Editing Petersen graph (section)

== Hamiltonian paths and cycles ==
[[Image:Petersen2 tiny.svg|class=skin-invert-image|thumb|right|The Petersen graph is hypo-Hamiltonian: by deleting any vertex, such as the center vertex in the drawing, the remaining graph is Hamiltonian. This drawing with order-3 symmetry is the one given by {{harvtxt|Kempe|1886}}.]]

The Petersen graph has a [[Hamiltonian path]] but no [[Hamiltonian cycle]]. It is the smallest bridgeless cubic graph with no Hamiltonian cycle. It is [[hypohamiltonian graph|hypohamiltonian]], meaning that although it has no Hamiltonian cycle, deleting any vertex makes it Hamiltonian, and is the smallest hypohamiltonian graph.

As a finite [[connectivity (graph theory)|connected]] vertex-transitive graph that does not have a Hamiltonian cycle, the Petersen graph is a counterexample to a variant of the [[Lovász conjecture]], but the canonical formulation of the conjecture asks for a Hamiltonian path and is verified by the Petersen graph.

Only five connected vertex-transitive graphs with no Hamiltonian cycles are known: the [[complete graph]] ''K''<sub>2</sub>, the Petersen graph, the [[Coxeter graph]] and two graphs derived from the Petersen and Coxeter graphs by replacing each vertex with a triangle.<ref name=foster>Royle, G. [http://www.cs.uwa.edu.au/~gordon/remote/foster/#census "Cubic Symmetric Graphs (The Foster Census)."] {{webarchive|url=https://web.archive.org/web/20080720005020/http://www.cs.uwa.edu.au/~gordon/remote/foster/ |date=2008-07-20 }}</ref> If ''G'' is a 2-connected, ''r''-regular graph with at most 3''r''&nbsp;+&nbsp;1 vertices, then ''G'' is Hamiltonian or ''G'' is the Petersen graph.<ref>{{citation|first1=D. A.|last1=Holton|first2=J.|last2=Sheehan|title-link= The Petersen Graph |title=The Petersen Graph|publisher=[[Cambridge University Press]]|year=1993|isbn=0-521-43594-3|page=32}}</ref>

To see that the Petersen graph has no Hamiltonian cycle, consider the edges in the cut disconnecting the inner 5-cycle from the outer one. If there is a Hamiltonian cycle ''C'', it must contain an even number of these edges. If it contains only two of them, their end-vertices must be adjacent in the two 5-cycles, which is not possible. Hence, it contains exactly four of them. Assume that the top edge of the cut is not contained in ''C'' (all the other cases are the same by symmetry). Of the five edges in the outer cycle, the two top edges must be in ''C'', the two side edges must not be in ''C'', and hence the bottom edge must be in ''C''. The top two edges in the inner cycle must be in ''C'', but this completes a non-spanning cycle, which cannot be part of a Hamiltonian cycle. Alternatively, we can also describe the ten-vertex [[Regular graph|3-regular graphs]] that do have a Hamiltonian cycle and show that none of them is the Petersen graph, by finding a cycle in each of them that is shorter than any cycle in the Petersen graph. Any ten-vertex Hamiltonian 3-regular graph consists of a ten-vertex cycle ''C'' plus five chords. If any chord connects two vertices at distance two or three along ''C'' from each other, the graph has a 3-cycle or 4-cycle, and therefore cannot be the Petersen graph. If two chords connect opposite vertices of ''C'' to vertices at distance four along ''C'', there is again a 4-cycle. The only remaining case is a [[Möbius ladder]] formed by connecting each pair of opposite vertices by a chord, which again has a 4-cycle. Since the Petersen graph has girth five, it cannot be formed in this way and has no Hamiltonian cycle.