Degree reduction of Bézier curves

Lizheng Lululz_zju@yahoo.com.cn

Mar. 8, 2006

Outline

Overview

Recent developments

Our work

Problem formulation

Problem I: Given a curve of degree n in ,to find a curve of degree m, such that,

( ), ( ) mind t t P Q

( )tP

( )tQ

sR

An example

Degree from 7 to 4

Two kinds of methods

Component-wise Vector decomposition Degree reduction at each decomposition Combining all the components

Euclidean [Brunnett et al., 1996] Consider all the components together ( ), ( ) ( ), ( ) for all ( )d t t d t t tP Q P Q Q

Unconstrained and constrained degree reduction

Constrained degree reduction

Problem II: Given a curve of degree n in ,to find a curve of degree m, such that,

I)

II)

( ), ( ) mind t t P Q

( )tP

( )tQ

sR

( ) ( )(0) (0), 0, ,j j j r P Q

( ) ( )(1) (1), 0, ,j j j s P Q

Metric choice

-norms on C[0,1]

Weighted -norms

Others Control points perturbing

pL

11

0d

pp

pf f t

pL

11

0d

pw p

pf w f t

Approximation Theory

L1-norm Chebyshev polynomials of second kind

L2-norm Legendre polynomials

L∞-norm Chebyshev polynomials of first kind

Lp-norms L1-norm

Kim and Moon, 1997 L2-norm

Ahn et al., 2004; Chen and Wang, 2002; Eck, 1995; Zheng and Wang, 2003;

Zhang and Wang, 2005; L∞-norm

Eck, 1993; Ahn, 2003

Present status

Unconstrained Solved and very mature

Constrained (Optimal approximation) Solved for L2-norm Unsolved for L1-norm and L∞-norm

Some methods have been proposed, but not optimal

Outline

Overview

Recent developments

Our work

Multiple degree reduction and elevation of Bézier curves using Jacobi-Bernstein

basis transformations

Rababah, A., Lee, B.G., Yoo, J.

Submitted to Applied Numerical Mathematics

Main contribution

Unified matrix representations An unconstrained component-wise metho

d Explicit Simple and efficient Explicit approximating error Include three previous methods

Lp-norm, p =1, 2, ∞

Jacobi polynomials

Orthonormal on [0,1] with

[Szegö, 1975]

Special kinds of orthonormal polynomials α=β=-1/2

Chebyshev polynomials of second kind α=β=0

Legendre polynomials α=β=1/2

Chebyshev polynomials of first kind

Jacobi-Bernstein basis transformation

( , ) ( , ) ( , )0 0, , , ,n n

n n nP P B B M

1( , ) ( , ) ( , )0 0, , , ,n n

n n nB B P P M

Jacobi Bernstein

[Rababah, 2004]

( , )nM

μ,ν= 0,1, … ,n

Lemma

Degree elevation

(1)

(2)

(3)

(4)

Theorem for degree elevation

Jacobi-weighted L2-norm

Jacobi-weighted L2-norm

Special cases forJacobi-weighted L2-norm

α=β=-1/2 L∞ -approximation

α=β=0 L2 -approximation

α=β=1/2 L1 -approximation

Degree reduction by degree elevation

0

( ) ( )m

m mi i

i

b x b B x

0

( ) ( )n

ni i

i

f x c B x

?

Degree reduction by degree elevation

0

( ) ( )m

m mi i

i

b x b B x

0

( ) ( )n

ni i

i

f x c B x

= 0

?

Degree reduction by Jacobi polynomials

0

( ) ( )m

m mi i

i

b x b B x

0

( ) ( )n

ni i

i

f x c B x

?

Degree reduction by Jacobi polynomials

0

( ) ( )m

m mi i

i

b x b B x

0

( ) ( )n

ni i

i

f x c B x

(1)

?

Degree reduction by Jacobi polynomials

0

( ) ( )m

m mi i

i

b x b B x

0

( ) ( )n

ni i

i

f x c B x

(1)

(2)

?

Degree reduction by Jacobi polynomials

0

( ) ( )m

m mi i

i

b x b B x

0

( ) ( )n

ni i

i

f x c B x

(1)

(2)

(3)

?

Degree reduction by Jacobi polynomials

0

( ) ( )m

m mi i

i

b x b B x

0

( ) ( )n

ni i

i

f x c B x

(1)

(2)

(4)

0

( ) ( )m

m mi i

i

b x b B x

(3)

Theorem for degree reduction

Error estimation

Example

Example

Summary:Advantages

Unified matrix representations Explicit approximating error Optimal approximation under uncons

trained degree reduction Include three previous methods

Lp-norm, p=1, 2, ∞

Summary:Disadvantages

Constrained degree reduction Unsolved

Challenge to unified representation?

Outline

Overview

Recent developments

Our work

Optimal multi-degree reduction of Bézier curves with G 1-continuity

Lizheng Lu and Guozhao Wang

To be published in JZUS

Motivation:parametric representations

4 3 219 1( ) 18 15 6

2 2f x x x x x 23 1

( )4 4

x t t t

Optimal approximation not unique

Motivation:Geometric Hermite Interpolation

Theorem. [Boor et al., 1987] If the curvature at one endpoint is not vanished, a planar curve can be interpolated by cubic splin

e with G 2-continuity and that the approximation order is 6.

BHS method. More methods about GHI. [Degen, 2005]

Main contributions

Multi-degree reduction G 1: position and tangent direction Minimize Euclidean distance

between control points Optimal approximation

An example

Ahn et al., 2004Ours

Problem

Problem III: Given a curve of degree n in ,to find a curve of degree m, such that,

I) G k-continuous:

II) ( ), ( ) mind t t P Q

( )tP( )tQ

sR

0,1 0,1

d ( ) d ( ( )), 0, ,

d d

i i

t ti i

t ti k

t t

P Q

Special cases

G 0-continuity Endpoint interpolation

G 1-continuity Position and tangent direction

G 2-continuity G 1 + curvature

Main challenges

Consider all the components together

Control points free moving How to be optimal?

Error estimating Numerical problems

Convergence, uniqueness, stability, etc.

Algorithm overview

G 1 condition Discrete coefficient norm

Through degree elevation Solution and improvement

Numerical methods

G 1-continuous

0

( ) ( )n

ni i

i

t B t

P p0

( ) ( )m

mi i

i

t B t

Q q

Given Goal

G 1-continuous

0 0 1 0 0 1 0

1 1 1

, ( ),

, ( ).m n m n n n

n

mn

m

q p q p p p

q p q p p p

G 1 continuous at both endpoints

0

( ) ( )n

ni i

i

t B t

P p0

( ) ( )m

mi i

i

t B t

Q q

Given Goal

G 1 condition

δ0=1

δ0>0

δ0<0

Influence of v

Singular case

Singular case

Singular case

Singular case

Solution condition

δ0=1

δ0>0

δ0<0

0

0,1v

v

Discrete coefficient norm

,

( )

ˆm m

n n m m n m

t

Q B Q

B T Q B Q

Given Goal

( ) n nt P B P

T

0 1, , ,n nP p p p T

0 1ˆ ˆ ˆ ˆ, , ,n nQ q q q

Discrete coefficient norm

,

( )

ˆm m

n n m m n m

t

Q B Q

B T Q B Q

Given Goal

( ) n nt P B P

2

,0

ˆ( ), ( )n

n n m m i ii

d t t

P Q P T Q p q

Discrete coefficient norm: (l2-norm)

T

0 1, , ,n nP p p p T

0 1ˆ ˆ ˆ ˆ, , ,n nQ q q q

Algorithm:Step 1

0 0 1 1 0 0 0 1 0

1 1 0 1 1

, ( ),

, ( ).m n m m n n n

n

mn

m

q p q q = p p p

q p q q p p p

(a)

Algorithm:Step 1

0 0 1 1 0 0 0 1 0

1 1 0 1 1

, ( ),

, ( ).m n m m n n n

n

mn

m

q p q q = p p p

q p q q p p p

2

0 0 1 1 1 12

( ) ( ) ( ) ( ) ( ) ( )m

m m m m mi m m m m

iit B t B t B t B t B t

Q q qq q q

(a)

(b)

Algorithm:Step 1

0 0 1 1 0 0 0 1 0

1 1 0 1 1

, ( ),

, ( ).m n m m n n n

n

mn

m

q p q q = p p p

q p q q p p p

2

0 0 1 1 1 12

( ) ( ) ( ) ( ) ( ) ( )m

m m m m mi m m m m

iit B t B t B t B t B t

Q q qq q q

0 1 2, , 2( , ),ii i m qq

(a)

(b)

(c)?

Algorithm:Step 1.c

2

0 0 1 1 1 12

( ) ( ) ( ) ( ) ( ) ( )m

m m m m mi m m m m

iit B t B t B t B t B t

Q q qq q q

22,( ), ( ) n n m mE d t t P Q P T QError:

Algorithm:Step 1.c

2

0 0 1 1 1 12

( ) ( ) ( ) ( ) ( ) ( )m

m m m m mi m m m m

iit B t B t B t B t B t

Q q qq q q

22,( ), ( ) n n m mE d t t P Q P T Q

,n m mT Q,c cn m mT Q

,f fn m mT Q

T0 1 1( , , , )c

m m mQ q q q q

Error:

where

Algorithm:Step 1.c

2

0 0 1 1 1 12

( ) ( ) ( ) ( ) ( ) ( )m

m m m m mi m m m m

iit B t B t B t B t B t

Q q qq q q

22,( ), ( ) n n m mE d t t P Q P T Q

,n m mT Q,c cn m mT Q

,f fn m mT Q

T0 1 1( , , , )c

m m mQ q q q q

2

, ,c c f f

n n m m n m mE P T Q T Q

Error:

where

Algorithm:Step 1.c

2

, ,c c f

n n m m m mnfE P Q QT TError:

T T

, , , ,f c c f f fn m n n m m n m n m m 0 T P T Q T T Q

We minimize E and write the results in vector-matrix form, which is,

Algorithm:Step 1.c

2

, ,c c f

n n m m m mnfE P Q QT TError:

T T

, , , ,f c c f f fn m n n m m n m n m m 0 T P T Q T T Q

We minimize E and write the results in vector-matrix form, which is,

1T T

0 1 , , , ,,f f f f f c cm m n m n m n m n n m m

Q Q T T T P T Q

Finally, we obtain

Algorithm:Step 1 (summary)

1T T

0 1 , , , ,,f f f f f c cm m n m n m n m n n m m

Q Q T T T P T Q

2

, ,c c f f

n n m m n m mE P T Q T Q

T0 1 0 1 1, ( , , , )c c

m m m m Q Q q q q q (1)

(2)

Algorithm:Step 1 (summary)

Convex Quadratic Implicit

1T T

0 1 , , , ,,f f f f f c cm m n m n m n m n n m m

Q Q T T T P T Q

2

, ,c c f f

n n m m n m mE P T Q T Q

0 1,E E

T0 1 0 1 1, ( , , , )c c

m m m m Q Q q q q q (1)

(2)

Algorithm:Step 2

0 1,E E

The solution of E can be achieved by many numerical methods. e.g. [Press, 1988]

0 1,

Algorithm:Step 2

0 1,E E

The solution of E can be achieved by many numerical methods. e.g. [Press, 1988]

0 1,

0 1

0

0 1

1

,0,

,0.

E

E

We solve it as follows.

Use solve of MATLAB

Efficient No initial values

Algorithm overview

Singular case

0, 0,1v v Condition:

Possible reasons:1). Edge length irregular; → Subdivision2). End edge length too small → Improved

Improved algorithm

2 2 2

, , 0 0 1 11 1c c f fn n m m n m m nE G G P T Q T Q p p

Modified error:

Improved algorithm

2 2 2

, , 0 0 1 11 1c c f fn n m m n m m nE G G P T Q T Q p p

Modified error:

is the standard Gaussian function.

2 2( ) exp 2G x x

Algorithm 2 overview

Algorithm 1

Algorithm 2

Ahn et al., 2004Ours

Degree from 5 to 4 Degree from 5 to 3

Subdivision

Summary

Multi-degree reduction G 1: position and tangent direction Optimal approximation With subdivision Better approximation

Summary:Questions

Lack theory support Error obtained afterward Numerical solution How to use Approximation Theory?

G 2-continuitious degree reduction

G 2 = G 1 + curvature

0 10 3

0

1n

n

p p

p

G k -continuous[Degen, 2005]

G k -continuous[Degen, 2005]

G 2 -continuous

G 1

G 2

G 2 condition

Example for G 2 -continuous

A conjecture

If we consider the problem of degree reduction

with G k -continuity (k >1), the approximating curve

will be much closer to the original curve than that

obtained by consider it with C k-continuity.

A conjecture

If we consider the problem of degree reduction

with G k -continuity (k >1), the approximating curve

will be much closer to the original curve than that

obtained by consider it with G k-continuity.

■ Very difficult■ Future work

Thank You!