The Consecutive-4
Digraphs are Hamiltonian
Gerard J. Chang, Frank K. Hwang, and Li–Da TongDEPARTMENT OF APPLIED MATHEMATICS NATIONAL CHIAO TUNG UNIVERSITY HSINCHU 30050, TAIWAN
Received March 14, 1997; revised September 1, 1998
Abstract: Du, Hsu, and Hwang conjectured that consecutive-d digraphs are Hamiltonian for d = 3, 4. Recently, we gave an infinite class of consecutive-3 digraphs, which are not Hamiltonian. In this article we prove the conjecture for
d = 4. c 1999 John Wiley & Sons, Inc. J Graph Theory 31: 1–6, 1999 Keywords: Hamiltonian circuit; consecutive-ddigraph; network; loop
1. INTRODUCTION
Define G(d, n, q, r), also known as a consecutive-d digraph, to be a digraph whose nnodes are labeled by the residues modulo n, and a link i → j from node i to node j exists if and only if j ∈ {qi + k (mod n): r ≤ k ≤ r + d − 1}, where 1 ≤ q ≤ n − 1, 1 ≤ d ≤ n − 1and 0 ≤ r ≤ n − 1 are given. Many computer networks and multiprocessor systems use consecutive-d digraphs for the topology of their interconnection networks. For example, q = 1 yields the multiloop networks [13], also known as circulant digraphs [14], with the skip set {r, r + 1, . . . , r + d − 1}. q = d and r = 0 yields the generalized de Bruijn digraphs [8, 12], and q = r = n − dyields the Imase–Itoh digraphs [9].
In some applications, it is important to know whether a consecutive-d digraph embeds a Hamiltonian circuit. This issue was first raised by Pradhan [11]. Neces-sary and sufficient conditions for generalized de Bruijn digraphs and the Imase–Itoh digraphs to be Hamiltonian were given by Du, Hsu, Hwang, and Zhang [5]. For
Contract grant sponsor: National Science Council Contract grant number: NSC86-2115-M009-002.
c
the case of gcd(n, q) ≥ 2, Du, Hsu, and Hwang [4] showed that G(d, n, q, r) is Hamiltonian if and only if d ≥ gcd(n, q). So, we may only consider the case when gcd(n, q) = 1. Necessary and sufficient conditions for consecutive-d di-graphs to be Hamiltonian were given by Hwang [7] for d = 1 and by Du and Hsu [3] (also see [2]) for d = 2. Furthermore, Du, Hsu, and Hwang [4] proved that consecutive-d digraphs are Hamiltonian for d ≥ 5, and conjectured they are also for d = 3, 4. Du and Hsu [3] gave partial support to this conjecture by proving its validity under the condition q ≤ d. Recently, we [1] gave an infinite class of ex-amples that consecutive-3 digraphs are not necessarily Hamiltonian. In this article, we prove that consecutive-4 digraphs are Hamiltonian, and thus, completely settle the conjecture.
2. SOME GENERAL REMARKS
Throughout this article, we assume that gcd(n, q) = 1. In this case, G(d, n, q, r) is a regular digraph of indegree and outdegree both d. In particular, G(1, n, q, r0) is the disjoint union of cycles.
Let G(4, n, q, r) denote the underlying consecutive-4 digraph. Consider the digraph G(1, n, q, r + 1). Suppose that G(1, n, q, r + 1) consists of c disjoint cycles C1, C2, . . . , Cc. If c = 1, then G(4, n, q, r) is Hamiltonian and we are done. Suppose that c > 1. A link-interchange method was introduced in [4] to merge two cycles. Since 0 < q < n, there exists a cycle with more than one node. Furthermore, this cycle remains to contain more than one node throughout merges. Let i be a node on this cycle such that i + 1 is not. Such an i always exists unless the cycle is Hamiltonian. Suppose that i0 → iand (i + 1)0 → i + 1are in G(1, n, q, r + 1), where i0 6= ibut (i + 1)0 could be i + 1. We replace these two links by the two links i0 → i + 1and (i + 1)0 → iand call this an {i, i + 1} interchange, which merges the two cycles i and i + 1 are on into one. Note that the link i0 → i + 1is in G(1, n, q, r + 2) and the link (i + 1)0→ iis in G(1, n, q, r).
Two interchanges {i, i + 1} and {j, j + 1} do not interfere with each other, if {i, i + 1} ∩ {j, j + 1} = ∅. But if the intersection is not empty, say, j + 1 = i, then doing the interchange {i − 1, i} after {i, i + 1} means replacing (i + 1)0 → i by (i + 1)0 → i − 1, which is in G(1, n, q, r − 1), but not in G(4, n, q, r). However, we can do {i, i + 1} after {i − 1, i}. This is because we are replacing (i − 1)0 → i and (i + 1)0 → i + 1by (i − 1)0 → i + 1and (i + 1)0 → i, where (i − 1)0 → i + 1is in G(1, n, q, r + 3) and (i + 1)0 → iis in G(1, n, q, r). Therefore, we can do two consecutive interchanges, if we do it in the right order, namely, do the smaller pair first. Similarly, if we start with decomposing G(1, n, q, r + 2) into cycles, then we can do two consecutive interchanges, if we do the larger pair first.
We will now represent two consecutive interchanges {i − 1, i} and {i, i + 1} by the set {i − 1, i, i + 1}. In defining an interchange, the two nodes involved are assumed to be on different cycles, and these are cycles updated to previous merges. For example, when the interchange {i, i + 1} is performed after the interchange
{i − 1, i}, then the three nodes i − 1, i, i + 1 are on different cycles originally (if i − 1and i + 1 are on the same cycle, we have no reason to perform the second interchange). A legitimate interchange set (without three consecutive interchanges) can be represented by a set S = {S1, S2, . . . , Ss}, where the Si’s are disjoint and each Siis a subset of two or three consecutive nodes. Note that after one or more interchanges in Siare performed, then all cycles intersecting Siare connected.
Let X and Y be two sets of subsets of {1, 2, . . . , m}. Define Bm(X, Y )to be the bipartite graph with vertex set X ∪Y , and there exists an edge between Xi ∈ X and Yj ∈ Y if and only if Xi∩ Yj 6= ∅. Let Cr+1 (respectively, Cr+2) denote the set of all disjoint cycles in G(1, n, q, r + 1) (respectively, G(1, n, q, r + 2)). Then we have the following.
Lemma 1. G(4, n, q, r)is Hamiltonian if gcd(n, q) = 1 and there exists a legiti-mate interchange set S such that either Bn(S, Cr+1)or Bn(S, Cr+2)is connected.
Proof. Since gcd(n, q) = 1, both G(1, n, q, r + 1) and G(1, n, q, r + 2) are disjoint unions of cycles. Applying the link-interchange method by using the legitimate interchange set S, we can merge the cycles into a Hamiltonian cycle of G(4, n, q, r). Q.E.D.
3. ALGORITHM FOR CONSTRUCTING S
We are unable to find an explicit legitimate interchange set S such that Bn(S, Cr+1) or Bn(S, Cr+2)is connected for all n and q. However, for each given set (n, q, r), we give an algorithm to construct such S. In fact, our construction applies to a more general setting where C1, C2, . . . , Cc do not have to come from G(1, n, q, r + 1) or G(1, n, q, r + 2), but merely a disjoint partition of {1, 2, . . . , n}.
Lemma 2. Let P = {P1, P2, . . . , Pp} be a partition of {1, 2, . . . , m} such that all |Pj| ≥ 2 except one part can be a singleton. Then there exists S = {S1, S2, . . . , Ss},where Si’s are disjoint consecutive subsets of {1, 2, . . . , m} with all |Si| = 2or 3 such that Bm(S, P )is connected and the Sicontaining m (if any) has |Si| = 2.
Proof. We shall prove the lemma by induction on m. It is trivially true for m ≤ 4. Assume m ∈ Pi and m − 1 ∈ Pj.
If |Pi| ≥ 3, then |Pi− {m}| ≥ 2. Let P0 be obtained from P by deleting m from Pi. By the induction hypothesis, there exists S such that Bm−1(S, P0) is connected. Clearly, Bm(S, P )is also connected.
Now, suppose that |Pi| ≤ 2. If i 6= j, let P0be obtained from P by replacing Pi and Pjby Pi0 = Pi∪ Pj− {m − 1, m}. Note that Pi0is nonempty, since Pior Pjis not a singleton. Also, P0
i is a singleton only when Pior Pjis. Thus, P0has at most one singleton. By the induction hypothesis, there exists S0such that B
m−2(S0, P0) is connected. Then Bm(S0∪ {{m − 1, m}}, P )is connected.
If i = j, i.e., Pi= {m−1, m}, let P0 = P −{Pi}. By the induction hypothesis, Bm−2(S0, P0)is connected for some S0. Set S = S0∪ {{m − 2, m − 1}}, if m − 2
is not in any Sk. Otherwise, assume m−2 ∈ Sk(then |Sk| = 2). Let S be obtained from S0by adding m − 1 to S
k. Then Bm(S, P )is connected. Q.E.D. Theorem 1. Suppose that gcd(n, q) = 1. Then G(4, n, q, r) is Hamiltonian.
Proof. We first note that a consecutive-1 digraph G(1, n, q, r0)has a loop i → i (i.e., i ≡ qi + r0(mod n)) if and only if gcd(n, q − 1) divides r0. In the affirmative case, the number of loops is gcd(n, q − 1), see [7].
If gcd(n, q − 1) > 1, then either G(1, n, q, r + 1) or G(1, n, q, r + 2) has no loop, as gcd(n, q − 1) cannot divide both r + 1 and r + 2. If gcd(n, q − 1) = 1, then both G(1, n, q, r + 1) and G(1, n, q, r + 2) have exactly one loop. In either case, since gcd(n, q) = 1, either G(1, n, q, r + 1) or G(1, n, q, r + 2) partitions the node-set into a set C of disjoint cycles with at most one singleton-cycle. By Lemma 2, there exists a legitimate interchange set S such that Bn(S, C)is connected. The
theorem then follows from Lemma 1. Q.E.D.
Note that the inductive proof of Lemma 2 implies a linear-time algorithm to construct S.
4. EXPLICIT CONSTRUCTION OF S
When gcd(n, q) = 1 and 3 divides n, we can give an explicit construction of S that works for all n and q. Throughout this section, S = {{3i − 2, 3i − 1, 3i} : i = 1, 2, . . . , n/3}.
Theorem 2. If gcd(n, q) = 1 and 3 divides n, then either Bn(S, Cr+1) or Bn(S, Cr+2)is connected.
Proof. It is now easier to consider S as a set E of links (a subset of size 3 corresponds to two consecutive links). To show Bn(S, Cr+1) or Bn(S, Cr+2)is connected, it suffices to show that E ∪ G(1, n, q, r + 1) or E ∪ G(1, n, r + 2) is connected. We first consider E ∪ G(1, n, q, r + 1). Note that for i → i + 1 in E, both i → qi + r + 1 and i + 1 → q(i + 1) + r + 1 are in G(1, n, q, r + 1). Hence, qi + r + 1and qi + r + 1 + q are connected in E ∪ G(1, n, q, r + 1). Let E ∪ Q be obtained from E ∪ G(1, n, q, r + 1) by replacing the two links i → qi + r + 1 and i + 1 → q(i + 1) + r + 1with the q-link qi + r + 1 → qi + r + 1 + q for every i such that i → i + 1 is in E. Then E ∪ G(1, n, q, r + 1) is connected if E ∪ Q is. We now explore the connectivity of E ∪ Q.
Partition the nodes into n/3 groups, where group i consists of nodes 3i − 2, 3i − 1, 3i. We will refer to them as the first, second, and third node of the group. We show that the groups are interconnected through the q-links. A q-link (i, j) will be called an (x, y) q-link if i is the xthnode of a group and j the ythnode of a group. Since gcd(n, q) = 1 and 3 divides n, we have that 3 does not divide q. Therefore, each group has two q-links going out and two q-links going in. The 2n/3 q-links contain two patterns of size n/3 each: one pattern corresponds to the (x, y) pattern of the q-link generated by the link (1, 2), the other by the link (2, 3). As 3 does not divide q, we have x 6≡ y (mod 3). So there are six permissible combinations for
these two patterns: (i) (1, 2), (2, 3); (ii) (1, 3), (3, 2); (iii) (2, 3), (3, 1); (iv) (2, 1), (1, 3); (v) (3, 1), (1, 2); (vi) (3, 2), (2, 1). The two q-links (i, j) and (i0, j0)going out from a group have different patterns (x, y) and (x0, y0). Note that i−j = i0−j0. Since i and i0 are in the same group, j and j0 are either in the same group or in consecutive groups. Furthermore, it is easily seen that j and j0 are in the same group if and only if (x − y)(x0− y0) > 0. Thus, for the middle four combinations, the two q-links from a group go to two consecutive groups. This implies that every pair of consecutive groups is connected; hence E ∪ Q is.
For the first and last pattern, the two q-links from a group go to the same group. So E ∪Qis not connected. However, let E ∪Q0be obtained from E ∪G(1, n, q, r +2) by replacing the two links i → qi + r + 2 and i + 1 → q(i + 1) + r + 2 with the q-link qi + r + 2 → qi + r + 2 + q for every i such that i → i + 1 is in E. Then the combination of the two patterns of q-links is (2, 3), (3, 1) for case (i), and (1, 3), (3, 2) for case (vi). In either case, the two q-links from a group go to two different groups. So, E ∪ Q0, consequently, E ∪ G(1, n, q, r + 2) is connected. Q.E.D. Unfortunately, E = {(3i − 2 → 3i − 1) ∪ (3i − 1 → 3i) : i = 1, 2, . . . , bn/3c} does not work when 3 does not divide n. A counterexample G(4, 25, 13, 7) was given by Xuding Zhu (group 1 and group 5 are not connected in G(1, 25, 13, 8) and group 4 and group 8 not connected in G(1, 25, 13, 9)).
5. CONCLUSIONS
It is known that consecutive-d digraph is Hamiltonian for d ≥ 5, but not necessarily so for d ≤ 3. In this article, we prove the conjecture that consecutive-4 digraphs are Hamiltonian, and thus completely settle the issue. Of course, our result for d = 4implies that for d ≥ 5. Our result also implies that there exist at least bd/4c disjoint Hamiltonian circuits for a consecutive-d digraph.
ACKNOWLEDGMENTS
The authors thank the referees for many constructive suggestions.
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