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Unformatted text preview: 2Tr 1 vertices. So the degree of every vertex is  2Tr 1 1 . Then we choose the vertex, say v , with the biggest number of edges go out, say x . So (  )= (  ) x 12 2Tr 1 1 T r 1 Thus v point to at least Tr 1 vertices, which can make sure of a transitive tournament on r 1 vertices. So this transitive tournament plus v is a transitive tournament on r vertices. That is, in any tournament on 2Tr 1 vertices, we can find a transitive tournament on r vertices. Because T(r) is the smallest number such that every tournament on T(r) vertices contains a transitive tournament on r vertices, we know ( ) 2Tr 1 T r . Thus we have:   =  = Tr 2Tr 1 22Tr 2 2r 2T2 2r 2 2 2r 1 Page 1...
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 Fall '09
 Steven
 Graph Theory

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