(d, k) digraph problem, that is, to maximize the number of vertices in a digraph of maximum out-degree d and diameter k. By line digraph iterations it is possible ...
400
IEEE
TRANSACTIONS ON COMPUTERS, VOL. c-33, NO. 5, MAY 1984
Line Digraph Iterations and the Digraph Problem MIGUEL A. FIOL, J. LUIS ANDRES YEBRA, Abstract -This paper studies the behavior of the diameter and the average distance between vertices of the line digraph of a given digraph. The results obtained are then applied to the so-called (d, k) digraph problem, that is, to maximize the number of vertices in a digraph of maximum out-degree d and diameter k. By line digraph iterations it is possible to construct digraphs with a number of vertices larger than (d2 - l)/d2 times the (nonattainable) Moore bound. In particular, this solves the (d, k) digraph problem for k = 2. Also, the line digraph technique provides us with a simple local routing algorithm for the corresponding networks.
Index Terms -Communication network, (d, k) graph problem, line digraph, Moore bound, routing algorithm.
AND
IGNACIO ALEGRE DE MIQUEL
k
Manuscript received November 22, 1982; revised May 31, 1983 and
October 10, 1983. The authors are with the Department of Mathematics, E.T.S. de Ing. de Telecomunicacion, Polytechnic University of Barcelona, Spain.
d(x, y).
N2x, YEV
To reduce the number of transit switching stages k and/or k must be small. In fact, sometimes k cannot exceed.a given value. In this context, the (d, k) digraph problem consists of determining the digraph with the largest number of vertices N for given values of the maximum out-degree d and of the diameter k. It is easy to see that N is related to d and k by the Moore bound N
1 + d + d2 +
...
I. INTRODUCTION
R ECENT advances in VLSI technology and microprocessors have had a great impact on the design of computer systems, stimulating research on interconnection networks. These networks can be modeled by graphs whose N vertices represent the processing elements or nodes of the network and whose edges represent the links between them. The graphs thus obtained can be directed or undirected. We are concerned here with directed graphs only, called digraphs for short. A digraph D = (V, E) consists of a set V of points called vertices and a set E of directed edges between vertices of D. If [x, y] is an edge from x to y, we say that x is adjacent to y and also that y is adjacent from x. (We shall assume that D has no parallel edges, that is, that there is at most one edge from each vertex x to any other vertex y.) The network must allow communication between any pair of processing elements. Of course, the complete symmetric digraph KN in which each vertex is adjacent to all others satisfies this condition. But for large values of N this solution is too expensive or even technically unfeasible. So, we are forced to use digraphs with the property that every vertex is adjacent to no more than a few, say d, vertices, but such that there is a (directed) path from any vertex to any other. The minimum number of edges needed to go from a vertex x to another vertex y is called the distance d(x, y) from x to y. (Note that in a digraph d(x,y) is not necessarily equal to d( y, x).) The maximum distance between any pair of vertices is the diameter k of the digraph, while the average distance between vertices of the digraph is defined as
(d, k)
+ d' = NM(d,k) ik+ 1 =
dk+l
d-
-
if d 1
=
1
if d> 1.
1
(1)' The Moore bound is attained for d = 1 by a ring digraph in which every vertex is connected to the following one, and for k = 1 by the complete symmetric digraph Kd+l. However, it is known [3] that this bound cannot be attained for d > 1 andk> 1. Using the results obtained in Section II about the diameter of a line digraph, we construct in Section III a family of digraphs that for any values of d > 1 and k > 1 comes close to the Moore bound. Other constructions are presented in Section IV. Another important requirement that should be given careful consideration is the existence of efficient routing algorithms to find short paths between any two nodes of the network. In Section V we present a simple local routing -algorithm that finds the shortest path between any two vertices of the digraph. IL. LINE DIGRAPH ITERATIONS In the line digraph L(D) of a digraph D = (V, E), each vertex represents an edge of D, that is V(L(D)) = {uv [u, V] E E(D)}; and the vertex uv is adjacent to the vertex wz iff v = w (i.e., when the edge [u, v] is adjacent to the edge [w, z] in D). Fig. 1 shows a digraph D and its line
digraph L(D). The number of vertices in L(D) is
NL
=
IV(L(D))I
=
|E(D)| = veV, d+(v) = E d-(v) VEV
(2)
where d+(v) and d-(v) are the out- and in-degree of vertex
v, respectively.
0018-9340/84/0500-0400$01.00 (© 1984 IEEE
FIOL et
al.: LINE DIGRAPH ITERATIONS
401
To prove (5), note that for any two vertices i,j E V(D) with d(i,j) = m there are at most d-(i) * d+(j) = d2 vertices at distance m + 1 in L(D). To carry out an exact computation we must leave out from this set one vertex each time that there is an edge to i that coincides with an edge from j; that is, for each edge [j, i]. Therefore,
i
\Or,,
L(D) I4
4
kL= NL2
24
40
When D is a d-regular digraph, that is, when d+(v) = d-(v) = d for all vertices v E V, it has |E| = dN edges. Then its line digraph L(D) is also d-regular with
(3)
vertices and d2N edges. And for d > 1 the sequence of line digraph iterations
L(D),L2(D)
=
L(L(D)), * Ln(D)
=
+
1I- E {d(i,j) [j, i]eE(D)
+
I}) (6)
and so
Fig. 1. A digraph and its line digraph.
NL = dN
d2{d(i,j)
i,jeV(D)
L(Ln-(D)), *
k- < >f(d2
i,jeV(D)
d(i,j) + d2N2) = k +
1.
Remarks: 1) This last bound may be improved upon if the girth (= length of the minimum cycle) q of D is known. Then d(i,j) q - 1 whenever [j, i] E E(D) and so >
e o +a(q d(i,j)
Er
[j, il]eE(D)
-[j, i]eE(D)
1)
+(
=
qdN
Therefore, we obtain from (6) is an infinite sequence of d-regular digraphs. We remark now that each vertex x in L2(D) represents an kL 1) therefore it must have far-reaching consequences in conof order N, diameter k, and average distance between vertices nection with the (d, k) digraph problem. To begin with, the maximum number of verticesf(d, k) in k. Then, the order NL, diameter kL, and average distance a d-regular digraph of diameter k must satisfy between vertices kL of L(D) satisfy N~dk+1~
1
f(d,k + 1) df(d,k) (8) (3) (4) because the line digraph L(D) of a d-regular digraph with f(d, k) vertices already has df(d, k) vertices. (5) In particular, from f(d, 1) = d + 1 we obtain Proof: (3) has already been deduced from (2). To prove d(d + 1) S f(d, 2) < NM(d, 2) = d2 + d + 1
NL
:
dN kL= k + 1 kL< k + 1. =
(4), note that to go from the vertex u = ij to the vertex v = pq, v + u, in L(D) is equivalent to going from the edge [i, j] to the edge [p, q ] in D. But the path of minimum length with these two terminal edges contains the shortest path between j and p. Therefore,
dL(D)(u, V) = dD(j,p) + 1 This will imply kL = k + 1 if there are at least two different edges [i, j], p, q ] E E(D) such that d(j, p) = k. But, as d-(j) = d 2, at least one of the d edges [.,j] must be different from [p, q]. ,
so that
f(d, 2)
=
d2 + d
(9)
and this value is attained by L(Kd+I). Also, the digraphs Dkd = Lkl- (Kd+l) are d-regular, with diameter k and number of vertices d2 lk+I1 dk+ld2~ d-1 N =dl(d + 1) = d2 d-l d> =
d2
1
d2NM(d, k)
(10)
402
IEEE TRANSACTIONS ON COMPUTERS, VOL.
c-33,
NO.
5, MAY 1984
D2
02
Fig. 3. A 2-regular digraph of diameter 4 on 25 vertices.
Fig. 2. The digraphs Dl = K3,D2 = L(K3) and D2 = L2(K3).
that is, N is larger than (d2 - 1)/d2 times the (nonattainable) Moore bound. This family of digraphs was already described in [9] as shift register state diagrams. Other constructions can be found in [6] and [11]. We show in Fig. 2 the first three members of the family corresponding to d = 2. We expect that f(d, k) must not be too far from N = (d + 1)dk . We should point out, however, that for d = 2, f(2, k) 25 * 2k4 because the graph of Fig. 3 (found by computer search) has 25 vertices and diameter 4. Finally, we mention that, for solutions of the (d, k) digraph problem, all vertices must have the same out-degree since otherwise there would be a vertex with out degree d+(v) > d - 1, and this implies -
N
-
