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Degree reduction of Bézier curves. Lizheng Lu lulz_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,. An example. - PowerPoint PPT Presentation
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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!
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