J. Comp. & Math. Sci. Vol.4 (6), 419-424 (2013)
Lower Bound of (r, 2, r ( r-1 )) Regular Graph N. R. SANTHI MAHESWARI1 and C. SEKAR2 1
Department of Mathematics, G. Venkataswamy Naidu College, Kovilpatti, INDIA. 2 Department of Mathematics, Aditanar College of Arts and Science, Tiruchendur, INDIA. (Received on: October 28, 2013) ABSTRACT A graph is called ( r , 2, r ( r - 1) )- regular graph if each vertex in the graph G is at a distance one away from exactly r number of vertices and each vertex in the graph G is at a distance two away from exactly r ( r - 1 ) number of vertices. That is, d (v) = r and d2 (v) = r (r-1 ) , for all v in G. In this paper, we have proved that for any r > 0, the order of ( r, 2, r ( r-1)) regular graph is r2 +1 if and only if diam ( G ) = 2 and we have listed out of all ( r, 2, r ( r - 1) ) regular graph of diameter two. We have seen that the lower bound of ( r , 2, r ( r-1 ) )- regular graph is r2 + 1. Also we have suggested a method to construct ( r ,2, r ( r-1 ) )- regular graph using Neil Robertson’s construction of Hoffman singleton graph. Keywords: Distance degree regular graph ; (d, k) - regular graph; (2, k)-regular; (r, 2 , k)-regular graph; girth; diameter; semiregular. Mathematics subject code classification: 05C12.
1. INTRODUCTION In this paper, we consider only finite, simple, connected graphs. We follows graph theoretic terminology proposed in the works of J.A. Bondy and U.S.R. Murty4 and Harary6.We denote the vertex set and edge set of a graph G by V(G) and E(G) respectively. The degree of a vertex v is the number of edges incident at v. A graph G is regular if all its vertices have the same degree.
For a connected graph G, the distance d (u, v) between two vertices u and v is the length of a shortest ( u, v ) path. Therefore, the degree of a vertex v is the number of vertices at a distance 1 from v, and it is denoted by d ( v ). This observation suggests a generalization of degree. That is, d d ( v ) is defined as the number of vertices at a distance d from v. Hence d1 ( v ) = d ( v ) and N d ( v ) denote the set of all vertices that are at a distance d away from v in a graph G. Hence N1( v ) = N ( v ).
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A graph G is distance d – regular3 if every vertex of G has the same number of vertices at a distance d from it. A graph G is said to be ( d , k ) - regular graph if d d ( v ) = k , for all v in G. The (1 , k) – regular graphs are nothing but our k-regular graphs. A graph G is (2 , k ) - regular if d 2 ( v ) = k , for all v € G . We observe that (2 , k) - regular graphs and k – semiregular graphs are same. The concept of the semiregular graph was introduced and studied by Alison pechin northup2. A graph is said to be k semiregular graph if each vertex of G is at a distance two away from exactly k vertices of G. The ( 2 , k ) - regular graph may be regular or non-regular. They are regular graphs which are ( 2 , k ) - regular and non regular graphs which are ( 2 , k ) – regular. Let us denote r- regular graphs which are (2, k) regular by (r, 2, k) - regular graphs13. Girth of a graph is the smallest cycle in the graph and diameter of graph G = max {d ( u , v ) / u ,v in G}. In this paper, we will discuss with lower bound of a (r, 2, r (r-1)) regular graphs and we have listed the (r, 2, r(r-1)) - regular graphs of order r2+1.
6. For any r > 1, a graph G is (r, 2, r(r-1)regular if G is r-regular with girth at least five13. 7. For any n ≥ 5, (n ≠ 6, 8) and any r > 1, there exists a (r, 2, r (r-1)) - regular graph on n x 2r-2 vertices with girth five13. 8. Let k € N be odd, k ≥ s3, k ≠ m2+m+3 and k ≠ m2+m-1 for each nonnegative integer m. Then there is no k-regular graph with girth 5 and k2+312. 9. For every prime power q ≥ 7, there exists a q+2 regular graph of girth 5 with 2(q2-1) vertices9. 10. Let q ≥ 7 be a prime power and let k ≤ q+2. Then there a exists a k-regular graph with girth 5 and with 2(k -1)(q -1) vertices9. 11. Let q ≥ 5 be a prime power and let k ≤ q+2. Then there exists a k - regular graph with girth 5 and with 2q (k-2) vertices9. 12. There is only graph with less than 20 vertices of valency 4 and girth 510.
2. PRELIMINARIES
( r - 1) ) - regular graph, then V (G ) ≥ 1+ r ( r – 1 )+ r = r2+1. That is, the order of (r, 2 , r ( r-1 ) ) - regular graph is at least r2 + 1. Let us look at (r, 2, r(r-1) )- regular graph of order precisely r2+1.
1. Any (r, 2, k) - regular graph has at least k+r+1 vertices[14]. 2. If r and k are odd, then (r, 2, k ) - regular graph has at least k+r+2 vertices14. 3. For any r ≥ 2 and k ≥ 1, G is a (r, 2, k)regular graph of order r+k+1 if and only if diam (G) = 214. 4. For any odd r ≥ 3, there is no (r, 2, 1)regular graph14. 5. For any r > 1, if G is a (r, 2, (r-1)(r-1))regular graph, then G has girth four14.
3. LOWER BOUND OF (r, 2, r ( r-1 )) – REGULAR GRAPHS In our observation , if G is (r, 2, r
Theorem 3.1 Let G be a (r, 2, r (r-1)) regular graph of order r2 +1 if and only if diam ( G ) = 2. Proof Let G be a (r, 2, r ( r - 1 )) regular graph of order r2 +1 if and only if d (v) = r
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N. R. Santhi Maheswari, et al., J. Comp. & Math. Sci. Vol.4 (6), 419-424 (2013)
and d2 (v) = r (r - 1) and V (G ) = r2 +1 if and only if V (G ) = d (v)+ d2 (v)+1, for all v € V(G) if and only if diam ( G ) = 2. 3.2 The (r, 2, r ( r-1 )) - regular graphs of order r2+1 Hoffman singleton11 proved that there exists a regular graph of degree r with r2+1 vertices and diameter two only for r = 2, 3 , 7 and possibly 579. The existence of a graph with r = 57 has not yet been established. Here, we will see three ( r, 2, r (r - 1 ) ) - regular graph with exactly r2 + 1 vertices. 3.3 List of (r, 2, r ( r-1)) - regular graphs of order r2+1 (i). For r = 2, pentagon C5 is unique graph with diameter 2 and girth five of order 5. It is (2, 2, 2 (2 - 1 ) ) regular of order 22+1. (ii). For r = 3, Peterson graph is unique graph with diameter 2 and girth 5 of order 10. It is (3, 2, 3( 3 -1 ) ) regular of order 32 + 1. (iii).For r = 7, Hoffman singleton graph is unique graph with diameter 2 and girth 5 of order 50. It is (7, 2,7 ( 7 - 1 ) ) - regular of order 72 + 1.
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vertex j of Ph is adjacent to vertices j-1, j+1 and vertex j of Qi is adjacent to vertices j-2 , j+2 of Qi . Now join vertex j of Ph to vertex hi+j of Qi .( All indices mod 5 ). We get Hoffman singleton graph. 4.1 Construction of ( r, 2, r ( r-1)) - regular graphs due to Neil Robertson Using the construction of Neil Robertson of Hoffman singleton graph5,11, we can construct r (= 3, 4, 5, 6, ) regular graph of girth 5. For r = 3 , 4 , 5 , 6 ,… we take (r- 2) pentagons Ph (0 ≤ i ≤ r-3) and (r-2) pentagrams Qi (0 ≤ i ≤ r-3) so that vertex j of Ph is adjacent to vertices j-1, j+1 and vertex j of Qi is adjacent to vertices j-2 , j+2 of Qi . Now join vertex j of Ph to vertex hi+j of Qi. (All indices mod 5 ). When r = 3 , we get Peterson graph which is ( 3 , 2, 3 ( 3 -1 ))- regular graph on 10 vertices with girth 5.
When r= 4 , we get ( 4 , 2, 4 ( 4 -1 ))- regular graph on 20 vertices with girth 5.
4. CONSTRUCTION OF HOFFMAN SINGLETON GRAPH9 DUE TO NEIL ROBERTSON We take five pentagons Ph (0 ≤ i ≤ 4) and five pentagrams Qi (0 ≤ i ≤ 4) so that
When r = 5, we get (5, 2, 5 ( 5 - 1 ) )- regular graph on 30 vertices with girth 5.
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N. R. Santhi Maheswari, et al., J. Comp. & Math. Sci. Vol.4 (6), 419-424 (2013)
Fig. (1)
When r = 6, we get (6,2, 6 ( 6 - 1 ) )- regular graph on 40 vertices with girth 5
Qi is adjacent to vertices j-2, j+2 of Qi . Now join vertex j of ph to vertex hi+j of Qi. (All indices mod 5). Joining the vertex 0 of P0 to vertex 0 of Q6 and joining the vertex 0 of p6 to 0 of Q0, we get a graph with four cycle. Therefore, here we cannot have a (8, 2, 8 ( 8-1 ))- regular using this construction.. When n = 5, degree r varies from 3 to 7 only . For any r ( 3 ≤ r ≤ 7 ) , the vertex labeled j ( with 0 ≤ j≤ 4) residing in a pentagon labeled Ph ( 0 ≤ h ≤ r-2 ) and pentagrams labeled Qi (0 ≤ i ≤ r-2 ) so that vertex j of Ph is adjacent to vertices j -1, j+1 and vertex j of Qi is adjacent to vertices j-2, j+2 of Qi ,. now join vertex j of ph to vertex hi+j of Qi.(All indices mod 5). For any r (3 ≤ r ≤ 7), we get (r, 2, r ( r-1 )) regular graph having 10 ( r-2 ) vertices and 5 r ( r-2) edges with girth 5. Theorem. 4.3 If n ≥ 5 (n ≠ 6, 8 ) is an integer such that (3 ≤ r ≤ n+2), then there is a (r, 2, r(r-1))regular graph of order 2n (r-2) vertices and n r(r-2) edges. Proof
Fig. (2)
Remark. 4.2. For r = 8, (construction due to Neil Robertson of Hoffman singleton graph5,11, we take six pentagons Ph and six pentagrams Qi so that vertex j of ph is adjacent to vertices j-1, j+1 and vertex j of
When n = 6, By the construction in 4.2, we have a graph with girth 3 and n = 8, we have a graph with girth 4. Let n ≥ 5 , (n ≠ 6, 8 ) for ( 3 ≤ r ≤ n+2 ), we take the vertex labeled i (with 0 ≤ i ≤ n-1) residing in a labeled Ph (0 ≤ h ≤ r-2) and labeled Qi (0 ≤ i ≤ r-2) so that vertex j of Ph is adjacent to vertices j-1 , j+1 and vertex j of Qi is adjacent to vertices j-2, j+2 of Qi.. Now join the vertex j of ph to vertex hi+j of Qi.(All indices mod n). It is r- regular graph with 2n(r-2) vertices and nr(r-2) edges with girth 5.
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Table 4.8 Order of (r, 2, r ( r - 1))- regular graphs Regular 3-regular 4-regular 5-regular 6-regular 7-regular 8-regular 9-regular 10-regular 11-regular 12-regular 13-regular 14-regular 15-regular 16-regular
Order of (r, 2, r(r1))-regular graph 10 19 30 40 50 80 96 126 156 216 240 288 312 336
First constructed by Peterson Neil Robertson10 Wegner9 O’Keefe and Wong8 Hoffman and singleton5,11 Royle9 LeifKjaer Jorgensen9 Exoo9 LeifKjaer Jorgensen9 LeifKjaer Jorgensen9 Exoo9 LeifKjaer Jorgensen9 LeifKjaer Jorgensen9 Leif Kjaer Jorgensen9
Theorem 4.4 If n ≥ 7 (n ≠9, 12, 15 ) is an integer such that (3 ≤ r ≤ n+2), then there is (r, 2, r(r-1) )regular graph of order 2n(r-2) vertices and nr(r-2) edges with girth 6. Proof When n = 9, 12. By construction in 4.2, we have a graph of girth 3, girth 4 respectively. When n = 15, we have (r, 2, r (r-1))regular graph of order 30 ( r-2) vertices and 15r(r-2) edges with girth 5. Let n ≥ 7 ( n ≠ 9 , 12 , 15 ) is an integer. For ( 3 ≤ r ≤ n+2 ), the vertex labeled i (with 0 ≤ i ≤ n-1) residing in a labeled Ph (0 ≤ h ≤ r-2) and labeled Qi (0 ≤i ≤ r-2) so that vertex j of Ph is adjacent to vertices j-1 , j+1 and vertex j of Qi is adjacent to vertices j-3, j+3 of Qi. now join vertex j of ph to vertex hi+j of Qi .(All indices mod n). It is ( r, 2, r ( r-1) ) - regular graph
Order of the graph in theorem 4.3 10 (n = 5) 20 (n = 5) 30 (n = 5) 40 (n = 5) 50 (n = 5) 84 (n =7) 98 (n = 7) 128 (n = 8), girth 6 162 (n = 9) 200 (n = 10) Smallest order 242 (n = 11) 288 (n = 12) same 338 (n = 13) 392 (n = 14)
of order 2n ( r-2 ) vertices and n r ( r-2) edges with girth 6. Theorem 4.5 If n ≥ 9 ( n ≠12, 16, 20, 24 ) is an integer such that ( 3 ≤ r ≤ n+2 ), then there is ( r, 2, r ( r – 1 )- regular graph of order 2 n ( r – 2 ) vertices and n r (r-2) edges with girth 7. Proof When n = 12, 16 . By construction in 4.2, we have a graph with girth 3 and girth 4 respectively. when n = 20, 24. we have (r, 2, r(r-1))- regular of order 2n(r-2) vertices and nr(r-2) edges with girth 5 and girth 6 respectively. Let n ≥ 9, ( n ≠ 12 ,16 , 20 , 24 ) is an integer. For (3 ≤ r ≤ n + 2), the vertex labeled i ( with 0 ≤ i ≤ n-1) residing in a labeled Ph (0 ≤ h ≤ r-2) and labeled Qi (0 ≤ i
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≤ r-2) so that vertex j of Ph is adjacent to vertices j-1 , j+1 and vertex j of Qi is adjacent to vertices j-4, j+4 of Qi. now join vertex j of ph to vertex hi+j of Qi.(All indices mod n). It is (r, 2, r(r-1))- regular of order 2n(r-2) vertices and nr(r-2) edges with girth 7. That is, If n ≥ 9 (n ≠ 12, 16, 20, 24) is an integer such that (3 ≤ r ≤ n+2), then there is (r, 2, r(r-1))- regular graph of order 2n(r-2) vertices and nr(r-2) edges with girth 7. From the theorem 4.3 , theorem 4.4 and theorem 4.5 we can generalize. Theorem 4.6 If n ≥ 5 and ( 2 ≤ k < n/2 ) and n ≠ ( 2 + t ) k , 1 ≤ t ≤ 2 and for any r ≥ 2, there is an ( r, 2, r ( r-1 ))- regular graph on 2 n (r-2) vertices and n r ( r - 2 ) edges with girth k + 3 . In particular , when n = ( 2 + t ) k , (3 ≤ t ≤ k ), we can get (r, 2, r ( r – 1 ))regular of order 2 n ( r-2 ) vertices and n r ( r - 2 ) edges with girth ( 2 + t ). Remark 4.7 If n = ( 2 + t ) k , ( t = 1, 2 ) we have graphs with girth 3,4 respectively. Therefore we cannot have ( r, 2, r(r-1))- regular for n = (2 + t) k, for (t = 1,2) using this construction. REFERENCES
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Journal of Computer and Mathematical Sciences Vol. 4, Issue 6, 31 December, 2013 Pages (403-459)