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Discrete Mathematics with Graph Theory (3rd Edition) 276

# Discrete Mathematics with Graph Theory (3rd Edition) 276 -...

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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 = n 2 2 other vertices. If n is odd, this means that both v and w are adjacent to at least n 2 1 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 8 1 be the set of such vertices) and w must be adjacent to the other half (denote this set 8 2 ).
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