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- 对于k=1和k=2,易知:Sumner猜想本文对奇数和偶数分别构造了k支配边临界图,给出了f(n,k)的一个上界:Easily we can get f(n,1) = , f(n,2) = , Sumner conjecturedf(n,3) . This paper gives out two classes of domination k-edge-critical graphrespectively when k is even and odd, proves that:f(n,k) (2k -2)
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