By Claude Flament

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Then G' has ordern+ (m+ 1-2n) = m + 1- n and size m + m + 1- 2n + 1 = 2(m + 1 - n). Furthermore, it is easily checked that j6(G; x, y)j = j6(G'; x', x')j. Hence it suffices to prove the theorem when m = 2n and x = y. 22. The construction of C'. Given a vertex z, recall that d+(z) is the number of edges starting at z and recall that d-(z) is the number of edges ending at z. Call d(z) = d+(z) + d-(z) the degree of z and f(z) = d+(z) - d-(z) theflux at z. We may assume that G contains an Euler circuit (an Euler trail from x to x; otherwise, there is nothing to prove.

Does our brief acquaintance with graphs help us tackle the problems? As it will transpire in the first section, the answer is yes; for after a short review of the basic ideas of electricity we make use of spanning trees to obtain solutions. Some of these results can be reformulated in terms of tilings of rectangles and squares, as we shall show in Section 2. The last section introduces elementary algebraic graph theory, which is then applied to electrical networks. It should be emphasized that in the problems we consider we use hardly more than the terminology of graph theory; virtually the only concept to be used is that of a spanning tree.

6. Let G be a graph of order n. Prove the equivalence of the following assertions. (i) G is a tree. (ii) G is connected and has at most n - 1 edges. (iii) G is acyclic and has at least n - 1 edges. (iv) G = Kn for n = 1,2, and if n :::: 3, then G =I- Kn and the addition of an edge to G produces exactly one new cycle. 7. - Show that every connected graph G of order at least two contains vertices x and y such that both G - x and G - y are connected. 8. In the puzzle of jealous husbands, three husbands and their wives wish to cross a river.

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