SearcharxivSearch

arXiv · 1712.10322

Proof of Kelly-Ulam Conjecture

Abstract

The deck of a graph $X$, $D(X)$, is defined as the multiset of all vertex-deleted subgraphs of $X$. Two graphs are said to be hypomorphic, if they have the same deck. Kelly-Ulam conjecture states that any two hypomorphic graphs on at least three vertices are isomorphic. In this paper, we first prove that for two finite simple hypomorphic graphs the number of $l$-paths between two arbitrary vertices are equal, where $1 \leq l \leq n - 2$. As a consequence, it is proved that the Kelly-Ulam conjecture is correct over the category of all finite simple graphs.

Explore related subjects

Keep this discovery

BibTeXRIS

Adel Tadayyonfar, Ali Reza Ashrafi. 2017-12-24. Proof of Kelly-Ulam Conjecture. https://arxiv.org/abs/1712.10322

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