运筹与管理 ›› 2023, Vol. 32 ›› Issue (10): 108-113.DOI: 10.12005/orms.2023.0327

• 理论分析与方法探讨 • 上一篇    下一篇

彩色Ore定理

高立青1,2, 王健3   

  1. 1.太原理工大学 经济管理学院,山西 太原 030024;
    2.电子科技大学 经济与管理学院,四川 成都 611731;
    3.太原理工大学 数学学院,山西 太原 030024
  • 收稿日期:2020-09-25 出版日期:2023-10-25 发布日期:2024-01-31
  • 通讯作者: 王健(1988-),男,山西长治人,副教授,博士,研究方向:图论及其应用,极值组合。
  • 作者简介:高立青(1987-),女,山东烟台人,副教授,博士,研究方向:运筹学,图论及其应用。
  • 基金资助:
    国家自然科学基金资助项目(72004154)

A Rainbow Version of Ore Theorem

GAO Liqing1,2, WANG Jian3   

  1. 1. School of Economics and Management, Taiyuan University of Technology, Taiyuan 030024, China;
    2. School of Economics and Management, University of Electronic Science and Technology of China, Chengdu 611731, China;
    3. School of Mathematics, Taiyuan University of Technology, Taiyuan 030024, China
  • Received:2020-09-25 Online:2023-10-25 Published:2024-01-31

摘要: 设G1,G2,···,Gn是在同样的顶点集合V上的n个图,且满足|V|=n。设C是包含V中所有顶点的一个圈,如果C的边集合是从G1,G2,···,Gn中每个图选择一条边得到的,则称C为{G1,G2,···,Gn}上的一个彩色哈密尔顿圈。设P是包含V中所有顶点的一条路,如果P中每条边均来自G1,G2,···,Gn-1中不同的图,那么称P为{G1,G2,···,Gn-1}上的一个彩色哈密尔顿路。最近,Joos和Kim证明了彩色的Dirac定理, 即如果G1,G2,···,Gn中每个图的最小度都大于等于n/2,则{G1,G2,···,Gn}中包含一个彩色哈密尔顿圈。在本文中,我们采用移位方法证明了如果G1,G2,···,Gn中每个图的边数都大于等于$\left(\begin{array}{c} n-1 \\ 2 \end{array}\right)+2$,则{G1,G2,···,Gn}中包含一个彩色哈密尔顿圈。进一步地,如果G1,G2,···,Gn-1中每个图的边数都大于等于$\left(\begin{array}{c} n-1 \\ 2 \end{array}\right)+1$,则{G1,G2,···,Gn-1}中包含一条彩色哈密尔顿路。

关键词: Ore定理, 彩色哈密尔顿圈, 彩色哈密尔顿路

Abstract: Let G1,G2,···,Gn be n graphs on the same vertex set V with |V|=n. If C is a cycle of length n such that each edge belongs to a different graph, then we call C a rainbow Hamilton cycle on {G1,G2,···,Gn}. Similarly, if P is a path of length n-1 such that each edge belongs to a different graph of G1,G2,···,Gn-1, then we call P a rainbow Hamilt on path on {G1,G2,···,Gn-1}. Recently, Joos and Kim have proved a rainbow version of Dirac theorem, i.e., if each of G1,G2,···,Gn has minimum degree at least n/2, then {G1,G2,···,Gn} contains a rainbow Hamilton cycle. In this paper, by the shifting method we show that if each of G1,G2,···,Gn has at least $\left(\begin{array}{c} n-1 \\ 2 \end{array}\right)+2$ edges, then {G1,G2,···,Gn} contains a rainbow Hamilton cycle. Moreover, if each of G1,G2,···,Gn-1 has at least $\left(\begin{array}{c} n-1 \\ 2 \end{array}\right)+1$ edges, then {G1,G2,···,Gn-1} contains a rainbow Hamilton path.
The shifting operator is a powerful tool in extremal set theory, which was invented by Erdös, Ko and Rado and further developed by Frankl. The shifting operator can be applied to graphs and hypergraphs and preserve the number of edges. It is well known that the shifting operator cannot increase the matching number of graphs and hypergraphs. In the present paper, we show that if a graph system {G1,G2,···,Gn} does not contain a rainbow Hamilton cycle, then one can apply the shifting operator to G1,G2,···,Gn simultaneously without producing a rainbow Hamilton cycle. By applying the shifting operator to G1,G2,···,Gn repeatedly, eventually we shall obtain a sequence of shifted graphs G1,G2,···,Gn.
For two pairs (x1,x2)and (y1,y2), define the shifted partial order<as: (x1,x2)<(y1,y2) if and only if either x1≤y1 and x2≤y2 or x1≤y2 and x2≤y1 hold. The shifted graphs have the following nice property: if x1x2 is an edge of G and G is shifted, then for any pair (x1,x2)<(y1,y2),y1y2 is also an edge of G. Let Hn,a,b be a graph with the vertex set {1,2,···,n} such that (i)the subgraph induced by {1,2,···,a} is a complete graph; (ii){a+1,a+2,···,n} forms an independent set; (iii)The neighbors of x are 1,2,···,b for each x in{a+1,a+2,···,n}. It is easy to see that Hn,a,b is a shifted graph. Moreover, Hn,a,b is non-Hamiltonian if n-a≥b and a≥b.
By applying a technique developed by Akiyama and Frankl, we show that some Gi has to be a subgraph of Hn,a,b for some a and b. We prove the result by distinguishing two cases: n is odd and n is even. For n=2k, let f1={1,2k},f2={2,2k-1},···,fk={k,k+1}and let fk+1={2,2k},fk+2={3,2k-1},···,f2k-1={k,k+2},f2k={1,k+1}. Then f1,f2,···,f2k form a Hamilton cycle. Since {G1,G2,···,Gn} does not contain a rainbow Hamilton cycle, we infer that there exists i such that fi$\notin$Gi. If i=2k, then {1,k+1}$\notin$E(Gn). By using the property of shifted graphs, we obtain that Gi has to be a subgraph of H2k,k,0. If i≤k, then we have fi={i,2k+1-i}$\notin$E(Gi). It follows that Gi has to be a subgraph of H2k,2k-i,i-1. If i≥k, set j=i-k, then fi={j,2k-j}$\notin$E(Gi). Since fi<fj={j,2k+1-j},fj$\notin$E(Gi). Then we can show that Gi has to be a subgraph of H2k,2k-j,j-1.
For n=2k+1, let f1={1,2k+1},f2={2,2k},···,fk={k,k+2}and let fk+1={2,2k+1},fk+2= {3,2k},···,f2k={k+1,k+2},f2k+1={1,k+1}. It is easy to see that these edges form a Hamilton cycle. Since {G1,G2,···,Gn}does not contain a rainbow Hamilton cycle, there exists i such that fi$\notin$Gi. If i=2k+1,then {1,k+1}$\notin$E(Gi). Then Gi has to be a subgraph of H2k+1,k,0. If i≥k+1, set j=i-k, then fi={j+1,2k+2-j}$\notin$E(Gi), implying that Gi has to be a subgraph of H2k+1,2k+1-j,j. If 1≤i≤k, then fi={i,2k+2-i}$\notin$E(Gi). As f′={i+1,2k+2-i} satisfies fi<f′,Gi has to be a subgraph of H2k+1,2k+1-i,i. By computations, we show that all these Hn,a,b graphs have at most $\left(\begin{array}{c} n-1 \\ 2 \end{array}\right)+1$ edges, which leads to a contradiction. Thus, we conclude that if each of G1,G2,···,Gn has at least $\left(\begin{array}{c} n-1 \\ 2 \end{array}\right)+2$ edges, then{G1,G2,···,Gn} contains a rainbow Hamiltonian cycle. By a similar argument, we establish the corresponding result for rainbow Hamiltonian paths as well.

Key words: Ore theorem, rainbow Hamiltonian cycle, rainbow Hamiltonian path

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