So 2 n vertices is 2 g x 2nn12 xn n 0 5 consider

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Unformatted text preview: (Note that we consider here actual graphs, not isomorphism classes.) SOLUTION. By definition of graphs, ￿ G ∈ Gn , then E (G) ⊆ C2,n := {S ⊆ ￿ if {1, . . . , n} | |S | = 2}. Note that |C2,n | = n , hence the number of possible graphs with 2 (n) = 2n(n−1)/2 . So 2 n vertices is 2 ΦG ( x ) = ￿ 2n(n−1)/2 xn . n ≥0 5. Consider now w(G) = |E (G)| on G100 . Find ΦG100 (x). SOLUTION. The number of vertices is fixed, so we know that any G ∈ G100 have E (G) ⊆ {{a, b} | 1 ≤ a < b ≤ n}. If P2...
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This document was uploaded on 03/07/2014.

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