Solutions for Homework 1 Graph Theory and Complex Networks ...

3 k-Regular GraphsFigure 1: Labelled graphs G 1 and G 3 .We observe that the the number of vertices n = 2k + 3 is odd, and constructthe Harary graph for odd n and odd k. So, Harary graph H k,2k+3 is G = (V, E)with V = {0, 1, 2, . . . , 2k + 2}. Its edge set E(G) can be constructed as follows:1. E 1 consists of the edges that connect every vertex i to its k/2 left-handneighbors and to its k/2 right-hand neighbors;2. E 2 = {〈0, k + 1〉, 〈1, k + 2〉, . . . , 〈k + 1, 2k + 2〉}.So, the resulting edge set is E(G) = E 1⋃E2 .4 Feestje!People at the party can be represented as a graph with 55 nodes, and each hasexactly 5 edges. By the sum-degree theorem, the total number of edges wouldhave to be 55×52, which is not an integer. Hence, such a graph does not exist.5 Mathematica, basic functionalitySee solutions-1.nb file.2