Linear and rational factorization of tropical polynomials Bo Lin 1 - - PowerPoint PPT Presentation

linear and rational factorization of tropical polynomials
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Linear and rational factorization of tropical polynomials Bo Lin 1 - - PowerPoint PPT Presentation

Background Tools Main result and algorithms Examples Linear and rational factorization of tropical polynomials Bo Lin 1 Ngoc Mai Tran 2 1 School of Mathematics, Georgia Institute of Technology 2 Dept. of Mathematics, University of Texas, Austin


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Background Tools Main result and algorithms Examples

Linear and rational factorization

  • f tropical polynomials

Bo Lin 1 Ngoc Mai Tran 2

1School of Mathematics, Georgia Institute of Technology

  • 2Dept. of Mathematics, University of Texas, Austin

Emory University Algebra Seminar February 25th, 2020 arXiv:1707.03332v3

  • B. Lin & N. Tran

Linear and rational factorization of tropical polynomials

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Outline

Background: factorization of tropical polynomials is hard; Tools: Cayley trick, signed Minkowski sum of polytopes; Main results and algorithms; Examples: homogeneous linear polynomials and more.

  • B. Lin & N. Tran

Linear and rational factorization of tropical polynomials

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Tropical algebra

Definition On the set R = R ∪ {−∞} we define two commutative binary

  • perations ⊕ and ⊙ as follows: for a, b ∈ R and c ∈ R,

a ⊕ b = max(a, b), a ⊙ b = a + b. c ⊕ −∞ = c, c ⊙ −∞ = −∞. The triple (R, ⊕, ⊙) is called the tropical semiring.

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Linear and rational factorization of tropical polynomials

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Tropical algebra

Definition On the set R = R ∪ {−∞} we define two commutative binary

  • perations ⊕ and ⊙ as follows: for a, b ∈ R and c ∈ R,

a ⊕ b = max(a, b), a ⊙ b = a + b. c ⊕ −∞ = c, c ⊙ −∞ = −∞. The triple (R, ⊕, ⊙) is called the tropical semiring. Remark We use the so-called max-plus operations for convenience. There is an equivalent way to define the tropical semiring: replace max by min and −∞ by ∞.

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Linear and rational factorization of tropical polynomials

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Tropical polynomials

Definition A tropical polynomial is a function f : Rn → R, such that for any x = (x1, x2, . . . , xn) ∈ Rn, f(x) =

  • a∈A
  • fa ⊙

n

  • i=1

x⊙ai

i

  • = max

a∈A

  • fa +

n

  • i=1

aixi

  • where A ⊆ Nn is finite and fa ∈ R for a ∈ A.
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Linear and rational factorization of tropical polynomials

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Background Tools Main result and algorithms Examples

Tropical polynomials

Definition A tropical polynomial is a function f : Rn → R, such that for any x = (x1, x2, . . . , xn) ∈ Rn, f(x) =

  • a∈A
  • fa ⊙

n

  • i=1

x⊙ai

i

  • = max

a∈A

  • fa +

n

  • i=1

aixi

  • where A ⊆ Nn is finite and fa ∈ R for a ∈ A.

Remark We ignore the ground field K, and directly take valuations as the coefficients of each term in the polynomial.

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Linear and rational factorization of tropical polynomials

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Roots of tropical polynomials

Definition Let f(x) be a tropical polynomial. A point x ∈ Rn is a root of f if the maximum is attained at least twice in the evaluation of f(x).

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Roots of tropical polynomials

Definition Let f(x) be a tropical polynomial. A point x ∈ Rn is a root of f if the maximum is attained at least twice in the evaluation of f(x). Definition Let f be a tropical polynomial. The tropical hypersurface T(f) is the set of all roots of f.

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Linear and rational factorization of tropical polynomials

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Roots of tropical polynomials

Definition Let f(x) be a tropical polynomial. A point x ∈ Rn is a root of f if the maximum is attained at least twice in the evaluation of f(x). Definition Let f be a tropical polynomial. The tropical hypersurface T(f) is the set of all roots of f. Example Suppose f(x, y) = max(x, y, 0). The graph

  • f T(f) is the right figure.

(0, 0)

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Equivalence of tropical polynomials

There are three levels of equivalence between two tropical polynomials f and g.

1 f =1 g: f and g have the same terms and coefficients; 2 f =2 g: f(x) = g(x) for all x ∈ Rn; 3 f =3 g: T(f) = T(g).

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Equivalence of tropical polynomials

There are three levels of equivalence between two tropical polynomials f and g.

1 f =1 g: f and g have the same terms and coefficients; 2 f =2 g: f(x) = g(x) for all x ∈ Rn; 3 f =3 g: T(f) = T(g).

Example max(2x, 0) = max(2x, x − 1, 0) for all x ∈ R. Then max(2x, 0) =2 max(2x, x − 1, 0). In addition, they both =3 max(3x + 2, x + 1, 2), as all of them have a unique root x = 0. In this project, we focus on =2, i.e. the equivalence of polynomial functions.

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Fundamental theorem of tropical algebra

Like ordinary polynomials, we also want to write tropical polynomials as (tropical) product of other tropical polynomials.

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Fundamental theorem of tropical algebra

Like ordinary polynomials, we also want to write tropical polynomials as (tropical) product of other tropical polynomials. In the univariate case, we have the following result. Theorem (Fundamental theorem of tropical algebra) Every tropical polynomial in one variable with rational coefficients equals to a product of linear tropical polynomials with rational coefficients as functions.

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Fundamental theorem of tropical algebra

Like ordinary polynomials, we also want to write tropical polynomials as (tropical) product of other tropical polynomials. In the univariate case, we have the following result. Theorem (Fundamental theorem of tropical algebra) Every tropical polynomial in one variable with rational coefficients equals to a product of linear tropical polynomials with rational coefficients as functions. Example f1(x) = max(4x, 3x + 2, 2x + 1, −3) =2 max(x, 2) + max(x, −1) + 2 max(x, −2).

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Multivariate: factorization is NP-complete

Deciding whether a general tropical polynomial is factorizable is hard.

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Multivariate: factorization is NP-complete

Deciding whether a general tropical polynomial is factorizable is hard. Theorem (Kim-Roush ’05, Grigg ’07) The factorization of multivariate tropical polynomials is NP-complete.

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Newton polytope and regular subdivision

Definition Let f(x) = maxa∈A (fa + n

i=1 aixi). The Newton polytope of f,

denoted by Newt(f), is the convex hull of {(a1, a2, . . . , an)|a ∈ A}. Newt(f) is a lattice polytope in Rn. It tells us what terms could appear in the polynomial.

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Newton polytope and regular subdivision

Definition Let f(x) = maxa∈A (fa + n

i=1 aixi). The Newton polytope of f,

denoted by Newt(f), is the convex hull of {(a1, a2, . . . , an)|a ∈ A}. Newt(f) is a lattice polytope in Rn. It tells us what terms could appear in the polynomial. Remark The regular subdivision of Newt(f) induced by the weights fa provides important information of T(f).

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Example: regular subdivision of Newt(f2)

Let f2(x1, x2) = max(2x1 − 3, 2x2 − 1, x1 + x2, x1, x2 + 1, 0). Then Newt(f2) = Conv((0, 0), (1, 0), (0, 1), (0, 2), (1, 1), (2, 0)). If we choose the weight vector as w = (−3, −1, 0, 0, 1, 0), the regular subdivision of Newt(f2) is

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Duality between T(f) and the regular subdivision of Newt(f)

Figure 1: Duality between T(f2) and ∆Newt(f2)

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Duality between T(f) and the regular subdivision of Newt(f)

Proposition Let ∆f be the regular subdivision of Newt(f) w.r.t. the vector fa. Then the tropical hypersurface T(f) is the polyhedral complex dual to ∆f.

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Duality between T(f) and the regular subdivision of Newt(f)

Proposition Let ∆f be the regular subdivision of Newt(f) w.r.t. the vector fa. Then the tropical hypersurface T(f) is the polyhedral complex dual to ∆f. Remark This result tells us that regular subdivision is a useful tool to study tropical polynomials.

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S-units and S-factorizable

Definition A tropical polynomial f is a unit if the induced regular subdivision

  • n Newt(f) is trivial. For a set of lattice polytopes S in Rn, an

S-unit f is a unit such that Newt(f) is a translation of some polytope in S.

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S-units and S-factorizable

Definition A tropical polynomial f is a unit if the induced regular subdivision

  • n Newt(f) is trivial. For a set of lattice polytopes S in Rn, an

S-unit f is a unit such that Newt(f) is a translation of some polytope in S. Definition A tropical polynomial f is called S-factorizable if it equals to a tropical product of S-units. And f is called S-rational if there exist a polynomial g and an S-factorizable polynomial h such that f ⊙ g = h. f is called strong S-rational if in addition g is also S-factorizable.

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Cayley trick and Newton polytope

The polyhedral version of the Cayley trick is the following Theorem (Sturmfels ’94 (Thm 5.1)) Let S be a set of polytopes. Then f is S-factorizable if and only if ∆f is a regular mixed subdivision of Newt(f) with respect to a sequence of possibly repeated polytopes in S.

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Cayley trick and Newton polytope

The polyhedral version of the Cayley trick is the following Theorem (Sturmfels ’94 (Thm 5.1)) Let S be a set of polytopes. Then f is S-factorizable if and only if ∆f is a regular mixed subdivision of Newt(f) with respect to a sequence of possibly repeated polytopes in S. So we can consider the decomposition of a polytope into Minkowski sums of other polytopes. But this is hard, too. Theorem (Gao-Lauder ’01) The decomposition problem of integral polygons is NP-complete.

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Results in special cases

There are results that work in special cases: Gritzmann-Sturmfels(’93): d-dimensional polytopes with up to n vertices; Fukuda(’04): zonotopes; Fukuda-Weibel(’05): V-polytopes.

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Our contribution

In this work, we present a large class of polytopes S such that the set of S-factorizable tropical polynomials has unique and local

  • factorization. Here local means that if each cell of ∆f is a

Minkowski sum of some polytopes in S, then f is S-factorizable.

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Our contribution

In this work, we present a large class of polytopes S such that the set of S-factorizable tropical polynomials has unique and local

  • factorization. Here local means that if each cell of ∆f is a

Minkowski sum of some polytopes in S, then f is S-factorizable. We also present algorithms to determine whether a tropical polynomials f is S-factorizable or S-rational. And if the answer is positive, we compute the decomposition (though with exponential time complexity).

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Example: SK3-rational but not SK3-factorizable

For each positive integer n, our first choice of S is the standard simplex Conv(ei | 1 ≤ i ≤ n) and its faces, denoted by SKn. which are the Newton polytopes of the tropical polynomials of the form max(x1 + c1, x2 + c2, . . . , xn + cn), where each ci ∈ R.

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Example: SK3-rational but not SK3-factorizable

For each positive integer n, our first choice of S is the standard simplex Conv(ei | 1 ≤ i ≤ n) and its faces, denoted by SKn. which are the Newton polytopes of the tropical polynomials of the form max(x1 + c1, x2 + c2, . . . , xn + cn), where each ci ∈ R. Example ˜ f2(x) = max(2x1 − 3, 2x2 − 1, x1 + x2, x1 + x3, x2 + x3 + 1, 2x3) is not SK3-factorizable, but ˜ f2 is SK3-rational. In fact ˜ f2 + max(x1 − 1, x2, x3) =2 max(x1 − 1, x2 − 2, x3) + max(x1 − 3, x2, x3) + max(x1, x2 + 1, x3) =1 max(3x3, x2 + 2x3 + 1, 2x2 + x3 + 1, 3x2 − 1, x1 + 2x3, x1 + x2 + x3, x1 + 2x2, 2x1 + x3 − 1, 2x1 + x2 − 1, 3x1 − 4).

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Example: ˜ f2

Note that ˜ f2 is the homogenization of f2. Figure 2 shows the rational factorization:

Figure 2: The rational factorization of ˜ f2

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Signed Minkowski sum

The S-rational examples motivate us to consider the following. Definition Let P, Q be two nonempty polytopes of Rn. If there exists a nonempty polytope R ⊂ Rn such that P = Q + R, then we define the Minkowski difference P − Q as R.

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Signed Minkowski sum

The S-rational examples motivate us to consider the following. Definition Let P, Q be two nonempty polytopes of Rn. If there exists a nonempty polytope R ⊂ Rn such that P = Q + R, then we define the Minkowski difference P − Q as R. Definition Let P1, P2, · · · , Pm be non-empty polytopes in Rn, c1, c2, · · · , cm ∈ Z with at least one being positive. If there exists a polytope P ′ such that

  • ci<0

(−ci)Pi + P ′ =

  • ci>0

ciPi, then the signed Minkowski sum m

i=1 ciPi is defined to be P ′.

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H-representations and b-vectors

We focus on the H-representation of polytopes. Definition For matrix H ∈ Zr×n whose rows are primitive vectors, and a vector b ∈ Rr, let PH,b denote the possibly empty polytope given by PH,b = {x ∈ Rn | Hx ≤ b}. Suppose H =

  • h1

h2 · · · hr T , where hi is the i-th row vector

  • f H. For any polytope P, let

v(H, P) =

  • maxx∈P h1 · x

maxx∈P h2 · x · · · maxx∈P hr · x T . And let b(H) = {b ∈ Rr | PH,b = ∅ and v(H, PH,b) = b}.

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Example

Let H be   −1 −1 1 1   . And b = (0, 0, 1)T . Then P(H, b) = {(x1, x2) ∈ R2 | x1 ≥ 0, x2 ≥ 0, x1 + x2 ≤ 1} = Conv((0, 0), (0, 1), (1, 0)). And v(H, P(H, b)) = b, so b ∈ b(H).

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H-representable polytopes

Proposition Let Pi be lattice polytopes and H ∈ Zr×n contains all primitive normal vectors of their Minkowski sum

i=1 Pi. Suppose vectors

bi are such that PH,bi = Pi. For y+, y− ∈ Nm, suppose the signed Minkowski sum m

i=1 (y+ i − y− i )PH,bi is well-defined. Then m

  • i=1

(y+

i − y− i )PH,bi = PH,m

i=1 (y+ i −y− i )bi.

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H-representable polytopes

Proposition Let Pi be lattice polytopes and H ∈ Zr×n contains all primitive normal vectors of their Minkowski sum

i=1 Pi. Suppose vectors

bi are such that PH,bi = Pi. For y+, y− ∈ Nm, suppose the signed Minkowski sum m

i=1 (y+ i − y− i )PH,bi is well-defined. Then m

  • i=1

(y+

i − y− i )PH,bi = PH,m

i=1 (y+ i −y− i )bi.

Remark This proposition enables us to decompose a polytope into a signed Minkowski sum of polytopes on the level of the b-vectors. Then linear algebra can be used.

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Basis and unique factorization

Let S be a finite set of lattice polytopes in Rn. Let H(S) ∈ Zr×n be a matrix whose row vectors are all distinct primitive normal vectors of the polytope

S∈S S, with coordinate-wise

lexicographic order. Then H(S) is uniquely defined. Definition B(S) = {b ∈ Zr ∩ b(H(S)) | PH(S),b ∈ S}, B(S) = {b ∈ Zr ∩ b(H(S)) | ∅ = PH(S),b ⊂ Znis a lattice polytope}. S is a basis if B(S) is a basis over Z for ZB(S). S is a full basis if B(S) is a basis over Z for B(S).

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The good polynomials and polytopes of S

Let S be a finite set of lattice polytopes. Definition Let N[S] be the set of S-factorizable polynomials, E[S] be the set

  • f S-rational polynomials and Z[S] be the set of strong S-rational

polynomials.

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The good polynomials and polytopes of S

Let S be a finite set of lattice polytopes. Definition Let N[S] be the set of S-factorizable polynomials, E[S] be the set

  • f S-rational polynomials and Z[S] be the set of strong S-rational

polynomials. Let f be a unit tropical polynomial in Rn and P = Newt(f). Then Proposition

(i)

f ∈ E[S] if and only if P = PH(S),b for some b ∈ B(S)

(ii)

f ∈ N[S] if and only if P = PH(S),b for some b ∈ NB(S).

(iii)

f ∈ Z[S] if and only if P = PH(S),b for some b ∈ B(S) ∩ ZB(S).

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Example: a basis without local factorization

Consider the regular subdivision of the square: (1, 1) (0, 0)

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Example: a basis without local factorization

Consider the regular subdivision of the square: (1, 1) (0, 0) If both triangles belong to S, then S does not have local

  • factorization. This motivates us to exclude such pairs of polytopes

in good S.

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Canonical, hierarchical and orientation

To define positive basis, we need the following: Definition A polytope S is canonical if for any proper face P, P ≤ S. A set

  • f polytopes S is canonical if S is canonical for all S ∈ S. S is

hierarchical if for any S ∈ S, all proper faces of S also belong to S.

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Canonical, hierarchical and orientation

To define positive basis, we need the following: Definition A polytope S is canonical if for any proper face P, P ≤ S. A set

  • f polytopes S is canonical if S is canonical for all S ∈ S. S is

hierarchical if for any S ∈ S, all proper faces of S also belong to S. Definition Let S be a hierarchical set of polytopes. An orientation τ is a map from the row vectors of H(S) to {1, −1}, such that τ(v) = −τ(−v). Let Hτ

+ = {v | τ(v) = 1}. S is positive with

  • rientation τ if for each v ∈ Hτ

+ and S ∈ S, face−v(S) is not a

proper face of S.

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Example: a non-positive basis

Let S ={Conv((0, 0), (1, 0), (0, 1)), Conv((1, 1), (1, 0), (0, 1)), Conv((0, 0), (1, 0))}. Then S is a basis (note that S is not hierarchical). However, S is not positive.

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Example: a non-positive basis

Let S ={Conv((0, 0), (1, 0), (0, 1)), Conv((1, 1), (1, 0), (0, 1)), Conv((0, 0), (1, 0))}. Then S is a basis (note that S is not hierarchical). However, S is not positive. Since Conv((1, 0), (0, 1)) is a facet for some polytope in S, the vectors (1, 1) and (−1, −1) are row vectors of H(S). Then for any

  • rientation τ, if τ((1, 1)) = 1, then

face(−1,−1)(Conv((1, 1), (1, 0), (0, 1))) = Conv((1, 0), (0, 1)) is a proper face of Conv((1, 1), (1, 0), (0, 1)); the other case is similar.

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Example: a positive basis

Let S consist of the polytope Conv((0, 0), (1, 0), (0, 1)) and its proper faces. Then S is a positive basis.

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Example: a positive basis

Let S consist of the polytope Conv((0, 0), (1, 0), (0, 1)) and its proper faces. Then S is a positive basis. If we homogenize, this S becomes the family SK3.

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Main Theorems

Theorem (L.-Tran ’17+) If S is a positive basis, then N[S] has unique and local factorizations.

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Main Theorems

Theorem (L.-Tran ’17+) If S is a positive basis, then N[S] has unique and local factorizations. Theorem (L.-Tran ’17+) If S is a positive basis, then Z[S] has unique and local

  • factorizations. In addition, Z[S] = E[S] if and only if S is a full

positive basis. In this case, f ∈ Z[S] = E[S] if and only if the edges of cells in ∆f are parallel to edges in S.

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Algorithms of factorizing f

Given a tropical polynomial f and a finite set S of lattice

  • polytopes. We have algorithms for each of the following purposes:

1 Decide whether S is a positive basis. 2 Given a positive basis S, decide whether f ∈ Z[S] \ N[S],

f ∈ N[S], or neither.

3 If f ∈ N[S], obtain the unique factorization for f. 4 If f ∈ Z[S] \ N[S], obtain a g ∈ N[S] such that f ⊙ g ∈ N[S].

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SKn revisited

The family SKn of simplices is important because the factorization into linear polynomials has applications in economics and combinatorics.

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SKn revisited

The family SKn of simplices is important because the factorization into linear polynomials has applications in economics and combinatorics. Proposition SKn is a full positive basis.

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A formula using generalized permutohedra

If f is strong SKn-rational, then each cell C in ∆f is a generalized

  • permutohedra. So there exist (zI)I⊆[n] ∈ R2n such that

C = {(t1, . . . , tn) ∈ Rn |

n

  • i=1

ti = z[n],

  • i∈I

ti ≥ zI ∀I ⊆ [n]}.

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A formula using generalized permutohedra

If f is strong SKn-rational, then each cell C in ∆f is a generalized

  • permutohedra. So there exist (zI)I⊆[n] ∈ R2n such that

C = {(t1, . . . , tn) ∈ Rn |

n

  • i=1

ti = z[n],

  • i∈I

ti ≥ zI ∀I ⊆ [n]}. The following result gives a M¨

  • bius inversion formula to

decompose C. Theorem (Ardila-Benedetti-Doker ’10) C =

  • I⊆[n]

yI · Conv({ei | i ∈ I}), where yI =

J⊆I (−1)|I|−|J|zJ.

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Factors of a SK3-factorizable h1

Let n = 3 and consider the finite set SK3. Let h1(x1, x2, x3) = max(x2 + 2x3 + 4, 2x2 + x3 + 6, 3x2 + 7, x1 + 2x3 + 5, x1 + x2 + x3 + 7, x1 + 2x2 + 8, 2x1 + x3 + 7, 2x1 + x2 + 8, 3x1 + 5).

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Factors of a SK3-factorizable h1

Let n = 3 and consider the finite set SK3. Let h1(x1, x2, x3) = max(x2 + 2x3 + 4, 2x2 + x3 + 6, 3x2 + 7, x1 + 2x3 + 5, x1 + x2 + x3 + 7, x1 + 2x2 + 8, 2x1 + x3 + 7, 2x1 + x2 + 8, 3x1 + 5). ∆h1 has 5 maximal cells:

1 Conv({(2, 1, 0), (2, 0, 1), (1, 2, 0), (1, 1, 1)}), 2 Conv({(1, 2, 0), (1, 1, 1), (0, 3, 0), (0, 2, 1)}), 3 Conv({(1, 1, 1), (1, 0, 2), (0, 2, 1), (0, 1, 2)}), 4 Conv({(1, 1, 1), (1, 0, 2), (2, 0, 1)}), 5 Conv({(2, 1, 0), (2, 0, 1), (3, 0, 0)}).

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signed Minkowski sum - M¨

  • bius inversion

Next we write the 5 maximal cells as Minkowski sums of polytopes in SK3. Both methods work here. Let’s decompose C1 = Conv({(2, 1, 0), (2, 0, 1), (1, 2, 0), (1, 1, 1)}) using the M¨

  • bius

inversion formula. C1 is a generalized permutohedron with parameters: ∅ {1} {2} {3} {1, 2} {1, 3} {2, 3} {1, 2, 3} z 1 2 1 1 3 .

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signed Minkowski sum - M¨

  • bius inversion

Next we write the 5 maximal cells as Minkowski sums of polytopes in SK3. Both methods work here. Let’s decompose C1 = Conv({(2, 1, 0), (2, 0, 1), (1, 2, 0), (1, 1, 1)}) using the M¨

  • bius

inversion formula. C1 is a generalized permutohedron with parameters: ∅ {1} {2} {3} {1, 2} {1, 3} {2, 3} {1, 2, 3} z 1 2 1 1 3 . Then the coefficients yI =

J⊆I (−1)|I|−|J|zJ are:

∅ {1} {2} {3} {1, 2} {1, 3} {2, 3} {1, 2, 3} y 1 1 1 . Thus C1 = S{1} + S{1,2} + S{2,3}.

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signed Minkowski sum - b-vector

H(SK3) is the transpose of (up to row permutation): 1 1 1 0 1 0 0 −1 −1 −1 0 −1 0

1 1 0 1 0 1 0 −1 −1 0 −1 0 −1 0 1 0 1 1 0 0 1 −1 0 −1 −1 0 0 −1

  • .
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signed Minkowski sum - b-vector

H(SK3) is the transpose of (up to row permutation): 1 1 1 0 1 0 0 −1 −1 −1 0 −1 0

1 1 0 1 0 1 0 −1 −1 0 −1 0 −1 0 1 0 1 1 0 0 1 −1 0 −1 −1 0 0 −1

  • .

The vectors v(H(SK3), P) for the simplices P ∈ SK3 are      

{1,2,3}: 1 1 1 1 1 1 1 −1 0 {1,2}: 1 1 1 1 1 1 0 −1 −1 0 {1,3}: 1 1 1 1 1 0 1 −1 0 −1 0 {2,3}: 1 1 1 1 0 1 1 −1 0 0 −1 0 {1}: 1 1 1 0 1 0 0 −1 −1 −1 0 −1 0 {2}: 1 1 0 1 0 1 0 −1 −1 0 −1 0 −1 0 {3}: 1 0 1 1 0 0 1 −1 0 −1 −1 0 0 −1

      . This 7 × 14 matrix has full rank.

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signed Minkowski sum - C2

Take C2 = Conv({(1, 2, 0), (1, 1, 1), (0, 3, 0), (0, 2, 1)}). Then v(H, C2) = (3, 3, 2, 3, 1, 3, 1, −3, −2, 0, −2, 0, −1, 0).

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signed Minkowski sum - C2

Take C2 = Conv({(1, 2, 0), (1, 1, 1), (0, 3, 0), (0, 2, 1)}). Then v(H, C2) = (3, 3, 2, 3, 1, 3, 1, −3, −2, 0, −2, 0, −1, 0). It turns out that v(H, C2) belongs to the span of the previous matrix and it is the sum of the 2-nd, 4-th, and 6-th rows. Thus C2 = S{2} + S{1,2} + S{2,3}.

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Factors of a SK3-factorizable h1

So we write the 5 maximal cells as Minkowski sums of polytopes in SK3 respectively: S{1} + S{1,2} + S{2,3}, S{2} + S{1,2} + S{2,3}, S{3} + S{1,2} + S{2,3}, S{1} + S{3} + S{1,2,3}, 2S{1} + S{1,2,3}.

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Factors of a SK3-factorizable h1

So we write the 5 maximal cells as Minkowski sums of polytopes in SK3 respectively: S{1} + S{1,2} + S{2,3}, S{2} + S{1,2} + S{2,3}, S{3} + S{1,2} + S{2,3}, S{1} + S{3} + S{1,2,3}, 2S{1} + S{1,2,3}. For each maximal cell C, we find a unique homogeneous linear function lC(x1, x2, x3) such that for a ∈ C, lC(a1, a2, a3) = −(h1)a. This is called the Legendre transform.

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Legendre transform of the 5 maximal cells

The coefficients of the linear functions are (8/3, 8/3, 5/3), (10/3, 7/3, 4/3), (11/3, 8/3, 2/3), (3, 3, 1), (5/3, 14/3, 11/3).

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Legendre transform of the 5 maximal cells

The coefficients of the linear functions are (8/3, 8/3, 5/3), (10/3, 7/3, 4/3), (11/3, 8/3, 2/3), (3, 3, 1), (5/3, 14/3, 11/3). The algorithm always chooses a polytope from SK3 with largest dimension that appears in the Minkowski sums, and fix a maximal cell σ. In this case, we choose S{1,2,3} and σ = Conv({(1, 1, 1), (1, 0, 2), (2, 0, 1)}). Then we get a linear factor from the Legendre transform max(x1 + 3, x2 + 3, x3 + 1).

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The remaining factors

Now there are two more linear factors. The important thing is to determine what are the contributions of the first factor in the Minkowski sums. For another maximal cell η, we find all vertices v ∈ S{1,2,3} such that lσ(v) − lη(v) is maximal. Write the convex hull of these vertices as a Minkowski sum of polytopes in SK3, which is the contribution of the first factor, and we want to delete them.

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The remaining factors

Now there are two more linear factors. The important thing is to determine what are the contributions of the first factor in the Minkowski sums. For another maximal cell η, we find all vertices v ∈ S{1,2,3} such that lσ(v) − lη(v) is maximal. Write the convex hull of these vertices as a Minkowski sum of polytopes in SK3, which is the contribution of the first factor, and we want to delete them. The five polytopes are S{1,2}, S{2}, S{3}, S{1,2,3}, S{1}.

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The remaining Minkowski sums

Now the Minkowski sums become

1 S{1} + S{1,2} + S{2,3}, 2 S{2} + S{1,2} + S{2,3}, 3 S{3} + S{1,2} + S{2,3}, 4 S{1} + S{3} + S{1,2,3}, 5 S{1} + S{1} + S{1,2,3}.

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The remaining Minkowski sums

Now the Minkowski sums become

1 S{1} + S{1,2} + S{2,3}, 2 S{2} + S{1,2} + S{2,3}, 3 S{3} + S{1,2} + S{2,3}, 4 S{1} + S{3} + S{1,2,3}, 5 S{1} + S{1} + S{1,2,3}.

Repeat the procedure, we can find the other two linear factors of h1: max(x1, x2 + 3, x3 + 2), max(x1 + 1, x2).

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˜ f2 revisited

Recall SK3 consists of polytopes Conv((1, 0, 0), (0, 1, 0), (0, 0, 1)) and its faces. For convenience denote these polytopes as S{1,2,3}, S{1,2}, S{3}, etc. Note that ∆Newt( ˜

f2) has four

2-dimensional cells: Conv((2, 0, 0), (1, 0, 1), (1, 1, 0)) and other two symmetric ones, plus Conv((1, 0, 1), (1, 1, 0), (0, 1, 1)).

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˜ f2 revisited

Recall SK3 consists of polytopes Conv((1, 0, 0), (0, 1, 0), (0, 0, 1)) and its faces. For convenience denote these polytopes as S{1,2,3}, S{1,2}, S{3}, etc. Note that ∆Newt( ˜

f2) has four

2-dimensional cells: Conv((2, 0, 0), (1, 0, 1), (1, 1, 0)) and other two symmetric ones, plus Conv((1, 0, 1), (1, 1, 0), (0, 1, 1)). Conv((2, 0, 0), (1, 0, 1), (1, 1, 0)) = S{1,2,3} + (1, 0, 0), but Conv((1, 0, 1), (1, 1, 0), (0, 1, 1)) = S{1,2}+S{1,3}+S{2,3}−S{1,2,3}. So this cell is the reason that ˜ f2 is not SK3-factorizable. And if Newt(g) = S{1,2,3}, g may suffice. To find the coefficients of g we need the Legendre transform and the steps are similar to the previous example.

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Another full positive basis S2

This example comes from tropical plane curves of degree 2. Let S2 consists of the following ten polytopes P1, . . . , P10 (and their faces).

1 2 3 4 6 7 8 9 10 5

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Another full positive basis S2

This example comes from tropical plane curves of degree 2. Let S2 consists of the following ten polytopes P1, . . . , P10 (and their faces).

1 2 3 4 6 7 8 9 10 5

H(S2) is the following 14 × 3 matrix: 1 −1 0 0 1 −1 1 −1 1 −1 2 −2 1 −1

0 0 1 −1 1 −1 −1 1 2 −2 1 −1 1 −1 0 0 0 0 0 0 0 0 0 0 0 1 −1

  • .
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Example: ∆Newt(f3)

Let f3(x1, x2, x3) = max(2x1 + 2x2, x1 + 3x2 − 2, x1 + x2 + 2x3 − 3, 3x1 + x3 − 1, x1 + 2x2 + x3 − 4, 4x1 − 3). Then Newt(f3) is Conv((2, 2, 0), (1, 3, 0), (1, 1, 2), (3, 0, 1), (1, 2, 1), (4, 0, 0)) and ∆Newt(f3) consists of three triangles C1, C2, C3: C1 C2 C3

Figure 3: The projection of ∆Newt(f3) onto coordinates x1 and x2

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Computing g3

Using the b-vectors, we can write C1 =P1 − P2 + P3 − P4 + P10 + Conv((1, 0, 1)), C2 = − P1 + 2P4 + P7 + Conv((1, 1, −2)), C3 = − P3 + 2P4 + P9 + Conv((2, 0, −2)). Then Newt(g) should be at least the Minkowski sum P1 + P2 + P3 + P4.

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Computing g3

Using the b-vectors, we can write C1 =P1 − P2 + P3 − P4 + P10 + Conv((1, 0, 1)), C2 = − P1 + 2P4 + P7 + Conv((1, 1, −2)), C3 = − P3 + 2P4 + P9 + Conv((2, 0, −2)). Then Newt(g) should be at least the Minkowski sum P1 + P2 + P3 + P4. In fact, the following polynomial works: g3(x1, x2, x3) = max(2x1, 2x3 − 10/3, x2 + x3 − 2) + max(x1 + x3, 2x3 − 5/3, x2 + x3 − 1/3) + max(2x3, 2x2 − 1, x1 + x3 − 2) + max(2x1, 2x3 − 5, x1 + x2 − 2).

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Computing h3

We take the product f3 ⊙g3 and apply the algorithms again, we get h3(x1, x2, x3) = x1 − 3x3 + 2 max(2x3 − 5 2, x1 + x3, x2 + x3 − 2) + max(2x3 − 8 3, x1 + x3 − 1, 2x2) + max(x2 + x3 − 2, 2x1) + 2 max(2x3, x1 + x3 − 2, x2 + x3 − 1 2) + max(2x3 − 10 3 , x1 + x2 − 1 3, 2x1).

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The subdivision of Newt(h3)

Figure 4: Two ways to decompose ∆h3: by writing h3 as a product of S2-units, or by writing h3 = f3 ⊙ g3.

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Open problems

Conjecture Let E be a finite set of primitive lattice edges in Rn. Then there exists a full positive basis S such that E corresponds to the 1-dimensional polytopes in S.

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Open problems

Conjecture Let E be a finite set of primitive lattice edges in Rn. Then there exists a full positive basis S such that E corresponds to the 1-dimensional polytopes in S. This conjecture implies the following: Conjecture Any tropical polynomial is S-rational for some full positive basis S.

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Some references

[ABD10] Federico Ardila, Carolina Benedetti, and Jeffrey Doker. Matroid polytopes and their volumes. Discrete & Computational Geometry, 43(4):841–854, 2010. [BGK16] Elizabeth Baldwin, Paul Goldberg, and Paul Klemperer. Tropical intersections and equilibrium (day 2 slides). Hausdorff School on Tropical Geometry and Economics, 2016. [GL01] Shuhong Gao and Alan G.B. Lauder. Decomposition of polytopes and

  • polynomials. Discrete & Computational Geometry, 26(1):89–104, 2001.

[Mur03] Kazuo Murota. Discrete convex analysis. SIAM, 2003. [Pos09] Alexander Postnikov. Permutohedra, associahedra, and beyond. International Mathematics Research Notices, 2009(6):1026–1106, 2009. [Stu94] Bernd Sturmfels. On the Newton polytope of the resultant. J. Algebraic Combin., 3(2):207–236, 1994.

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The End Thank you!

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