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The Discussion About The Total Irredundance Number Of Graphs

Posted on:2005-11-07Degree:MasterType:Thesis
Country:ChinaCandidate:F WangFull Text:PDF
GTID:2120360122491742Subject:Operational Research and Cybernetics
Abstract/Summary:PDF Full Text Request
Because the search on the domination numbers is more and more attached importance to people get more deeply aware of it and put forward many domination parameters, such as: total domination number, minus domination number, negative dominationnumber, and so on. Those type of domination number play an important role in structure of graphs.In the long-term studying it is felt by some researchers that a deeper understanding of the concept of domination in graphs can best be obtained by studying the more general concept of irredundance in graphs. To date, the resaerch about the irredundance in graphs is more and more deeper. Resaerchers have got a lot of properties about the irredundant set in graphs. Cockayne and Mynhardt is motivated by the belief that the study of total irredundance will shed light on the concept of total domination. In fact, Hedetniemi et al. in [8] have given the definition of the total irredundant set of a graph, and discussed the irredundance number of some special graphs. Odile favaron et al.in [9] have gotten the sufficient and essencial condition of the graph G satisfying irt(G) =0 and the structure of a tree T satisfying irt(T)= 1.This paper mainly discusses some structure characterization of the graph G satisfying irt(G) = 1 and the condtitution of the graph G satisfying ir(G) = irt(G) = 1. The mainly results are:(1)The graph G - (V, E) is gotten by the clone contraction ,u0 V, consider v0 is the zero layer, divide the vertices of G into k layers , if k > 4,then irt(G) > 2;(2)The graph G is a connected graph ,v0 V(G), satisfying N[v0] = V(G) .if G1,G2,..., Gk are all the connected component of G - u0, then there exist k0 in {1,2, ..., k},s.t.irt (Gk0) =1,irt(Gi) = 0,i= 1,2, ..., k0-1,k0+ 1, ...,k.The discussing of the structure of the graph G satisfying ir(G) = irt(G) =1 has partly answered the open problem (2) in [9].
Keywords/Search Tags:domination number, irredundance number, total irredundant set, total irredundance number, construction characteration
PDF Full Text Request
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