Computing Nearest Gcd with Certification G. Chze, A. Galligo, B. - - PowerPoint PPT Presentation

Computing Nearest Gcd with Certification

• G. Chèze, A. Galligo,
• B. Mourrain, J.-C. Yakoubsohn

Institut de Mathématiques de Toulouse, Université de Nice, Inria.

RAIM 2011

• G. Chèze (IMT)

Computing Nearest Gcd with Certification 1 / 31

Plan

1

The approximate gcd problem

2

A bisection approach

3

Complexity results

4

Some examples

• G. Chèze (IMT)

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The problem

Given f(z) and g(z) in Cd[z], find p(z) and q(z) in Cd[z], which are solutions of the minimization problem min

p, q∈Cd[z], Resultant(p,q)=0 ||f − p||2 + ||g − q||2.

• G. Chèze (IMT)

Computing Nearest Gcd with Certification 3 / 31

Bibliography

1

Algebraic approach via Euclid’s Algorithm, resultant and subresultant : Schönhage, Noda-Sasaki, Corless-Gianni-Trager-Watt, Hribernig-Stetter, Emiris-Galligo-Lombardi, Rupprecht, Beckermann-Labahn.

2

Padé approximation and structured matrices approach : Bini-Pan, Bini-Boito, Zhi, Yang-Zhi.

3

Rootfinding and cluster root approach : Pan, Zeng, Graillat-Langlois.

4

Optimization approach : Karmarkar-Lakshman, Kaltofen-Yang-Zhi, Nie-Demmel-Gu, Terui.

• G. Chèze (IMT)

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Karmarkar and Lakshman’s formula

The approximate gcd problem can be reduced to the minimization problem : min

z∈C

f(z)f(z) + g(z)g(z) d

k=0 zkzk

. If z0 is a minimum we get p and q with this formula : p(z) = f(z) − f(z0) d

k=0 z0 kzk

×

d

• k=0

z0

kzk

q(z) = g(z) − g(z0) d

k=0 z0 kzk

×

d

• k=0

z0

kzk

• G. Chèze (IMT)

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Our goal

What do we want ?

1

Give an algorithm to solve the Karmarkar and Lakshman’s minimization problem in a certified way.

2

Study the complexity of this algorithm.

• G. Chèze (IMT)

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Some notations

We set : X + iY = z, F(X, Y) = f(z)f(z) + g(z)g(z) d

k=0 zkzk

. G(X, Y) =

• G1(X, Y), G2(X, Y)
• is the system of the numerators
• f the gradient ∇F(X, Y).
• G. Chèze (IMT)

Computing Nearest Gcd with Certification 7 / 31

Some notations

E is an exclusion test : E(Gi, S) is true if Gi has (in a certified way) no zero in the square S otherwise it returns false. I is an inclusion test : I(G, S) is true if G has (in a certified way) one zero in the square S otherwise it returns false.

• G. Chèze (IMT)

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An exclusion test

Proposition

Let S(x, y, r) be a square. We set : M(Gi, x, y, r) = |Gi(x, y)| −

• k≥1

||DkGi(x, y)|| k! r k. If M(Gi, x, y, r) > 0 then S(x, y, r) does not contain any zero of Gi(X, Y).

Definition

E

• Gi, S
• is true if M(Gi, x, y, r) > 0 else E(Gi, S) is false.
• G. Chèze (IMT)

Computing Nearest Gcd with Certification 9 / 31

An exclusion test

Proposition

Let S(x, y, r) be a square. We set : M(Gi, x, y, r) = |Gi(x, y)| −

• k≥1

||DkGi(x, y)|| k! r k. If M(Gi, x, y, r) > 0 then S(x, y, r) does not contain any zero of Gi(X, Y).

Definition

E

• Gi, S
• is true if M(Gi, x, y, r) > 0 else E(Gi, S) is false.
• G. Chèze (IMT)

Computing Nearest Gcd with Certification 9 / 31

An exclusion test

Proposition

Let S(x, y, r) be a square. We set : M(P, x, y, r) = |P(x, y)| −

• k≥1

||DkP(x, y)|| k! r k. If M(P, x, y, r) > 0 then S(x, y, r) does not contain any zero of P(X, Y).

Definition

E+ P, S

• is true if M(P, x, y, r) > 0 and P(x, y) > 0 else

E+(P, S) is false.

• G. Chèze (IMT)

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An inclusion test

Definition

β := β(G; x, y) = ||DG(x, y)−1G(x, y)|| γ := γ(G; x, y) = supk≥2 1

k!||DG(x, y)−1DkG(x, y)||

1/(k−1) α := α(G; x, y) = βγ.

• G. Chèze (IMT)

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An inclusion test

Theorem (Smale α-Theorem)

If α(G, x, y) < (13 − 3 √ 17)/4 then G has one and only one zero (x∗, y ∗) in the open ball B(x, y, σ(x, y)) with σ(x, y) = 1 + α − √ 1 − 6α + α2 4γ and the Newton’s iteration from (x, y) converges quadratically to (x∗, y∗).

• G. Chèze (IMT)

Computing Nearest Gcd with Certification 12 / 31

An inclusion test

Definition

I(G, S) is true if :

1

α(G; x, y) < 13 − 3 √ 17 4 ,

2

S(x, y, r) ⊂ B

• x, y, σ(x, y)
• , i.e. r < σ(x, y).
• G. Chèze (IMT)

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A bisection approach

S0 := S(x0, y0, r0) a square containing all the global minima of F := N/D. L := {S0}, µ = F(x0, y0). While L is not empty do For each square of L do E+(N − µD, S), E(G1, S), E(G2, S). If at least one of these tests is true then remove S from L, Else do I(G, S) : If I(G, S) is false then divide S in 4 equal squares S1, . . . , S4 and substitute S by {S1, . . . , S4} in L. If I(G, S) is true then we have an approximate zero of G.

• G. Chèze (IMT)

Computing Nearest Gcd with Certification 14 / 31

A bisection approach

If I(G, S) is true then compute the local minimum (x∗, y∗)

• f F in S, µ∗ := F(x∗, y∗).

If µ∗ > µ then remove S. Else µ∗ ≤ µ then remove the bad squares, and µ := µ∗.

• G. Chèze (IMT)

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“Proof of the algorithm”

Computation of an initial square

Lemma (Karmarkar-Lakshman)

Let F be the rational bivariate function corresponding to the approximate gcd problem. Let (x, y) be a global minimum of F then (x, y) ≤ 5 max(f2, g2). ⇒ S0 = S

• (0, 0), 5 max(f2, g2)
• G. Chèze (IMT)

Computing Nearest Gcd with Certification 16 / 31

“Proof of the algorithm”

We can satisfy the inclusion test.

Proposition

Let (x∗, y∗) be a zero of G and suppose that DG(x∗, y ∗) is

• inversible. If

(x, y) − (x∗, y ∗) ≤ 0.08 γ(G, x∗, y∗) then α(G, x, y) ≤ 13 − 3 √ 17 4 .

• G. Chèze (IMT)

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Complexity results

Theorem

Assume that all the global minima of G(X, Y) = 0 are regular and are contained in a square S0 := S(x0, y0, r0). Let Cd := 1 + √ 2 log 2(4d − 2) and J :=

• log2

3r0Cdγ(G,AF ) K δ0

• , where 0.07 ≤ δ0 ≤ 0.08, and

K is a constant related to the geometry of the problem, AF is the set of global minima of F. Denote by Rj the set of retained squares at step j. Then, for j ≥ J, 1- The Newton’s iteration applied to G from a point in Rj converges quadratically to a minimum of F.

• G. Chèze (IMT)

Computing Nearest Gcd with Certification 18 / 31

Complexity results

Theorem

2- When the approximate gcd problem has a unique solution the number of exclusion steps belongs to O

• A d3 log3(d)

minAF dF

• DG(x, y)−1, Σ

+ B d3 , where A, B are constants related to the geometry of the problem. G1 G2 G1 G2 Small A, B. Big A, B.

• G. Chèze (IMT)

Computing Nearest Gcd with Certification 19 / 31

An example

Input : f(z) = 9x3 − 18x2 + 3x + 1, g(z) = −37x3 + 63x2 − 30x + 3. Output : z = 0.479, p(z) = 9.059x3 − 17.875x2 + 3.26x + 1.544, q(z) = −36.916x3 + 63.174x2 − 29.636x + 3.758.

• G. Chèze (IMT)

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An example

Graph of F(X, Y)

0.2 0.4 0.6 0.8 1 0.05 0.1 5 10 15 20 x y N/D

• G. Chèze (IMT)

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An example

z = 0.479

• G. Chèze (IMT)

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An example

z = 0.479

• G. Chèze (IMT)

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An example

z = 0.479

• G. Chèze (IMT)

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An example

z = 0.479

• G. Chèze (IMT)

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An example

z = 0.479

• G. Chèze (IMT)

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An example

z = 0.479

• G. Chèze (IMT)

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Random examples

5 10 15 20 25 30 2 4 6 8 10 12

cpu d

FIG.: f = (z − 1).f1 + 10−3ǫ1, g = (z − 1).g1 + 10−3ǫ2

Random means that we use the Gaussian distribution with mean 0 and variance 1. Our algorithm gives an approximate gcd near z − 1.

• G. Chèze (IMT)

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Solution at infinity

f(x) = z2 − 4z + 3, and g(z) = z2 + 4z + 3. The global minimum is reached at infinity ! lim

(x,y)→∞ F(x, y) = F(∞) = 2,

N(X, Y) − 2D(X, Y) = 16 + 42X 2 + 18Y 2 > 0. Remark : We compute F(∞) in the following way : F(∞) = Fr(0), where Fr(z) = zdeg(F)F(1/z).

• G. Chèze (IMT)

Computing Nearest Gcd with Certification 29 / 31

Solution at infinity

FIG.: F has no global minima in C.

• G. Chèze (IMT)

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Solution at infinity

Input : f(x) = z2 − 4z + 3, and g(z) = z2 + 4z + 3. We exclude the intial square S0 = [−130, 130]2 after 10 iterations, i.e. the size of the smallest exluded square is 130/210. Output : p = −4z + 3 and q = 4z + 3.

• G. Chèze (IMT)

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