talk: a polyhedral approach to computing border bases (2010)

8/14/2019 Talk: A polyhedral approach to computing border bases (2010)

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Motivation Algebra 101 - a quick recap Border bases General characterization Hardness Computational results

A polyhedral approach to

computing border bases

Sebastian Pokutta

Technische Universitat DarmstadtDepartment of Mathematics

(joint work with Gabor Braun)

Feb 2010 / Darmstadt

Sebastian Pokutta Polyhedral border bases Darmstadt 2010 1 / 30
http://find/http://goback/


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Why polynomial systems?

Gian-Carlo Rota (Summer 1985): The one contribution

of mine that I hope will be remembered has consisted in

just pointing out that all sorts of problems of combinatoricscan be viewed as problems of location of the zeros of

certain polynomials and in giving these zeros a

combinatorial interpretation. This is now called the critical

problem.



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Polynomial systems naturally occur in discrete optimization.

0/1 solutions to a linear system of equations Ax = b are given by

{x {0, 1}n | Ax = b, x2i xi = 0 i [n]}.

System of polynomial equations.

Graph isomorphism, graph coloring Modeling dynamical systems

(Very successful application in oil exploration)



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{x {0, 1}n | Ax = b, x2i xi = 0 i [n]}.






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{x {0, 1}n | Ax = b, x2i xi = 0 i [n]}.





G C


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{x {0, 1}n | Ax = b, x2i xi = 0 i [n]}.





M i i Al b 101 i k B d b G l h i i H d C i l l


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Polynomials

Our objects of interest: polynomials

p(X) =mNn

finite support

amxm

Set of indeterminates: X := {x1, . . . , xn}

Set of monomials : Tn := {xm | m Nn}

Degree of p: deg(p) := maxmNn,am=0

||m||1

Leading form of p: maximal-degree terms

Polynomials with coefficients in K: K[X]

Example: p(X) = 3x21 x12 x

75 + 1x

91 x

12 4x

33 x

24 1

deg(p) = 10 and LF(p) = 3x21 x12 x

75 + 1x

91 x

12


M ti ti Al b 101 i k B d b G l h t i ti H d C t ti l lt


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Polynomials


p(X) =mNn

finite support

amxm




||m||1




75 + 1x

91 x

12 4x

33 x

24 1

deg(p) = 10 and LF(p) = 3x21 x12 x

75 + 1x

91 x

12



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Polynomials


p(X) =mNn

finite support

amxm




||m||1




75 + 1x

91 x

12 4x

33 x

24 1

deg(p) = 10 and LF(p) = 3x21 x12 x

75 + 1x

91 x

12


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Motivation Algebra 101 a quick recap Border bases General characterization Hardness Computational results

Polynomials


p(X) =mNn

finite support

amxm




||m||1




75 + 1x

91 x

12 4x

33 x

24 1

deg(p) = 10 and LF(p) = 3x21 x12 x

75 + 1x

91 x

12


http://find/


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Motivation Algebra 101 a quick recap Border bases General characterization Hardness Computational results

Polynomials


p(X) =mNn

finite support

amxm




||m||1




75 + 1x

91 x

12 4x

33 x

24 1

deg(p) = 10 and LF(p) = 3x21 x12 x

75 + 1x

91 x

12


http://find/


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g q p p

Polynomials


p(X) =mNn

finite support

amxm




||m||1




75 + 1x

91 x

12 4x

33 x

24 1

deg(p) = 10 and LF(p) = 3x21 x12 x

75 + 1x

91 x

12


http://find/


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Polynomials


p(X) =mNn

finite support

amxm




||m||1




75 + 1x

91 x

12 4x

33 x

24 1

deg(p) = 10 and LF(p) = 3x21 x12 x

75 + 1x

91 x

12


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Vector spaces vs. rings

Polynomials generate a vector space (isomorphic to K):

p, q K[X], K p+ q K[X] and p K[X]

Polynomials generate a ring:

p, q K[X] p+ q K[X] and pq K[X]

K[X] is an associative K-algebra.


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Ideals and factors

Ideal:

I K[X] subring with p K[X], q I then pq I.

Ideals are absorbing.

Factor/epimorphic images: I K[X] ideal, then:

K[X]/I := {p+ I | p K[X]}.

A fundamental result that (almost always) holds:

Every object is an epimorphic image of a free one.

Question: What does that mean?


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Ideals and factors

Ideal:




K[X]/I := {p+ I | p K[X]}.





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Ideals and factors

Ideal:




K[X]/I := {p+ I | p K[X]}.





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Ideals and factors

Ideal:




K[X]/I := {p+ I | p K[X]}.





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Ideals and factors

One thing to remember: Considering the factor means adding relations .

The result restated:

Every object arises as an unrestricted object plus some relations.

Example 1:Consider:x2i xi = 0 if and only if xi {0, 1}.

Define: I :=

x2i xi | i [n]K[X]

and let V := K[X]/I.

In V we enforce x2i xi = 0 for all i [n].

(Meta-) Example 2:Polytopes P [0, 1]n arise by taking [0, 1]n and adding inequalities.


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Ideals and factors





Define: I :=

x2i xi | i [n]K[X]


In V we enforce x2i xi = 0 for all i [n].



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Ideals and factors





Define: I :=

x2i xi | i [n]K[X]


In V we enforce x2

i xi = 0 for all i [n].



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Ideals and factors





Define: I :=

x2i xi | i [n]K[X]


In V we enforce x2





Id l d f
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Ideals and factors





Define: I :=

x2i xi | i [n]K[X]


In V we enforce x2





Id l d f t
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Ideals and factors





Define: I :=

x2i xi | i [n]K[X]


In V we enforce x2





Ideals and factors
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Ideals and factors





Define: I :=

x2i xi | i [n]K[X]


In V we enforce x2





Degree filtrations and zero dimensional ideals
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Degree filtrations and zero-dimensional ideals

Let I ideal and define Ii := {p I | deg(p) i}.

Degree filtration of I:

I =i


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Degree filtrations and zero-dimensional ideals



I =i


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Degree filtrations and zero dimensional ideals



I =i


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g



I =i


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Definition (Order ideal)Let O Tn such that whenever m1 O, m2 Tn with m2|m1 itfollows m2 O. Then O is an order ideal.

Definition (Border of O)

Let O be an order ideal. Then

O := {xim | m O, i [n]} \ O

is the border of O.



Order ideals and borders
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Definition (Order ideal)Let O Tn such that whenever m1 O, m2 Tn with m2|m1 itfollows m2 O. Then O is an order ideal.

Definition (Border of O)

Let O be an order ideal. Then

O := {xim | m O, i [n]} \ O

is the border of O.



Order ideals and borders
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x

y

An order ideal and its border in T2



Border bases
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Definition (O-border basis)

Let O = {t1, . . . , t} be an order ideal with O := {b1, . . . , b}.Further, let I K[X] be a zero-dimensional ideal andG = {g1, . . . , g} K[X] a set of polynomials.Then G is an O-border basis of I if:

(i) G is of the form: gj = bj

i[] ijti with ij K;

(ii) GK[X] = I;

(iii) K[X] = I OK as vector spaces.



Border bases
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i[] ijti with ij K;

(ii) GK[X] = I;




Border bases
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i[] ijti with ij K;

(ii) GK[X] = I;




Border bases
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i[] ijti with ij K;

(ii) GK[X] = I;




Border bases
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i[] ijti with ij K;

(ii) GK[X] = I;




Border bases
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A necessary and sufficient condition.

Lemma

Let O be an order ideal and I a zero-dimensional ideal. Then Osupports a border basis of I if and only if

K[X] = I OK as vector spaces.

Order ideals with this property are admissible.



The catch: by far, not every order ideal supports a border basis.
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First, even if O Tn has the right cardinality, i.e.,

|O| = dim(K[X]/I)

it does not necessarily support a border basis:

ExampleLet R := {(1, 1);(1, 1); (0, 0);(1, 0); (0; 1)} A2(Q). Let

I := {p K[X] | p(r) = 0 r R}.

Then O := {1, x, x2

, x3

, x4

} cannot support a border basis:

OK I = {0} as x3 x I.



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|O| = dim(K[X]/I)



I := {p K[X] | p(r) = 0 r R}.

Then O := {1, x, x2

, x3

, x4


OK I = {0} as x3 x I.



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|O| = dim(K[X]/I)



I := {p K[X] | p(r) = 0 r R}.

Then O := {1, x, x2

, x3

, x4


OK I = {0} as x3 x I.



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|O| = dim(K[X]/I)



I := {p K[X] | p(r) = 0 r R}.

Then O := {1, x, x2

, x3

, x4


OK I = {0} as x3 x I.



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Let be a term-ordering. Then

K[X] = I Tn \ LT(I)K.

Question: Is every order ideal induced by a term-ordering?

Answer: No, this isnt the case either!

Example

Consider the ideal I :=

x2 + xy + y2, xy2, y4K[X]

. Then

LTDegLex(I) = {x2, xy2, y4} and ODegLex = {1, x, y, xy, y

2, y3}.

Via substitution: O = {1, x, y, x2, y2, y3}.If now x


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Example


x2 + xy + y2, xy2, y4K[X]

. Then


2, y3}.



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Example


x2 + xy + y2, xy2, y4K[X]

. Then


2, y3}.



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Example


x2 + xy + y2, xy2, y4K[X]

. Then


2, y3}.



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Example


x2 + xy + y2, xy2, y4K[X]

. Then


2, y3}.



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Finding admissible order ideals - a combinatorial problem.

The problem in finding an admissible order ideal is twofold.

(i) We have to choose dim(K[X]/I) monomials so that weobtain an order ideal;

(ii) The chosen monomials have to form a basis of K[X]/I.



The combinatorial problem and a polyhedral description
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Th d id l l
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The order ideal polytope.

Let z denote the characteristic vector of an order ideal O, i.e.,zm = 1 if and only if m O.

zm1 zm2 m1, m2 Tnd1 : m1 | m2

mTnd1

zm = d

mU

zm dim U IK/I U Tnd1 : |U| = d

zm [0, 1] m Tnd1

Figure: Order ideal polytope P(I) with d := dim (K[X]/I)




Th d id l l
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zm1 zm2 m1, m2 Tnd1 : m1 | m2

mTnd1

zm = d

mU


zm [0, 1] m Tnd1





Th d id l l t
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zm1 zm2 m1, m2 Tnd1 : m1 | m2

mTnd1

zm = d

mU


zm [0, 1] m Tnd1





The order ideal pol tope
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zm1 zm2 m1, m2 Tnd1 : m1 | m2

mTnd1

zm = d

mU


zm [0, 1] m Tnd1





The order ideal polytope
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zm1 zm2 m1, m2 Tnd1 : m1 | m2

mTnd1

zm = d

mU


zm [0, 1] m Tnd1





A polyhedral characterization
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A polyhedral characterization.

Theorem (Braun, Pokutta 2009)

Let I be a zero-dimensional ideal. There is a bijection between itsorder ideals and integral points of the order ideal polytope of I.The bijection is given by

: z P(I) ZTn

O(z) := {m Tn | zm = 1}.




Sketch of proof
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Sketch of proof.

First, observe that the condition

zm1 zm2 m1, m2 Tnd1 : m1 | m2

indeed ensures that O(z) is an order ideal.

Second, the condition

m

Tn

d1

zm = d

ensures that O(z) is of the right size, i.e., |O(z)| = dim(K[X]/I).




Sketch of proof
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Sketch of proof.

First, observe that the condition

zm1 zm2 m1, m2 Tnd1 : m1 | m2

indeed ensures thatO

(z) is an order ideal.

Second, the condition

m

Tn

d1

zm = d

ensures that O(z) is of the right size, i.e., |O(z)| = dim(K[X]/I).




Sketch of proof
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Sketch of proof.

It remains to show thatmU


is equivalent toI O(z)K = {0},

i.e., the image of O(z) is linearly independent in the factorK[X]/I.

together with |O(z)| = dim(K[X]/I) it follows

K[X] = I OK.




Sketch of proof.
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Sketch of proof.

It remains to show thatmU


is equivalent toI O(z)K = {0},

i.e., the image of O(z) is linearly independent in the factorK[X]/I.

together with |O(z)| = dim(K[X]/I) it follows

K[X] = I OK.




Sketch of proof.
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Sketch of proof.

With our interpretation of z being the characteristic vectormU


can be rewritten as

|U O(z)| dim(U IK/I).

The size of U O(z) is at most the dimension of the vector spacegenerated by the image of U in the factor K[X]/I .

Obviously necessary for directness.




Sketch of proof.
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p

With our interpretation of z being the characteristic vectormU


can be rewritten as

|U O(z)| dim(U IK/I).

The size of U O(z) is at most the dimension of the vector spacegenerated by the image of U in the factor K[X]/I .

Obviously necessary for directness.




Sketch of proof.
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p

For sufficiency choose U := O(z) in

|U O(z)| dim(U IK/I),

so that we obtain

|O(z)| dim(O(z) IK/I).

The dimension of the vector space generated bythe image of O(z) in the factor K[X]/I is at least |O(z)|.

Sufficient for directness.




Sketch of proof.
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p

For sufficiency choose U := O(z) in

|U O(z)| dim(U IK/I),

so that we obtain

|O(z)| dim(O(z) IK/I).

The dimension of the vector space generated bythe image of O(z) in the factor K[X]/I is at least |O(z)|.

Sufficient for directness.



A complexity theoretic consideration

A common problem.
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Given a zero-dimensional ideal I, find a nice basis of K[X]/I, i.e.,one that contains a favorable mix of monomials.

Question: How hard is it to choose an order ideal with respect to a(linear) preference?




A common problem.
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Given a zero-dimensional ideal I, find a nice basis of K[X]/I, i.e.,one that contains a favorable mix of monomials.

Question: How hard is it to choose an order ideal with respect to a(linear) preference?




Choosing nice bases is NP-hard
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Theorem (Braun, Pokutta 2009)

Let I K[X] be a zero-dimensional ideal and let c ZTn .Computing an order ideal O that is admissible for I and maximizesc is NP-hard.




A zero-dimensional ideal that solves k-Clique.
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We will establish hardness by a reduction from k-Clique.

For this, for any graph , we will construct a zero-dimensional ideal

I whose order ideals are in one-to-one with the graphs cliques.

We defineFn,k := {vj | j [n k]} T

n=3

with vj :=

i[n] ij

xi and define In,k := Fn,kK[X].




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n=3

with vj :=

i[n] ij





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n=3

with vj :=

i[n] ij





Characterizing the order ideals of In,k.
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Lemma

Let n N and k [n]. Then In,k is a zero-dimensional ideal suchthat an order ideal O is admissible if and only if

(i) O=1 Tn=1 with |O=1| = k;

(ii) O=2 = {xy | x, y O=1};

(iii) O= = for 3.

Therefore, the order ideals of In,k are in one-to-one correspondence

with the choices of k vertices and the associated edges on acomplete graph Kn.




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Lemma


(i) O=1 Tn=1 with |O=1| = k;

(ii) O=2 = {xy | x, y O=1};

(iii) O= = for 3.






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Lemma


(i) O=1 Tn=1 with |O=1| = k;

(ii) O=2 = {xy | x, y O=1};

(iii) O= = for 3.






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Lemma


(i) O=1 Tn=1 with |O=1| = k;

(ii) O=2 = {xy | x, y O=1};

(iii) O= = for 3.






Sketch of proof.
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Observe, that the polynomials vj are homogeneous of degree oneand the coefficient matrix A := (vj)j[nk] is a Vandermondematrix, i.e., every square submatrix is invertible.

If we thus remove any k columns (belonging to k variables of{x1, . . . , xn}) from A we obtain an invertible matrix. (without lossof generality we choose {x1, . . . , xk}).

Therefore, {x1, . . . , xk, v1, . . . , vnk} is a basis for thehomogeneous polynomials of degree one. So K[X] is also apolynomial ring in those variables.

Sebastian Pokutta Polyhedral border bases Darmstadt 2010 26 / 30 Motivation Algebra 101 - a quick recap Border bases General characterization Hardness Computational results


Sketch of proof.
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Sketch of proof.
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Sketch of proof.
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We can therefore conclude

OK= K[x1, . . . , xn]Fn,kK[X]

= K[x1, . . . , xk]Tk=3K[X] .

Substitution preserves degrees, homogeneity, etc.Thus for any order ideal: |O=1| = k, O=2 = {xy | x, y O=1},and O= = for all 3.

The other direction follows similary.



Sketch of proof.
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We can therefore conclude

OK= K[x1, . . . , xn]Fn,kK[X]

= K[x1, . . . , xk]Tk=3K[X] .

Substitution preserves degrees, homogeneity, etc.Thus for any order ideal: |O=1| = k, O=2 = {xy | x, y O=1},and O= = for all 3.

The other direction follows similary.



Proving hardness.
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Let = (V,E) with n := |V| and k [n] be an instance ofk-Clique.

We consider In,k and define c ZTn3 via

cm =

1, if m = xuxv and either (u, v) E or u = v;

0, otherwise.

Observe that there exists an admissible order ideal O with scorek(k+1)

2 if and only if contains a clique of size k



Proving hardness.
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cm =


0, otherwise.





Proving hardness.
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cm =


0, otherwise.




l l d d l [ ] [ ] # d d l
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polynomial system order ideal signature optimization [s] counting [s] # order ideals

x3, xy2 + y3 (1, 3, 1, 1, 1) < 0.01 0.02 3

vanishing ideal of the points(0, 0, 0, 1), (1, 0, 0, 2),(3, 0, 0, 2), (5, 0, 0, 3),

(1, 0, 0, 4), (4, 4, 4, 5),(0, 0, 7, 6)).

(1, 4, 2) < 0.01 0.02 45

x + y + z u v, x2 x,y2 y, z2 z, u2 u, v2 v

(1, 4, 5) < 0.01 0.35 1,260

x + y + z u v, x3 x,

y

3

y, z

2

z, u

2

u, v

2

v

(1, 4, 7, 6) 0.02 51.50 106,820


(1, 4, 8, 9) 0.02 53.00 108,900


(1, 4, 9, 12, 9) 0.08 300.00* > 1,349,154

x + y + z u v + a,x2 x, y2 y, z2 z,u2 u, v2 v, a2 a

(1, 5, 9) < 0.01 8.68 30,030

Table: Computational results (performed with SCIP and CoCoA).

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Thank you!

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talk: a polyhedral approach to computing border bases (2010)

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