Periods in missing lengths of rainbow cycles
A cycle in an edge‐colored graph is said to be rainbow if no two of its edges have the same color. For a complete, infinite, edge‐colored graph G, define \documentclass{article}\usepackage{amssymb}\usepackage[mathscr]{euscript}\footskip=0pc \pagestyle{empty}\begin{document}\begin{eqnarray*} {\mathsc...
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Veröffentlicht in: | Journal of graph theory 2009-06, Vol.61 (2), p.98-110 |
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Sprache: | eng |
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Zusammenfassung: | A cycle in an edge‐colored graph is said to be rainbow if no two of its edges have the same color. For a complete, infinite, edge‐colored graph G, define
\documentclass{article}\usepackage{amssymb}\usepackage[mathscr]{euscript}\footskip=0pc \pagestyle{empty}\begin{document}\begin{eqnarray*} {\mathscr{G}}({G}) = \{ {n} \ge {2} | {\rm {no}}\, {n{-}{\rm {cycle \, of}}} \,{{G}}\, {{\rm {is}\, {\rm {rainbow}}}\}.\end{eqnarray*}\end{document}
Then G(G) is a monoid with respect to the operation n∘m=n+ m−2, and thus there is a least positive integer π(G), the period of G(G), such that G(G) contains the arithmetic progression {N+ kπ(G)|k⩾0} for some sufficiently large N. Given that n∈G(G), what can be said about π(G)? Alexeev showed that π(G)=1 when n⩾3 is odd, and conjectured that π(G) always divides 4. We prove Alexeev's conjecture: Let p(n)=1 when n is odd, p(n)=2 when n is divisible by four, and p(n)=4 otherwise. If 2 |
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ISSN: | 0364-9024 1097-0118 |
DOI: | 10.1002/jgt.20371 |