An edge of a relaxation either is an edge of the polytope or exits it at a neighbouring vertex
ProvedHirsch.relaxation_exit_vertexhirsch-conjecturelinear-programmingpolytopes
Let , let be a set of row indices, and let be a set on which the inequalities indexed by hold. Let and let be an edge of (that is, : and the segment is an extreme subset of ). Then:
- if , the segment is an edge of : ;
- if , there is a point with or , at which some inequality with index outside is tight:
In words: walking from along an edge of the relaxation, one either stays inside along an edge of , or leaves through one of the dropped inequalities, and the exit point is itself or a neighbour of in . This is the geometric content of the Kalai--Kleitman sentence "a shorter path in could not be a path in and thus must meet a facet not in "; no simplicity or general-position hypothesis is used.
Formalization Note The exit point is the last point of inside ; is an extreme subset of because is extreme in .
Preamble
import Mathlib import Definitions.Def_Hirsch_model open scoped RealInnerProductSpace
Formal statement
namespace Hirsch
theorem relaxation_exit_vertex (d n : ℕ)
(a : Fin n → EuclideanSpace ℝ (Fin d)) (b : Fin n → ℝ)
(T : Finset (Fin n)) (Q : Set (EuclideanSpace ℝ (Fin d)))
(hPQ : Hpoly a b ⊆ Q) (hQ : ∀ z ∈ Q, ∀ j ∈ T, ⟪a j, z⟫ ≤ b j)
(x y : EuclideanSpace ℝ (Fin d)) (hx : x ∈ Hpoly a b) (hadj : Adj Q x y) :
(y ∈ Hpoly a b → Adj (Hpoly a b) x y) ∧
(y ∉ Hpoly a b → ∃ z ∈ segment ℝ x y, (z = x ∨ Adj (Hpoly a b) x z) ∧
∃ j, j ∉ T ∧ ⟪a j, z⟫ = b j) := by sorry
end HirschSource
G. Kalai, D. J. Kleitman, A quasi-polynomial bound for the diameter of graphs of polyhedra, Bull. AMS 26 (1992) 315-316, proof of the Lemma (the relaxed polyhedron cut out by the touched facets has no shorter paths), https://arxiv.org/abs/math/9204233; M. J. Todd, arXiv:1402.3579, Lemma 1. Prove2Me analogue for the full relaxation: Hirsch.relaxation_no_new_neighbours (ccfa0184-461a-49d1-b15d-1c6668e2d911).