Matching preclusion for some interconnection networks

Matching Preclusion for Some Interconnection Networks

Eddie Cheng and László LiptákDepartment of Mathematics and Statistics, Oakland University, Rochester, Michigan

The matching preclusion number of a graph is the mini-mum number of edges whose deletion results in a graphthat has neither perfect matchings nor almost-perfectmatchings. In this paper, we find this number for variousclasses of interconnection networks and classify all theoptimal solutions. © 2007 Wiley Periodicals, Inc. NETWORKS,Vol. 50(2), 173–180 2007

Keywords: interconnection networks; perfect matching; Cayleygraphs; (n, k)-star graphs

1. INTRODUCTION AND PRELIMINARIES

A perfect matching in a graph is a set of edges such thatevery vertex is incident with exactly one edge in this set. Analmost-perfect matching in a graph is a set of edges such thatevery vertex except one is incident with exactly one edge inthis set, and the exceptional vertex is incident to none. So, ifa graph has a perfect matching, then it has an even number ofvertices; and if a graph has an almost-perfect matching, thenit has an odd number of vertices. The matching preclusionnumber of a graph G, denoted by mp(G), is the minimumnumber of edges whose deletion leaves the resulting graphwithout a perfect matching or almost-perfect matching. Anysuch optimal set of edges of size mp(G) is called an optimalmatching preclusion set. We define mp(G) = 0 if G has nei-ther a perfect matching nor an almost-perfect matching. Thisconcept of matching preclusion was introduced by Brighamet al. [4]. They introduced this concept as a measure of robust-ness in the event of edge failure in interconnection networks,as well as a theoretical connection to conditional connectiv-ity, “changing and unchanging of invariants,” and extremalgraph theory. We refer the readers to [4] for details and addi-tional references. In [4], the matching preclusion number wasdetermined for three classes of graphs, namely, the completegraphs, the complete bipartite graphs, and the hypercubes; inaddition, all the optimal solutions were found. In this article,we do the same for some well-known classes of intercon-nection networks. All these interconnection networks have

Received June 2006; accepted December 2006Correspondence to: E. Cheng; e-mail: [email protected] 10.1002/net.20187Published online in Wiley InterScience (www.interscience.wiley.com).© 2007 Wiley Periodicals, Inc.

an even number of vertices, so only perfect matchings areconsidered. The following proposition is obvious.

Proposition 1.1. Let G be a graph with an even number ofvertices. Then mp(G) ≤ δ(G) where δ(G) is the minimumdegree of G.

Proof. Deleting all edges incident to a single vertexwill give a graph with no perfect matchings and the resultfollows. ■

We call an optimal solution of the form given in the proofof Proposition 1.1 a trivial optimal matching preclusion set.

Useful distributed processor architectures offer the advan-tage of improved connectivity and reliability. An importantcomponent of such a distributed system is the system topol-ogy, which defines the interprocessor communication archi-tecture. The first such important topology is the well-studiedclass of hypercubes. In the late 1980s, Akers et al. [1] pro-posed the star graphs as an alternative to the hypercubesand this work has since generated a considerable amount ofresearch including fault tolerant routings, strong connectivityproperties, various Hamiltonian properties, broadcasting, ori-entation, and embedding. In certain applications, every vertexrequires a special partner at any given time and the matchingpreclusion number measures the robustness of this require-ment in the event of link failures as indicated in [4]. Hencein these interconnection networks, it is desirable to have theproperty that the only optimal matching preclusion sets arethe trivial ones. In this article, we investigate this propertyfor Cayley graphs generated by transpositions in Section 2and (n, k)-star graphs in Section 3. Each of these classes is ageneralization of the popular star graphs.

2. CAYLEY GRAPHS GENERATEDBY TRANSPOSITIONS

As mentioned earlier, one of the most popular and funda-mental interconnection networks is the class of star graphs.Stars graphs are Cayley graphs generated by a special set oftranspositions. In fact, many of their properties are shared byCayley graphs generated by any set of transpositions.

Let � be a finite group and let S be a set of elements of �

such that the identity of the group does not belong to S. The

NETWORKS—2007—DOI 10.1002/net

FIG. 1. Star graph.

Cayley graph for (�, S) is the directed graph with vertex setbeing the set of elements of � and in which there is an arc fromu to v if there is an s ∈ S such that u = vs. The Cayley graphfor (�, S) is strongly connected if and only if S is a generatingset for �. In this article, we choose the finite group to be �n,the symmetric group on {1, 2, . . . , n}, and the generating set Sto be a set of transpositions. The vertices of the correspondingCayley graph are permutations, and since S only has trans-positions, there is an arc from vertex u to vertex v if and onlyif there is an arc from v to u. Hence, we can regard theseCayley graphs as undirected graphs. (We use round bracketssuch as (a1, a2, . . . , an) to denote a permutation written in thecycle notation, and square brackets such as [a1, a2, . . . , an] orsimply a1a2 . . . an, to write the permutation as a rearrange-ment of the symbols.) With transpositions as the generatingset, a simple way to depict S is via a graph with vertex set{1, 2, . . . , n} where there is an edge between i and j if and onlyif the transposition (i, j) belongs to S. This graph is called thetransposition generating graph of (�n, S) or simply (trans-position) generating graph if it is clear from the context. Infact, the star graphs were introduced via the generating graphK1,n−1, where the center is 1 and the leaves are 2, 3, . . . , n.Hence they are called star graphs, though perhaps the moredescriptive term should be “star-generated graphs.”

We note that the Cayley graph generated by the transpo-sitions in S is connected if and only if the generating graphcorresponding to S is connected. Since an interconnectionnetwork need to be connected, we require the transpositiongenerating graph to be connected. Suppose the transposi-tion generating graph has m edges. Then it is clear that theCayley graph generated by it is m-regular and bipartite withbipartition sets being the odd and even permutations. If thetransposition generating graph is a tree, we call it a (trans-position) generating tree. This includes the star graph whosegenerating tree is K1,n−1 and the bubble-sort graph (Akers andKrishnamurthy [2]), whose generating tree is a path. A treeon three vertices must be a path of length two and hence itgenerates a 6-cycle. A tree on four vertices is either K1,3 or apath of length three. The corresponding graphs are shown in

Figures 1 and 2, respectively. (We note that in Figure 1, thedashed lines with the same symbol indicate the same edge.)

Since a tree with at least two vertices has at least twoleaves, without loss of generality we may assume that verticesn and n−1 are leaves; this will help in describing the structureof the corresponding Cayley graph.

Our first goal is to find the matching preclusion number forCayley graphs generated by transposition trees and classifythe optimal solutions. We will accomplish this via a Hamil-tonian property. A cycle (path) in a graph is a Hamiltoniancycle (path) if it contains every vertex in the graph. We needthe following results.

Theorem 2.1 ([7]). Let G be a Cayley graph obtained froma connected transposition generating graph on {1, 2, . . . , n}where n ≥ 4. For every pair of vertices u and v in differentbipartition sets, there is a Hamiltonian path between u and v.

Theorem 2.1 is proved by Tchuente [7], so we will notgive a proof here. (For other related properties of these Cay-ley graphs, we refer the reader to Araki [3].) This result isneeded for our proof of the next result. In fact, one can proveTheorem 2.1 using similar techniques.

Theorem 2.2. Let G be a Cayley graph obtained from a con-nected transposition generating graph on {1, 2, . . . , n} wheren ≥ 4. Suppose u and v are adjacent in G. Then G − {u, v}has a Hamiltonian cycle.

Proof. Since a connected transposition generating graphhas a spanning tree which generates a Cayley graph that isa subgraph of G, it is enough to prove the result for gen-erating trees. If n = 4, then G is the graph in Figure 1 orthe graph in Figure 2. Since the graph in Figure 1 is edge-transitive, there is only one case to check and a solution is

FIG. 2. Bubble-sort graph.

174 NETWORKS—2007—DOI 10.1002/net

FIG. 3. The base case for the star graph in Theorem 2.2.

given in Figure 3. Although the graph in Figure 2 is not edge-transitive, it has at most four edge-transitive classes. Sincethere is an edge between u and v if and only if there is a trans-position s ∈ {(12), (23), (34)} such that u = vs, the mappingπ �→ ρπ is an automorphism for every permutation ρ. Wefirst consider the edges of the 6-cycle in the upper left sideof Figure 2. By choosing ρ = 3214, we have 1234 �→ 3214and 2134 �→ 2314, and hence the edges (1234, 2134) and(3214, 2314) are in the same edge-transitive class. By choos-ing ρ = 1324, we have 1234 �→ 1324 and 2134 �→ 3124, andhence the edges (1234, 2134) and (1324, 3124) are in the sameedge-transitive class. Similarly, (2134, 2314), (3214, 3124),and (1234, 1324) are in one edge-transitive class. We now

FIG. 4. A base case for the bubble-sort graph in Theorem 2.2.


note that the representation given in Figure 2 is symmetricalong the center vertically and horizontally. Thus, there is anautomorphism that maps an edge on the other three 6-cycles(upper right, lower right, and lower left) to some edge on theupper left 6-cycle. So checking the edges (1234, 1324) and(1234, 2134) covers the edges in these four 6-cycles. Thesolutions for them are given in Figures 4 and 5. In additionto the symmetries mentioned, one can also see from this rep-resentation that both a rotation of 90 degrees and a reflectionalong the line y = x give automorphisms. Hence, check-ing the edges (1234, 1243) and (2314, 2341) will cover theremaining edges. The solutions for them are given in Figures 6and 7.


NETWORKS—2007—DOI 10.1002/net 175


We proceed with induction on n. Assume that n ≥ 5 andthe result is true for n − 1. Pick an edge uv. Let Hi be thesubgraph of G with i in the last position for 1 ≤ i ≤ n.Since n is a leaf in the generating tree, each Hi is isomor-phic to a Cayley graph generated by a tree on n − 1 vertices,and every vertex in Hi has exactly one neighbor not in Hi.Hence the edges whose endpoints are in different Hi’s forma perfect matching in G. Let’s call this perfect matching M.The edges in M are precisely the edges generated by theleaf edge incident to vertex n in the generating tree. If uvis an edge in M, then we redefine the Hi’s via the otherleaf, that is, via the (n − 1)th position. Note that since nand n−1 are leaves in the generating tree and n ≥ 5, they arenot adjacent. Hence, we may assume that uv is not an edgein M.

For notational convenience, we may assume that uv isin H1. We need to find a Hamiltonian cycle in G − {u, v}.By the induction hypothesis, there is a Hamiltonian cycle C1

in H1 − {u, v}. There is an edge u1v1 on C1 such that u1’sunique neighbor not in H1 and v1’s unique neighbor not inH1 are in different Hi’s. This is true since there are exactly(n − 2)! edges in M that are between Hi and Hj for everypair of distinct i and j and n ≥ 5. For notational conve-nience, we may assume that u1’s unique neighbor not in H1

is v2 in H2 and v1’s unique neighbor not in H1 is un in Hn.Now without loss of generality, we may assume that u1 is anodd permutation and v1 is an even permutation. Pick an edgebetween Hi and Hi+1, say uivi+1, where ui in Hi is odd andvi+1 in Hi+1 is even for i = 2, 3, . . . , n − 1. Now by Theo-rem 2.1, there is a Hamiltonian path Pi between ui and vi inHi for i = 2, 3, . . . , n. Now C1 − {u1v1}, P2, . . . , Pn and theedges u1v2, u2v3, . . . , un−1vn, unv1 form a Hamiltonian cyclein G − {u, v}. ■

We are now ready for the main theorem in this section.

Theorem 2.3. Let G be a Cayley graph obtained from agenerating tree on {1, 2, . . . , n} where n ≥ 3. Then mp(G) =δ(G) = n − 1. Moreover, if n ≥ 4 then the only optimalsolutions are the trivial matching preclusion sets.

Proof. If n = 3, then G is a 6-cycle and hence the state-ment is correct. (We note that the additional classificationof optimal solutions is not correct in this case.) If n = 4,then G is the graph in Figure 1 or the graph in Figure 2. Itis easy though tedious to check that our claim is correct forthese graphs. We proceed with induction on n. Assume thatn ≥ 5 and the result is true for n − 1. Let Hi be the subgraphof G with i in the last position for 1 ≤ i ≤ n. Since n is aleaf in the generating tree, each Hi is isomorphic to a Cayleygraph generated by a tree on n − 1 vertices, and every vertexin Hi has exactly one neighbor not in Hi. Hence, the edgeswhose endpoints are in different Hi’s form a perfect matchingin G. Let’s call this perfect matching M. Let F be an optimalsolution attaining mp(G). Then mp(G) ≤ δ(G) = n − 1 byProposition 1.1. Since G − F does not have a perfect match-ing, |F ∩ M| ≥ 1. If |F ∩ E(Hi)| ≤ n − 3 for every i, thenby the induction hypothesis, Hi − F has a perfect match-ing for every i and hence G − F has a perfect matching, acontradiction. Therefore, for notational convenience, we mayassume |F ∩ E(H1)| ≥ n − 2. But we now have |F| ≤ n − 1,|F ∩ M| ≥ 1 and |F ∩ E(H1)| ≥ n − 2. Hence |F| = n − 1,|F ∩ M| = 1, |F ∩ E(H1)| = n − 2 and |F ∩ E(Hi)| = 0 fori �= 1. So mp(G) = n − 1.

Since H1−F does not have a perfect matching, F ∩ E(H1)

is the set of all edges incident to a single vertex, say v1, by theinduction hypothesis. If the unique edge in F ∩ M is incidentto v1, then we are done. Suppose not; let u1 be the uniqueneighbor of v1 not in H1. Consider G − {u1, v1}. It containsat most one edge from F, namely, the unique edge in F ∩ M.Then by Theorem 2.2, G − {u1, v1} has a Hamiltonian cycle.Since G − {u1, v1} has an even number of vertices, this cyclecan be partitioned into two disjoint perfect matchings of G −{u1, v1}. At least one of them does not contain the uniqueedge in F ∩ M. Hence this matching together with the edgeu1v1 form a perfect matching of G − F, a contradiction. ■

We note that the existence of a Hamiltonian cycle is notnecessary in the proof of Theorem 2.3. The proof remainsthe same if we have a perfect 2-factor (that is, a spanningsubgraph where each vertex has degree 2, which in this caseimplies that it is a collection of even cycles as the graph isbipartite) in G−{u1, v1}. However, the proof of the existenceof such 2-factors is not easier than the existence of a Hamil-tonian cycle. We are now ready to improve Theorem 2.3 toCayley graphs generated by arbitrary connected transpositiongraphs.

Theorem 2.4. Let G be a Cayley graph obtained from a con-nected transposition generating graph on {1, 2, . . . , n} with medges where n ≥ 3. Then mp(G) = δ(G) = m. Moreover, ifn ≥ 4 then the only optimal solutions are the trivial matchingpreclusion sets.


Proof. If n = 3, then the connected transposition gen-erating graph is either a path of length 3, which generatesa 6-cycle, or it is K3 which generates K3,3, and it is easyto see that the result is correct. (In fact, the additional clas-sification of optimal solutions is correct for K3,3.) We nowassume n ≥ 4. Let Q be the connected transposition gener-ating graph for G. Pick a spanning tree T in Q. Let H bethe Cayley graph generated by T . Then H is a subgraphof G. We note that any edge in E(Q) \ E(T) will gener-ate a perfect matching in G. Let the edges in E(Q) \ E(T)

be e1, e2, . . . , em−n+1 and let Mi be the perfect matchingof G induced by ei. Then E(H), M1, M2, . . . , Mm−n+1 par-tition E(G). Let F be an optimal solution to the matchingpreclusion problem of G. Then |F ∩ E(H)| ≥ n − 1 (by The-orem 2.3) and |F ∩ Mi| ≥ 1 for i = 1, 2, . . . , m − n + 1.But |F| ≤ m. Hence |F ∩ E(H)| = n − 1, |F ∩ Mi| = 1 fori = 1, 2, . . . , m − n + 1 and |F| = m. So mp(G) = m asrequired.

We now classify the optimal solutions. Let F be such anoptimal solution. Then |F| = m. By Theorem 2.3, F ∩ E(H)

consists of precisely all the edges incident to a vertex in H,say v. So we have identified n − 1 elements of F. We wantto show that F consists of precisely all the edges incidentto v in G. We note that the n − 1 edges incident to v in Hare generated by the n − 1 edges in T in a one-to-one cor-respondence. We now want to show that the unique edge inF ∩ Mi is incident to v for every i = 1, 2, . . . , m − n + 1.Now add ei to T to create a cycle in the resulting graph.Next, we delete an edge fi �= ei from the cycle. Call theresulting tree Ti. Let Gi be the Cayley graph generated byTi. Then Gi is a subgraph of G. By repeating the same argu-ment, |F ∩ E(Gi)| = n − 1 and the elements in F ∩ E(Gi)

are precisely all the edges incident to a single vertex, sayvi. Now F ∩ E(H) and F ∩ E(Gi) share exactly n − 2 ele-ments, namely, those generated by the edges common toboth T and Ti. These n − 2 edges are incident to both vand vi. Since n − 2 ≥ 2, v = vi. But i is arbitrary, henceF consists of precisely the edges incident to v, and we aredone. ■

3. (n, k )-STAR GRAPHS

Although the star graph has proven to be an attrac-tive alternative to the hypercube, one drawback it hasis the restriction on the number of vertices. (The hyper-cube also has this drawback though not as severe.) Sincethe star graph has n! vertices, anyone wanting to builda multiprocessor network using this topology is forcedto build one with n! vertices for some value of n. Thisled in part to the introduction of the class of (n, k)-stargraphs in [5], which is a generalization of the class of stargraphs.

The (n, k)-star graph Sn,k with 1 ≤ k < n is governedby the two parameters n and k. The vertex set of Sn,k con-sists of all the permutations of k elements chosen fromthe ground set {1, 2, . . . , n}. Two vertices [a1, a2, . . . , ak]

FIG. 8. The graph S4,2.

and [b1, b2, . . . , bk] are adjacent if one of the followingholds:

1. There exists 2 ≤ r ≤ k such that a1 = br , ar = b1 andai = bi for i ∈ {1, 2, . . . , k} \ {1, r}.

2. ai = bi for i ∈ {2, . . . , k} and a1 �= b1.

Hence given a vertex [a1, a2, . . . , ak], it has k − 1 neigh-bors via the adjacency rule (1) by exchanging a1 with eachof ai, i ∈ {2, 3, . . . , k}, and it has n − k neighbors via theadjacency rule (2) by exchanging a1 with each element in{1, 2, . . . , n} \ {a1, a2, a3, . . . , ak}. We note that adjacencyrule (1) is precisely the rule for star graphs. In keeping withthe terminology for star graphs, an edge corresponding tothis rule is a star edge. An edge corresponding to the secondrule is a residual edge. Figure 8 gives S4,2. (For convenience,we write the (n, k)-permutation [i, j] as ij, for example, [1, 4]as 14.) Note that given an edge in Sn,k with the labellings ofits two endpoints, one can immediately determine whether itis a star edge or a residual edge. The family of Sn,k graphsgeneralizes the star graph, as Sn,n−1 is isomorphic to the stargraph with n! vertices. Since the graph reduces to the com-plete graph if k = 1, we assume k ≥ 2 for the rest of thearticle.

As in the previous section, we let Hi be the subgraph ofSn,k induced by vertices whose labelling has i in the last posi-tion. Then Hi is isomorphic to Sn−1,k−1 if k ≥ 2. Moreover,the edges not in any of the Hi’s, namely the star edges viaexchanging of symbols in the first and last (kth) position,form a perfect matching in Sn,k . We call them cross edges.

Since we already know the matching preclusion sets forstar graphs, we will only consider Sn,k with n−k ≥ 2. We willfirst consider Sn,2. We note that in this case Hi is isomorphicto Kn−1, and the graph obtained by contracting each Hi intoa vertex is isomorphic to Kn. From Figure 8, one can checkthat mp(S4,2) = 3, but there is a nontrivial matching preclu-sion set: for example, the set containing the edge between 14and 41, the edge between 24 and 42, and the edge between34 and 43 (as their deletion leaves a component of odd size).In general, a set of edges with exactly one end in Hi has sizen − 1 and is a matching preclusion set for k = 2 whenevern − 1 is odd. We call these semitrivial matching preclusion


sets. We need the following result for complete graphs in ourdiscussion.

Theorem 3.1 ([4]). Let r be even. Then mp(Kr) = r − 1.Moreover, if r �= 4 then the only optimal solutions are thetrivial matching preclusion sets; an optimal solution for K4

is either the edge set of a 3-cycle or a trivial matchingpreclusion set.

We also need a result similar to Theorem 2.2 for Sn,k :

Theorem 3.2. Let n − k ≥ 2, n ≥ 5, and k ≥ 2. Suppose uand v are adjacent in Sn,k. Then Sn,k−{u, v}has a Hamiltoniancycle.

We will not give a proof of this theorem. Instead, we referthe readers to a stronger result. It is proved by Hsu et al. [6]that if n − k ≥ 2, k ≥ 2, and at most n − 3 vertices aredeleted from Sn,k , then the resulting graph has a Hamiltoniancycle. The proof of this result is rather involved and we donot need the full strength of the statement here. The proof ofTheorem 3.2 is much simpler and can be done in a way similarto the proof of Theorem 2.2. We will also need the followingadditional results to aid us in proving our main theorem of thissection. We begin with a definition. A graph is Hamiltonianconnected if there is a Hamiltonian path between every pairof vertices.

Proposition 3.3. Let n ≥ 4 and 1 ≤ p ≤ n. If all thevertices that have p in the last position are deleted from Sn,2,then the resulting graph is Hamiltonian.

Proof. For notational convenience, we consider Sn,2 −V(H1) where the Hi’s are defined as usual. Choose the uniquecross edges v2u3, v3u4, . . . , vn−1un, vnu2 where ui, vi ∈ V(Hi)

for i = 2, 3, . . . , n. Since these edges are independent andthere is a Hamiltonian path between ui and vi in Hi for i =2, 3, . . . , n as Hi is isomorphic to Kn−1, we are done. ■

Proposition 3.4. Let n ≥ 5 and let u be a vertex in Sn,2.Then Sn,2 and Sn,2 − {u} are Hamiltonian connected.

Proposition 3.4 can be proved in a similar way as in theproof of Proposition 3.3 but with a tighter analysis. We willnot give the proof explicitly. Instead, we refer the readers to astronger result whose proof is quite complicated. It is provedin [6] that if n − k ≥ 2, k ≥ 2, and at most n − 4 verticesare deleted from Sn,k , then the resulting graph is Hamiltonianconnected.

Proposition 3.5. Let n ≥ 6 and 1 ≤ p ≤ n. Suppose all thevertices with the last symbol being p and an additional vertexw are deleted from Sn,3. Let G1 be the resulting graph. If u andv are two vertices in G1 with different symbols in the last posi-tion, then there is a Hamiltonian path between u and v in G1.

Proof. We define the Hi’s as usual. For notational con-venience, we may assume p = 1, u ∈ V(H2), and v ∈ V(Hn).

Since there are (n − 2)!/(n − 3)! = n − 2 ≥ 4 edgesbetween every pair of Hj’s, we can find cross edges viui+1

avoiding u, v, w where vi ∈ V(Hi) and ui+1 ∈ V(Hi+1) fori = 2, 3, . . . , n − 1. Let u2 = u and vn = v. We now applyProposition 3.4 to obtain an appropriate path from ui to vi inHi for i = 2, 3, . . . , n to complete the proof. ■

Before we prove the main result for Sn,k , we need to firstconsider the boundary case where k = 2.

Lemma 3.6. Let n ≥ 4. Then mp(Sn,2) = n−1. Moreover, ifn is odd, then the only optimal solutions are the trivial match-ing preclusion sets; if n is even, then the only optimal solutionsare the trivial and semi-trivial matching preclusion sets.

Proof. Consider the Hi’s defined at the beginning of thissection, and let M denote the perfect matching correspondingto the cross edges. Let F be an optimal matching preclusionset. One can check that the result is correct for n = 4.

Suppose n > 4 is odd. Then each Hi isomorphic to Kn−1

has n − 1 (even) number of vertices. Then |F ∩ M| ≥1, and F ∩ E(Hi) is a matching preclusion set for Hi (so|F ∩ E(Hi)| ≥ n − 2) for at least one i since F is a match-ing preclusion set. But |F| ≤ n − 1 since it is optimal.Therefore, |F ∩ M| = 1 and, for notational convenience,|F ∩ E(H1)| = n − 2 and |F ∩ E(Hi)| = 0 for i �= 1. Somp(Sn,2) = n−1. We now classify the optimal solutions. Sup-pose n �= 5. Then F ∩E(H1) consists of all the edges incidentto a single vertex, say v. Let vw be the unique cross edge at v.If vw is the unique element in F ∩ M, then we are done. Sup-pose not. Consider the graph Sn,2 −{v, w}. It contains at mostone element from F, namely, the unique element in F ∩ M.By Theorem 3.2, Sn,2 −{v, w} has a Hamiltonian cycle. SinceSn,2 −{v, w} has an even number of vertices, this cycle can bepartitioned into two disjoint perfect matchings of Sn,2−{v, w}.At least one of them does not contain the unique element inF ∩ M. Hence this matching together with the edge vw forma perfect matching of Sn,2 − F, a contradiction. We now con-sider n = 5. If F ∩E(H1) is a trivial matching preclusion set,then we are done via the same argument. So we may assumeF ∩ E(H1) is the edge set of a 3-cycle (by Theorem 3.1).For notational convenience, assume that the vertices in H1

are y2, y3, y4, y5 and the edges in H1 − F are y2y3, y2y4, y2y5.There is only one additional edge in F, and it is in F∩M. Nowconsider the cross edges incident to vertices in H1; at most oneof them is in F. Hence we may assume y3z3 and y4z4 are not inF and that z3 ∈ V(H3) and z4 ∈ V(H4). Now we obtain a per-fect matching in S5,2−F to arrive at a contradiction as follows:y2y5, y3z3, y4z4, a perfect matching in H2, a perfect matchingin H5 and a perfect matching in the subgraph induced by(V(H3) ∩ V(H4)) \ {z3, z4}. This last matching exists if theunique cross edge between H3 and H4 is not in F. If it does,then cross edges incident to y2 and y3 are not in F. Hence,we can repeat the construction with these two edges instead.

Now suppose n > 4 is even. We first note that, as before,|F ∩ M| ≥ 1. Let G′ be the graph obtained from Sn,2 bycontracting each Hi to a vertex. Then G′ is isomorphic to Kn.


If G′−(F∩M) has no perfect matching, then |F∩M| ≥ n−1by Theorem 3.1. But |F| ≤ n − 1. Hence |F| = n − 1 andF ∩ M is a trivial matching preclusion set for G′. So F isa semitrivial matching preclusion set for Sn,2. Now supposeG′ −(F ∩M) has a perfect matching. Then exactly one vertexfrom each Hi, say vi, is matched by the edges in Sn,2 corre-sponding to this perfect matching in G′ − (F ∩ M). Hence atleast one of Hi−{vi}, isomorphic to Kn−2, does not have a per-fect matching after F is deleted. For notational convenience,say i = 1. By Theorem 3.1, |F ∩ E(H1 − {v1})| ≥ n − 3. Weconsider two separate cases.

Case 1 (n ≥ 8). We know that |F ∩E(H1 −{v1})| ≥ n −3.Suppose |F ∩E(H1 −{v1})| = n−3. Then F ∩E(H1 −{v1})must be a trivial preclusion set for H1 −{v1} by Theorem 3.1.For notational convenience, assume that the vertices in H1

are v1, y2, y3, . . . , yn−1 and the edges y2y3, y2y4, . . . , y2yn−1

are in F. Since |F| ≤ n − 1, we have |F ∩ M| ≤ 2 and|F ∩E(Hi)| ≤ 2 for i �= 1. Our next goal is to show y2v1 ∈ F.Assume not. Consider the n − 3 edges in M that are incidentto y3, y4, . . . , yn−1. At least one of these edges is not in F, saythe one incident to y3. Now it follows from Theorem 3.1 thatG′ − (F ∩ M) has a perfect matching containing this edge as|F ∩ M| ≤ 2 and n ≥ 6. We can obtain a perfect matchingin H1 − {y3} − F as follows: y2v1 and a perfect matching M1

from (H1 −{v1, y2, y3})−F. (We note that M1 exists because|F ∩ E(H1 − {v1, y2, y3})| = 0.) This easily extends to a per-fect matching in Sn,2 − F, which gives a contradiction. So alledges in H1 incident to y2 are in F. Let y2z be the unique crossedge incident to y2. To complete the proof, we claim y2z ∈ F.We first note Sn,2 − {y2, z} has at most one element from F,namely, the unique element in F ∩ M. Now, by Theorem 3.2,Sn,2 −{y2, z} has a Hamiltonian cycle containing at most oneedge from F. This gives a contradiction as we have a perfectmatching in Sn,2 − F.

We will now consider the case when |F ∩ E(H1 − {v1})|is n − 1 or n − 2. Since |F ∩ M| ≥ 1, we must have|F ∩ E(H1 −{v1})| = n − 2 and |F ∩ M| = 1. Now the graphwith vertex set V(H1) \ {v1} and edge set F ∩ E(H1 − {v1})has n − 2 vertices and n − 2 edges. So it must contain acycle, and hence at least two vertices are of degree 2 in thisgraph. For at least one of them, the unique cross edge that itis incident with is not the edge in F ∩ M. Let this vertex be z.Clearly there is a perfect matching in G′ that uses this crossedge (incident with z) and does not use the unique edge inF ∩M. Now we can extend this to a perfect matching for Sn,2

as |F ∩ E(H1 − {z})| ≤ n − 2 − 2 = n − 4.

Case 2 (n = 6). This case is very similar to the case n ≥ 8except for the possibility of non-trivial optimal matchingpreclusion sets for K4. We omit the details. ■

We need two technical lemmas before we are ready forthe main result of this section.

Lemma 3.7. Let n ≥ 6. Let W be the set of vertices in Sn,3

and 1 ≤ p �= q ≤ n, where the third position is q and the

second position is p. Let w be an (additional) vertex in Sn,3

where the third position is not q. Let G1 be the graph obtainedby deleting W and w from Sn,3. Then G1 is Hamiltonian.

Proof. We define the Hi’s as usual. We may assumeq = 1. Now each Hi is isomorphic to Sn−1,2. We note thatthere are n − 2 cross edges between every pair of Hi’s. NowW is a set of vertices in H1. In fact W is the vertex set ofone of the “substructures” (isomorphic to Kn−2) in H1. ByProposition 3.3 (since n − 1 ≥ 4), H1 − W is Hamiltonian.Consider a Hamiltonian cycle C on vertices v1, v2, . . . , vN inH1 − W where N = (n − 2)2. We want to find an edge onthis cycle whose endpoints have cross edges in Sn,3 that avoidw and go to different Hi’s. Now, each of these vertices has across edge in Sn,3. So there is at least one edge on the cyclewhose endpoints have cross edges going to different Hi’s.(If not, there will be at least (n − 2)2 > n − 2 cross edgesbetween H1 and another Hi, a contradiction.) Assume v1v2 issuch an edge. If the cross edges of v1 and v2 avoid w, thenwe have found the edge. Assume not; we may assume that v1

is adjacent to w. Now examine each of the following edges:v2v3, v3v4, . . . , vN−1vN . At least one has the desired property.(If not, there will be at least (n−2)2 −1 > n−2 cross edgesbetween H1 and another Hi, a contradiction.) Let this edgebe xy. Let xx1 and yy1 be the cross edges in Sn,3. Now byProposition 3.5, there is a Hamiltonian path between x1 andy1 in Sn,3 − (V(H1) ∪ {w}) as n ≥ 6. This path together withthe Hamiltonian path between x and y in H1 − W , and theedges xx1 and yy1 give a Hamiltonian cycle in G1, so we aredone. ■

Lemma 3.8. Consider the graph S5,3. If the edge between145 and 415, the edge between 345 and 435, the edgebetween 245 and 425 and one additional cross edge f inthis graph are deleted, then the resulting graph has a perfectmatching.

Proof. Let F be the set of deleted edges and define theHi’s as usual. Three specific edges are deleted from H5. Wewill construct a matching M in S5,3 −F. Consider the 3 crossedges that are incident with 345, 245 and 145, respectively.At least one of them is not f . Pick this cross edge e = u1v1

between H5 and Hp, where u1 ∈ {345, 245, 145} and v1 ∈V(Hp). Put e in M. Since 345, 245, 145 induce a K3, pick theedge that is not incident with e and put it in M. Now there are3 cross edges between H5 and Hp, and one of them is e. Fromthe remaining two, pick the one that is not f , say u2v2 whereu2 ∈ V(H5) and v2 ∈ V(Hp); put u2v2 in M. To complete thematching, simply pick a perfect matching in each of Hi wherei �∈ {1, p}, a perfect matching in H5 −{345, 245, 145, u2}, anda perfect matching in Hp − {v1, v2}. Since Hi is isomorphicto S4,2, it is easy to see that the last two matchings exist.Hence we are done. ■

We note that Lemma 3.8 can also be easily checked (andwe did) using CPLEX [8] via the standard integer program-ming formulation for the matching problem as one only has


to solve the perfect matching problem on 60 vertices 30 timesas there are 30 possible choices for f .

Theorem 3.9. Let n ≥ 4. Then mp(Sn,2) = n − 1. More-over, if n is odd, then the only optimal solutions are the trivialmatching preclusion sets; if n is even, then the only opti-mal solutions are the trivial and the semitrivial matchingpreclusion sets. If n − k ≥ 2 and k ≥ 3, then mp(Sn,k) =n − 1 and the only optimal solutions are the trivial matchingpreclusion sets.

Proof. The first statement is Lemma 3.6. We proceedwith induction on n. If n = 4, then we only have to con-sider S4,2, in which case the validity of the statement is alreadyestablished. Consider Sn,k with n−k ≥ 2 and n ≥ 5. If k = 2,we are done. So we may assume k ≥ 3. Again, we let Hi bethe subgraph of Sn,k induced by vertices whose labelling has iin the last position. Then Hi is isomorphic to Sn−1,k−1. Recallthat the edges not in any of the Hi’s, namely the cross edges,form a perfect matching in Sn,k . Let M denote this perfectmatching.

Since each Hi is isomorphic to Sn−1,k−1, it has (n−1)!/(n−k)! vertices, which is even as k ≥ 3. (So unlike the proofof Lemma 3.6, there is only one main case.) Let F be anoptimal matching preclusion set. Now we use the inductionhypothesis to conclude that |F ∩ M| ≥ 1 and |F ∩ E(Hi)| ≥n − 2 for at least one i since F is a matching preclusion set.But |F| ≤ n − 1, since it is optimal. Therefore, |F| = n − 1,|F ∩ M| = 1, and without loss of generality, |F ∩ E(H1)| =n−2 and |F ∩E(Hi)| = 0 for i �= 1. So mp(Sn,k) = n−1. Wenow classify the optimal solutions. We consider two cases.

Case 1 (k ≥ 4). In this case, k − 1 ≥ 3. By the inductionhypothesis, F ∩ E(H1) is a trivial matching preclusion set.So F ∩ E(H1) must consist of all edges incident to a singlevertex, say v. Let vw be the unique cross edge at v. If vw is theunique element in F ∩ M, then we are done. Otherwise, byTheorem 3.2, Sn,k −{v, w} (a graph with at most one elementfrom F) has a Hamiltonian cycle. Since Sn,k − {v, w} has aneven number of vertices, this cycle can be partitioned intotwo disjoint perfect matchings of Sn,k − {v, w}. At least oneof them does not contain the edge vw, hence this matchingtogether with the edge vw form a perfect matching of Sn,k −F,a contradiction.

Case 2 (k = 3). If F ∩ E(H1) is a trivial matching preclu-sion set for H1, then the proof is exactly as in Case 1. So we

assume that F ∩ E(H1) is a semitrivial matching preclusionset for H1 (which is isomorphic to Sn−1,2). This means thatn − 1 is even. Moreover, we have a complete graph in H1

on n − 2 vertices, say u1, u2, . . . , un−2, such that the edges inH1 with exactly one endpoint in {u1, u2, . . . , un−2} is the setF ∩ E(H1).

The case n = 5 is done as Lemma 3.8 provides a per-fect matching in S5,3 − F, a contradiction. Hence assumen ≥ 6. Now, each of u1, u2, . . . , un−2 is incident to a crossedge in Sn,3, that is, an element of M. Since n ≥ 6, atleast one of them is not in F, say u1z. Now we will con-struct a perfect matching in Sn,3 − F to get a contradiction.We use u1z and a perfect matching on u2, u3, . . . , un−2 (aneven number of vertices). By Lemma 3.7, there is a Hamil-tonian cycle in Sn,3 − {u1, u2, . . . , un−2, z}, which containstwo disjoint perfect matchings on Sn,3 −{u1, u2, . . . , un−2, z}.As usual, at most one contains the element from F ∩ M asSn,3 −{u1, u2, . . . , un−2, z} contains at most one element fromF, so we are done. ■

Acknowledgments

The authors thank Professor Douglas Shier, the editor whohandled our paper, and the three anonymous referees for anumber of helpful comments and suggestions.

REFERENCES

[1] S.B. Akers, D. Harel, and B. Krishnamurthy, The star graph:An attractive alternative to the n-cube, Proc Int Conf ParallelProcess (1987), 393–400.

[2] S.B. Akers and B. Krishnamurthy, A group-theoretic modelfor symmetric interconnection networks, IEEE Trans Comput38 (1989), 555–565.

[3] T. Araki, Hyper hamiltonian laceability of Cayley graphsgenerated by transpositions, Networks 48 (2006), 121–124.

[4] R.C. Brigham, F. Harary, E.C. Violin, and J. Yellen, Perfect-matching preclusion, Congressus Numerantium 174 (2005),185–192.

[5] W.K. Chiang and R.J. Chen, The (n, k)-star graph: A general-ized star graph, Inform Proc Lett 56 (1995), 259–264.

[6] H.C. Hsu, Y.L. Hsieh, J. Tan, and L.H. Hsu, Fault hamiltonicityand fault hamiltonian connectivity of the (n, k)-star graphs,Networks 42 (2004), 189–201.

[7] M. Tchuente, Generation of permutations by graphicalexchanges, Ars Combinatoria 14 (1982), 115–122.

[8] ILOG CPLEX, webpage. http://www.ilog.com/products/cplex/.


Documents

Matching preclusion for some interconnection networks