By A. F. Horadam, W. D. Wallis

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**Extra info for Combinatorial Mathematics VI**

**Sample text**

Let d = (d1 , . . , dn ) be a sequence of positive integers, where n ≥ 2. Then d is a degree sequence of a tree iff ni=1 di = 2n − 2. Proof. If d is a degree sequence of a tree of size m then ni=1 di = 2m = 2n − 2. For the converse, if n = 2 then d = (1, 1), the degree sequence of the tree K2 . Suppose then that n > 2 and the result holds for sequences of length n − 1. We may assume d1 ≥ · · · ≥ dn . Then d1 > 1, since otherwise n n i=1 di = n < 2n − 2, and dn = 1, since otherwise i=1 di ≥ 2n.

Then G − e is a connected plane graph of order n and size m − 1 having r − 1 regions (the two regions of G adjacent to e lying in a single region of G − e). The result follows. 3. Let G be a plane graph of order n and size m having r regions and k components. Then n − m + r = k + 1. 4. If G is a planar graph of order n ≥ 3 and size m then m ≤ 3n − 6. Proof. We may assume that G is connected, since otherwise we may add edges to get a connected planar graph of order n and size greater than m. Embed G in the plane and let the number of regions be r.

Applying part (a) to each component and adding, we have m = n − k, so k = 1, as required. (c) Suppose that (1) and (3) hold. Then G has a spanning tree T . By part (a), T has size m, so T = G, giving (2). 4. Let G be an acyclic graph of order n and size m with k components. Then m = n − k. 5. Let u and v be vertices of a graph G, and suppose there are distinct u-v paths P and Q of lengths k and l. Then there is a cycle of length at most k + l in G, and if P · Qr is not a cycle, there is one of length less than k + l.