Graph families defined by their automorphisms
distance-transitive distance-regular strongly regular
symmetric (arc-transitive) t-transitive, t ≥ 2 skew-symmetric
(if connected)
vertex- and edge-transitive
edge-transitive and regular edge-transitive
vertex-transitive regular (if bipartite)
biregular
Cayley graph zero-symmetric asymmetric

In graph theory, a regular graph is a graph where each vertex has the same number of neighbors; i.e. every vertex has the same degree or valency. A regular directed graph must also satisfy the stronger condition that the indegree and outdegree of each vertex are equal to each other. A regular graph with vertices of degree k is called a k‑regular graph or regular graph of degree k. Also, from the handshaking lemma, a regular graph contains an even number of vertices with odd degree.

Regular graphs of degree at most 2 are easy to classify: a 0-regular graph consists of disconnected vertices, a 1-regular graph consists of disconnected edges, and a 2-regular graph consists of a disjoint union of cycles and infinite chains.

A 3-regular graph is known as a cubic graph.

A strongly regular graph is a regular graph where every adjacent pair of vertices has the same number l of neighbors in common, and every non-adjacent pair of vertices has the same number n 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 Km is strongly regular for any m.

A theorem by Nash-Williams says that every k‑regular graph on 2k + 1 vertices has a Hamiltonian cycle.

• 0-regular graph
• 1-regular graph
• 2-regular graph
• 3-regular graph

## Existence

It is well known that the necessary and sufficient conditions for a $k$ regular graph of order $n$ to exist are that $n\geq k+1$ and that $nk$ is even.

Proof: As we know a complete graph has every pair of distinct vertices connected to each other by a unique edge. So edges are maximum in complete graph and number of edges are ${\binom {n}{2))={\dfrac {n(n-1)}{2))$ and degree here is $n-1$ . So $k=n-1,n=k+1$ . This is the minimum $n$ for a particular $k$ . Also note that if any regular graph has order $n$ then number of edges are ${\dfrac {nk}{2))$ so $nk$ has to be even. In such case it is easy to construct regular graphs by considering appropriate parameters for circulant graphs.

## Algebraic properties

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.

There is also a criterion for regular and connected graphs : a graph is connected and regular if and only if the matrix of ones J, with $J_{ij}=1$ , is in the adjacency algebra of the graph (meaning it is a linear combination of powers of A).

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

$D\leq {\frac {\log {(n-1))){\log(\lambda _{0}/\lambda _{1})))+1.$ ## Generation

Fast algorithms exist to enumerate, up to isomorphism, all regular graphs with a given degree and number of vertices.