数学季刊 ›› 2007, Vol. 22 ›› Issue (3): 359-363.

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Km V Kn的最小直径定向

  


  1. 1. Shanghai Gaoqiao High School  2. School of Mathematics and Computer Science Nanjing Normal University 

  • 收稿日期:2004-12-01 出版日期:2007-09-30 发布日期:2023-10-26
  • 作者简介: MIAO Xiao-yan(1979-), female, native of Shanghai, a lecturer of Shanghai Gaoqiao High School, engagea in graph theory; SUN Zhi-ren(1964-), male, native of Rugao, Jiangsu, a profensor of Nanjing Normal University, engages in graph theory.
  • 基金资助:
     Supported by the National Natural Science Foundation of China(10671095)

Minimum Diameter Orientations of Km V Kn


  1. 1. Shanghai Gaoqiao High School  2. School of Mathematics and Computer Science Nanjing Normal University 
  • Received:2004-12-01 Online:2007-09-30 Published:2023-10-26
  • About author: MIAO Xiao-yan(1979-), female, native of Shanghai, a lecturer of Shanghai Gaoqiao High School, engagea in graph theory; SUN Zhi-ren(1964-), male, native of Rugao, Jiangsu, a profensor of Nanjing Normal University, engages in graph theory.
  • Supported by:
     Supported by the National Natural Science Foundation of China(10671095)

摘要: For a graph G,let D denote an orientation of G having minimum diameter. Define f(G)=diamD.In this paper,we concentrate on exploring the minimum diameter of Km V Kn(m≥1,n≥1).Some special cases are known: f(Km V Kn)=∞,2,3, where m=1 and n≥1,m=2 or m≥4 and n=1,m=3 and n=1,respectively. So we only consider the case when m≥2 and n≥2.The following results are obtained. (1) f(Km V Kn)=3,where m=2,3,n≥2 and m=n=4.(2) f(Km V Kn)=2, where m≥5 and m is odd,2≤n≤m-m.(3) f(Km V Kn)=2,where m≥4 and m≡0(mod4),2≤n≤m-(m/2+1).(4) f(Km V Kn)=2,where m≥6 and m≡2(mod4),2≤n≤m-m/2.(5) f(Km V Kn)=3,where m≥4,n>m. 

关键词: minimum diameter, orientation, containment-free

Abstract: For a graph G,let D denote an orientation of G having minimum diameter. Define f(G)=diamD.In this paper,we concentrate on exploring the minimum diameter of Km V Kn(m≥1,n≥1).Some special cases are known: f(Km V Kn)=∞,2,3, where m=1 and n≥1,m=2 or m≥4 and n=1,m=3 and n=1,respectively. So we only consider the case when m≥2 and n≥2.The following results are obtained. (1) f(Km V Kn)=3,where m=2,3,n≥2 and m=n=4.(2) f(Km V Kn)=2, where m≥5 and m is odd,2≤n≤m-m.(3) f(Km V Kn)=2,where m≥4 and m≡0(mod4),2≤n≤m-(m/2+1).(4) f(Km V Kn)=2,where m≥6 and m≡2(mod4),2≤n≤m-m/2.(5) f(Km V Kn)=3,where m≥4,n>m. 

Key words: minimum diameter, orientation, containment-free

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