arXiv:1801.08345v1 [math.CO] 25 Jan 2018 rmo h ae-aiicieinieaiey odtrieteexis the determine to iteratively, criterion Havel-Hakimi the or orem 15)adHkm 16)dsrb loihsfrsligtegraph the solving for algorithms problem. describe tion Tripat (1962) Erd˝os-Gallai theore and the Hakimi (1994) of and Triesch proofs (1955) and different provide Aigner (2003) (1986), Vijay Choudum graphic. be h rp elzto rbe tde h xsec fagahon graph a of existence the studies problem realization graph The Introduction 1 rvddncsayadsffiin odtosfrtesequence the for conditions sufficient and necessary provided ihagvndge sequence given a with ∗ If E-Mail: xsec fcnetdrglradnearly and regular connected of Existence d,teei once nearly connected a is there odd, see,teei connected a is there even, is nay 00,6E0 01,91D30. 90B15, 62E10, 60C05, Classification: ondary: Subject 2000 AMS existence. graphs, words: Key d i = o integers For [email protected] k o l 1 all for e okUiest,AuDhabi Abu University, York New once eua rps once eryregular nearly connected graphs, regular Connected hrmrhnGanesan Ghurumuruhan k ≥ ≤ and 2 i eua graphs regular ≤ n, d n 1 k ≥ ti osbet s h Erd˝os-Gallai The- the use to possible is it ≥ Abstract − k eua rp on graph regular d k 1 + 2 1 − ≥ eua rp on graph regular , epoetefloig If following: the prove we . . . rmr:6J0 03;Sec- 60K35; 60J10, Primary: ≥ d n . r˝ sadGla (1960) Gallai Erd˝os and n etcs If vertices. n ∗ vertices. n Havel and m { n d · i n n } k 1 · vertices ≤ realiza- is k iand hi i ≤ tence n to of k−regular graphs on n vertices. In this paper, we directly prove the exis- tence of connected k−regular graphs on n vertices for all permissible values of n ≥ k +1, using induction. We first have some definitions. Let n ≥ 2 be any integer and let Kn the with vertex set V = {1, 2,...,n} and containing edges between all pairs of vertices. The edge between vertices i and j in Kn is referred to as (i, j). Let G =(V, E) be any subgraph of Kn with edge set E. Vertices u and v are adjacent in G if the edge (u, v) ∈ E. A path P = (v1,...,vt) in G is a sequence of distinct vertices such that vi is adjacent to vi+1 for 1 ≤ i ≤ t − 1. The vertices v1 and vt are then connected by a path in G. The graph G is said to be connected if any two vertices in G are connected by a path in G (Bollobas (2002)). The degree of a vertex v in the graph G is the number of vertices adjacent to v, in G. The graph G is called a k−regular graph if the degree of each vertex in G is exactly k. If G is a k−regular graph on n vertices, then the product n · k is sum of degrees of the vertices in G which in turn is twice the number of edges in G. Therefore it is necessary that either n or k is even. Finally, the graph G is said to be a nearly k−regular graph on n vertices if n − 1 vertices have degree k and one vertex has degree k − 1. The following is the main result of the paper.

Theorem 1. The following statements hold. (a) For all even integers k ≥ 2 and all integers n ≥ k +1, there is a con- nected k−regular graph on n vertices. (b) For all odd integers k ≥ 3 and all even integers n ≥ k +1, there is a connected k−regular graph on n vertices. (c) For all odd integers k ≥ 3 and all odd integers n ≥ k +2, there is a connected nearly k−regular graph on n vertices.

The paper is organized as follows. In Section 2, we prove Theorem 1.

2 Proof of Theorem 1

The following fact is used throughout. (a1) If the minimum degree of a vertex in a graph G is δ ≥ 2, then there exists a path in G containing δ edges. Proof of (a1): Let P =(v1,...,vt) be the longest path in G; i.e., P is a path containing the maximum number of edges. Since P is the longest path, all

2 the neighbours of v1 in G belong to P. Since v1 has at least δ neighbours in G, we must have t ≥ δ +1 and so P has at least δ edges.

We prove (a), (b) and (c) in Theorem 1 separately below. Proof of (a) : For n = k +1, the complete graph Kn is a connected k−regular graph on n vertices, with k even. Suppose Gn(k) is a connected k−regular graph with vertex set {1, 2,...,n}, for some n ≥ k +1. Use property (a1) and let Pn(k) be a path in Gn(k) containing k edges. Since k is even, k k there are 2 vertex disjoint edges ei = (ui, vi), 1 ≤ i ≤ 2 in Pn(k); i.e., k the set k {ui, vi} has k distinct vertices. Remove the edges ei, 1 ≤ i ≤ S1≤i≤ 2 2 and add k new edges

k 2 [{(n +1,ui)} [{(n +1, vi)}. i=1 The resulting graph is a connected k−regular graph with vertex set {1, 2,...,n +1} and is defined to be Gn+1(k).

Proof of (b) : The proofis analogous as in the case of(a). For n = k+1, the complete graph Kn is a connected k−regular graph on n vertices with k odd. Suppose Gn(k) is a connected k−regular graph with vertex set {1, 2,...,n}, for some even n ≥ k +1. We use Gn(k) to construct a connected k−regular graph Gn+2(k) with vertex set {1, 2,...,n +2} as follows. By property (a1), the graph Gn(k) contains a path Qn(k)=(q1,...,qk+1) consisting of k edges {(qj, qj+1)}1≤j≤k. Remove the k−1 edges {(qj, qj+1)}1≤j≤k−1 and add the following edges: (i) For 1 ≤ j ≤ k − 1, j odd, add the edges {(n +1, qj), (n +1, qj+1)}. (ii) For 1 ≤ j ≤ k − 1, j even, add the edges {(n +2, qj), (n +2, qj+1)}. (iii) Add the edge (n +1, n + 2). Since k is odd, the total number of edges added in step (i) is k − 1 and so there are k − 1 edges with n + 1 as an endvertex after step (i). Similarly, after step (ii) there are k − 1 edges with n + 2 as an endvertex. Finally, the resulting graph after step (iii) is a connected k−regular graph with vertex set {1, 2,...,n +1, n +2} and is defined to be Gn+2(k).

Proof of (c) : Let k ≥ 2 be odd and let n ≥ k + 2 be odd. We use the graph Gn−1(k) with vertex set {1, 2,...,n − 1} obtained in (b)

3 above to construct the graph Gn(k) with vertex set {1, 2,...,n}. From prop- erty (a1), the graph Gn−1(k) contains a path Sn−1(k) consisting of k edges. k−1 k−1 Since k is odd, there are 2 vertex disjoint edges fi = (xi,yi), 1 ≤ i ≤ 2 in Sn−1(k); i.e., the set k−1 {xi,yi} has k − 1 distinct vertices. Remove S1≤i≤ 2 k−1 the edges fi, 1 ≤ i ≤ 2 and add k − 1 new edges

k−1 2 [{(n, xi)} [{(n, yi)} i=1 and define the resulting graph to be Gn(k). By construction, Gn(k) is con- nected, the vertex n has degree k − 1 and the rest of all the vertices have degree k.

Acknowledgement I thank Professors Rahul Roy and Federico Camia for crucial comments and for my fellowships.

References

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