On the Face Lattice of the Metric Polytope

A counterexample to the dominating set conjecture

“…The remaining 738 orbits providing counterexamples to the dominating set conjecture are made of vertices satisfying with equality at least 38 inequalities. Some of these counterexamples have a relatively large incidence and adjacency: For example, the vertex 1 9 (1, 2, 3, 3, 4, 4, 4, 6, 3, 4, 4, 3, 3, 5, 7, 5, 5, 6, 6, 2, 4, 6, 3, 3, 3, 3, 3, 3, 3, 3, 6, 6, 4, 6, 4, 2) satisfies with equality 44 inequalities and has 84 fractional adjacent vertices, and the vertex 1 12 (3,3,3,6,7,7,7,7,6,6,3,4,4,4,4,6,3,4,4,4,4,9,4,4,8,8,7,7,7,7,8,4,4,4,4,8) satisfies with equality 43 inequalities and has 202 fractional adjacent vertices. See [2] for a complete list of the known counterexamples.…”

“…The vertex given in Proposition 1 is adjacent to the following 37 fractional vertices of met 9 : (2, 2, 2, 3, 3, 3, 5, 5, 4, 2, 5, 5, 5, 3, 3, 4, 5, 1, 3, 3, 3, 5, 3, 3, 5, 5, 4, 2, 2, 4, 4, 2, 2, 4, 2, 2) 1 9 (1, 2, 4, 2, 5, 5, 4, 4, 3, 3, 3, 6, 6, 3, 3, 6, 4, 3, 3, 2, 6, 6, 3, 3, 6, 6, 5, 3, 2, 6, 6, 3, 3, 5, 3, 4) 1 9 (2, 2, 3, 3, 4, 4, 5, 3, 4, 3, 5, 6, 6, 3, 3, 5, 5, 2, 4, 3, 5, 6, 3, 3, 6, 6, 5, 3, 2, 6, 6, 3, 3, 5, 3, 4) 1 9 (2, 2, 3, 3, 4, 4, 5, 5, 4, 3, 5, 6, 6, 3, 3, 5, 5, 2, 4, 3, 3, 6, 3, 3, 6, 6, 5, 3, 2, 6, 6, 3, 3, 5, 3, 4) 1 9 (2, 2, 3, 3, 4, 4, 5, 5, 4, 3, 5, 6, 6, 3, 3, 5, 5, 2, 4, 3, 5, 6, 3, 3, 6, 6, 5, 3, 2, 6, 6, 3, 3, 5, 3, 6) 1 9 (2, 2, 3, 3, 4, 4, 5, 5, 4, 3, 5, 6, 6, 3, 3, 5, 5, 2, 4, 3, 5, 6, 3, 3, 6, 6, 3, 3, 2, 6, 6, 3, 3, 5, 3, 4) 1 9 (2, 2, 3, 3, 4, 4, 5, 5, 4, 3, 5, 6, 6, 3, 3, 5, 5, 2, 4, 3, 5, 6, 3, 3, 6, 6, 5, 3, 2, 6, 6, 3, 3, 3, 3, 4) 1 9 (2, 2, 3, 3, 4, 4, 5, 5, 4, 3, 5, 6, 6, 3, 3, 5, 5, 2, 6, 3, 5, 6, 3, 3, 6, 6, 5, 3, 2, 6, 6, 3, 3, 5, 3, 4) 1 9 (2, 2, 3, 3, 4, 6, 5, 5, 4, 3, 5, 6, 6, 3, 3, 5, 5, 2, 4, 3, 5, 6, 3, 3, 6, 6, 5, 3, 2, 6, 6, 3, 3, 5, 3, 4) 1 9 (3, 2, 2, 4, 3, 3, 6, 6, 5, 3, 5, 6, 6, 3, 3, 4, 6, 1, 5, 4, 4, 6, 3, 3, 6, 6, 5, 3, 2, 6, 6, 3, 3, 5, 3, 4) 1 10 (2, 2, 4, 3, 5, 4, 5, 5, 4, 4, 5, 7, 6, 3, 3, 6, 5, 3, 4, 3, 7, 7, 3, 4, 7, 7, 6, 3, 2, 6, 7, 4, 4, 5, 3, 4) 1 10 (2, 2, 4, 3, 5, 5, 5, 6, 4, 4, 5, 7, 7, 3, 4, 6, 5, 3, 3, 3, 6, 7, 3, 3, 7, 6, 6, 4, 2, 7, 6, 4, 3, 6, 3, 5) 1 10 (3, 3, 2, 4, 4, 3, 7, 7, 6, 3, 7, 7, 6, 4, 4, 5, 7, 1, 6, 4, 4, 6, 4, 3, 7, 7, 6, 3, 3, 7, 7, 3, 3, 6, 4, 4) 1 12 (3,3,3,5,5,5,7,7,6,4,8,8,8,4,4,6,8,2,6,4,6,8,4,4,8,8,8,4,4,8,8,4,4,…”

“…The conjecture was further substantiated by the computation of met 7 and met 8 . The 275,840 vertices of met 7 and the 1,550,825,600 vertices of met 8 are adjacent to a cut, see [3][4][5].…”

Symmetry in Integer Linear Programming

“…These algorithms are based on clever enumeration procedures using the symmetry group of the polyhedron and can be carried out for problems of small dimension. For example, a complete linear description is known for the TSP polytope on a complete graph with up to 10 nodes [27], the Linear Ordering polytope with up to 8 items [27], the Cut polytope on a complete undirected graph with up to 9 nodes [27], the Metric cone and Metric polytope on a complete graph with up to 8 nodes [35,36].…”

The hypermetric cone and polytope on eight vertices and some generalizations

“…Also, the number of vertices and facets in the hypermetric polytope HY P P n , see Section 4, is given for 3 ≤ n ≤ 8, with number of orbits under Sym(n) and 2 n−1 switchings C n = 3 n = 4 n = 5 n = 6 n = 7 n = 8 CU T n , e 3(1) 7(2) 15 (2) For CU T 8 and CU T P 8 , sets of facets were found in [7]; completeness of these sets was shown in [14]. The enumeration of orbits of extreme rays of M ET n for n ≤ 8 was done in [24,9,10]. In Section 2 the facets of the hypermetric cone HY P 8 are determined with the help of the connection with geometry of numbers and the list of simplices of dimension 7.…”