<p>Kirkman schedule is one of the typical single round-robin (abbrev. SRR) tournaments. The partial team swap (abbrev. PTS) is one of the typical procedures of changing from an SRR tournament to another SRR tournament, which is used in local search for solving the traveling tournament problem. An SRR of <i>n</i> teams (of even number) can be represented by a 1-factorization of the complete graph <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq1.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="24" /> </InlineMediaObject> <EquationSource Format="TEX">\(K_n\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>K</mi> <mi>n</mi> </msub> </math></EquationSource> </InlineEquation>. It is known that the 1-factorization of any Kirkman schedule is “perfect” when <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq2.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="70" /> </InlineMediaObject> <EquationSource Format="TEX">\(n=p+1\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>=</mo> <mi>p</mi> <mo>+</mo> <mn>1</mn> </mrow> </math></EquationSource> </InlineEquation> for prime numbers <i>p</i>, meaning that any pair of 1-factors in the 1-factorization forms a Hamilton cycle <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq3.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="22" /> </InlineMediaObject> <EquationSource Format="TEX">\(C_n\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>C</mi> <mi>n</mi> </msub> </math></EquationSource> </InlineEquation> in <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq1.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="24" /> </InlineMediaObject> <EquationSource Format="TEX">\(K_n\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>K</mi> <mi>n</mi> </msub> </math></EquationSource> </InlineEquation>, called a 2-edge-colored Hamilton cycle. We are concerned with the cycle structure after applying the PTS to Kirkman schedules, that is, how a 2-edge-colored Hamilton cycle <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq3.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="22" /> </InlineMediaObject> <EquationSource Format="TEX">\(C_n\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>C</mi> <mi>n</mi> </msub> </math></EquationSource> </InlineEquation> is decomposed into two 2-edge-colored cycles of length 2<i>d</i> and <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq6.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="52" /> </InlineMediaObject> <EquationSource Format="TEX">\(n-2d\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> <mi>d</mi> </mrow> </math></EquationSource> </InlineEquation>, say, <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq7.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="28" /> </InlineMediaObject> <EquationSource Format="TEX">\(C_{2d}\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>C</mi> <mrow> <mn>2</mn> <mi>d</mi> </mrow> </msub> </math></EquationSource> </InlineEquation> and <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq8.gif" Format="GIF" Height="17" Rendition="HTML" Resolution="72" Type="Linedraw" Width="44" /> </InlineMediaObject> <EquationSource Format="TEX">\(C_{n-2d}\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>C</mi> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> <mi>d</mi> </mrow> </msub> </math></EquationSource> </InlineEquation> for some number <InlineEquation ID="IEq9"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq9.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="67" /> </InlineMediaObject> <EquationSource Format="TEX">\(d\in [n/2]\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>d</mi> <mo>∈</mo> <mo stretchy="false">[</mo> <mi>n</mi> <mo stretchy="false">/</mo> <mn>2</mn> <mo stretchy="false">]</mo> </mrow> </math></EquationSource> </InlineEquation>. We characterize the numbers <i>d</i> such that any cycle <InlineEquation ID="IEq10"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq7.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="28" /> </InlineMediaObject> <EquationSource Format="TEX">\(C_{2d}\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>C</mi> <mrow> <mn>2</mn> <mi>d</mi> </mrow> </msub> </math></EquationSource> </InlineEquation> is not generated by <i>any</i> PTS. Moreover, in case that a cycle <InlineEquation ID="IEq11"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq7.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="28" /> </InlineMediaObject> <EquationSource Format="TEX">\(C_{2d}\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>C</mi> <mrow> <mn>2</mn> <mi>d</mi> </mrow> </msub> </math></EquationSource> </InlineEquation> is generated, we show that the number of <InlineEquation ID="IEq12"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq7.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="28" /> </InlineMediaObject> <EquationSource Format="TEX">\(C_{2d}\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>C</mi> <mrow> <mn>2</mn> <mi>d</mi> </mrow> </msub> </math></EquationSource> </InlineEquation> for any <InlineEquation ID="IEq13"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq13.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="59" /> </InlineMediaObject> <EquationSource Format="TEX">\(d\ne n/4\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>d</mi> <mo>≠</mo> <mi>n</mi> <mo stretchy="false">/</mo> <mn>4</mn> </mrow> </math></EquationSource> </InlineEquation> generated by any PTS is at most <InlineEquation ID="IEq14"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq14.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="41" /> </InlineMediaObject> <EquationSource Format="TEX">\(n-2\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>-</mo> <mn>2</mn> </mrow> </math></EquationSource> </InlineEquation>. For the case of <InlineEquation ID="IEq15"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq15.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="59" /> </InlineMediaObject> <EquationSource Format="TEX">\(d=n/4\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>d</mi> <mo>=</mo> <mi>n</mi> <mo stretchy="false">/</mo> <mn>4</mn> </mrow> </math></EquationSource> </InlineEquation> (i.e., <InlineEquation ID="IEq16"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq16.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="32" /> </InlineMediaObject> <EquationSource Format="TEX">\(C_{n/2}\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>C</mi> <mrow> <mi>n</mi> <mo stretchy="false">/</mo> <mn>2</mn> </mrow> </msub> </math></EquationSource> </InlineEquation>), the number of <InlineEquation ID="IEq17"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq16.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="32" /> </InlineMediaObject> <EquationSource Format="TEX">\(C_{n/2}\)</EquationSource> <EquationSource Format="MATHML"><math> <msub> <mi>C</mi> <mrow> <mi>n</mi> <mo stretchy="false">/</mo> <mn>2</mn> </mrow> </msub> </math></EquationSource> </InlineEquation> generated by any PTS is at most <InlineEquation ID="IEq18"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="10878_2025_1329_Article_IEq18.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="61" /> </InlineMediaObject> <EquationSource Format="TEX">\(2(n-2)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>2</mn> <mo stretchy="false">(</mo> <mi>n</mi> <mo>-</mo> <mn>2</mn> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation>, and there is some PTS to achieve the upper bound.</p>

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On the initial transition of graphs of Kirkman schedules by the partial team swap

  • Yusuke Kashiwagi,
  • Masaki Yamamoto,
  • Takamasa Yashima

摘要

Kirkman schedule is one of the typical single round-robin (abbrev. SRR) tournaments. The partial team swap (abbrev. PTS) is one of the typical procedures of changing from an SRR tournament to another SRR tournament, which is used in local search for solving the traveling tournament problem. An SRR of n teams (of even number) can be represented by a 1-factorization of the complete graph \(K_n\) K n . It is known that the 1-factorization of any Kirkman schedule is “perfect” when \(n=p+1\) n = p + 1 for prime numbers p, meaning that any pair of 1-factors in the 1-factorization forms a Hamilton cycle \(C_n\) C n in \(K_n\) K n , called a 2-edge-colored Hamilton cycle. We are concerned with the cycle structure after applying the PTS to Kirkman schedules, that is, how a 2-edge-colored Hamilton cycle \(C_n\) C n is decomposed into two 2-edge-colored cycles of length 2d and \(n-2d\) n - 2 d , say, \(C_{2d}\) C 2 d and \(C_{n-2d}\) C n - 2 d for some number \(d\in [n/2]\) d [ n / 2 ] . We characterize the numbers d such that any cycle \(C_{2d}\) C 2 d is not generated by any PTS. Moreover, in case that a cycle \(C_{2d}\) C 2 d is generated, we show that the number of \(C_{2d}\) C 2 d for any \(d\ne n/4\) d n / 4 generated by any PTS is at most \(n-2\) n - 2 . For the case of \(d=n/4\) d = n / 4 (i.e., \(C_{n/2}\) C n / 2 ), the number of \(C_{n/2}\) C n / 2 generated by any PTS is at most \(2(n-2)\) 2 ( n - 2 ) , and there is some PTS to achieve the upper bound.