Teresa Cortadellas, Carlos D’Andrea, Eulàlia Montoro - UB · Solving the Rational Interpolation Problem Teresa Cortadellas, Carlos D’Andrea, Eulàlia Montoro FoCM 2017 Barcelona

Solving the Rational Interpolation Problem

Teresa Cortadellas, Carlos D’Andrea, Eulàlia Montoro

FoCM 2017 Barcelona



Lagrange Interpolation

Given (x1, y1), . . . , (xN , yN) ∈ K2

compute P ∈ K[x ] of minimal degreesuch that P(xi) = yi , i = 1, . . . ,N




Given (x1, y1), . . . , (xN , yN) ∈ K2





Given (x1, y1), . . . , (xN , yN) ∈ K2




“Easy” Solution

1 x1 x2

1 . . . xN−11

1 x2 x22 . . . xN−1

2... ... ... . . ....

1 xN x2N . . . xN−1

N

a0

...aN−1

=

y1...yN

P =N−1∑j=0

ajxj =

N∑j=1

yj

∏i 6=j

x − xixj − xi



“Easy” Solution

1 x1 x2

1 . . . xN−11

1 x2 x22 . . . xN−1

2... ... ... . . ....

1 xN x2N . . . xN−1

N

a0

...aN−1

=

y1...yN

P =N−1∑j=0

ajxj =

N∑j=1

yj

∏i 6=j

x − xixj − xi



“Easy” Solution

1 x1 x2

1 . . . xN−11

1 x2 x22 . . . xN−1

2... ... ... . . ....

1 xN x2N . . . xN−1

N

a0

...aN−1

=

y1...yN

P =N−1∑j=0

ajxj

=N∑j=1

yj

∏i 6=j

x − xixj − xi



“Easy” Solution

1 x1 x2

1 . . . xN−11

1 x2 x22 . . . xN−1

2... ... ... . . ....

1 xN x2N . . . xN−1

N

a0

...aN−1

=

y1...yN

P =N−1∑j=0

ajxj =

N∑j=1

yj

∏i 6=j

x − xixj − xi



Hermite Interpolation

Given x1, . . . , xL,y10, . . . , y1N1−1, . . . , yL0, . . . , yLNL−1

compute P ∈ K[x ] of minimal degreesuch that P (j)(xi) = yij













How to solve it?

1 x1 x2

1 . . . xN−11

0 1 2x1 . . . (N − 1)xN−21

......

... . . ....

0 0 0 . . . ∗xN−NL−1L

a0

...aN−1

=

y10...

yLNL−1

P =N−1∑j=0

ajxj =

L∑j=0

Pj(x)



How to solve it?

1 x1 x2

1 . . . xN−11

0 1 2x1 . . . (N − 1)xN−21

......

... . . ....

0 0 0 . . . ∗xN−NL−1L

a0

...aN−1

=

y10...

yLNL−1

P =N−1∑j=0

ajxj =

L∑j=0

Pj(x)



How to solve it?

1 x1 x2

1 . . . xN−11

0 1 2x1 . . . (N − 1)xN−21

......

... . . ....

0 0 0 . . . ∗xN−NL−1L

a0

...aN−1

=

y10...

yLNL−1

P =N−1∑j=0

ajxj

=L∑

j=0

Pj(x)



How to solve it?

1 x1 x2

1 . . . xN−11

0 1 2x1 . . . (N − 1)xN−21

......

... . . ....

0 0 0 . . . ∗xN−NL−1L

a0

...aN−1

=

y10...

yLNL−1

P =N−1∑j=0

ajxj =

L∑j=0

Pj(x)



Computing

Linear Algebra O(N3)

Chinese Remainder Theorem(Hermite)Divided Differences

O(N log2(N))



Computing


Chinese Remainder Theorem(Hermite)

Divided Differences

O(N log2(N))



Computing



O(N log2(N))



Computing



O(N log2(N))



Rational Interpolation


compute R ∈ K(x) such that

R (j)(xi) = yij

of minimal degree






R (j)(xi) = yij

of minimal degree






R (j)(xi) = yij

of minimal degree






R (j)(xi) = yij

of minimal degree



Sub-Problem (Cauchy/osculatory)

Given 0 ≤ n ≤ N − 1 computeA ∈ K[x ]≤n, B ∈ K[x ]≤N−n−1 suchthat R := A

B solves the interpolationproblem




Given 0 ≤ n ≤ N − 1

computeA ∈ K[x ]≤n, B ∈ K[x ]≤N−n−1 suchthat R := A





Given 0 ≤ n ≤ N − 1 computeA ∈ K[x ]≤n, B ∈ K[x ]≤N−n−1

suchthat R := A





Given 0 ≤ n ≤ N − 1 computeA ∈ K[x ]≤n, B ∈ K[x ]≤N−n−1 suchthat R := A




Problem with the Sub-Problem

It may be unsolvable!

N = 3, n = 1, x1 = 1, x2 = 2y1,0 = 1, y1,1 = 0, y2,0 = 0

��a0+a1x

b0+b1x





N = 3, n = 1, x1 = 1, x2 = 2y1,0 = 1, y1,1 = 0, y2,0 = 0

��a0+a1x

b0+b1x





N = 3, n = 1, x1 = 1, x2 = 2y1,0 = 1, y1,1 = 0, y2,0 = 0

��a0+a1x

b0+b1x





N = 3, n = 1, x1 = 1, x2 = 2y1,0 = 1, y1,1 = 0, y2,0 = 0

��a0+a1x

b0+b1x



The “Problem” is the denominator

One needs B(xi) 6= 0∀i = 0, . . . , LFor “generic” input, there is a unique

solutionDescription of the set of “bad points”

given in Cortadellas-D-MontoroarXiv:1704.08563




One needs B(xi) 6= 0∀i = 0, . . . , L

For “generic” input, there is a uniquesolution

Description of the set of “bad points”given in Cortadellas-D-Montoro

arXiv:1704.08563





solution

Description of the set of “bad points”given in Cortadellas-D-Montoro

arXiv:1704.08563





solutionDescription of the set of “bad points”

given in Cortadellas-D-MontoroarXiv:1704.08563



The Weak Sub-Problem

Given the data xi , yij computeA ∈ K[x ]≤n, B ∈ K[x ]≤N−n−1 such

that A(xi) = yiB(xi)

A(j)(xi) =

j∑s=0

(j)syisB(j−s)(xi)




Given the data xi , yij

computeA ∈ K[x ]≤n, B ∈ K[x ]≤N−n−1 such


A(j)(xi) =

j∑s=0

(j)syisB(j−s)(xi)




Given the data xi , yij computeA ∈ K[x ]≤n, B ∈ K[x ]≤N−n−1

suchthat A(xi) = yiB(xi)

A(j)(xi) =

j∑s=0

(j)syisB(j−s)(xi)






A(j)(xi) =

j∑s=0

(j)syisB(j−s)(xi)






A(j)(xi) =

j∑s=0

(j)syisB(j−s)(xi)



Linear Algebra solves it!

The Weak Problem is alwayssolvableThe solution is “unique”The denominator of “the” solutiondoes not vanish in the xi ’s ⇐⇒the sub-problem can be solved




The Weak Problem is alwayssolvable

The solution is “unique”The denominator of “the” solutiondoes not vanish in the xi ’s ⇐⇒the sub-problem can be solved




The Weak Problem is alwayssolvableThe solution is “unique”

The denominator of “the” solutiondoes not vanish in the xi ’s ⇐⇒the sub-problem can be solved




The Weak Problem is alwayssolvableThe solution is “unique”The denominator of “the” solutiondoes not vanish in the xi ’s ⇐⇒

the sub-problem can be solved




The Weak Problem is alwayssolvableThe solution is “unique”The denominator of “the” solutiondoes not vanish in the xi ’s ⇐⇒the sub-problem can be solved



How to solve it

Linear AlgebraSymmetric functions/SubresultantsEuclidean AlgorithmBarycentric CoordinatesJacobi’s Method. . .



How to solve it

Linear Algebra

Symmetric functions/SubresultantsEuclidean AlgorithmBarycentric CoordinatesJacobi’s Method. . .



How to solve it

Linear AlgebraSymmetric functions/Subresultants

Euclidean AlgorithmBarycentric CoordinatesJacobi’s Method. . .



How to solve it

Linear AlgebraSymmetric functions/SubresultantsEuclidean Algorithm

Barycentric CoordinatesJacobi’s Method. . .



How to solve it

Linear AlgebraSymmetric functions/SubresultantsEuclidean AlgorithmBarycentric Coordinates

Jacobi’s Method. . .



How to solve it

Linear AlgebraSymmetric functions/SubresultantsEuclidean AlgorithmBarycentric CoordinatesJacobi’s Method

. . .



How to solve it

Linear AlgebraSymmetric functions/SubresultantsEuclidean AlgorithmBarycentric CoordinatesJacobi’s Method. . .



Euclidean Algorithm revisited

f :=∏L

i=1(x − xi)Ni

g ∈ K[x ]≤N−1, g(j)(xi) = yij

γ(`)f + β(`)g = α(`), ` = 1, . . .




f :=∏L

i=1(x − xi)Ni

g ∈ K[x ]≤N−1, g(j)(xi) = yij

γ(`)f + β(`)g = α(`), ` = 1, . . .




f :=∏L

i=1(x − xi)Ni

g ∈ K[x ]≤N−1, g(j)(xi) = yij

γ(`)f + β(`)g = α(`), ` = 1, . . .




f :=∏L

i=1(x − xi)Ni

g ∈ K[x ]≤N−1, g(j)(xi) = yij

γ(`)f + β(`)g = α(`), ` = 1, . . .




f :=∏L

i=0(x − xi)Ni

g ∈ K[x ]≤N−1, g(j)(xi) = yij

γ(`)`−1f + β

(`)` g = α

(`)N−1−`, ` = 1, . . .

` = N − n − 1, A = α(`), B = β(`)

work!




f :=∏L

i=0(x − xi)Ni

g ∈ K[x ]≤N−1, g(j)(xi) = yij

γ(`)`−1f + β

(`)` g = α

(`)N−1−`, ` = 1, . . .

` = N − n − 1, A = α(`), B = β(`)

work!



But...

you may have α(`) = β(`) = 0 !

However picking “the closest” nonzero` to N − n − 1 works...



But...





But...


However

picking “the closest” nonzero` to N − n − 1 works...



But...





Complexity?

Computing f , g : O(N log2 N)

One single step of the EEA costs

O(M(N) log(N)) = O(N log2 N)



Complexity?






Complexity?



O(M(N) log(N))

= O(N log2 N)



Complexity?






Can you improve this?

Linear Algebra based methods:O(N3)Barycentric Coordinates:more stable numerically but yetO(N3)Jacobi’s Method:single roots+ mild conditions O(N2)




Linear Algebra based methods:O(N3)

Barycentric Coordinates:more stable numerically but yetO(N3)Jacobi’s Method:single roots+ mild conditions O(N2)




Linear Algebra based methods:O(N3)Barycentric Coordinates:

more stable numerically but yetO(N3)Jacobi’s Method:single roots+ mild conditions O(N2)




Linear Algebra based methods:O(N3)Barycentric Coordinates:more stable numerically but yetO(N3)

Jacobi’s Method:single roots+ mild conditions O(N2)




Linear Algebra based methods:O(N3)Barycentric Coordinates:more stable numerically but yetO(N3)Jacobi’s Method:single roots+ mild conditions O(N2)



Barycentric coordinates

(Schneider - Werner 86)

Any choice of a1, . . . , aN 6= 0 makes

R =

∑Ni=1

aiyi(x−xi )∑N

i=1ai

(x−xi )

a “solution” of the WRIPRIP is solvable ⇐⇒ ai 6= 0






R =

∑Ni=1

aiyi(x−xi )∑N

i=1ai

(x−xi )

a “solution” of the WRIP

RIP is solvable ⇐⇒ ai 6= 0






R =

∑Ni=1

aiyi(x−xi )∑N

i=1ai

(x−xi )

a “solution” of the WRIPRIP is solvable ⇐⇒

ai 6= 0






R =

∑Ni=1

aiyi(x−xi )∑N

i=1ai

(x−xi )

a “solution” of the WRIPRIP is solvable ⇐⇒ ai 6= 0



Jacobi’s method

(Egecioglu-Koc 89)

From a Bézout type identity:C f + B g = A

we pass to xkC + xk Bf g = xkAf If

k + deg(A) ≤ N − 2, Res∞(xkAf ) = 0

Separated equations for thecoefficients of A and B!



Jacobi’s method

(Egecioglu-Koc 89)



k + deg(A) ≤ N − 2, Res∞(xkAf ) = 0




Jacobi’s method

(Egecioglu-Koc 89)


we pass to xkC + xk Bf g = xkAf

Ifk + deg(A) ≤ N − 2, Res∞(x

kAf ) = 0




Jacobi’s method

(Egecioglu-Koc 89)



k + deg(A) ≤ N − 2, Res∞(xkAf ) = 0




Jacobi’s method

(Egecioglu-Koc 89)



k + deg(A) ≤ N − 2, Res∞(xkAf ) = 0




Our contribution

(Cortadellas-D-Montoro)

Extend Jacobi’s method to:the case with multipliciteseliminate the “conditions”improve the complexity



Our contribution


Extend Jacobi’s method to:

the case with multipliciteseliminate the “conditions”improve the complexity



Our contribution


Extend Jacobi’s method to:the case with multiplicites

eliminate the “conditions”improve the complexity



Our contribution


Extend Jacobi’s method to:the case with multipliciteseliminate the “conditions”

improve the complexity



Our contribution


Extend Jacobi’s method to:the case with multipliciteseliminate the “conditions”improve the complexity



Moreover...

The matrix of residues iscomputable, of size n × n, ofHankel typeComputing an element in thekernel of this matrix can be done atcost O(n log2 n)

B can be interpolated a posteriori



Moreover...

The matrix of residues iscomputable, of size n × n, ofHankel type

Computing an element in thekernel of this matrix can be done atcost O(n log2 n)




Moreover...





Moreover...


B can be interpolated a posterioriTeresa Cortadellas, Carlos D’Andrea, Eulàlia Montoro


Cost of the Algorithm

(Cortadellas-D-Montoro)O(n log2(n) + N log(N))

Euclides approach: O(N log2(N))



Cost of the Algorithm

(Cortadellas-D-Montoro)O(n log2(n) + N log(N))

Euclides approach: O(N log2(N))



What about the minimal problem?

Given (x1, y1), . . . , (xN , yN) ∈ K2

compute a rational functionR ∈ K(x) such that R (j)(xi) = yij of

minimal degree



What about the minimal problem?

Given (x1, y1), . . . , (xN , yN) ∈ K2

compute a rational functionR ∈ K(x) such that R (j)(xi) = yij of

minimal degree



Weak Version (single root)

(Ravi 97)

Any pair (A,B) solves the weak RIP

⇐⇒

x − x1 0 . . . 0 1 −y1

0 x − x2 . . . 0 1 −y2...

......

......

0 0 . . . x − xN 1 −yN

∗...∗AB

=

0




(Ravi 97)


⇐⇒

x − x1 0 . . . 0 1 −y1

0 x − x2 . . . 0 1 −y2...

......

......

0 0 . . . x − xN 1 −yN

∗...∗AB

=

0




(Ravi 97)


⇐⇒

x − x1 0 . . . 0 1 −y1

0 x − x2 . . . 0 1 −y2...

......

......

0 0 . . . x − xN 1 −yN

∗...∗AB

=

0



Hilbert’s Syzygy’s Theorem

The kernel of this matrix is free ofrank 2There are two generators ofdegrees µ ≤ N − µµ should be the minimal degree!




The kernel of this matrix is free ofrank 2

There are two generators ofdegrees µ ≤ N − µµ should be the minimal degree!




The kernel of this matrix is free ofrank 2There are two generators ofdegrees µ ≤ N − µ

µ should be the minimal degree!




The kernel of this matrix is free ofrank 2There are two generators ofdegrees µ ≤ N − µµ should be the minimal degree!



Don’t bother computing that!


µ is closest degree of a remainderin the EEA to N

2The “µ-basis” are just twoconsecutive rows of the EEA!



Don’t bother computing that!


µ is closest degree of a remainderin the EEA to N

2The “µ-basis” are just twoconsecutive rows of the EEA!



From the Weak to the Strong...


One uses the stratification of “badpoints” given in

Cortadellas-D-MontoroarXiv:1704.08563



From the Weak to the Strong...

(Cortadellas-D-Montoro)One uses the stratification of “bad

points” given inCortadellas-D-MontoroarXiv:1704.08563



To conclude...

ThanksAgnes, Marta and Thorsten!



To conclude...

ThanksAgnes, Marta and Thorsten!



And Thank you!



Documents

Teresa Cortadellas, Carlos D’Andrea, Eulàlia Montoro - UB · Solving the Rational Interpolation Problem Teresa Cortadellas, Carlos D’Andrea, Eulàlia Montoro FoCM 2017 Barcelona