Note on Ward-Horadam H(x) - binomials' recurrences and related interpretations, II

We deliver here second new $\textit{H(x)}-binomials'$ recurrence formula, were $H(x)-binomials'$ array is appointed by $Ward-Horadam$ sequence of functions which in predominantly considered cases where chosen to be polynomials . Secondly, we supply a review of selected related combinatorial interpretations of generalized binomial coefficients. We then propose also a kind of transfer of interpretation of $p,q-binomial$ coefficients onto $q-binomial$ coefficients interpretations thus bringing us back to $Gy{\"{o}}rgy P\'olya$ and Donald Ervin Knuth relevant investigation decades ago.Comment: 57 pages, 8 figure

Why Delannoy numbers?

This article is not a research paper, but a little note on the history of combinatorics: We present here a tentative short biography of Henri Delannoy, and a survey of his most notable works. This answers to the question raised in the title, as these works are related to lattice paths enumeration, to the so-called Delannoy numbers, and were the first general way to solve Ballot-like problems. These numbers appear in probabilistic game theory, alignments of DNA sequences, tiling problems, temporal representation models, analysis of algorithms and combinatorial structures.Comment: Presented to the conference "Lattice Paths Combinatorics and Discrete Distributions" (Athens, June 5-7, 2002) and to appear in the Journal of Statistical Planning and Inference

On the expected number of perfect matchings in cubic planar graphs

A well-known conjecture by Lov\'asz and Plummer from the 1970s asserted that a bridgeless cubic graph has exponentially many perfect matchings. It was solved in the affirmative by Esperet et al. (Adv. Math. 2011). On the other hand, Chudnovsky and Seymour (Combinatorica 2012) proved the conjecture in the special case of cubic planar graphs. In our work we consider random bridgeless cubic planar graphs with the uniform distribution on graphs with $n$ vertices. Under this model we show that the expected number of perfect matchings in labeled bridgeless cubic planar graphs is asymptotically $c\gamma^n$, where $c>0$ and $\gamma \sim 1.14196$ is an explicit algebraic number. We also compute the expected number of perfect matchings in (non necessarily bridgeless) cubic planar graphs and provide lower bounds for unlabeled graphs. Our starting point is a correspondence between counting perfect matchings in rooted cubic planar maps and the partition function of the Ising model in rooted triangulations.Comment: 19 pages, 4 figure

The Complexity of Simultaneous Geometric Graph Embedding

Given a collection of planar graphs $G_1,\dots,G_k$ on the same set $V$ of $n$ vertices, the simultaneous geometric embedding (with mapping) problem, or simply $k$-SGE, is to find a set $P$ of $n$ points in the plane and a bijection $\phi: V \to P$ such that the induced straight-line drawings of $G_1,\dots,G_k$ under $\phi$ are all plane. This problem is polynomial-time equivalent to weak rectilinear realizability of abstract topological graphs, which Kyn\v{c}l (doi:10.1007/s00454-010-9320-x) proved to be complete for $\exists\mathbb{R}$, the existential theory of the reals. Hence the problem $k$-SGE is polynomial-time equivalent to several other problems in computational geometry, such as recognizing intersection graphs of line segments or finding the rectilinear crossing number of a graph. We give an elementary reduction from the pseudoline stretchability problem to $k$-SGE, with the property that both numbers $k$ and $n$ are linear in the number of pseudolines. This implies not only the $\exists\mathbb{R}$-hardness result, but also a $2^{2^{\Omega (n)}}$ lower bound on the minimum size of a grid on which any such simultaneous embedding can be drawn. This bound is tight. Hence there exists such collections of graphs that can be simultaneously embedded, but every simultaneous drawing requires an exponential number of bits per coordinates. The best value that can be extracted from Kyn\v{c}l's proof is only $2^{2^{\Omega (\sqrt{n})}}$