On the structure of almost Moore digraphs containing selfrepeats

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1 On the structure of almost Moore digraphs containing selfrepeats E.T. Baskoro 1, Y.M. Cholily 1, M. Miller 2 1 Department of Mathematics Institut Teknologi Bandung Jl. Ganesa 10 Bandung 40132, Indonesia {ebaskoro,yus}@dns.math.itb.ac.id 2 School of Information Technology and Mathematical Sciences University of Ballarat P.O. Box 663. Ballarat, Vic. 3353, Australia m.miller@ballarat.edu.au Abstract. An almost Moore digraph G of degree d > 1, diameter k > 1 is a diregular digraph with the number of vertices is one less than the Moore bound. If G is an almost Moore digraph, then for each vertex u V (G) there exists a vertex v V (G), called repeat of u and denoted by r(u) =v, Such that there are two walks of lenght k from u to v. The smallest positive integer p such that the composition r p (u) =u is called the order of u. If the order of u is 1 then u is called a selfrepeat. It is known that if G is an almost Moore digraph then G contains exactly k selfrepeats or none. In this paper, we present the possible vertex orders of an almost digraph containing selfrepeats for d 4, k 3. 1 Introduction Let G be a directed graph (digraph). Unless otherwise stated in this paper, all digraphs G in this paper are finite and connected. We use standard notation V (G) for the set of vertices and A(G) for the set of edges of G. An important problem in graph theory is the degree/ diameter problem. Determine the largest number of vertices n of G, for given maximum out-degree d and diameter at most k. It is easy to see that a natural upper bound, called the Moore bound and denoted by Permanent address : Department of Mathematics, Muhammadiyah University of Malang. Jl. Tlogomas 246 Malang Indonesia. yus@umm.ac.id

2 2 E.T. Baskoro, Y.M. Cholily, M. Miller M d,k, for n can be found by counting the maximum possible number of vertices in spanning tree argument, that is, n M d,k =1+d + d d k. It is well known that digraphs with the number of vertices equal to M d,k do not exists for d>1andk>1 (for a proof see [6] or [15]). The non existence of Moore digraphs for d>1 and k>1 motivates us to study the existence of digraphs with orders close to the Moore bound for d 2, k 2. Since Moore digraphs do not exist for d 2andk 2theexistence of a digraph of maximum out-degree d 2, diameter k 2 and order M d,k 1 becomes an interesting problem. Such digraphs (if any) will be denoted by (d, k)-digraphs. By an easy counting argument we can show that every vertex of (d, k)-digraphs must have the same out-degree d. Furthermore, Miller et al. [14] showed that any vertex in a (d, k)-digraph must also have the same in-degree d. Therefore a (d, k)-digraph (if any) must be diregular of degree d, for k 2. Hovewer, the order of a (d, k)-digraph G one less than the Moore bound then every vertex u V (G) there exists exactly one vertex v V (G) such that there are two walks of length k from u to v. Such a vertex v is called the repeat of u, denoted by r(u) =v. In case r(u) =u, vertex u is called a selfrepeat (the two walks, in this case, have lengths 0 and k). Moreover, Baskoro, Miller and Plesník [3] showed that the function r is an automorphism on V (G). For any integer p 1, the repeat function defines a composition r p (v) =r(r p 1 (v)) with r 0 (v) =v. Then, for every vertex v of G, there exists a smallest natural number ω(v), called the order of v, such that r ω(v) (v) =v. Hence, ω(v) is the length of the permutation cycle containing v. The study of the existence of (d, k)-digraphs has received much attention. Some partial results on the existence of (d, k)-digraphs have been obtained. Fiol, Allegre and Yebra pointed out in [8] that (d, 2)-digraphs for any d 2 exist, for example Kautz digraphs (line digraphs of complete digraphs). In particular, for d = 2, Miller and Fris [12] showed that there are exactly three non-isomorphic (2, 2)- digraphs. Furthermore, Gimbert [11] showed that there is only one

3 On the structure of almost Moore digraphs containing selfrepeats 3 (d, 2)-digraph, namely the line digraph L(K d+1 ) of the complete digraph K d+1, for d 3. For diameter k 3, it is known that there are no (2,k)-digraphs [13]. Recently, it has been proved that there are no (3,k)-digraphs with k 3 [4]. Thus, the remaining case still open to investigate is the existence of (d, k)-digraphs when d 4 and k 3. Some necessary conditions for the existence of (d, k)-digraphs have been obtained (see [2],[3],[10]). One such condition is that any (d, k)-digraph contains at most one C k, k 3[3].Gimbertused sophisticated algebraic methods to prove that the number of permutation cycles of even length must be a multiple of k [10]. Some new conditions for the existence of (d, k)-digraphs in terms of the orders of their vertices we presented in [7]. In addition, in [1] the structure of a (d, k)-digraph which contains selfrepeats has been determined for diameter k = 3. In this paper we deal with the case k 4. We present an explicit formula for enumerating vertices of all orders present in a (d, k)-digraph based on the given information of the repeat structure of out-neighbours of any one selfrepeat. This is given as Theorem 2 in section 3. 2 Preliminaries Let v be a vertex of a (d, k)-digraph G. For integer i, letn i (v) be a set of vertices at distance i from v, ifi 0, and at distance i to v, ifi 0. Hence, for particular cases i =0andi = ±1 wehave N 0 (v) ={v}, N 1 (v) =N + (v) isasetsofout-neighbourhood of v, and N 1 (v) =N (v) in-neighbourhood of v. Let u, v be two vertices of a (d, k)-digraph G. Since the diameter of G is k then there exists at least one walk of length k from u to v. The following lemmas give the upper bound on the orders of internal vertices in a walk connecting u and v, in terms of the orders of u and v. Moreprecisely,ifω(u) =m and ω(v) =n then the least common multiple of m and n, denoted by lcm(m, n), is an upper bound. Lemma A.[7] Let u, v be two vertices of a (d, k)-digraph G with ω(u) =m, ω(v) =n. IfW is a walk of length <kconnecting u to v in G then the orders of each internal vertex in W must divide

4 4 E.T. Baskoro, Y.M. Cholily, M. Miller lcm(m, n). If u is not a repeat of v, then the following lemma give the results as Lemma A, except that the walk W is longer than in Lemma A. Lemma B.[7] Let u, v be two vertices of a (d, k)-digraph G where ω(u) =m, ω(v) =n and r(u) v. IfW is a walk of length k connecting u to v in G then the order of each internal vertex in W must divide lcm(m, n). Considering both Lemma A and Lemma B, it is easy to see that for a special case when u and v have the same order and ω(u) isthe smallest order in V (G) then all vertices in a walk connecting u and v also have the same order as u. From Lemma B we shall derive the following corollary. Let v be a selfrepeat of a (d, k)-digraph G. Hence, there exist d pairwise internally disjoint walks (v, z 1,,y 1,v), (v, z 2,,y 2,v),, (v, z d,,y d,v)oflengths k+1. Applying Lemma B on (v, z 1,,y 1 ), we get ω(z 1 ) ω(y 1 ). Applying Lemma B to (z 1,,y 1,v)weobtain ω(y 1 ) ω(z 1 ). Thus ω(z 1 )=ω(y 1 ). Similarly, we will get ω(z i )=ω(y i ) for any i =2, 3,,d. Hence the following corollary holds. Corollary A. [7] For any selfrepeat v in a (d, k)-digraph, the permutation r of N + (v) has the same cycle structure as the permutation r of N (v). Let v be a selfrepeat of a (d, k)-digraph G. Recently, [7] has pointed out that if N + (v) contains a vertex of order m and a vertex of order n then G must also contains a vertex of order lcm(m, n). In addition, [1] gives the structure of orders of all vertices in digraph G for d 4andk = 3 based on assuming that the repeat structure of an out-neighborhood of any selfrepeat is given. In this paper we shall generalize this result for any diameter k 3. 3 The Results A method to enumerate the orders of vertices of a (d, 3)-digraph, d 4, containing selfrepeats was given in [1]. In this section we shall

5 On the structure of almost Moore digraphs containing selfrepeats 5 discuss the extension of the digraph for degree d 4 and arbitrary diameter k 3. Therefore, from now on, we consider G to be a (d, k)-digraph, d 4, k 3 containing selfrepats. Let v 0 be a selfrepeat of G and let N + (v 0 ) consist of permutation cycles with lengths 1=s 0,s 1,s 2,,s t and multiplicities 1 = m 0,m 1,m 2,,m t.from this, it is clear that d =1+ t m i s i. i=1 Before continuing our discussion, we shall introduce some definitions of sets and their notations. We denote S 1, the set of all vertices orders of N + (v 0 ). Thus, S 1 = {s 0,s 1,,s t }. Next, we define a set, denoted by S 2, as the set of all lcm(s i,s j )wheres i,s j S 1 and lcm(s i,s j ) / S 1.Later,ifS 2 then the i th element of S 2 will be denoted by s 2,i. In general, we can continue to define a set S m, 3 m k, as the set of the least common multiples of any m vertex orders of S 1 but including only those lcm wich are not already members of any s i, i<m. Similarly, if S m then i th element of S m will be denoted by s m,i. Since the diameter of G is k and v 0 is a selfrepeat, then by a spanning tree argument it is easy to see that V (G) = k i=0 N i (v 0 ). Clearly, for i = 0, the set S 0 = N 0 (v 0 )={v 0 } consists of a selfrepeat vertex, v 0,only.Fori =1,S 1 = N 1 (v 0 )=N + (v 0 ) consists of a selfrepeat vertex and m i s i vertices of order s i, i =0, 1, 2,,t. From this information, we shall provide a general formula giving the number of vertices in G order of vertices of G. Henceforth, we use the notation n(p, q) to denote the number of vertices of order p in N q (v 0 ), q 0. Clearly, N i (v 0 ) contains exactly one selfrepeat vertex for every i<kbut N k (v 0 ) does not. Hence, we can summarize using our notations: { 1 for i =0, 1,,k 1, n(1,i)= (1) 0 for i = k. Apart from selfrepeat vertices, we also give the orders of vertices of N + (v 0 ), i.e. n(s i, 1) = m i s i, for i =1, 2,,t. (2) Later in this paper, we will consider orders in N i (v 0 ), 2 i k. The possible vertex orders in N i (v 0 ), 2 i k, are as follows.

6 6 E.T. Baskoro, Y.M. Cholily, M. Miller Theorem 1. If u N i (v 0 ), 2 i k, thenω(u) Proof. We shall use induction on i, i =2, 3,,k.Letu N 2 (v 0 ) then there exists x N + (v 0 ) such that u N + (x). By Lemma B on walk (v 0,x,u), ω(u) must be a multiple of ω(x) S 1.Sincethe diameter of G is k then there exists a walk (u,,y,v 0 )oflength k from u to v. By Corollary A, ω(y) S 1. Therefore, Lemma B on walk (u,,y,v) we conclude that ω(u) must be a multiple of ω(y). On the other hand, applying Lemma B to walk (x, u,,y) of length k, ω(u) must divide lcm(ω(x),ω(y)). Thus, ω(u) must divide the least common multiple of ω(x) S 1 and ω(y) S 1.Thus ω(u) S 1 S 2. To complete the proof, assume the theorem holds for i, where 2 <i<k.letubeavertex of N i+1 (v 0 ). Similarly to the proof for i = 2, there exists a walk (x, u,,y)oflength k from x N i (v 0 )toy N (v 0 )andω(u) =lcm(ω(x),ω(y)). Since x N i (v 0 ) then, by our assumption, ω(x) i j=1 i j=1 S j. S j. By Corollary A, ω(y) S 1. Therefore, we have ω(u) =lcm(ω(x),ω(y)) where ω(x) ω(y) S 1. This concludes the proof. i j=1 S j and Now, we shall enumerate the number of vertices of G according to their possible orders. Since V (G) = k N i (v 0 ), then the enumeration process will be done for every level i of N i (v 0 ). The number of vertices of vertex orders present in levels 0, 1, and 2 was given in [1]. Equation 1 enumerate the number of selfrepeat vertex present in every levels i =1, 2,,k. Equation 2 enumerate the number of non selfrepeat vertices of orders s i for i =1, 2,,t. The non-selfrepeat vertices of N 2 (v 0 )are i=0 n(s i, 2) = m i s i + n(s i, 1) + n(s i, 1)s i + n(s i, 1) m i s i s j s i,j i n(s j, 1) + s j s i,j i m j s j + lcm(s j,s l )=s i n(s j, 1)m l s l. (3)

7 On the structure of almost Moore digraphs containing selfrepeats 7 n(s 2,i, 2) = n(s j, 1)m l s l. (4) lcm(s j,s l )=s 2,i Next, we shall enumerate vertex orders of N i (v 0 ) for 3 i k. Let u N i (v 0 ) and consider a walk (x, u,,y)oflength k from x N i 1 (v 0 )toy N (v 0 ). Furthermore, ω(u) =lcm(ω(x),ω(y)). By Theorem 1 ω(x) i 1 S l and by Corollary A, ω(y) S 1. Hence, l=1 we have i 1cases,thatis,ω(x) S l and ω(y) S 1 for l = 1, 2,,i 1. Since 3 i k then N i (v 0 ) usually contains vertices of orders s j for each j {0, 1, 2,,t}.Asweknowthatforj =0,s 0 is order of selfrepeat vertex, has given in Eq.(1) then let we consider for j {1, 2,,t}. Therefore, ω(u) =s j for some j {1, 2,,t} can occur from exactly the following six cases (i) ω(x) = 1 and ω(y) =s j, (ii) ω(x) =s j and ω(y) = 1, (iii) ω(x) =s j and ω(y) =s j, (iv) ω(x) =s j and ω(y) =s l s j, l j, (v)ω(x) =s l s j, l j and ω(y) =s j and the last (vi) ω(x) s j, ω(y) s j but lcm(ω(x),ω(y)) = s j. From the six cases, then the number of vertices of order s j, j = 1, 2,,t,inN i (v 0 ), i {3, 4,,k} is : n(s j,i)=m j s j +n(s j,i 1)+n(s j,i 1)s j +n(s j,i 1) s l s j,l j m l s l + m j s j n(s l,i 1) + n(s l,i 1)m p s p. (5) s l s j,l j lcm(s l,s p)=s j If S m, m =2, 3,,i then, by Theorem 1, N i (v 0 )contains vertices of order s m,j S m. Consider again the walk (x, u,,y)of length k and, for m =2, 3,,i 1, let ω(u) =s m,j, for some s m,j S m. Then the order s m,j can occur in four different ways, namely, ω(x) =s m 1,q and ω(y) =s l where lcm(s m 1,q,s l )=s m,j, special case for m = 2 the order s m 1,j means s j, ω(x) =s m,j and y is a selfrepeat, ω(x) =s m,j and ω(y) =s l,wheres l s m,j,and

8 8 E.T. Baskoro, Y.M. Cholily, M. Miller ω(x) =s m,p and ω(y) =s l where lcm(s m,p,s l )=s m,j. From the four cases, the number of vertices of order s m,j on N i (v), i =3, 4,,k and m =2, 3,,i 1is: n(s m,j,i)= lcm(s m 1,q,s l )=s m,j n(s m 1,q,i 1)m l s l + n(s m,j,i 1)+ n(s m,j,i 1) s l s m,j m l s l + lcm(s m,p,s l )=s m,j n(s m,p,i 1)m l s l. (6) In addition to this, if S i then the set N i (v 0 ) contains vertices of order s i,j. This order will appear if there exists s i 1,p S i 1 and s l S 1 such that lcm(s i 1,p,s l )=s i,j. Hence the number of vertices of order s i,j S i is n(s i,j,i)= lcm(s i 1,p,s l )=s i,j n(s i 1,p,i 1)m l s l. (7) From the above investigation, we have the following theorem. j=1 Theorem 2. Let G be a (d, k)-digraph, d 4 and k 3, containing selfrepeats. Let v be a selfrepeat of G where N + (v) consists of permutation cycles with lengths s 0 =1,s 1,s 2,,c t and multiplicities 1,m 1,m 2,,m t.theng consists of exactly k selfrepeats, k n(s i,j) vertices of order s i, i {1, 2,,t} and k n(s l,i,j) vertices of orders s l,i S l for l =2, 3,,k. Proof. The proof follows directly by summing up the number of vertices of the same order from Eqs.(1)-(7). Next, let α be an order of vertex in G. Ifx V (G) andr α (x) =x then it is easy to see that ω(x) α. Letu, v be two vertices of G where r α (u) =u and r α (v) =v. The following theorem show that u and v have the same number of out-neighbour (in-neighbour) which have order dividing α. The following theorem was first given in [2]. It is included here since we have found a shorter, more elegance proof. j=l

9 On the structure of almost Moore digraphs containing selfrepeats 9 Theorem 3. Let G be (d, k)-digraph containing selfrepeats, d, k 2. Letα be an order of some vertices of G. LetV α = {x V (G) r α = x}. Then, every vertex of V α has the same number of out-neighbours and in-neighbours wich are in V α. Proof. Let α =1,thenV α is C k. Therefore the conclusion is hold. Let α > 1. It is easy to see that V 1 and V 1 V α.letv be a selfrepeat of G. Clearly that v V α.letn + (v) containing 1 d 1 d vertices which are in V α. We shall show that every vertex of V α has the same number of out-neighbour (in-neighbour) which are in V α as v. Letx V α and ω(x) 1. Since the diameter of G is k and v is a selfrepeat then there exists d internally disjoint walks,(x, x i,,v i,v) for i = 1, 2,,d,oflength k + 1 from x to v. By corollary A, N (v) containing d 1 vertices which are in V α. Hence, by using Lemma B on walks, (x, x i,,v i ) for each i = 1, 2,,d, of length k then ω(x i ) must divide lcm(ω(x),ω(v i )). Since x V α, its mean ω(x) α, thenω(x i )must divide lcm(α, ω(v i )) for each i {1, 2,,d}. Therefore N + (x) containing at least d 1 vertices have order divide α. Assuming that N + (x) containing more than d 1 vertices of V α.by considering again on walks (x, x i,,v i ) for each i {1, 2,,d} then N (v) also containing more than d 1 vertices of V α. By Corollary A, it is a contradiction that N (v) andn + (v) contain sam permutation cycles structure. Thus N + (x) contains the same number of vertices which are in V α as N + (v). Similar argument for N (v) contains the same number of vertices which are in V α. Thus the theorem has been proven. As the end of this section, we list some open problem : Problem 1 Do there exist (d, k)-digraphs containing selfrepeats, for d 4, k 3? Problem 2 Do there exist (d, k)-digraphs, d 4 and k 3, without selfrepeats? References 1. E.T. Baskoro, Y.M. Cholily, M. Miller, Structure of repeat cycles in almost Moore digraphs with selfrepeats and diameter 3, Accepted to Bulletin ICA

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