Slide No. 1

Shape Interrogation IIShape Interrogation IITakis SakkalisTakis Sakkalis†‡†‡

Nicholas M.Nicholas M. PatrikalakisPatrikalakis††

††Massachusetts Institute of TechnologyMassachusetts Institute of Technology‡‡Agricultural Univ. of AthensAgricultural Univ. of Athens

International Summer School onComputational Methods for Shape Modeling and Analysis

Genova, 14-18 June 2004, Area della Ricerca, CNR, Genova

Slide No. 2

IntroductionIntroduction

• Polynomials are used in various branches of computational science.

• They can be found in mathematics, computer science, engineering and many other fields.

• There are two basic reasons for that: – Most functions can be approximated by polynomial functions, and – They are rather easy to use in a computer code.

• Thus, they serve as good substitutes for functions that are difficult to deal with.

Slide No. 3

• In this talk we will discuss some of their applications in Computer Aided Geometric Design and Geometric Modeling.

• In particular, we will discuss:

– Polynomial systems and their solutions – Elements of elimination theory – Polynomial maps – Some Problems of this Area.

Slide No. 4

A Strange ExampleA Strange Example

• As an indication of the difference in moving from one dimension to the next, even for simple functions–like polynomials–let us consider the following:

• Example 1. Every polynomial function y = p(x) with p(x) > 0, ∀x ∈ R has at least one (real) critical point.

• Example 2. The polynomial

• has the property that, for every (x,y) ∈ R2, p(x,y) > 0, • but the function p(x,y) does not have any (real) critical point.

2 2 2( ) ( 1)p x y x y x x, = − − +

lim ( )x

p x| |→∞

= ∞.

( )lim ( ) Does not exist

x yp x y

| , |→∞

, .

Slide No. 5

Slide No. 6

Slide No. 7

Polynomial Systems

• Polynomials are popular in curve and surface representation.

• Thus, many critical problems in CAGD, such as surface interrogation, are reduced to finding the zero set of a system of polynomial equations

where and each is a polynomial of independent variables .

1( )= , ,L nf f fif

1( )mx x x= , ,Lm

( ) 0=f x

Slide No. 8

Polynomial Systems

• Several root-finding methods for polynomial systems have been used in practice.

• These can be categorized as: – Algebraic and hybrid methods, – Homotopy methods, and – Subdivision methods.

• Among those types, the subdivision methods have been widely used in practice.

• The Interval Projected Polyhedral (IPP) algorithm is one example, and it has successfully been applied to various problems.

Slide No. 9

MotivationMotivation

• Difficulties in handling roots with high multiplicity- Performance deterioration- Lack of robustness in numerical computation- Round-off errors during floating point arithmetic

• Limited research on root multiplicity of a system of equations- Heuristic approaches are needed for practical

purposes.

Slide No. 10

ObjectivesObjectives

• Develop practical algorithms to isolate and compute roots and their multiplicities.

• Improve the Interval Projected Polyhedron (IPP) algorithms.

Slide No. 11

Multiplicity of RootsMultiplicity of Roots

• Univariate Case- A root a of f(x)=0 has multiplicity k if

• Bivariate Case- Define

- Suppose that z0 is the only common point of Vf and Vg lying above x0. Consider h(x)=Resy(f,g), the resultant of f,g with respect to y. Then the multiplicity of z0=(x0,y0) as a root of the system is the multiplicity of x0 as a zero of h(x).

0)(,0)()()( )()1(' ≠==== − afandafafaf kkL

}0),(|),{(

}0),(|),{(

=∈=

=∈=

yxgyxV

yxfyxV

g

f

C

C

Slide No. 12

Degree of the Gauss MapDegree of the Gauss Map• Let p(x,y), q(x,y) be polynomials with rational coefficients

without common factors, of degrees n1 and n2, and let F=(p, q).

• Let A be a rectangle in the plane defined by so that no zero of F lies its

boundary , and does not vanish at its vertices.• Gauss map where S1 is the unit

circle.• G is continuous ( ).• and S1 carry the counterclockwise orientation.

• Degree d of G : an integer indicating how many times is wrapped around S1 by G.

,, 4321 ayaaxa ≤≤≤≤

4,3,2,1,,, 4321 =∈<< iaaaaa i QA∂ qp ⋅

,/,: 1 FFGSAG =→∂

AonF ∂≠ 0

A∂

A∂

Slide No. 13

Illustration of the Gauss MapIllustration of the Gauss Map

a1x

S1

y

a3

a4

a2

z=(x0,y0)

(0,0)

p(x,y)

q(x,y)

(0,0)

G=F/||F||

F1

X

Y

F

F1 / | F1|

Slide No. 14

• Preliminaries– R(x) : a rational function q(x)/p(x), where p, q are polynomials.– [a,b] : a closed interval, a < b. R does not become infinite at the

end points.

• Definition of the Cauchy indexBy the Cauchy index, of R over [a,b], we mean

where denotes the number of points in (a,b) at which

R(x) jumps from , respectively, as x is

moving from a to b. Notice that from the definition.

RI ba

−+

+− −= NNRI b

a

)( −+

+− NN

)( ∞−+∞∞+∞− toto

RIRI ab

ba −=

The The Cauchy Cauchy IndexIndex

Slide No. 15

• Preliminaries– A : a rectangle defined by [ 1, 2] x [ 3, 4] which encloses a

zero.– F = (p,q) does not vanish on the boundary of A, – is not zero at each vertex of A.– Let

Then, we set (for counterclockwise traversal of )

• Proposition*

IAF is an even integer and the multiplicity

.),(),(

,),(),(

,),(),(

,),(),(

4

44

3

33

2

22

1

11 axp

axqR

axpaxq

Ryapyaq

Ryapyaq

R ====

.43211

2

2

1

4

3

3

4RIRIRIRIFI a

aaa

aa

aaA +++=

.21

FId A−=

qp ⋅.A∂

A∂

•T. Sakkalis, “The Euclidean Algorithm and the Degree of the Gauss Map”,

SIAM J. Computing. Vol. 19, No. 3, 1990.

a a a a

The The Cauchy Cauchy Index (continued)Index (continued)

Slide No. 16

• p(x) = (x-1/2)5 = 0• A root of p(x), [ ] = [0.49,0.51].• P(z); (z = x+iy)

• Create

• Calculate the Cauchy index– Roots of f(x, 3) = 0– Calculation of

• Roots No. 2, 3, and 4 are selected since they lie within the interval [ ].

01.0,01.0,51.0,49.0 , 4321 =−===×= aaaa01] [-0.01,0.1] [0.49,0.5A

[0.530776834861365, 0.530776835861365]5[0.507265424645288, 0.507265426808589]4[0.499999997363532, 0.500000001889623]3[0.49273457408967, 0.492734576204823]2[0.46922316412099, 0.46922316512099]1

Roots of f(x,a3) = 0 in [0,1] (from the IPP)No.

),(),()21

()( 5 yxigyxfiyxzp +=−+=

332

1−=RI a

a

a

a

a

Illustrative Example for Multiplicity Illustrative Example for Multiplicity Computation Using the Computation Using the Cauchy Cauchy IndexIndex

Slide No. 17

• Similarly,

• Calculate

• The multiplicity m of the root is

2,3,2 1423

4

1

2

4

3==−= RIRIRI a

aaa

aa

1043211

2

2

1

4

3

3

4−=+++= RIRIRIRIFI a

aaa

aa

aaA

521

=−= FId A

..

assumedisAofnorientatiockwiseCountercloRIRI

Noteab

ba

∂−−=−

Illustrative Example (Continued)Illustrative Example (Continued)

Slide No. 18

• Use the map F directly.

Direct Computation MethodDirect Computation Method

Slide No. 19

-5

-4

-3

-2

-1

0

1

2

3

4

5

-6 -4 -2 0 2 4 6

f(x,y)

g(x

,y) IR3 (f(x,a3)=0)

IR2 (f(a2,y)=0)IR4 (f(x,a4)=0)IR1 (f(a1,y)=0)

)).,(),,((),(,: yxgyxfyxFF =→ 22 RR ,/,: 1 FFGSAG =→∂

Direct Computation MethodDirect Computation Method

Slide No. 20

• Univariate polynomial in complex variable z. (Substitute x with a complex variable z = x+iy)

• Input :– initial domain : – a complex polynomial : p(z)– tolerance, number of sample points

• Output– real and complex roots, multiplicities

• Algorithm– Quadtree decomposition – Direct degree computation method : complex interval

arithmetic.

],[],[ 2211 babaS ×=

0)( 01

11

1 =++++= −− azazazazp n

nn

n L

s1s2

s3 s4

b2

a1 b1

a2

Bisection Algorithm for Solving Bisection Algorithm for Solving Univariate Univariate Polynomial EquationsPolynomial Equations

Slide No. 21

4323

422

)4)(2()1(

)1()1()(

−−+++

−++=

ttttt

ttttp

ExamplesExamples

• Wilkinson polynomial • Complicated Polynomial (degree 22)

Slide No. 22

Solving aSolving a BivariateBivariate Polynomial Polynomial SystemSystem

• Change of Coordinates- CR : f and g are regular in y.- CU : whenever two points (x0,y0) and (x1,y1) satisfy f=g=0,

then y0=y1.

• Solving a Bivariate Polynomial System- Let f,g satisfy CR and CU and let h(x)=Resy(f,g). Then the

roots of the system f=g=0 are in a one to one correspondence with the roots of h(x). Moreover, zi=(xi,yi) is a real root if and only if xi is a real root of h(x).

- Let h(x)=Resy(f,g) and l(y)=Resx(f,g) and aij=[ti,ti+1]x[sj,sj+1]where in each subinterval [ti,ti+1] or [sj,sj+1] there exist precisely one root of h(x) and l(y), respectively. If aij

encloses a real root of f=g=0, then the following must be true

]),[],,([]),[],,([0 1111 ++++ ×∈ jjiijjii ssttgssttf

Slide No. 23

Solving aSolving a BivariateBivariate Polynomial Polynomial System : ExampleSystem : Example

Slide No. 24

Elimination TheoryElimination Theory

I. Resultants • Sylvester Resultant • Macaulay Resultant • Sparse Resultant • D-Resultant

II. Groebner Bases

III. Symbolic System Solving

Slide No. 25

Elements of Resultant TheoryElements of Resultant Theory

• Let:

• non zero polynomials, with complex coefficients.

• The resultant of a,b wrt t (or the t -resultant), is

• Observe that .

1 0

1 0

( )

( )

= + + +

= + + +

L

L

nn

mm

a t a t a t a

b t b t b t b

Res ( )t a b R, =01

1 0

0

0

0

n n

n

n

m

m

a a aa a a

R a ab b

b b

−

=

LLO O

L LL L LO O

L L L

Res ( ) Ct a b, ∈

Slide No. 26

Properties of the ResultantProperties of the Resultant

• Let us see some well known properties of the resultant: • Property 1. There exist polynomials of degrees

respectively, so that (1)

• Property 2. and have a common factor of positive degree.

• Property 3.• Let ,

• with an or bm∈C∗, and consider p(x)=Resy(a,b). If x0 is a root of p(y), then there exists y0∈C with the property

( ) ( ) C[ ]A t B t t, ∈

( ) ( ) ( ) ( ) ( )Resta t A t b t B t a b+ = , .

n m m n′ ′< , <

Res ( ) 0 ( )t a b a t, = ⇐⇒ ( )b t

11 0( ) ( ) ( )n n

n na x y a y a x y a x−−, = + + +L

0( ) ( ) [ ][ ]−

−=, = ∈∑m m i

m iib x y b x y k y x

0 0 0 0( ) ( ) 0a x y b x y, = , =

Slide No. 27

Cramer’s RuleCramer’s Rule

• Let f(x,y), g(x,y) ∈ C [x,y] two nonconstant polynomials, and let be indeterminates.

• Consider

• with F(0,0) = (0,0).

2 2C C ( )( ) Res ( )

( ) Res ( )

: → , = ,, , = − , − ,

, , = − , −

y

x

F F f gA x u v f u g v

B y u v f u g v

Slide No. 28

Cramer’s RuleCramer’s Rule

• Theorem[Cramer’s Rule] F has a polynomial inverse if and only if:

• and

• Moreover, if

• Then G is the inverse of F.• In addition,

0( ) ( , ),, , = +A x u v ax A u v

0( ) ( , ), 0, , = + ≠B y u v by B u v with ab

0 0( , ) ( , )( , ) : , , = − −

A x y B x yG x y

a b

1deg deg −=F F