By M. Peski n, D. Schroeder [RUSSIAN]

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Since G x has in d(x) edges and degree sum < 2m 2d(x), we have a' = a > 0. Hence deleting a vertex of maximum degree in nontrivial graph reduces the average degree and cannot increase it. 18. If k >- 2, then a k-regular bipartite graph has no cut-edge. Since components of k-regular graphs are k-regular, it suffices to consider a connected k-regular X, Y-bigraph. Let uc be a cut-edge, and let G and H be the components formed by deleting uv. Let in = V(G) n and a = V(G) P By symmetry, we may assume that u a V(G) P Y and v a V(H) P X.

Construction 2a (inductive). 26 (or in Construction 1). This graph is 3-regular with 10 vertices and cut-edge xy; note that 10 = 4• 1 + 6. From a (2k 1)-regular graph Gk_1 with 4k + 2 vertices such that Gk_1 xy has two components of order 2k + 1, we form Gk. Add two vertices for each component of Gk_1 xy, adjacent to all the vertices of that component. This adds degree two to each old vertex, gives degree 2k + 1 to each new vertex, and leaves xy as a cut-edge. The result is a (2k + 1)-regular graph Gj< of order 4k + 6 with cut-edge xy.

In the first case, its length is at least g, since H has girth g. In the second case, the copies of H that C visits correspond to a cycle in G, so C visits at least fg/21 such copies. For each copy, C must enter on one edge and then move to another vertex before leaving, since the copy is entered by only one edge at each vertex, Hence the length of such a cycle is at least 2 ig/21. 17. Deleting a vertex of maximum degree cannot increase the average degree, but deleting a vertex of minimum degree can reduce the average degree.