【摘要】 目的　寻找循环图 Cn<a1 ,a2 ,… ,ak>中的最大团 .方法　利用组合算法并结合图的特性 .结果求出了循环图 Cn<a1 ,a2 ,… ,ak>满足下列条件 :1 ai=a1 +(i-1 ) d(i=1 ,2 ,… ,k) ;2 d∈ Z+ 且 d≠ 1 ;3a1 ∈Z+且 a1 ≠md,m∈ Z+ ;4ak<(n+1 ) /2时的最大团的阶及其个数 ,n=2 ak时 ,最大团的阶为 2 ,个数为(2 k-1 ) n/2 ;n=2 ak+a1 +ld(l=0 ,1 ,… ,k-1 )时 ,最大团的阶为 3 ,个数为 (k-l) (k-l+1 ) n/6;n为其它数时 ,最大团的阶为 2 ,个数为 kn.结论　循环图 Cn<a1 ,a2 ,… ,ak>在满足一定邻接条件下 ,最大团是可求的更多还原

【Abstract】 Aim\ To find the maximum clique of circulant graph \%C\-n<a\-1,a\-2,\:,a\-k>\%. Methods By means of characteristics of the graph, the combinatorial algorithm is used. Results\ The order and the number of the maximum clique of circulant graph \%C\-n<a\-1,a\-2,\:,a\-k>\% are presented when it satisfies the following conditions: ① \%a\-i=a\-1+(i-1)d(i=1,2,\:,k); ② d\%∈Z\++ and \%d≠l\%; ③ \%a\-1∈Z\++ and \%a\-1≠md, m∈Z\++; ④ \%a\-k<n+12\%. When \%n=2a\-k+a\-1+ld(l=0,1,\:,k-1)\%, the order and the number are 3 and \%(k-l)(k-l+1)n/6\% respectively. When \%n=2a\-k\%, the order and the number are 2 and \%(2k-1)n/2\% respectively. In other cases, the order and the number are 2 and \%kn\% respectively. Conclusion\ The maximum clique problem is solvable when the adjacent condition is constrained.更多还原