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Topics in Computer GraphicsChap 7: Polynomial Curve Constructions

fall, 2011

University of Seoul

School of Computer Science

Minho Kim


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Table of contents

Aitkens Algorithm

Lagrange PolynomialsThe Vandermonde Approach

Limits of Lagrange Interpolation

Cubic Hermite Interpolation

Quintic Hermite Interpolation

Point-Normal Interpolation

Least Squares Approximation

Smoothing Equations

Designing with Bezier Curves

The Newton Form and Forward Differencing


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Motivation

Given n points in 2D/3D, Interpolation: find a (polynomial) curve that passes through

the data exactly. Aitkens algorithm (7.1) Lagrange polynomials (7.2) The Vandermonde approach (7.3) The Newton form and forward differencing (7.11)

Approximation: find a (polynomial) curve that goes throughthe data closely in a reasonable way.

Least squares approximation (7.8) Smoothing equations (7.9)

Normal/tangent data in addition to position data. Cubic/quintic Hermite interpolation (7.5&7.6) Point-normal interpolation (7.7)


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Outline

Aitkens Algorithm







Smoothing Equations




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Aitkens Algorithm

Problem: Given pointst pi u

ni

0

with corresponding

parameter values t ti un

i0

, find a curve p p tq such that

p p ti q pi, i 0, . . . , n .


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Aitkens Algorithm (contd)

1. Base case: Let

p1

ip

tq

ti1 t

ti1 tip

i

t ti

ti1 tip

i 1, i

0, . . . , n

1.

p1i p t q is a line segment for ti t ti 1. p1i p t q interpolates pi and pi 1: p

1

i p ti q pi and

p

1

ip

ti 1 q

pi 1

.2. Let pn 1

0p tq is a polynomial of degree n 1 that

interpolates p0, . . . ,pn 1 and pn

1

1p

tq

is one that

interpolates p1, . . . ,pn.

3. Then

pn0 p tq tn t

tn t0pn 10

p tq t

t0

tn t0pn 11

p tq

is a polynomial curve of degree n and interpolates p0, . . . ,pn.


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Aitkens Algorithm (contd)

With p0i p tq : pi, the interpolating curve pn0 p

tq

is defined

recursively as

pri p tq ti

r t

tir ti

pr 1i p tq t

ti

tir ti

pr 1i1

p tq ,

5

r 1, . . . , n

i 0, . . . , n r.

Properties

Affine invariance

Linear precision

No convex hull property

Is it possible for a smooth interpolating curve has a convex

hull property?

No variation diminishing property

Does not answer if such polynomial curve is unique.

Not a closed form


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Outline

Aitkens Algorithm







Smoothing Equations




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Lagrange PolynomialsThe interpolating curve is given by

p p tq

n

i0

piLn

i

pt

q

where Lni p tq are Lagrange polynomialsdefined as

Ln

i

pt

q :

nj

0

j$i

pt

tj q

nj 0j

$i

p ti tj q

Lagrange polynomials are cardinal:

Lni p tj q i,j

where i,j is the Kronecker delta.

t Lni p tq uni

0

form a basis of all polynomials of degree n Interpolating polynomial (of degree n) is unique

n

i 0

Lni

p tq 1p tq


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Outline

Aitkens Algorithm







Smoothing Equations




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The Vandermonde Approach

What is the monomial form

pn

p

tq

n

j0ajt

j

of the interpolating polynomial?


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The Vandermonde Approach (contd)

1.

pnp

t0 q p0 a0 a1t0 antn0

pn p t1 q p1 a0 a1t1 antn1

...

pnp

tn q pn a0 a1tn antnn

2.

"

"

"

!

p0p1

...

pn

(

0

0

0

)

"

"

"

!

1 t0 tn0

1 t1 tn1

... ... . . . ...

1 tn tnn

(

0

0

0

)

"

"

"

!

a0a1

...

an

(

0

0

0

)

p

Ta

3. Ift

tj unj

0

are distinct, detT $ 0. (Why?)

detT is the Vandermonde of the interpolation problem

4. Solution: a T

1p


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Generalized Vandermonde

For some basis t Fnj p tq unj

0

, (e.g. Bernstein basis) the

interpolating polynomial is expressed as

pnp

tq

n

j0

cjFnj p tq

and the coefficientstcj u

nj

0

can be found as

"

"

"

!

c1c2

...

cn

(

0

0

0

)

"

"

"

!

Fn0 p

t0 q Fn1 p

t0 q Fnn p t0 q

Fn0 p t1 q Fn1 p t1 q F

nn p t1 q

... ... . . . ...

Fn0 p

tn q Fn1 p

tn q Fnn p tn q

(

0

0

0

)

1

"

"

"

!

p0p1

...

pn

(

0

0

0

)


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Outline

Aitkens Algorithm

Lagrange Polynomials







Smoothing Equations




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Small change of input results in large change in output

The process is ill-conditioned. Polynomial interpolation is not shape preserving.

Runge phenomonen Expensive to solve

for construction (solving dense linear system) for evaluation (for high n)

Piecewise schemes are better! (e.g. spline interpolation)

O li


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Outline

Aitkens Algorithm








Smoothing Equations



C bi H i I l i


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Find the cubic polynomial that interpolates the following:

pp

0q

p0

E

3

9p p 0q m0 R3

9pp 1q m1 R

3

pp 1q p1 E

3,

Easier to work with Bezier form

p p tq

3

j 0

bjB3j p tq

1. b0 p0 and b3 p1

2. 9pp 0q 3b0 m0 and 9p p 1q 3b2 m1

b1 p0 1

3m0 and b2 b1

1

3m1

C bi H it I t l ti C di l F


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Cubic Hermite Interpolation: Cardinal Form

pp tq p0B30 p tq

p0 13m0

B31 p tq

p1 13m1

B32 p tq p1B33 p tq

pp tq p0H30 p tq m0H

31 p tq m1H

32 p tq p1H

33 p tq

H30 p tq B30 p tq B

31 p tq

H31 p tq 1

3B31 p tq

H32 p tq 1

3B32 p tq

H33 p tq B32 p tq B

33 p tq

C bi H it I t l ti C di l F


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Cubic Hermite Interpolation: Cardinal Form

H30 p 0q 1ddt

H30 p 0q 0ddt

H30 p 1q 0 H30 p 1q 0

H31 p 0q 0ddt

H31 p 0q 1ddt

H31 p 1q 0 H31 p 1q 0

H32 p 0q 0

ddt

H32 p 0q 0

ddt

H32 p 1q 1 H

32 p 1q 0

H33

p

0q

0ddtH

33

p

0q

0ddtH

33

p

1q

0 H33

p

1q

1

H30 p

tq

H33 p

tq 1p tq (Why?)

Not invariant under affine domain transformations

Not symmetric

O tline


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Outline

Aitkens Algorithm








Smoothing Equations



Outline


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Outline

Aitkens Algorithm








Smoothing Equations



Outline


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Outline

Aitkens Algorithm








Smoothing Equations





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P 1 number of data points can be interpolated by apolynomial of degree n if d P.

What if n

P?

approximation required

How to measure the closeness of the approximating curve

and the data points? least squares approximation: To find a polynomial curve

xp

tq

that minimizes

fp

c0, . . . , cnq

P

j 0x

p

tjq

pj2

.

Least Squares Approximation (contd)


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Least Squares Approximation (cont d)

1. The approximating curve is expressed as

xp

tq

c0Cn0 p tq cnC

nn p tq

for some basist

Cnj unj

0

.

2. Assuming xp tq interpolates all the points t pj uPj 0, we get a

linear system

"

!

Cn0 p t0 q Cnn p t0 q...

. . ....

Cn0 p tP q Cnn p tP q

(

0

)

"

!

c0...

cn

(

0

)

"

!

p0...

pP

(

0

)

MC P

where

MR

pP

1

q pn

1

q ,C R p n 1q 3,P R p P 1q 3, P n

Linear systems for each component

underdetermined (no solution in general)

Least Squares Approximation (contd)


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Least Squares Approximation (cont d)

3. Can be converted to a solvable linear system by

multiplying MT

:MTMC MTP

MTM is always invertible. (Why?) The solution minimizes

fp c0, . . . , cn q

P

j

0

xp tj q pj

2.

(Why?)

The solution satisfies

ff

fcdk

0, k

0, . . . , n , d

1, 2, 3.

Outline


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Outline

Aitkens Algorithm








Smoothing Equations



Smoothing Equations


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Smoothing Equations

Minimizing the least square error does not guarantee agood curve. (Fig 7.8)

We want a curve that does not wiggle much. (How?)

If the control polygon behaves nicely, then so does the

curve.

How to minimizes wiggling (of the polygon)?

minimize

the 2nd differences 2bi

b0 2b1 b2 0...

bn 2 2bn 1 bn 0

SB

0

Smoothing Equations (contd)


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Smoothing Equations (cont d)

Combined with the least squares problem,

M

S

&

B

P

0

&

Balancing the error and shape

p 1 q M

S

&

B

p 1 q P

0

&

0 1

Larger for noisier data

Outline


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Outline

Aitkens Algorithm








Smoothing Equations





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What if we want to move a point on the curve to change

the shape?

Given a Bezier curve

xp tq

n

i0

biBni p tq ,

if we change a point xp

tq

to y, how should we changetbi u

to t bi u such that the curve passes through y at t t?

y

n

i0

biBn

i

pt

q

Assume b0 and bn are unchanged. Underdetermined: 1 equation and n 1 unknowns

Designing with Bezier Curves (contd)


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Designing with Bezier Curves (cont d)

1.

Bn1 p tq Bnn 1 p tq$

"

!

b1

...

bn 1

(

0

)

y b0Bn0 p tq bnBnn p tq : z

AB

z

2. Assume all the points b1, . . . ,bn1 move along (unknown)

c direction:

B B ATc AB AB AATc z AB AATc

3. Since AAT is a scalar, with a : AAT

c

1

apy

x

q B

B

1

aAT

py

x

q.

Outline


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Aitkens Algorithm








Smoothing Equations



The Newton Form of Interpolating Polynomial


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p g y For equally spaced parameter intervals: ti 1 ti h

For cubic case,

pp

tq

p0 1h

pt

t0 q p0 1

2!h2p

t

t0 q p t t1 q 2p0

1

3!h3p t t0 q p t t1 q p t t2 q

3p0

where

ipj :

i1pj

1

i1pj and

0pj pj .

With g : 1{ h,

p0

p1 gp0

p2 gp1 g22p0

p3 gp2 g22p1 g

33p0

Fast Evaluation for Plotting


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g Efficient way to evaluate p

pt2 hq ,p p t2 2hq , . . .

p0

p1 gp0

p2 gp1 g

22

p0p3 gp2 g

22p1 g33p0

p4 gp3 g22p2 g

33p1

p5 gp4 g22p3 g

33p2

.

.

....

.

.

....

t g3pj u are all equal. (Why?) From right to left

g22p2 g33p1 g

22p1 gp3 g

22p2 gp2

p4 gp3 p3

In general, with qij : giipj ,

qij qi

1

j1

qij1, i 2, 1, 0

and pj

q0

j .

Fast for Plotting (contd)


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g ( )

No multiplication involved! (after start-up phase)

extremely fast

Given a polynomial in cubic Bezier form,

1. Compute initial four points.2. Evaluate sequence of points and plot using line segments.

Roundoff error is accumulated The curve may deviates

more and more...

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