Fleischner's theorem

An undirected graph ${\displaystyle G}$ is Hamiltonian if it contains a cycle that touches each of its vertices exactly once. It is 2-vertex-connected if it does not have an articulation vertex, a vertex whose deletion would leave the remaining graph disconnected. Not every 2-vertex-connected graph is Hamiltonian; counterexamples include the Petersen graph and the complete bipartite graph ${\displaystyle K_{2,3}}$.

The square of ${\displaystyle G}$ is a graph ${\displaystyle G^{2}}$ that has the same vertex set as ${\displaystyle G}$, and in which two vertices are adjacent if and only if they have distance at most two in ${\displaystyle G}$. Fleischner's theorem states that the square of a finite 2-vertex-connected graph with at least three vertices must always be Hamiltonian. Equivalently, the vertices of every 2-vertex-connected graph ${\displaystyle G}$ may be arranged into a cyclic order such that adjacent vertices in this order are at distance at most two from each other in ${\displaystyle G}$.

In Fleischner's theorem, it is possible to constrain the Hamiltonian cycle in ${\displaystyle G^{2}}$ so that for given vertices ${\displaystyle v}$ and ${\displaystyle w}$ of ${\displaystyle G}$ it includes two edges of ${\displaystyle G}$ incident with ${\displaystyle v}$ and one edge of ${\displaystyle G}$ incident with ${\displaystyle w}$. Moreover, if ${\displaystyle v}$ and ${\displaystyle w}$ are adjacent in ${\displaystyle G}$, then these are three different edges of ${\displaystyle G}$.[1]

In addition to having a Hamiltonian cycle, the square of a 2-vertex-connected graph ${\displaystyle G}$ must also be Hamiltonian connected (meaning that it has a Hamiltonian path starting and ending at any two designated vertices) and 1-Hamiltonian (meaning that if any vertex is deleted, the remaining graph still has a Hamiltonian cycle).[2] It must also be vertex pancyclic, meaning that for every vertex ${\displaystyle v}$ and every integer ${\displaystyle G}$k with ${\displaystyle 3\leq k\leq |V(G)}$, there exists a cycle of lengt ${\displaystyle k}$ containing ${\displaystyle v}$.[3]

If a graph ${\displaystyle G}$ is not 2-vertex-connected, then its square may or may not have a Hamiltonian cycle, and determining whether it does have one is NP-complete.[4]

An infinite graph cannot have a Hamiltonian cycle, because every cycle is finite, but Carsten Thomassen proved that if ${\displaystyle G}$ is an infinite locally finite 2-vertex-connected graph with a single end then ${\displaystyle G^{2}}$ necessarily has a doubly infinite Hamiltonian path.[5] More generally, if ${\displaystyle G}$ is locally finite, 2-vertex-connected, and has any number of ends, then ${\displaystyle G^{2}}$ has a Hamiltonian circle. In a compact topological space formed by viewing the graph as a simplicial complex and adding an extra point at infinity to each of its ends, a Hamiltonian circle is defined to be a subspace that is homeomorphic to a Euclidean circle and covers every vertex.[6]

The Hamiltonian cycle in the square of an ${\displaystyle n}$-vertex 2-connected graph can be found in linear time,[7] improving over the first algorithmic solution by Lau[8] of running time ${\displaystyle O(n^{2})}$. Fleischner's theorem can be used to provide a 2-approximation to the bottleneck traveling salesman problem in metric spaces.[9]

A proof of Fleischner's theorem was announced by Herbert Fleischner in 1971 and published by him in 1974, solving a 1966 conjecture of Crispin Nash-Williams also made independently by L. W. Beineke and Michael D. Plummer.[10] In his review of Fleischner's paper, Nash-Williams wrote that it had solved "a well known problem which has for several years defeated the ingenuity of other graph-theorists".[11]

Fleischner's original proof was complicated. Václav Chvátal, in the work in which he invented graph toughness, observed that the square of a ${\displaystyle k}$-vertex-connected graph is necessarily ${\displaystyle k}$-tough; he conjectured that 2-tough graphs are Hamiltonian, from which another proof of Fleischner's theorem would have followed.[12] Counterexamples to this conjecture were later discovered,[13] but the possibility that a finite bound on toughness might imply Hamiltonicity remains an important open problem in graph theory. A simpler proof both of Fleischner's theorem, and of its extensions by Chartrand et al. (1974), was given by Říha (1991),[14] and another simplified proof of the theorem was given by Georgakopoulos (2009a).[15]