Regular graph

Regular graphs of degree at most 2 are easy to classify: a graph consists of disconnected vertices, a graph consists of disconnected edges, and a graph consists of a disjoint union of cycles and infinite chains. In analogy with the terminology for polynomials of low degrees, a or graph often is called a cubic graph or a quartic graph, respectively. Similarly, it is possible to denote k-regular graphs with k=5,6,7,8,\ldots as quintic, sextic, septic, octic, et cetera. A strongly regular graph is a regular graph where every adjacent pair of vertices has the same number of neighbors in common, and every non-adjacent pair of vertices has the same number of neighbors in common. The smallest graphs that are regular but not strongly regular are the cycle graph and the circulant graph on 6 vertices. The complete graph is strongly regular for any . Image:0-regular_graph.svg|0-regular graph Image:1-regular_graph.svg|1-regular graph Image:2-regular_graph.svg|2-regular graph Image:3-regular_graph.svg|3-regular graph ==Properties==

By the degree sum formula, a -regular graph with vertices has \frac{nk}2 edges. In particular, at least one of the order and the degree must be an even number. A theorem by Nash-Williams says that every graph on vertices has a Hamiltonian cycle. Let A be the adjacency matrix of a graph. Then the graph is regular if and only if \textbf{j}=(1, \dots ,1) is an eigenvector of A. Its eigenvalue will be the constant degree of the graph. Eigenvectors corresponding to other eigenvalues are orthogonal to \textbf{j}, so for such eigenvectors v=(v_1,\dots,v_n), we have \sum_{i=1}^n v_i = 0. A regular graph of degree k is connected if and only if the eigenvalue k has multiplicity one. The "only if" direction is a consequence of the Perron–Frobenius theorem. Let G be a k-regular graph with diameter D and eigenvalues of adjacency matrix k=\lambda_0 >\lambda_1\geq \cdots\geq\lambda_{n-1}. If G is not bipartite, then : D\leq \frac{\log{(n-1)}}{\log(\lambda_0/\lambda_1)}+1. == Existence ==

There exists a k-regular graph of order n if and only if the natural numbers and satisfy the inequality n \geq k+1 and that nk is even. Proof: If a graph with vertices is -regular, then the degree of any vertex v cannot exceed the number n-1 of vertices different from v, and indeed at least one of and must be even, whence so is their product. Conversely, if and are two natural numbers satisfying both the inequality and the parity condition, then indeed there is a -regular circulant graph C_n^{s_1,\ldots,s_r} of order (where the s_i denote the minimal `jumps' such that vertices with indices differing by an s_i are adjacent). If in addition is even, then k = 2r, and a possible choice is (s_1,\ldots,s_r) = (1,2,\ldots,r). Else is odd, whence must be even, say with n = 2m, and then k = 2r-1 and the `jumps' may be chosen as (s_1,\ldots,s_r) = (1,2,\ldots,r-1,m). If n=k+1, then this circulant graph is complete. == Generation ==

Fast algorithms exist to generate, up to isomorphism, all regular graphs with a given degree and number of vertices. == See also ==