By Claude Berge

ISBN-10: 0444103996

ISBN-13: 9780444103994

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**Extra resources for Graphs and hypergraphs**

**Example text**

D , - 1, d, To each tree R correspond trees H of graph G. Y (~z,)~P different spanning Hp) = = nl n, ... n,(nl + n2 + + np)P-2. D. Corollary 4 (Cayley [1889]). , x, belong to p different trees is T'(n ;p ) = pnn-P-' . , x, such that dH(xo)= p . , n } and I P I = p , let VPdenote the set of trees in V such that, for all i E P,the vertex x, is joined to x,. Then Hence Therefore, T'(n ; p ) = . (n - l ) ! P ! ( n - P I ! ,,n-p = n! ( p - 1) ! ( n - p ) ! p,tn-~-l . , x,} be a set of n vertices, and let E c P,(x) be a set of q edges that join pairs of vertices in X.

Is a connected subgraph of G* because we can always go from one face to another inside cycle p. p is also a connected subgraph of G*. Thus o ( A * ) is an elementary cocycle. The other part of the theorem may be proved similarly. D. Corollary. For a connected topological planar graph G, v(G*) = 1(G), 1(G*) = v ( G ) . The corollary is evident from Theorem 3, but it can also be shown using Euler’s Formula, which implies that v(G*)=rn-f+l =m-(2+rn-n)+l=n-l=A(G), 1(G*) = f - 1 = m‘- n 1 = v(G).

1) Let G be a connected graph without cocircuits. We shall assume that G is not strongly connected and produce a contradiction. Since G is not strongly connected, it has more than one strongly connected component. Since G is connected, there exist two distinct strongly connected components that are joined by an arc 29 TREES AND ARBORESCENCES (a, 6). Arc (a, b) is not contained in any circuit because otherwise a and b would be in the same strongly connected component. By the Arc Colouring Lemma, arc (a, b) is contained in some cocircuit.

### Graphs and hypergraphs by Claude Berge

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