The complexity of some families of cycle-related graphs

Abstract

In this paper, we derive new formulas for the number of spanning trees of a specific family of graphs-gear graphs, flower graphs, sun graphs and sphere graphs-using techniques from linear algebra, Chebyshev polynomials and matrix theory.

Edge even graceful labelling of new families of graphs

“…The quadrilateral friendship graph Fr (4) n is a planer undirected graph with 3n+1 vertices and 4n edges constructed by joining n copies of the cycle graph C 4 with a common vertex. u 1 , u 2 , .…”

“…Recently graph labelling attracted many brilliant researches in different disciplines, as coding theory, optimal circuits layouts, communication networks, astronomy, radar and graph decomposition problems. See [1][2][3][4][5] for some details and concerning applications.…”

“…0 ) = (6n − 2) + (6n − 8) + (6n − 12) + • • • + (2n + 4) + 2n = (6n − 2) + n−1 2 (8n − 8) = 4n 2 − 2n + 2), f (v 0 ) = (6n − 4) + (6n − 6) + (6n − 10) + • • • + (2n + 6) + (2n + 2) = (6n − 4) + n−1 2 (8n − 4) = 4n 2 − 2.Since n ≡ 5mod6, n = 6m + 5 for some positive integer m. Therefore, p = 6m + 7, q = k = 18m + 14, and 6n − 2 = 36m + 28.So f (v 0 ) = 4(6m + 5) 2 − 2(6m + 5) + 2 ≡ 8m + 8 =4 3 (n + 1), and f (v 0 ) = 4(6m + 3) 2 − 2 ≡ 20m + 14 = 2 3 (5n − 4).…”

“…Case (2) n = 4.n(n + 1), f (v 0 ) = (6n − 2) + (6n − 4) + • • • + 4n = n(5n − 1).Since n ≡ 0mod6, n = 6m for some positive integer m. Therefore, p = 6m + 2, q = k = 18m − 1, and 6n − 2 = 36m − 2. So f (v 0 ) = 6m(6m + 1) = 36m 2 + 6m ≡ 2m + 6m = 8m =4 3 n, and f (…”

Number of spanning trees of some families of graphs generated by a triangle

“…Another class of graphs for which an explicit formula has been derived is based on a prism [13,14]. Many works have conceived techniques to derive the number of spanning tree of a graph, which can be found at [15][16][17][18][19][20][21][22][23]. Now we introduce following Lemma which describes a way to calculate the number of spanning trees by an extension of Kirchhoff formula.…”