Teresa Cortadellas, Carlos D’Andrea, Eulàlia Montoro - UB · Solving the Rational Interpolation...

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

“Easy” Solution

1 x1 x2

1 . . . xN−11

1 x2 x22 . . . xN−1

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

1 xN x2N . . . xN−1

...aN−1

y1...yN

P =N−1∑j=0

ajxj =

N∑j=1

∏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

...aN−1

y1...yN

P =N−1∑j=0

ajxj =

N∑j=1

∏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

...aN−1

y1...yN

P =N−1∑j=0

=N∑j=1

∏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

...aN−1

y1...yN

P =N−1∑j=0

ajxj =

N∑j=1

∏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

...aN−1

y10...

yLNL−1

P =N−1∑j=0

ajxj =

L∑j=0

How to solve it?

1 x1 x2

1 . . . xN−11

0 1 2x1 . . . (N − 1)xN−21

......

... . . ....

0 0 0 . . . ∗xN−NL−1L

...aN−1

y10...

yLNL−1

P =N−1∑j=0

ajxj =

L∑j=0

How to solve it?

1 x1 x2

1 . . . xN−11

0 1 2x1 . . . (N − 1)xN−21

......

... . . ....

0 0 0 . . . ∗xN−NL−1L

...aN−1

y10...

yLNL−1

P =N−1∑j=0

How to solve it?

1 x1 x2

1 . . . xN−11

0 1 2x1 . . . (N − 1)xN−21

......

... . . ....

0 0 0 . . . ∗xN−NL−1L

...aN−1

y10...

yLNL−1

P =N−1∑j=0

ajxj =

L∑j=0

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(`)

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(`)

But...

you may have α(`) = β(`) = 0 !

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

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?

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

∑Ni=1

aiyi(x−xi )∑N

(x−xi )

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

∑Ni=1

aiyi(x−xi )∑N

(x−xi )

a “solution” of the WRIP

RIP is solvable ⇐⇒ ai 6= 0

∑Ni=1

aiyi(x−xi )∑N

(x−xi )

a “solution” of the WRIPRIP is solvable ⇐⇒

ai 6= 0

∑Ni=1

aiyi(x−xi )∑N

(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...

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

(Ravi 97)

⇐⇒

x − x1 0 . . . 0 1 −y1

0 x − x2 . . . 0 1 −y2...

......

0 0 . . . x − xN 1 −yN

∗...∗AB

(Ravi 97)

⇐⇒

x − x1 0 . . . 0 1 −y1

0 x − x2 . . . 0 1 −y2...

......

0 0 . . . x − xN 1 −yN

∗...∗AB

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!

Teresa Cortadellas, Carlos D’Andrea, Eulàlia Montoro - UB · Solving the Rational Interpolation...

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