaining terms, giving us

d b' dt =lim

 v 0 nb[t 〈n−1〉 , 

v ∣ v∣] n 2b[t

〈n−2〉 , 

v ∣ v∣ , v]...  v ∣ v∣= 1 b'=nb[t

〈n−1〉,

1]

High Order Derivative

• Using a similar pattern and proof, we arrive at

b

r=

n n−r! b[t

〈n−r〉 ,

1

〈r〉]

Bezier Blossom Subdivision

• Subdivision of a bezier curve at t=c

ci , f =b[0

〈n−i〉 ,c 〈i〉]

ci ,b=b[c

〈n−i〉 ,1 〈i 〉]

b[0,0,0] , b[0,0,1] , b[0,1,1], b[1,1,1] b[0,0,c], b[0,c ,1] , b[c ,1,1] b[0,c ,c] , b[c ,c ,1] b[c ,c ,c]

Bezier Degree Raising

• The new control points can be found by
• This is the average of all the points, with the

argument

• mitted from the sum

b

1= 1

n1∑

j=0 n1

b[t1,...t n1│t j] t j

Degree Raising Cont.

• Need Diagram of Example

Blossoms and B-Splines

• The control points for a quadratic Bezier curve are

b[0,0], b[0,1], and b[1,1].

– Successive pairs from the sequence: 0,0,1,1.

• We generalize

0,0,1,1  u0,u1,u2,u3

b[u ,u]=... u2−u u2−u1 b[u1,u] u−u1 u2−u1 b[u ,u2]=... u2−u u2−u1  u2−u u2−u0 b[u0 ,u1] u−u0 u2−u0 b[u1 ,u2]... u−u1 u2−u1  u3−u u3−u1 b[u1 ,u2] u−u1 u3−u1 b[u2 ,u3]

• We use the following identities:
• We can do the recursion

u= ui−u ui−u j u j u−u j ui−u j ui

b[u0,u1] , b[u1,u2] , b[u2,u3] b[u1,u] , b[u ,u2] b[u ,u]

B-Spline Blossoms

• Let

, an interval in a sequence of knots

• Ordered sets of successive knots,

such that each set contains

• r
• The set

is defined such that

– Consists of r+1 successive knots – Is the (r-i)th element in

U=[uI ,uI 1] U i

r

U i

r

uI uI1 U i

r

U i

r

uI U i

r

U i

r

• The control points for a B-Spline curve can be

defined as

• The point on the curve

are computed recursively

• The derivative is, similar to a Bezier derivative,

d i=b[U i

n−1]

b[u

〈n〉]

d i

ru=b[u 〈r〉 ,U i n−1−r] ,

r=1,... ,n; i=0,... ,n−r b' [u

〈n〉]=nb[u 〈n−1〉,

1] b

r[u 〈n〉]=

n! n−r! b[u

〈n−r〉 ,

1

〈r〉]

B-Spline Blossom Example 1

B-Spline Example 2

• Given a portion of a knot sequence and interval:
• If n=3, the evaluation is:
• k=...u3,u4,u5,u6,u7,u8,...;

U =[u5,u6] b[u3,u4,u5] , b[u4,u5,u6], b[u5,u6,u7] , b[u6,u7,u8] b[u ,u4,u5] , b[u ,u5,u6] , b[u ,u6,u7] b[u ,u ,u5] , b[u ,u ,u6] , b[u ,u ,u]

• We can now write this as:

b[u3,u4,u5] , b[u4,u5,u6], b[u5,u6,u7] , b[u6,u7,u8] b[u ,u4,u5] , b[u ,u5,u6] , b[u ,u6,u7] b[u ,u ,u5] , b[u ,u ,u6] , b[u ,u ,u] b[U 0

2] ,

b[U 1

2] ,

b[U 2

2] ,

b[U 3

2]

b[u ,U 0

1],

b[u ,U 1

1],

b[u ,U 2

1]

b[u ,u ,U 0

0] ,

b[u ,u ,U 1

0],

b[u ,u ,u]

Knot Insertion

• A B-Spline defined over the interval, U, with control

points

• every n-tuple of successive knots
• If we insert the knot

into the knot vector, the new control points are

– These are all n-tuples of the new knots containing at

least 1 of the original endpoints of U

 u d [U i

n−1]

d [  U i

n−1]

Knot Insertion

Degree Raising

• Let

be a sequence of n+1 real numbers

• Let

be the sequence with the value removed

• V

n│vi

V

n

vi  b[V

n1]= 1

n1∑

i=1 n1

b[V

n1│vi]

Degree Raising Example

• Say we define a B-Spline over
• The interval

becomes the interval

• Knowing that

u0=u1=u2 ,u3,... [u4,u5] [  u7,  u8]  d 4=  d 7[  u4,  u5,  u6,  u7]=... 1 4 d 4[  u4,  u5,  u6]d 4[  u4,  u5,  u7]d 4[  u4,  u6,  u7]d 4[  u5,  u7,  u7]  u4=  u5=u3 ,  u6=  u7=u4  d 4=1 2 d 4[u3,u3,u4]d 4[u3,u4,u4]

Smoothness

• In general, if a knot u has multiplicity r then a B-

Spline of degree n has smoothness at that knot

• The highest order osculant at a given knot is

C

n−r

• n−ru=b[ 

u

〈r〉,u 〈n−r〉]

 u

Smoothness Examples