John M. Harris, Jeffry L. Hirst, Michael J. Mossinghoff's Combinatorics and Graph Theory PDF

By John M. Harris, Jeffry L. Hirst, Michael J. Mossinghoff (auth.)

ISBN-10: 1475748035

ISBN-13: 9781475748031

ISBN-10: 1475748051

ISBN-13: 9781475748055

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The result is certainly true if n = 2, since T = K 2 in this case. Suppose the result is true for all orders from 2 ton- 1, and consider a tree T of order n :::-_ 3. We know that T has n - 1 edges, and since we can assume n :::-_ 3, T has at least 2 edges. If every edge of T is incident with a leaf, then T has at least two leaves; and the proof is complete. So assume that there is some edge of T that is not incident with a leaf, and let us say that this edge is e = u v. The graph T - e is a pair of trees, T1 and T2 , each of order less than n.

We now proceed to label the vertices of Gin preparation for the greedy algorithm. Let v 1 and v 2 be as labeled. Let v be labeled lin. Now choose a vertex adjacent to Vn that is not v 1 or v2 (such a vertex exists, since deg( Vn) :=:: 3 ). Label this vertex Vn-·t· Next choose a vertex that is adjacent to either V 11 or v" .. 1 and is not v 1, V;>, V 11 , or V11 _ 1• Call this vertex Vn-l· We continue this process. Since G - {v 1 , v2 } is connected, then for each i E {3, ... , n - I}, there is a vertex vi E V(G).

If n :::: 6, then the result is immediate, so we will suppose that n > 6. If we let D be the sum of the degrees of the vertices of G, then we have D = 2q :::: 2(3n - 6) = 6n - 12. If each vertex had degree 6 or more, then we would have D 2:: 6n, which is impossible. Thus there must be some vertex with degree less than or equal to 5. 0 Exercises 1. G is a planar graph of order 24, and it is regular of degree 3. How many regions are in a planar representation of G? 2. Let G be a connected planar graph of order less than 12.

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Combinatorics and Graph Theory by John M. Harris, Jeffry L. Hirst, Michael J. Mossinghoff (auth.)


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