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Extra info for An Introduction to the Khasia Language, Comprising Grammar, Selections for Reading, and a Vocabulary
Hence the average of the neighborhood avuv∈E(G) d(v) d(u) erage degrees is at least the average degree, and the pigeonhole principle yields the desired vertex. It is possible that every average neighborhood degree exceeds the average degree. Let G be the graph with 2n vertices formed by adding a matching between a complete graph and an independent set. Since G has n + n edges and 2n vertices, G has average degree (n + 1)/2. For each 2 vertex of the n -clique, the neighborhood average degree is n − 1 + 1/n .
Let q be the number of copies of k in d . If q > k + 1, then d has k s and (k − 1)s. If q = k + 1, then d has only (k − 1)s. If q < k + 1, then d has (k − 1)s and (k − 2)s. Also, if k = n − 1, then the first possibility cannot occur. Thus d has length n − 1, its largest value is less than n − 1, and its largest and smallest values differ by at most 1. Thus d ∈ A(n − 1), and we can apply the induction hypothesis to d . The induction hypothesis (d ∈ A(n − 1)) ⇒ (d ∈ B(n − 1)) tells us that d is graphic.
The degree list of G in nonincreasing order is n − 1 − dn , . . , n − 1 − d1 . Since G ∼ = G , the lists are the same. Since n is odd, the central elements in the list yield d(n+1)/2 = n − 1 − d(n+1)/2 , so d(n+1)/2 = (n − 1)/2. 62. When n is congruent to 0 or 1 modulo 4, there is an n -vertex simple graph G with 12 2n edges such that (G) − δ(G) ≤ 1. 31. More generally, let G be any 2k -regular simple graph with 4k + 1 vertices, where n = 4k + 1. Such a graph can be constructed by placing 4k + 1 vertices around a circle and joining each vertex to the k closest vertices in each direction.