274 Solutions to Exercises with u or v), we are left with a graph gf with 1£1-degu -degv edges. (By £, we mean the set of edges of g.) Since gf has n -2 vertices, gf has at most (n22) edges, so (n -1) (n -2) 2 + 2 -degu -degv::; 1£1-degu -degv::; 2 and (n -1) (n -2) deg u + deg v ~ 2 -2 + 2 _ (n -1)(n -2) (n -2)(n -3) 2 -2 -2 + (n -2)[(n -1) -(n -3)] = +2=n-2+2=n 2 and the result follows. 18. [BB] Add a new vertex adjacent to all existing vertices and apply the result of Exercise 16. 19. (thanks to Michael Watson) Let v and w be two adjacent vertices. Each of these vertices is adjacent to at least ~ -1 = n22 other vertices. If n is odd, this means that both v and w are adjacent to at least n21 other vertices. Since there are only n -2 other vertices in all, it follows that they must both be adjacent to a common vertex, and we have a triangle. So assume n is even and that 9 does not contain a triangle. Then v must be adja-cent to exactly half of the remaining n -2 vertices (let 81 be the set of such vertices) and w must be adjacent to the other half (denote this set 82).
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vertices, Graph theory objects, Michael Watson, Gabriel Andrew Dirac