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

ISBN-10: 1931233985

ISBN-13: 9781931233989

This ebook supplies a few new sorts of Fuzzy and Neutrosophic types which could learn difficulties in a innovative manner. the recent notions of bigraphs, bimatrices and their generalizations are used to construct those types that allows you to be beneficial to research time based difficulties or difficulties which want stage-by-stage comparability of greater than specialists. The types expressed right here could 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**

**Example text**

33 We illustrate this by the following example. 11: G = G1 ∪ G2 be a bigraph given by the following figure. 11 Here G = G1 ∪ G2 = {v1, v 2, v 3, …, v8} ∪ { v'1, v'2, v'3, …, v'8}. The graphs associated with G1 and G2 are given by the following figure. 11b This bigraph has a subgraph in common given by the following figure. 11c Thus we have the following interesting result. 3: Let G = G1 ∪ G2 be a bigraph which is strong subgraph glued bigraph, then G is a vertex glued graph and a edge glued graph.

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.

Thus if G = G1 ∪ G2 is a disjoint bigraph it has no bidegree associated with it. 44 Now we give an example of a bigraph G = G1 ∪ G2 which has a vertex with bidegree associated with it. 18. 18 Here G = = G1 ∪ G2 {u1, u2, u3, u4, u5} ∪ {u'1, u'2, u'3, u'4}. 18b. 18b Clearly G1 is 2-regular and G2 is 3-regular but the bigraph G = G1 ∪ G2 is not regular. Further G is not K1 + K2 biregular for we have 3 bivertices u3 = u'1, u4 = u'4 and u5 = u'3 and only the vertex u4 = u'4 has degree 2 + 3 and the other vertices are just of degree 4.

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

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