نتایج جستجو برای: strongly regular graphs
تعداد نتایج: 425818 فیلتر نتایج به سال:
In this paper we obtain an upper bound and also a lower bound for maximum edges of strongly 2 multiplicative graphs of order n. Also we prove that triangular ladder the graph obtained by duplication of an arbitrary edge by a new vertex in path and the graphobtained by duplicating all vertices by new edges in a path and some other graphs are strongly 2 multiplicative
A factor 2 and a factor 3 approximation algorithm are given for the isoperimetric number of Strongly Regular Graphs. One approach involves eigenvalues of the combinatorial laplacian of such graphs. In this approach, both the upper and lower bounds involve the spectrum of the combinatorial laplacian. An interesting inequality is proven between the second smallest and the largest eigenvalue of co...
The real monomial representations of Clifford algebras give rise to two sequences of bent functions. For each of these sequences, the corresponding Cayley graphs are strongly regular graphs, and the corresponding sequences of strongly regular graph parameters coincide. Even so, the corresponding graphs in the two sequences are not isomorphic, except in the first 3 cases. The proof of this nonis...
By means of an exhaustive computer search we have proved that the strongly regular graphs with parameters (v, k, λ, μ) = (105, 32, 4, 12), (120, 42, 8, 18) and (176, 70, 18, 34) are unique upto isomorphism. Each of these graphs occurs as an induced subgraph in the strongly regular McLaughlin graph. We have used an orderly backtracking algorithm with look-ahead and look-back strategies, applying...
A subset of the vertex set of a graph G, S ⊆ V (G), is a (k, τ)-regular set if it induces a k-regular subgraph of G and every vertex not in the subset has τ neighbors in it. This paper is a contribution to the given problem of existence of (k, τ)-regular sets associated with all distinct eigenvalues of integral strongly regular graphs. The minimal idempotents of the Bose-Mesner algebra of stron...
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