Applications of Bimatrices to Some Fuzzy and Neutrosophic by W. B. Vasantha Kandasamy, Florentin Smarandache, K.

By W. B. Vasantha Kandasamy, Florentin Smarandache, K. Ilanthenral

This e-book provides a few new sorts of Fuzzy and Neutrosophic versions that can examine difficulties in a innovative approach. the recent notions of bigraphs, bimatrices and their generalizations are used to construct those types with a view to be precious to research time based difficulties or difficulties which desire stage-by-stage comparability of greater than specialists. The versions expressed right here will be regarded as generalizations of Fuzzy Cognitive Maps and Neutrosophic Cognitive Maps.

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Additional resources for Applications of Bimatrices to Some Fuzzy and Neutrosophic Models

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31. 31 61 v'4 v"6 v"5 G3 The vertex set of G has only 14 points and the individual graphs of the trigraph is given by the following figures. 31c Infact G = G1 ∪ G2 ∪ G3 is a trigraph which is connected. Infact G is not a disjoint trigraph. 62 Now we proceed on to give an example of just a disjoint trigraph. 32. 32c Clearly this trigraph is a disjoint trigraph and the bigraph in it is a edge glued bigraph. Thus we have seen three types of trigraphs. In fact all properties of graphs can be proved with suitable and necessary modification for every trigraph is also a graph.

A bigraph can also be a bipartite bigraph. e. V G1C ( ) V G2C = V (G) and = V (G) and making two adjacent vertices u and v adjacent in GC if and only if they are non adjacent in G. 16a. 16b. 16b 43 We see if G is a disjoint bigraph so is its complement. 5: If G = G1 ∪ G2 be a disjoint bigraph such that G1 = G2C then G is a self complementary bigraph. Proof: Follows directly by the definition and the fact the bigraph is disjoint. We need to work more only when the bigraphs are not disjoint. 10: Let G = G1 ∪ G2 be a bigraph and v ∈ V (G1) ∪ V (G2).

21b. 21b Clearly G1 has both an isolated vertex u1 and a pendent vertex u6 but the bigraph G = G1 ∪ G2 has no pendent vertex or isolated vertex. 48 It is easily seen that the Euler’s theorem is true for all bigraphs. Let G = G1 ∪ G2 be a bigraph. The join of G1 ∨ G2 = G is different from the bigraph G. This is illustrated by the following example. 22. 22a. 22a Clearly G and G1 are distinct. 22b. 22b 49 Thus G = G1 ∪ G2 ≠ G1. 22c. 22c Clearly G = G1 ∪ G2 ≠ G1. Thus the join of two graphs G1 ∨ G2 is not the same as the bigraph given by G = G1 ∪ G2.

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