Subscribe to the gold package and get unlimited access to Shamra Academy
Register a new userThe Martin Gardner Polytopes
147
0
0.0
(
0
)
Ask ChatGPT about the research
No Arabic abstract
In the chapter Magic with a Matrix in emph{Hexaflexagons and Other Mathematical
rate research
Read More
A two-step model for generating random polytopes is considered. For parameters $d$, $m$, and $p$, the first step is to generate a simple polytope $P$ whose facets are given by $m$ uniform random hyperplanes tangent to the unit sphere in $mathbb{R}^d$, and the second step is to sample each vertex of $P$ independently with probability $p$ and let $Q$ be the convex hull of the sampled vertices. We establish results on how well $Q$ approximates the unit sphere in terms of $m$ and $p$ as well as asymptotics on the combinatorial complexity of $Q$ for certain regimes of $p$.
Neighborly cubical polytopes exist: for any $nge dge 2r+2$, there is a cubical convex d-polytope $C^n_d$ whose $r$-skeleton is combinatorially equivalent to that of the $n$-dimensional cube. This solves a problem of Babson, Billera & Chan. Kalai conjectured that the boundary $partial C^n_d$ of a neighborly cubical polytope $C^n_d$ maximizes the $f$-vector among all cubical $(d-1)$-spheres with $2^n$ vertices. While we show that this is true for polytopal spheres for $nle d+1$, we also give a counter-example for $d=4$ and $n=6$. Further, the existence of neighborly cubical polytopes shows that the graph of the $n$-dimensional cube, where $nge5$, is ``dimensionally ambiguous in the sense of Grunbaum. We also show that the graph of the 5-cube is ``strongly 4-ambiguous. In the special case $d=4$, neighborly cubical polytopes have $f_3=f_0/4 log_2 f_0/4$ vertices, so the facet-vertex ratio $f_3/f_0$ is not bounded; this solves a problem of Kalai, Perles and Stanley studied by Jockusch.
In Martin Gardners October, 1976 Mathematical Games column in Scientific American, he posed the following problem: What is the smallest number of [queens] you can put on a board of side n such that no [queen] can be added without creating three in a row, a column, or a diagonal? We use the Combinatorial Nullstellensatz to prove that this number is at least n, except in the case when n is congruent to 3 modulo 4, in which case one less may suffice. A second, more elementary proof is also offered in the case that n is even.
A neighborliness property of marginal polytopes of hierarchical models, depending on the cardinality of the smallest non-face of the underlying simplicial complex, is shown. The case of binary variables is studied explicitly, then the general case is reduced to the binary case. A Markov basis for binary hierarchical models whose simplicial complexes is the complement of an interval is given.
Considering $ntimes ntimes n$ stochastic tensors $(a_{ijk})$ (i.e., nonnegative hypermatrices in which every sum over one index $i$, $j$, or $k$, is 1), we study the polytope ($Omega_{n}$) of all these tensors, the convex set ($L_n$) of all tensors in $Omega_{n}$ with some positive diagonals, and the polytope ($Delta_n$) generated by the permutation tensors. We show that $L_n$ is almost the same as $Omega_{n}$ except for some boundary points. We also present an upper bound for the number of vertices of $Omega_{n}$.
Log in to be able to interact and post comments
comments
Fetching comments
Sign in to be able to follow your search criteria
