SearcharxivSearch

arXiv · 0904.3431

Algorithmic proof of Barnette's Conjecture

Abstract

In this paper we have given an algorithmic proof of an long standing Barnette's conjecture (1969) that every 3-connected bipartite cubic planar graph is hamiltonian. Our method is quite different than the known approaches and it rely on the operation of opening disjoint chambers, bu using spiral-chain like movement of the outer-cycle elastic-sticky edges of the cubic planar graph. In fact we have shown that in hamiltonicity of Barnette graph a single-chamber or double-chamber with a bridge face is enough to transform the problem into finding specific hamiltonian path in the cubic bipartite graph reduced. In the last part of the paper we have demonstrated that, if the given cubic planar graph is non-hamiltonian then the algorithm which constructs spiral-chain (or double-spiral chain) like chamber shows that except one vertex there exists (n-1)-vertex cycle.

Explore related subjects

Keep this discovery

BibTeXRIS

I. Cahit. 2009-04-22. Algorithmic proof of Barnette's Conjecture. https://arxiv.org/abs/0904.3431

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

Average Chord Lengths in a Triangle

Let $P$ be a point inside a triangle $T$. We consider the average length of the chords of $T$ through $P$, where the direction of the chord is chosen uniformly. An elementary formula is obtained in terms of the distances from $P$ to the sides and vertices of the triangle. Several classical triangle centers give especially simple specializations. For example, if $I$ is the incenter, then \[ M_T(I)=\frac{2r}{\pi} \log\left(\cot\frac A4\cot\frac B4\cot\frac C4\right). \] Our main result is the sharp inequality \[ M_T(P)\le \frac{p}{\pi\sqrt3}\log(2+\sqrt3), \] valid simultaneously for every triangle of perimeter $p$ and every interior point $P$. Thus, among all such pairs $(T,P)$, the largest possible average chord length occurs only when $T$ is equilateral and $P$ is its center. The proof is an elementary symmetrization argument. We close with brief remarks relating the problem to the radial center of a convex body, the electrostatic potential center of a triangle, and dual quermassintegrals.

math.GM

A Proof of Liu's Conjecture on the Fundamental Triangle Inequality

Let $a,b,c$ be the side lengths of a triangle, and let $R$ and $r$ denote its circumradius and inradius, respectively. We prove a conjecture of Liu stating that \[\sum_{\mathrm{cyc}} \left(\frac{a(b+c-a)}{bc}\right)^k \geq 2+\left(\frac{2r}{R}\right)^k,~~k>1, \] with the reverse inequality for $0<k<1$. The proof reduces the problem to three positive variables with fixed sum and product. We also determine the equality cases.

math.GM