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ЧЕБЫШЕВСКИЙ СБОРНИК
Том 20. Выпуск 2.
УДК 511.3 DOI 10.22405/2226-8383-2019-20-2-169-177
Гиперметрический конус и многогранник на графах
М. Дьютор
Дютур-Сикирич Мэтью — старший научный сотрудник Института Рудьер Боскович Бие-
ница, Загреб, Хорватия
e-mail: mathieu. dutour@gmail. com
Аннотация
Гиперметрический конус был определен в [9] и был широко изучен Мишелем Дезой и его сотрудниками. Еще одним ключевым его интересом был отрезной и метрический многогранник, который он рассматривал в своих последних работах в случае графов.
Здесь мы объединяем оба интереса, рассматривая гиперметрию на графах. Мы определяем их для любого графа и даем алгоритм вычисления экстремальных лучей и граней гиперметрического конуса на графах. Мы вычисляем гиперметрический конус для первого нетривиального случая К7 — {е}. Мы также вычисляем гиперметрический конус в случае графов без К5 минора.
Ключевые слова: алгебраические решётки, алгебраические сетки, тригонометрические суммы алгебраических сеток с весами, весовые функции.
Библиография: 21 название. Для цитирования:
М. Dutour. The hvpermetric cone and polvtope on graphs // Чебышевский сборник, 2019, т. 20, вып. 2, с. 169-177.
CHEBYSHEVSKII SBORNIK Vol. 20. No. 2.
UDC 511.3 DOI 10.22405/2226-8383-2019-20-2-169-177
The hypermetric cone and polytope on graphs
M. Dutour
M. Dutour — Senior Associate Researcher, Institut Rudjer Boskovic Bijenicka, Zagreb, Croatia e-mail: mathieu. dutour@gmail. com
Abstract
The hypermetric cone was defined in [9] and was extensively studied by Michel Deza and his collaborators. Another key interest of him was cut and metric polytope which he considered in his last works in the case of graphs.
Here we combine both interest by considering the hypermetric on graphs. We define them for any graph and give an algorithm for computing the extreme rays and facets of hypermetric cone on graphs. We compute the hypermetric cone for the first non-trivial case of K7 — {e}. We also compute the hypermetric cone in the case of graphs with no K5 minor.
Keywords: algebraic lattices, algebraic net, trigonometric sums of algebraic net with weights, weight functions.
Bibliography: 21 titles.
For citation:
M. Dutour, 2019, "The hvpermetric cone and polvtope on graphs Chebyshevskii sbornik, vol. 20, no. 2, pp. 169-177.
1. Introduction
Given an integer n and a vector b £ Zn such that h = 1 + 2s with s £ Z the hypermetric inequality is defined as
H(b,d)= ^ bzb3d(i, j) < s(s + 1)
1 <i<j<n
The hypermetric cone HYP(Kn) is the set of functions d : {1,..., n}2 ^ R, such that H(b, d) < 0 is satisfied for all be Zn with bi = 1- The elements of HYP(Kn) are named hypermetric.
A priori the hypermetric cone is not polyhedral since it is defined by an infinity of inequalities. However, the polvhedralitv of the hypermetric cone was proved in [19, 9, 10]. The hypermetric cone is interpreted in term of parameter space of Delaunav polvtopes and this viewpoint was introduced first in [3]. A complete description of the facets of the hypermetric cone was achieved in [4] for n = 6, [5, 20, 7] for n = 7 and [8].
Another viewpoint for the parameter space of Delaunav polvtopes is the Erdahl cone. It is the set of quadratic functions on Rn such that f(x) > 0 for x £ Zn. This viewpoint is used and developed in [18, 17, 14].
The hypermetric polytope HYPP(Kn) is the set of functions d : {1,...,n}2 ^ R, such that H (b, d) < s(s + 1) is satisfied for all be Zn with h = 1 + 2s an d s £ Z. It was defined in [8, 15] and it is related to centrally symmetric Delaunav polvtopes (see 2.2 for some summary).
In this work we define the hypermetric cone HYP(G) and polytope HYPP(G) of a graph G. This extends the construction of cut and metric cone and polvtopes of graphs (see [19, 11]). We provide algorithms for checking if an hypermetric belong to those.
G
We also define the hypermetric cone and polvtopes for the complete graph Kn.
G
algorithms to test if a distance function on the edge set of a graph is an hypermetric or not.
In Section 4 we compute the facets and extreme rays of the first non-trivial case K7 — {e}. We also prove that for graphs with k edges removed, the facet defining inequalities are obtained as sum of at most k hypermetric inequalities. We also characterize the facets of the hypermetric cone of graphs without K5 minor.
Characterizing the facet inequalities of other graphs is an interesting problem. In particular one good question is characterize the graphs G for which MET(G) = HYP(G) or HYP(G) = CUT(G).
2. Preliminary definitions
2.1. Cut and metric cones and poly topes
Given a graph G = (V,E) with n = IVfor a vertex subset S C V = {1,...,n}, the cut semimetric 5s(G) is a vector (actually, a symmetric {0,1}-matrix) defined as
Ss(x v) = i 1 lí{x,y}eE and |S n{x,y}I = 1
s( x, ) = 0
If G is connected, which will be the case in our work, there are exactly 2n 1 distinct cut semimetrics. The cut polytope CUTP(G) and the cut cone CUT(G) are defined as the convex hull of all such
semimetrics and the positive span of all non-zero ones among them, respectively. Their number of vertices, respectively extreme rays is 2n-1, respectively 2n-1 — 1 and their dimension is i.e., the number of edges of G.
The metric cone MET(Kn) is the set of all semimetrics on n points, i.e., the functions d : |1,...,n}2 ^ R^o satisfyi ng d(i,i) = 0 d(i,j) = d(j, i) for 1 < i,j < n and the triangle inequalities
d(i,j) < d(i, k) + d(j, k) for 1 < i,j,k < n. (1)
The metric polytope METP(Kn) is defined as the elements of MET(Kra) satisfying the perimeter inequalities
d(i, j) + d(j, k) + d(k, i) < < i, j, k < n. (2)
For a graph G = (V, E) of order IV| = n, let MET(G) and METP(G) denote the projections of MET(Kra) and METP(Kra), respectively, on the subspace RE indexed by the edge set of G. Clearly, CUT(G) and CUTP(G) are the projections of, respectively, CUT(Kra) and CUTP(Kra) on RE. We have the relaxation property:
CUT(G) C MET(G) and CUTP(G) C METP(G).
Definition 9. Let G = (V, E) be a graph.
(i) Given an edge e £ E, the edge inequality is
x(e) > 0.
(ii) For a cycle C, and an odd size set F C C the cycle inequality is
x(F) — x(C\F) <IF| — 1
where x(U) = ueu x(u)-
METP(G) is defined by all edge, bounding inequality x(e) < 1 and s-cvcle inequalities, while MET(G) is defined by all edge inequalities and s-cvcle inequalities with |F| = 1 (see [2] and [12, Section 27.31).
2.2. Hypermetric cone and Delaunay polytope
By a distance matrix D = )0<i,j<n we mean a matrix with Diti = ^d Dij = Djti. We define eo = 0 ei = (1,0,..., 0^ '., en = (0,..., 0,1).
We can associate to this an x n symmetric matrix Q, a vector v £ R^, a sealar c £ R and a function / defined on Rra by
f (x) = Q[x] + {v,x) + c
and which satisfies f (e0) = f (e1) = ■ ■ ■ = f (en) = 0. The matrix D satisfies Dij = Q[ei — ej].
This correspondence relates the hvpermetric cone HYP(Kra) with the Erdahl cone. See for example [14]. In other words we have that D £ HYP(Kra+1) if and onlv if f (x) > 0 to all x £ Zn.
If we express the problem purely in term of geometry of numbers what we have is that a distance matrix D £ HYP(Kn+1) if and only there exist a fc-dimensional lattice L with k < n, a Delaunay polvtopes P of L and Dij = \\vi — Vj ||2 for some vert ices Vi oi P (see [12, Chapter 2]).
Given a D £ HYP(Kn+1) which correspond to a Delaunay polytope P of dimension n. Then the set of vertices corresponds to the set of vector b £ Zn+1 with H(b, D) = 0. In this respect the set v0, v1, ..., vn form an affine basis. The set of vertices of P is then expressed as Y^i=0 b^i for H(b, D) = 0. This can be used to describe the Delaunay polvtopes and this method was used in [13] for describing the Delaunay polvtopes of dimension six.
Unfortunately, while powerful the method of hypermetric does not work out completely because there are Delaunay polvtopes which are not basic (see [16]). We found out that the Erdahl cone provides a better replacement in many contexts (see [18, 17, 14]).
The symmetries of the hvpermetric cone HYP(Kra) is Sym(n) for n = 4 (see [6]).
2.3. Hypermetric polytope and centrally symmetric Delaunay polytope
We cite following [8, Theorem 6]:
Theorem 1. A distance function d belongs to HYPP(Kn) if and only if there exist a centrally symmetric n-dimensional Delaunay polytope of center c, circumradius 1 and vertices Vi, 2c — Vi for 1 < i < n with ||Vi — Vj||2 = 4dij.
The correspondence can be made more precise. Consider the lattice
l® e Zn+1 s.t. ^
L ={x £ Zn+1 s.t. > vxi = 0 (mod 2) and the point b0 = (1, 0n_ 1). We define the matrix
A(d) = (Aij)i<i,j<n with Aij = I
1 if i = j
lj> ^ 1 1 — 2dj otherwise.
Then we have that d £ HYPP(Kn) if and only if we have (see [8]):
bfA(d)b > 1 for al 1 be b0 + L. (3)
The set of b e b0 + L such that blA(d)b = 1 correspond to the vertices of the centrally symmetric Delaunay polytope of center 60 for the lattice L. The points ±e» for 1 < i <n will always be among those.
It is important to point out that it is unlikely that all centrally symmetric Delaunay polvtopes could be expressed in this way because of the negative result [16] but we do not know any counterexample. However, we could construct a variant of the Erdahl cone for this centrally symmetric setting
ErdahlCent(n) = {Q e Sn such that Q[x — e1/2] > 1}
with Sn the set of n x n quadratic forms. The center of the Delaunay polvtope will be e1 /2 and the circumradius 1. The vertices of the Delaunay polytope will be the set of x e Zn such that Q[ x— 1/2] = 1
3. Definition of hypermetric cone on graphs
The hypermetric cone HYP(Kn) is defined as the set of metrics satisfying all hypermetric inequalities, that is
HYP(Kn) = i d e with H (b, d) < 0 for be Zn, V h = 1
Definition 10. Given a graph G on n vertices the hypermetric cone HYP(G) is defined as the projection of HYP(Kn) on RE(G).
Since we know that HYP(Kn) = CUT(Kn) for n < 6 we have HYP(G) = CUT(G) for G on
graph on at most 6 vertices.
Another elementary property is that HYP( G) is polyhedral since it is the projection of a polyhedral cone.
Theorem 2. Given a graph G on n vertices and ad e HYP(G). The set of possible distances D e HYP(Kn) such that proj(D) = d is bounded if and only G is a connected graph.
Proof. Given ad £ HYP(G) with G being connected. Given two vertices v and w there exist a path v = v0, v1, ..., vm = w with Vi adjacent to Vj. By iterating the triangle inequality one obtains
D(v, w) < d(v0, V1) +-----+ d(vm-1,vm) = C(d)
The value C(d) does not depend on D and so the set of possible D is connected.
If d £ HYP(G) then there exist at least one D with Proj(D) = d. Since G is not connected, there exist a subset S c {1,..., n} with vertices in S and {1,... ,n} — S not being adjacent. As a consequence we have Proj (5 s) = 0. Sinee 5 s £ HYP(Kn) we have that for all a > 0 the relation Proj(D + a5s) = d with D + a5s £ HYP(Kra). ■
Theorem 3. Given a graph G on n vertices, testing if a given d £ RE(G) belongs to HYP(G) can be done by iteratively solving linear programs.
Proof. We take a distance function d £ RE(G) and we want to find a matrix D with Proj(D) = d. Thus we need to find the possible values Dij with {i,j} £ E(G). We need to solve following linear program:
minimize E{i,j}/E(G) Did
satisfying to H(b, D) < 0
for b £ Zn with bi = 1
In other words what we have is an infinite linear program. WThat we can solve only is finite linear system.
The algorithm for solving that is to start from a finite set S of vectors b and then gradually expand it until a conclusion is reached. The finite starting point is the triangular inequalities of the metric cone. Then we iterate:
(i) We solve the program for a fixed set S.
(ii) If the program is unfeasible then this means that the elements d does not belong to HYP(Kra). The problem is resolved.
(in) If the optimal solution D0 of the linear program be longs to HYP(Kra) then the problem is resolved.
(iv) On the other hand if D0 is not an hvpermetric then there exist a d such that H(b, D0) > 0. We add b'rnD and reiterate.
Since the hvpermetric cone is polyhedral, after a finite set of addition one will eventually obtain a solution of the problem. ■
Theorem 4. If f (x) is a linear function defined on
rE(G)
then we can check whether f is valid
on HYP(G) by a sequence of linear program.
Proof. Since f is defined on RE(G) we can trivially extend it to by setting f (e) = 0 for all
e £ E(G).
The idea is to consider the linear program
minimize f (D)
satisfying to H(b, D) < 0
for b £ Zn with bi = 1 and Y,1<i,j<n Did < 1
This infinite linear programming is very similar to the one of Theorem 3 and the same iterative strategy can be used. Let's denote D0 the optimal solution which is an hvpermetric. If f(D0) < 0 then we have proved that f is not valid on HYP(G). If on the other hand f(D0) > 0 then the inequality is valid. ■
In practice the implementation of the above algorithms can be fairly complex. The linear programs are large and hard to solve. In our implementation we use cdd which uses exact arithmetic and provides both primal and dual solution in exact rational arithmetic. However, cdd uses the simplex algorithm and is very small in some cases. The idea is then to use floating point arithmetic and the glpk program which has better algorithm and can approximately solve linear programs. From the approximate solution we can derive the incidence and from the incidence get an exact solution in most cases. If this approach fails, then we fall back to the more expensive in time cdd. We only accept an approximate solution if we can derive a primal and dual solution. In any case of failure we fall back to cdd.
If we have a distance matrix D checking if it belongs to HYP(Kn) is done in the following way. This defines an x n-rational quadratic form Q, a vecto r v e Qn and a scalar C such that
f(x) =Q[x] + (v ,x) +C
with /(0) = f(e 1) = ■ ■ ■ = /(en) = 0. What we need is check if there is a x e Zn such that ( x) < 0 Q
eigenvector. By approximating the eigenvector with a rational vector and multiplying with some factor we can find ax e Zn with f(x) < 0. If Q is positive semidefinite then take the kernel K = {x e Zn with Q[x] = 0} and L a subspace of Zn such that K ®Z L = Zn. By restricting the L
problem of finding x e Zn with f(x) < 0 is a Closest Vector Problem and we can solve it bv using a [19].
The interest of above two algorithms is that they give an algorithm for computing HYP( G). We can start with a list of hypermetric of full dimension in HYP( G). This is not difficult to obtain: We can for example take the cuts and project them on RE(G). Then we iterate the following:
(i) We compute the facets of the convex body defined by those hvpermetrics.
(ii) For each facet we check if the corresponding facet defining inequality f (x) is also valid on HYP(G).
(iii) If all inequalities are also valid on HYP( G) then we have computed the list of extreme rays and facets of HYP(G).
If not then we insert the hvpermetrics that were found to be counterexample to the initial list of hvpermetrics and reiterate.
Each insertion will increase the hypermetric cone until one has the complete description of HYP(G).
We haven't implemented the algorithms of this section and it would be hard to do so. The main measure of the complexity should be the number of edges of the graph because it is a direct measure of the complexity of the problem. A way to speed up the process is to use the symmetries of the graph for the computation. Based on that, the first interesting case would likely be HYP(P etersen).
All of the above is for hypermetric cone. But we can just as well define the hypermetric polytope for a graph: Take HYPP(G) be the projection of HYPP(Kn) on The algorithms can be
adapted just as well to this case.
4. Computing HYP(G) for some graphs
Theorem 5. The hypermetric cone HYP(K7 — {e}) has 8782 extreme rays and 7210 facets.
Proof. The hvpermetric cone HYP(K7) is defined bv 3773 facet inequalities in 14 orbits and it has 37170 extreme rays in 29 orbits (see [7]). Let us denote e = {1,2}. We consider the projection obtained by eliminating the component de. The normal to the equation de = 0 is the distance function determined by d?j = 1 if {i,j} = e and 0 otherwise.
The facets Fi of the cone HYP(K7) are defined by inequalities fi(d) > 0. The facets of HYP(K7 — {e}) are obtained in two ways:
(i) The facets fi of HYP(K7) with fi(de) = 0. The corresponding facet defining inequality of HYP(^7 — {e}) is fi(x) > 0.
(ii) The ridges of HYP(K7) obtained as intersection Fi n Fj of two facets with fi(de) x fj(de) < 0. We can find a > 0 and ft > 0 with (afi + ftfj)(de) = 0. The corresponding facet defining inequality of HYP(^ — {e}) is afi(x) + ftfj(x) > 0.
Bv using this result and computing the ridges of HYP(K7) we can get the facets of the projection in The symmetries of HYP(K7 — {e}) are induced by the symmetries of the graph K7 — {e} and the group is Sym({1, 2}) x Sym({3, 4, 5, 6, 7}) of order 240. The total number of orbits of facets is thus 7210 in 80 orbits.
The extreme rays of HYP(K7 — {e})) are the projection of extreme rays of HYP(K7). We check
8782
in 73 orbits. ■
Theorem 6. If G is a complete graph with k edges removed then the facets of HYP(G) are determined as a sum aiH(b%, d) for m < k, a.i > 0 and b% £ Zn with = 1-
proof. The method of 5 can be generated to any n. It implies that the facets of HYP(Kn — {e}) are induced by hypermetric inequalities and sum with positive coefficient of two hypermetric inequalities. The method extends to any number of edges and give us that the facets of HYP(Kn — {e1,..., ek}) are formed by hvpermetric inequalities and sums with positive coefficients of at most k hvpermetric inequalities. ■
Above theorem is in a sense a relatively negative result. For graphs with few edges it gives the facets as sums of too many hypermetric inequality to be practical.
Theorem 7. If G is a graph without K5 minor then the facets of HYP(G) and HYPP(G)) are induced by the cycle inequalities and the non-negative inequalities.
Proof. By the result of [21, 1] we have that MET(G) = CUT(G) and METP(G) = CUTP(G) if G has no K5 minor. Since we have
CUT(Kra) c HYP(Kra) c MET(Kra) and CUTP(Kra) c HYPP(Kra) c METP(Kn).
As a consequence we get that the facets of MET(G) = HYP(G) and METP(G) = HYPP(G). Thus the facets of MET(G) and METP(G) are induced by the cycle and facet inequalities. ■
Seymour's theorem [21] is even stronger and states that CUT(G) = MET(G) occurs if and only if G has no K5 minor. Could it be that MET(G) = HYP(G) occurs for other graphs that have K5 as minor?
Also interesting would be to characterize cases where HYP(G) = CUT(G). If G has at most 6 vertices then equality holds. The example of K7 — {e} shows that we cannot characterize the equality with K6 minor.
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