Opposite vertices of base polyhedra

Definitions and mathematical statement

Let $V=\{1,\ldots,n\}$ be finite and let $f:2^V\to\mathbb R$ be normalized and submodular: \[ f(\varnothing)=0,\qquad f(X)+f(Y)\ge f(X\cap Y)+f(X\cup Y)\quad(X,Y\subseteq V). \] For $x\in\mathbb R^V$, write $x(X)=\sum_{i\in X}x_i$. The associated base polyhedron is \[ B(f)=\{x:x(X)\le f(X)\ (X\subseteq V),\quad x(V)=f(V)\}. \] It is a bounded polyhedron in this finite setting. A vertex is an extreme point, one that cannot be expressed as a nontrivial convex combination of two other points of the polyhedron.

Suppose $0\in B(f)$ and every vertex of $B(f)$ belongs to $\{-1,0,1\}^V$. Must there be a vertex $v$ such that $-v$ is also a vertex?

The hypothesis $0\in B(f)$ in particular forces $f(V)=0$ and $f(X)\ge0$ for every $X$. Central symmetry of the entire polyhedron is not assumed or requested; a single opposite pair is sufficient. The degenerate polyhedron $\{0\}$ satisfies the conclusion with $v=0$. Merely expressing the origin as a convex combination of vertices is weaker: the question asks for an exact pair of additive opposites among the vertices themselves.

Short English statement

Must a base polyhedron containing the origin and having only $-1,0,1$ vertices contain a pair of opposite vertices?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3000058 (AMR-029-0058), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Egres Open - List of open problems, Open-problem page: Opposite vertices of base polyhedra. Original source indexed by AMR-029-0058. https://oldlemon.cs.elte.hu/egres/open/List_of_open_problems.
• S3 EGRES Open, Opposite vertices of base polyhedra. Individual source page and explanatory remarks. https://lemon.cs.elte.hu/egres/open/Opposite_vertices_of_base_polyhedra.