A Textbook of Graph Theory by R. Balakrishnan, K. Ranganathan

By R. Balakrishnan, K. Ranganathan

 This moment variation comprises new chapters: one on domination in graphs and the opposite at the spectral homes of graphs, the latter including a dialogue on graph energy.  The bankruptcy on graph colours has been enlarged, overlaying extra subject matters resembling homomorphisms and colors and the individuality of the Mycielskian as much as isomorphism. 

This ebook additionally introduces numerous fascinating issues reminiscent of Dirac's theorem on k-connected graphs, Harary-Nashwilliam's theorem at the hamiltonicity of line graphs, Toida-McKee's characterization of Eulerian graphs, the Tutte matrix of a graph, Fournier's facts of Kuratowski's theorem on planar graphs, the evidence of the nonhamiltonicity of the Tutte graph on forty six vertices, and a concrete software of triangulated graphs.

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4. 11 Conversely, suppose that e = uv is a cut edge of G. Then the deletion of u results in the deletion of the edge u v. Since G is cubic, G-u is disconnected. Accordingly, u is a cut vertex of G. 2. 3 Prove or disprove: Let G be a simple connected graph with n(G) ::: 3. Then G has a cut edge if, and only if, G has a cut vertex. 4 Show that in a graph, the number of edges common to a cycle and an edge cut is even. 2. Connectivity and Edge-Connectivity We now introduce two parameters of a graph, which, in a way, measure the connectedness of the graph.

1 (Redel [107]) path . Proof (By induction on the number of vertices n of the tournament. ) The result can be directly verified for all tournament s up to three vertices. Hence suppose that the result is true for all tournaments on n ~ 3 vertices. Let T be a tournament on (n + 1) vertices VI , V2 , . , Vn+l ' Now delete Vn+ 1 from T . The resulting digraph T' is a tournament on n vertices and hence by the induction hypothesis contains a directed Hamilton path. Assume that the Hamilton path is VI V2 .

Let u and v be the end vertices of an edge of E . For each edge of E that does not have both u and v as end vertices, remove an end vertex that is different from u and v. If there are t such edges , at most t vertices have been removed. If the resulting graph is disconnected, then K :5 t < A. Otherwise, there will remain a subset of edges of E having u and v as end vertices, the removal of which would disconnect the graph. Hence, in addition to the already removed vertices, the removal of one of u and v will result in either a disconnected graph or a trivial graph.

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