Tight Upper Bound on the Number of Vertices of Polyhedra with $0,1$-Constraint Matrices

polyhedron $P(A,b)=\{x\in \mathbb{R}^d~:~Ax\leq b\}$ when the
$m\times d$ constraint matrix $A$ is subjected to certain
restriction. For instance, if $A$ is a 0/1-matrix, then there can be
at most $d!$ vertices and this bound is tight, or if the entries of
$A$ are non-negative integers so that each row sums to at most $C$,
then there can be at most $C^d$ vertices. These bounds are
consequences of a more general theorem that the number of vertices
of $P(A,b)$ is at most $d!\cdot W/D$, where $W$ is the volume of the
convex hull of the zero vector and the row vectors of $A$, and $D$
is the smallest absolute value of any non-zero $d\times d$
subdeterminant of $A$.


Joint wotk with Khaled Elbassioni and Raimund Seidel