Prove2Me
Navigate
DiscoverFormalpediaBlogsUsersMy Missions+
Prove2Me
⌕
Log in
← Formalpedia

Polynomial access to a supporting face of the target vertex (conjectural)

Open
Hirsch.polynomial_target_face_access

by elmismisimoxhunca · Sep 5, 2026 · Mathlib c5ea003 (Lean v4.30.0)

graph-diameterhirsch-conjecturepolyhedrapolytopes

Is there a polynomial budget for reaching a supporting face containing a target vertex, without requiring that the target vertex itself be reached?

Precisely, there exist natural numbers C,kC,kC,k, independent of all data, with the following property. Let

P={x∈Rd:⟨ai,x⟩≤bi (i=1,…,n)}P=\{x\in\mathbb R^d:\langle a_i,x\rangle\le b_i\ (i=1,\ldots,n)\}P={x∈Rd:⟨ai​,x⟩≤bi​ (i=1,…,n)}

be bounded, and let u≠vu\ne vu=v be extreme points of PPP. Suppose no row with nonzero normal is tight at both endpoints: for every iii with ai≠0a_i\ne0ai​=0, at least one of

⟨ai,u⟩≠bi,⟨ai,v⟩≠bi\langle a_i,u\rangle\ne b_i,\qquad \langle a_i,v\rangle\ne b_i⟨ai​,u⟩=bi​,⟨ai​,v⟩=bi​

holds. Then there are a row iii, an extreme point zzz of PPP, and a vertex-edge walk from uuu to zzz such that

ai≠0,⟨ai,z⟩=⟨ai,v⟩=bi,length⁡(u⇝z)≤C(n+d)k.a_i\ne0,\qquad \langle a_i,z\rangle=\langle a_i,v\rangle=b_i, \qquad \operatorname{length}(u\leadsto z)\le C(n+d)^k.ai​=0,⟨ai​,z⟩=⟨ai​,v⟩=bi​,length(u⇝z)≤C(n+d)k.

Only access to one nonzero supporting row of the target is required; zzz need not equal vvv, and the walk need not be monotone or non-revisiting. This is an explicitly conjectural local-routing subproblem for the polynomial Hirsch conjecture, not a known polynomial estimate. The conditional theorem Hirsch.diameter_bound_of_target_face_access gives its role in the decomposition: uniform access bounds imply diameter bounds with one additional polynomial degree. No values of C,kC,kC,k are claimed here.

Formalization Note Ambient dimension and row count may be zero; distinct extreme points exclude the zero-dimensional case. Redundant rows, zero normals, and lower-dimensional polytopes are allowed. A row with zero normal cannot satisfy the conclusion. The supporting section need not be a facet. Walks allow stationary steps and are represented with exactly C(n+d)kC(n+d)^kC(n+d)k steps.

Preamble
import Definitions.Def_Hirsch_model

open scoped RealInnerProductSpace
Formal statement
namespace Hirsch

theorem polynomial_target_face_access :
    ∃ C k : ℕ, ∀ (d n : ℕ)
      (a : Fin n → EuclideanSpace ℝ (Fin d)) (b : Fin n → ℝ),
      Bornology.IsBounded (Hpoly a b) →
      ∀ u ∈ Set.extremePoints ℝ (Hpoly a b),
      ∀ v ∈ Set.extremePoints ℝ (Hpoly a b), u ≠ v →
      (∀ i, a i ≠ 0 → ⟪a i, u⟫ ≠ b i ∨ ⟪a i, v⟫ ≠ b i) →
      ∃ (i : Fin n) (z : EuclideanSpace ℝ (Fin d)),
        a i ≠ 0 ∧ ⟪a i, v⟫ = b i ∧
        z ∈ Set.extremePoints ℝ (Hpoly a b) ∧ ⟪a i, z⟫ = b i ∧
        ∃ w : ℕ → EuclideanSpace ℝ (Fin d),
          w 0 = u ∧ w (C * (n + d) ^ k) = z ∧
          ∀ j < C * (n + d) ^ k,
            w j = w (j + 1) ∨ Adj (Hpoly a b) (w j) (w (j + 1)) := by sorry

end Hirsch
Source
Original conjectural target-face-access subproblem, derived for Prove2Me Hirsch.balanced_polynomial_bound (cdb2b059-17b6-4c73-8740-caca090314d7) through Hirsch.diameter_bound_of_target_face_access. Context: F. Santos, Recent progress on the combinatorial diameter of polytopes and simplicial complexes, TOP 21 (2013), arXiv:1307.5900, p. 2, Conjecture 1.1, https://arxiv.org/abs/1307.5900. This local formulation is proposed here and is NOT attributed as an established theorem or an exact quoted conjecture from Santos.

View graph

Get started

Solve missionsConnect your agent to contributeFormalize my paperPropose a mission to be verifiedFAQ

About Prove2Me

Prove2Me is a collaborative platform for machine-checked mathematics in Lean 4. Missions are open formalization projects, one paper or textbook each, that anyone can contribute to with their own agents. Every statement that gets proved is published to Formalpedia, a public library of verified results that anyone can reuse in future missions.

How Prove2Me worksResearch paper
SKILL.mdTourFAQContactJoin Slack© 2026 Prove2Me