Strong matching preclusion for augmented cubes

Theoretical Computer Science 491 (2013) 71–77 Contents lists available at SciVerse ScienceDirect Theoretical Computer Science journal homepage: www.elsevier.com/locate/tcs Strong matching preclusion for augmented cubes Eddie Cheng a,∗ , Shalin Shah b , Vyom Shah c , Daniel E. Steffy a a Department of Mathematics and Statistics, Oakland University, Rochester, MI 48309, United States b University of Michigan, Ann Arbor, MI 48109, United States c University of California, San Diego, 9500 Gilman Dr., La Jolla, CA 92093, United States article info Article history: Received 7 November 2012 Received in revised form 15 March 2013 Accepted 2 May 2013 Communicated by S.-y. Hsieh Keywords: Interconnection networks Perfect matching Augmented cubes abstract The strong matching preclusion number of a graph is the minimum number of vertices and edges whose deletion results in a graph that has neither perfect matchings nor almost perfect matchings. The concept was introduced by Park and Son. In this paper, we study the strong matching preclusion problem for the augmented cube graphs. As a result, we find smp(AQn ) and classify all optimal solutions. © 2013 Elsevier B.V. All rights reserved. 1. Introduction A perfect matching in a graph is a set of edges such that every vertex is incident with exactly one edge in this set. An almost perfect matching in a graph is a set of edges such that every vertex except one is incident with exactly one edge in this set, and the exceptional vertex is incident to none. If a graph has a perfect matching, then it has an even number of vertices; if a graph has an almost perfect matching, then it has an odd number of vertices. The matching preclusion number of a graph G, denoted by mp(G), is the minimum number of edges whose deletion leaves the resulting graph without a perfect matching or an almost perfect matching. Any such optimal set is called an optimal matching preclusion set. If G has neither a perfect matching nor an almost perfect matching, then mp(G) = 0. This concept of matching preclusion was introduced by [1] and further studied by [4–10,23,24,26]. They introduced this concept as a measure of robustness in the event of edge failure in interconnection networks, as well as a theoretical connection to conditional connectivity, ‘‘changing and unchanging of invariants’’ and extremal graph theory. We refer the readers to [1] for details and additional references. In [25], the concept of strong matching preclusion was introduced. The strong matching preclusion number of a graph G, denoted by smp(G), is the minimum number of vertices and edges whose deletion leaves the resulting graph without a perfect matching or an almost perfect matching. Any such optimal set is called an optimal strong matching preclusion set. Useful distributed processor architectures offer the advantages of improved connectivity and reliability. An important component of such a distributed system is the system topology, which defines the inter-processor communication architecture. Such system topology forms the interconnection network. We refer the readers to [16] for recent progress in this area and the references in its extensive bibliography. In certain applications, every vertex requires a special partner at any given time and the matching preclusion number measures the robustness of this requirement in the event of link failures as indicated in [1]. Hence in these interconnection networks, it is desirable to have the property that the only optimal matching preclusion sets and optimal strong matching preclusion sets are those whose deletion gives an isolated vertex in ∗ Corresponding author. Tel.: +1 248 370 4024. E-mail addresses: [email protected] (E. Cheng), [email protected] (S. Shah), [email protected] (V. Shah), [email protected] (D.E. Steffy). 0304-3975/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.tcs.2013.05.002 72 E. Cheng et al. / Theoretical Computer Science 491 (2013) 71–77 the resulting graph. Since interconnection networks are usually even, we only consider even graphs in this paper, that is, graphs with even numbers of vertices. Proposition 1.1. Let G be a graph with an even number of vertices. Then smp(G) ≤ mp(G) ≤ δ(G), where δ(G) is the minimum degree of G. Proof. Since G is even, mp(G) is the minimum number of edges whose deletion leaves a graph with no perfect matchings. Since deleting all edges incident to a single vertex will give a graph with no perfect matchings, mp(G) ≤ δ(G). The claim smp(G) ≤ mp(G) is obviously true as every matching preclusion set is a strong matching preclusion set.  An optimal solution of the form given in the proof of Proposition 1.1 is a trivial (optimal) matching preclusion set. Let F be an optimal strong matching preclusion set of a graph G = (V , E ). Suppose F = FV ∪ FE where FV consists of vertices and FE consists of edges. We may assume that no element in FE is incident to an element in FV since F is optimal. (If e ∈ FE is incident to an element of FV , then G − F = G − (F − {e}).) We call F a basic (optimal) strong matching preclusion set if F is an optimal strong matching preclusion set of G and G − F has an isolated vertex, that is, there exists a vertex v such that every vertex in FV is a neighbor of v and every edge in FE is incident to v . This includes the following scenario: F is a basic optimal matching preclusion set and G − F is odd without almost perfect matchings. We can further restrict this class as follows. If G − F is even and there is a vertex v such that every vertex in FV is a neighbor of v and every edge in FE is incident to v , then F is a trivial (optimal) strong matching preclusion set. For r-regular even graphs we have the following relationship between these classes of preclusion sets. Proposition 1.2 ([2]). Let r ≥ 2. Let G be an r-regular even graph. Suppose that smp(G) = r. Then every basic strong matching preclusion set is trivial. Hypercubes are the most basic class of interconnection networks. However, they have shortcomings and a number of their variants were introduced to address some of the issues. One such popular variant is the class of augmented cubes introduced in [11]. As an improvement upon the hypercubes, the augmented cube graphs are designed to be superior in many aspects. Not only do they retain some of the favorable properties of the hypercubes but also possess some embedding properties that the hypercubes do not have. For instance, a hypercube of the nth dimension contains cycles of all lengths from 3 to 2n whereas the hypercube contains only even cycles. As shown in [25], bipartite graphs are poor interconnection networks with respect to the strong matching preclusion property. However, augmented cubes are not bipartite and we will show in this paper that they have good strong matching preclusion properties. We now define the n-dimensional augmented cube AQn as follows. Let n ≥ 1, the graph AQn has 2n vertices, each labeled by an n-bit binary string u1 u2 · · · un such that ui ∈ {0, 1} for all i. AQ1 is isomorphic to the complete graph K2 where one vertex is labeled by the digit 0 and the other by 1. For n ≥ 2, AQn is defined recursively by using two copies of (n − 1)dimensional augmented cubes with edges between them. We first add the digit 0 to the beginning of the binary strings of all vertices in one copy of AQn−1 , which will be denoted by AQn0−1 , and add the digit 1 to the beginning of all the vertices of the second copy, which will be denoted by AQn1−1 . We now describe the edges between these two copies. Let u = 0u1 u2 · · · un−1 and v = 1v1 v2 · · · vn−1 be vertices in AQn0−1 and AQn1−1 , respectively. Then u and v are adjacent if and only if one of the following conditions holds: (1) ui = vi for every i ≥ 1. In this case, we call the edge (u, v) a cross edge and say u = v x and v = ux . (2) ui ̸= vi for every i ≥ 1. In this case, we call (u, v) a complement edge and denote u = v c and v = uc . Throughout this paper, we denote the set of cross edges and complement edges in AQn by Xn and Cn respectively. Clearly, AQn is (2n − 1)-regular, |Cn | = |Xn | = 2n−1 and the edges in Cn (Xn ) are independent. It is well-known that AQn is vertextransitive. Another important fact is that the connectivity of AQn is 2n − 1 for n ≥ 4. Some recent papers on augmented cubes include [3,6,13–15,17,21,22]. A few examples of augmented cubes are shown in Fig. 1.1. 2. Preliminaries Our objective is to show that smp(AQn ) = 2n − 1, which is the best possible result, and that all optimal solutions are trivial. In this section, we present some results that will be useful in our quest. Since the strong matching preclusion problem is a generalization of the matching preclusion problem and the latter problem has been solved for AQn , we state the corresponding result. Theorem 2.1 ([6]). Suppose n ≥ 4. Then mp(AQn ) = 2n − 1. Moreover, every optimal matching preclusion set is trivial. Given that a Hamilton cycle in an even graph induces two edge-disjoint perfect matchings, the following result uses the ‘‘fault Hamiltonian’’ property as a sufficient condition in determining the strong matching preclusion number. Proposition 2.2 ([2]). Let G be an r-regular even graph with the property that G − F is Hamiltonian for every F ⊆ V (G) ∪ E (G) where |F | ≤ r − 2. Then smp(G) = mp(G) = r. However, we are unaware of any relationship between such ‘‘fault Hamiltonian’’ property and the classification of optimal strong matching preclusion sets. In order to apply Proposition 2.2, we need Hamiltonian results for AQn . Fortunately, such a result is known. E. Cheng et al. / Theoretical Computer Science 491 (2013) 71–77 73 Fig. 1.1. Augmented cubes of dimensions 1 through 4. Theorem 2.3 ([15]). Let n ≥ 4. Suppose F ⊆ V (AQn ) ∪ E (AQn ). If |F | ≤ 2n − 4, then AQn − F is Hamiltonian connected1 ; if |F | ≤ 2n − 3, then AQn − F is Hamiltonian. 3. Main result It follows from Proposition 2.2 and Theorem 2.3 that smp(AQn ) = 2n − 1 for n ≥ 4. It remains to classify all optimal solutions. We claim that all optimal solutions are trivial. Given the recursive structure of augmented cubes, the natural method is to use induction. The first step is to check the base case. Lemma 3.1. smp(AQ4 ) = 7. Moreover every optimal strong matching preclusion set is trivial. Proof. This result was verified by a computer program. The check was done using a Python program and the NetworkX package [12] for graph representation. The program verified that for every 7-element fault sets F , unless F is trivial, AQ4 − F has a perfect matching or an almost perfect matching. In order to reduce the number of cases that had to be checked we note that Theorem 2.1 implies that we may assume F contains at least one vertex. Moreover, since AQ4 is vertex-transitive, one vertex in F can be fixed. Additionally, it can be assumed that no fault edge is incident with a fault vertex. In wall clock time, the computer verification took a couple of days on a modern desktop computer.  Before we present the proof of our main result, we need a number of easy technical results. We start with the following useful observation of augmented cubes which we will apply without explicitly referencing it. Proposition 3.2. Let n ≥ 3. Let u be a vertex of AQn . Then ux is adjacent to uc . Moreover, there is a unique vertex v such that u and v are adjacent, v c = ux and v x = uc . In other words, u, v, ux , uc form a complete graph on four vertices. We need two more facts regarding matchings which we will now state without proof. Proposition 3.3. Let G be a graph with no isolated vertices. Suppose that G has an almost perfect matching M that misses vertex v . Then there exists an almost perfect matching in G which misses a vertex other than v . Proposition 3.4. Let G be a graph with no isolated vertices. Suppose that G has an almost-perfect matching M that misses vertex w. If G does not contain a 2-path v − u − w in which v and w have degree 1, then there exist almost-perfect matchings M1 and M2 in G such that M, M1 and M2 miss different vertices. Our main result is that every optimal conditional strong matching preclusion set in AQn is trivial. Before proceeding with the proof we give some comments on the general strategy that will be applied. Due to the recursive structure of AQn it is natural to establish the result by induction. In particular, given any fault set F that is not a trivial strong matching preclusion set, we must show how to construct a perfect matching or almost perfect matching in AQn − F . We will consider several cases regarding how the faults are distributed among AQn0−1 , AQn1−1 and the set of cross edges. If many faults are concentrated within one of these two subgraphs, the induction hypothesis cannot be directly applied to recover a perfect matching or almost perfect matching in that subgraph. In such cases we will remove a set A from the fault set so that induction can be applied, building a perfect matching or almost perfect matching in each of the subgraphs and finally using the structural properties of AQn to augment the matchings to form a perfect matching or almost perfect matching in the entire graph, removing any dependence on the set A. Finally, even if the faults are distributed more evenly between AQn0−1 and AQn1−1 some care must still be taken; if an odd number of fault vertices appear in each half then induction will only provide an 1 A graph is Hamiltonian connected if there is a Hamiltonian path between every pair of vertices. 74 E. Cheng et al. / Theoretical Computer Science 491 (2013) 71–77 almost perfect matching in each, the union of these must still be augmented to produce a perfect matching in the entire graph. The art is to find the right balance in dividing the cases. This type of case analysis is the frequent method of choice in this area as seen in [18,20,24–26] among others for matching preclusion. Proofs for other properties on interconnection networks are equally involved; for example, see [13–15,19,21]. Theorem 3.5. Let n ≥ 4. Then smp(AQn ) = 2n − 1. Moreover, every optimal strong matching preclusion set is trivial. Proof. The claim that smp(AQn ) = 2n − 1 follows from Proposition 2.2 and Theorem 2.3. We now classify the optimal solutions. The proof is via induction. We first note that the statement is true if n = 4 by Lemma 3.1. Let n ≥ 5 and assume that the result is true for AQn−1 . Let F ⊆ V (AQn ) ∪ E (AQn ) be an optimal strong matching preclusion set. As remarked earlier, we may assume that no edge in F is incident to a vertex in F . Then, we show that either AQn − F contains a perfect matching or an almost perfect matching, or that F is a trivial strong matching preclusion set of AQn . Let F = FX ∪ FC ∪ F0 ∪ F1 where F0 and F1 denote the fault sets of AQn0−1 and AQn1−1 respectively. Similarly, FX is the set of faulty cross edges while FC denotes the set of faulty complement edges. We may assume that |F0 | ≥ |F1 |. We now divide the proof into four cases: Case 1: |F0 | = 2n − 1. Then |F1 ∪ FC ∪ FX | = 0. Note that AQn − F has no isolated vertices, so we will show that it has either a perfect matching or an almost perfect matching. We may assume that F = F0 contains vertices; otherwise, the result follows from Theorem 2.1. We pick two elements from F0 to form A. We either pick two vertices or one vertex together with an edge so that F0 − A contains an even number of vertices. Let F0′ = F0 − A. By construction, AQn0−1 − F0′ is an even graph. Suppose there is an isolated vertex v1 in AQn0−1 − F0′ . So every vertex in F0′ is adjacent to v1 and every edge in F0′ is incident to v1 . Since F0′ has an even number of vertices and the degree of v1 in AQn0−1 is odd, F0′ contains at least one edge say, (v1 , u1 ). If A consists of a vertex and an edge e. Then let A′ = (A − {e}) ∪ {(v1 , u1 )} and it is easy to see that AQn0−1 − (F0 − A′ ) does not have an isolated vertex, and we may choose A′ instead of A. Now suppose that A consists of two vertices y1 and y2 . (Recall that F0′ has an even number of vertices.) If F0′ has a vertex z, then we may choose A′ = {y1 , z } and it is easy to see that AQn0−1 − (F0 − A′ ) does not have an isolated vertex. Thus we assume that F0′ consists of edges only. Then F has two vertices and 2n − 3 edges. We claim that AQn − F has a perfect matching. Now by the induction hypothesis, AQn0−1 − {y1 , y2 , v1 } = AQn0−1 − (F0 ∪ {v1 }) has an almost perfect matching M0 missing, say, w . Consider the two cross edges (v1 , v1x ) and (w, w x ). By the induction hypothesis, AQn1−1 − {v1x , w x } has a perfect matching M1 . Now M0 ∪ M1 ∪ {(v1 , v1x ), (w, w x )} is a perfect matching in AQn − F , as required. Henceforth, we may assume that AQn0−1 − F0′ has no isolated vertices. Recall that by construction, AQn0−1 − F0′ is an even graph. So by the induction hypothesis, AQn0−1 − F0′ has a perfect matching MP . We consider two subcases. Subcase 1a: A contains distinct vertices v1 , v2 in AQn0−1 . So F has an even number of vertices and we want to find a perfect matching in AQn − F . If v1 and v2 are adjacent then (v1 , v2 ) ∈ MP and it is easy to extend it to a perfect matching in AQn − F . So we may assume that MP matches v1 and v2 to the vertices v1′ and v2′ , respectively in AQn0−1 − F0′ . Consider the cross edges (v1′ , v1′x ) and (v2′ , v2′x ). By the induction hypothesis, AQn1−1 − {v1′x , v2′x } has a perfect matching M1 . Now, (MP − {(v1 , v1′ )(v2 , v2′ )}) ∪ M1 ∪ {(v1′ , v1′x ), (v2′ , v2′x )} is a perfect matching in AQn − F , as required. Subcase 1b: A contains an edge (v, v ′ ) and a vertex u. (By assumption, u ̸∈ {v, v ′ } and v, v ′ ̸∈ F .) So F has an odd number of vertices and we want to find an almost perfect matching in AQn − F . Now let (u, u′ ) ∈ MP . We consider whether the edge (v, v ′ ) is part of the matching MP or not. If not, then MP − {(u, u′ )} is an almost perfect matching in AQn0−1 − F0 missing u′ , which can be extended to an almost perfect matching in AQn − F missing u′ , by using a perfect matching in AQn1−1 = AQn1−1 − F1 . Now assume instead that (v, v ′ ) ∈ MP . So MP − {(v, v ′ ), (u, u′ )} matches every vertex in AQn0−1 − F0 except v , v ′ , and u′ . Since each vertex has a complement and cross edge incident with it, we simply choose the cross edges and match v and v ′ to the vertices v x and v ′x in AQn1−1 − F1 . Since |F1 | = 0, it follows from the induction hypothesis that AQn1−1 − {v x , v ′x } has a perfect matching M1 . Furthermore, (MP − {(v, v ′ ), (u, u′ )}) ∪ M1 ∪ {(v, v x )(v ′ , v ′x )} is an almost perfect matching in AQn − F missing u′ , so we are done. Case 2: |F0 | = 2n − 2. Then |F1 ∪ FC ∪ FX | = 1. Note that AQn − F has no isolated vertices, so we will show that it has either a perfect matching or an almost perfect matching. Case 2a: F0 contains only vertices. We consider two possibilities. The first possibility is that the unique element in F1 ∪ FC ∪ FX is an edge. Then let A be a set containing two elements of F0 , say u and v . By the induction hypothesis, AQn0−1 − (F0 − A) has a perfect matching M0 . If (u, v) is an edge and it is in M0 , then it is easy to extend M0 to a perfect matching in AQn − F . So we may assume otherwise, and that (u, u′ ), (v, v ′ ) ∈ M0 . Since F1 ∪ FC ∪ FX contains exactly one edge, either {(u′ , u′x ), (v ′ , v ′x )} ∩ F = ∅ or {(u′ , u′c ), (v ′ , v ′c )} ∩ F = ∅. We may assume that (u′ , u′x ) and (v ′ , v ′x ) are not in F . Now by the induction hypothesis, AQn1−1 − (F1 ∪ {u′x , v ′x }) has a perfect matching M1 . Then (M0 − {(u, u′ ), (v, v ′ )}) ∪ M1 ∪ {(u′ , u′x ), (v ′ , v ′x )} is a perfect matching in AQn − F . The second possibility is the unique element in F1 ∪ FC ∪ FX is a vertex y in AQn1−1 . We consider the following scenarios. • Suppose yc , yx ∈ F0 . Let A = {yc , yx }. By the induction hypothesis, AQn0−1 − (F0 − A) has a perfect matching M0 . If (yc , yx ) is in M0 , then it is easy to extend M0 to an almost perfect matching in AQn − F . So we may assume otherwise, and that (yc , u′ ), (yx , v ′ ) ∈ M0 . Clearly neither u′ nor v ′ is adjacent to y. So we may assume that (u′ , u′x ) and (v ′ , v ′x ) are in AQn − F . Now by the induction hypothesis, AQn1−1 − (F1 ∪ {u′x , v ′x }) = AQn1−1 − ({y} ∪ {u′x , v ′x }) has an almost perfect matching M1 . Then (M0 − {(yc , u′ ), (yx , v ′ )}) ∪ M1 ∪ {(u′ , u′x ), (v ′ , v ′x )} is an almost perfect matching in AQn − F . E. Cheng et al. / Theoretical Computer Science 491 (2013) 71–77 75 • Suppose exactly one of yc and yx is in F0 . Without loss of generality, we may assume that yc ∈ F0 and yx ̸∈ F0 . Let v be a vertex in F0 that is neither yc nor yx . Let A = {yc , v}. By the induction hypothesis, AQn0−1 − (F0 − A) has a perfect matching M0 . If (yc , v) is an edge, and it is in M0 , then it is easy to extend M0 to an almost perfect matching in AQn − F . So we may assume that (yc , u′ ), (v, v ′ ) ∈ M0 . By construction, at most one of u′ and v ′ is adjacent to y. So we may apply the usual argument. • Suppose yc , yx ̸∈ F0 . Then let A be a set containing two elements of F0 , say u and v . By the induction hypothesis, AQn0−1 − (F0 − A) has a perfect matching M0 . If (u, v) is an edge, and it is in M0 , then it is easy to extend M0 to an almost perfect matching in AQn − F . So we may assume otherwise, and that (u, u′ ), (v, v ′ ) ∈ M0 . If at most one of u′ and v ′ is adjacent to y, we may apply the usual argument. Otherwise {u′ , v ′ } = {yc , yx } and hence u′ is adjacent to v ′ . Thus, (M0 − {(u, u′ ), (v, v ′ )}) ∪ {(u′ , v ′ )} is a perfect matching in AQn0−1 − F0 and it is easy to extend it to an almost perfect matching in AQn − F . Case 2b: F0 has at least one edge. If F0 has an even number of vertices, then let A be a set containing an edge from F0 ; otherwise let A be a set containing a vertex from F0 . Let F0′ = F0 − A. By construction, AQn0−1 − F0′ is an even graph. Suppose there is an isolated vertex v1 in AQn0−1 − F0′ . So every vertex in F0′ is adjacent to v1 and every edge in F0′ is incident to v1 . Since F0′ has an even number of vertices and the degree of v1 in AQn0−1 is odd, F0′ contains at least one edge say, (v1 , u1 ). If A consists of an edge e, then let A′ = (A − {e}) ∪ {(v1 , u1 )} and it is easy to see that AQn0−1 − (F0 − A′ ) does not have isolated vertices, and we may choose A′ instead of A. Now suppose that A consists of a vertex y. By construction, F0′ has an even number of vertices. If F0′ has a vertex z, then we may choose A′ = {z } and it is easy to see that AQn0−1 − (F0 − A′ ) does not have any isolated vertices. Thus we assume that F0′ consists of edges only. Then F0 has one vertex, namely, y. By the induction hypothesis, AQn0−1 − {y, v1 } = AQn0−1 − (F0 ∪ {v1 }) has a perfect matching M0 . If the unique element in F1 ∪ FC ∪ FX is an edge, then F has one vertex, and we claim that AQn − F has an almost perfect matching. Now M0 together with a perfect matching in AQn1−1 − F1 will be a desired almost perfect matching in AQn − F missing v1 . Now suppose the unique element in F1 ∪ FC ∪ FX is a vertex (in AQn1−1 ), say w1 . Then F has two vertices, and we claim that AQn − F has a perfect matching. Clearly w1 cannot be both v1x and v1c , so we may assume that it is not v1x . Let M1 be a perfect matching in AQn1−1 − {v1x , w1 }. Then M0 ∪ M1 ∪ {(v1 , v1x )} is a perfect matching in AQn − F , as required. Henceforth, we may assume that AQn0−1 − F0′ has no isolated vertices. Recall that by construction, AQn0−1 − F0′ is an even graph. So by the induction hypothesis, AQn0−1 − F0′ has a perfect matching MP . We consider two subcases. Subcase 2b(i): The element in A is the edge (v1 , v2 ) in AQn0−1 . We note that if (v1 , v2 ) ̸∈ MP , then it is easy to find a perfect matching or an almost perfect matching in AQn − F . So we may assume that (v1 , v2 ) ∈ MP . Since |F1 ∪ FC ∪ FX | = 1, we claim that we can match v1 and v2 to vertices in the graph AQn1−1 − F1 . If our claim is correct, then we may assume that (v1 , v1c ) and (v2 , v2c ) are edges in AQn − F . Let M1 be a perfect matching or an almost perfect matching in AQn1−1 −(F1 ∪{v1c , v2c }) depending on whether F1 contains a vertex. Then (MP − {(v1 , v2 )}) ∪ M1 ∪ {(v1 , v1c ), (v2 , v2c )} is the desired perfect matching or almost perfect matching in AQn − F . Now, our claim is clearly true if the unique element in F1 ∪ FC ∪ FX is an edge. Suppose the unique element in F1 ∪ FC ∪ FX is the vertex y in AQn1−1 . Then the claim is still true (by using either cross edges or complement edges) unless y is adjacent to both v1 and v2 . In this case, we may assume that y = v1x and y = v2c . Since the element in A is an edge, F0 must contain an even number of vertices. Thus F0 has an even number of edges and hence at least two edges. The natural argument is to choose another edge from F0 to form A instead of using (v1 , v2 ). However, we have already assumed that such an edge is chosen so that AQn0−1 − (F0 − A) has no isolated vertices. So care must be taken. If such an exchange produces an isolated vertex, then we may assume that the isolated vertex is v1 . If there is another edge (v1 , v3 ) belonging to F0 , then we can use A′ = {(v1 , v3 )} instead. Now it is easy to see that AQn0−1 − (F0 − A) has no isolated vertices and v3 is not adjacent to y. So we can match v1 and v3 to vertices in the graph AQn1−1 − F1 . So if our claim is not correct, then F0 consists of two edges, one of them is (v1 , v2 ), and 2n − 4 neighbors of v1 , except v2 . Moreover, the unique element in F1 ∪ FC ∪ FX is the vertex y in AQn1−1 , and y is adjacent to v1 and v2 . We may assume that v1c ̸∈ F . (So y = v1x .) We will use a different construction. Let w, z ∈ F0 , both neighbors of v . Let F0′′ = F0 − {w, z }. By the induction hypothesis, AQn0−1 − F0′′ has a perfect matching M0 . So we may assume that (v1 , w), (z , z ′ ) ∈ M0 . By the induction hypothesis, there is a perfect matching M1 in AQn1−1 − {y, v c }. Now (M0 − {(v1 , w), (z , z ′ )}) ∪ M1 ∪ {(v1 , v1c )} is an almost perfect matching in AQn − F missing z ′ . Subcase 2b (ii): The element in A is the vertex v1 in AQn0−1 . Let (v1 , v2 ) ∈ MP . Since |F1 ∪ FC ∪ FX | = 1 and we may use either cross edges or complementary edges, we may assume that (v2 , v2c ) is in AQn − F . Let M1 be a perfect matching or an almost perfect matching in AQn1−1 − (F1 ∪ {v2c }) depending on whether F1 contains a vertex. Then (MP − {(v1 , v2 )}) ∪ M1 ∪ {(v2 , v2c )} is the desired perfect matching or almost perfect matching in AQn − F . Case 3: |F0 | = 2n − 3. First, assume that F0 is not a trivial strong matching preclusion set of AQn0−1 − F0 . Then, by the induction hypothesis, each of AQn0−1 − F0 and AQn1−1 − F1 contains a perfect matching or an almost perfect matching. If at least one of AQn0−1 − F0 or AQn1−1 − F1 contains a perfect matching, we are done. So, we assume that both AQn0−1 − F0 and AQn1−1 − F1 have almost perfect matchings. Thus F1 contains only one vertex. (It may contain another edge.) In particular, M0 is an almost perfect matching of AQn0−1 − F0 missing w . If w is isolated in AQn0−1 − F0 , then either (w, w x ) or (w, w c ) is in AQn − F ; otherwise w is isolated in AQn − F and F is a basic optimal strong matching preclusion set of AQn , which implies 76 E. Cheng et al. / Theoretical Computer Science 491 (2013) 71–77 F is a trivial optimal strong matching preclusion set of AQn by Proposition 1.2. So, for convenience, we assume that (w, w c ) is in AQn − F . Let M1 be a perfect matching in AQn1−1 − (F1 ∪ {w c }). Then M0 ∪ M1 ∪ {(w, w c )} is a perfect matching in AQn − F . Therefore, we may assume that w is not isolated in AQn0−1 − F0 . This implies AQn0−1 − F0 has no isolated vertices. Suppose |{w c , w x , (w, w c ), (w, w x )} ∩ F | ≤ 1. Then we may assume that w c , (w, w c ) ̸∈ F . Thus AQn1−1 − (F1 ∪ {w c }) contains a perfect matching M1 , and hence M0 ∪ M1 ∪ {(w, w c )} is a perfect matching in AQn − F . So we may assume that |{w c , wx , (w, w c ), (w, w x )} ∩ F | = 2 and neither (w, wc ) nor (w, wx ) are in AQn − F . So, without loss of generality, we may assume that w c , (w, w x ) ∈ F1 . We apply Proposition 3.3 to find another almost perfect matching in AQn0−1 − F0 missing y ̸= w . Now |{yc , yx , (y, yc ), (y, yx )} ∩ F | ≤ 1 and we can repeat the argument. Henceforth, we may assume that F0 is a trivial strong matching preclusion set of AQn0−1 . So there exists an isolated vertex v in AQn0−1 − F0 and F0 contains vertices that are adjacent to v or edges that are incident to v . Moreover, F0 contains an even number of vertices. If neither (v, v c ) nor (v, v x ) is in AQn − F , then F is basic in AQn and hence trivial in AQn − F by Proposition 1.2. So, for convenience, we may assume that (v, v c ) is in AQn − F . Let f1 and f2 be two elements of F0 and let F0′ = F0 − {f1 , f2 }. Clearly we can pick f1 and f2 to either both be vertices or both be edges. By Theorem 2.3, AQn0−1 − F0′ contains a Hamiltonian cycle C since |F0′ | = 2n − 5. If f1 and f2 are both edges, then both are incident to v and they are in C ; thus C − {v} is a path P in AQn0−1 − F0 with an odd number of vertices. If f1 and f2 are both vertices, then both are adjacent to v and they are on C ; thus C − {v, f1 , f2 } is a path P in AQn0−1 − F0 with an odd number of vertices. Now P contains at least 2n−1 − (2n − 3) − 1 = 2n−1 − 2n + 2 vertices. For at least (2n−1 − (2n − 3) − 1)/2 = 2n−2 − n + 1 vertices, its deletion will separate P into two paths, each with an even number of vertices. Since 2n−2 − n + 1 ≥ 4 as n ≥ 5, there is one such vertex z such that (z , z c ) is in AQn − F . Thus C induces a matching M0 in AQn0−1 − F0 missing z. Now by the induction hypothesis, AQn1−1 − (F1 ∪ {v c , z c }) has a perfect matching or an almost perfect matching M1 . Then M0 ∪ M1 ∪ {(v, v c ), (z , z c )} is either a perfect matching or an almost perfect matching in AQn − F . Case 4: |F0 | < 2n − 3. By the induction hypothesis, AQn0−1 − F0 has a perfect matching or an almost perfect matching. Moreover, |F1 ∪ FX ∪ FC | > 2. Since |F0 | ≥ |F1 |, AQn−1 − F1 also contains a perfect matching or an almost perfect matching. If at least one of AQn0−1 − F0 and AQn1−1 − F1 has a perfect matching, we are done. So, assume both AQn0−1 − F0 and AQn1−1 − F1 have an odd number of vertices. We consider two subcases. Subcase 4a: |F0 | ≤ 2n − 5. Since there are 2n−1 cross edges and 2n−1 complement edges, we may assume that there is at least one cross edge and one complement edge not on F as 2n−1 > 2n − 1 for all n ≥ 5. We consider such a fault-free complement edge (v, v c ) between AQn0−1 − F0 and AQn1−1 − F1 where v is in AQn0−1 − F0 . Note that by assumption, both AQn0−1 − (F0 ∪ {v}) and AQn1−1 − (F1 ∪ {v c }) contain an even number of vertices. Now, |F0 ∪ {v}|, |F1 ∪ {v c }| ≤ 2n − 4. So AQn0−1 − (F0 ∪ {v}) and AQn1−1 − (F1 ∪ {v c }) have perfect matchings M0 and M1 , respectively. Thus M0 ∪ M1 ∪ {(v, v c )} is a perfect matching in AQn − F . Subcase 4b: |F0 | = 2n − 4. Hence |F1 ∪ FX ∪ FC | = 3. We note that F1 contains either one vertex or three vertices. Thus |FX ∪ FC | ≤ 2. We start with an almost perfect matching M0 missing w in AQn0−1 − F0 . Suppose |{w c , wx , (w, w c ), (w, w x )} ∩ F | ≤ 1. Then we may assume that w c , (w, w c ) ̸∈ F . Thus AQn1−1 − (F1 ∪ {w c }) contains a perfect matching M1 , and hence M0 ∪ M1 ∪{(w, w c )} is a perfect matching in AQn − F . (We note that |F1 ∪{w c }| ≤ 4 < 2n − 3 as n ≥ 5. So we may assume that |{wc , wx , (w, w c ), (w, w x )} ∩ F | ≥ 2. Thus the construction will not work. One may want to apply Proposition 3.3 to find another almost perfect matching in AQn0−1 − F0 missing y ̸= w . However, it is possible that |{yc , yx , (y, yc ), (y, yx )} ∩ F | ≥ 2. (To be precise, this happens when w c = yc and w x = yc .) Instead, we apply Proposition 3.4. Clearly AQn0−1 − F0 has no isolated vertices. If there is a forbidden 2-path w − u − v in AQn0−1 − F0 where both v and w are of degree 1, then we can completely determine F0 . Since |F0 | = 2n − 4 and F0 contains an odd number of vertices, F0 must contain exactly one edge (w, v) and 2n − 5 vertices, each adjacent to both w and v . But such a configuration is impossible in AQn0−1 . (Otherwise, deleting these 2n − 5 vertices together with u will disconnect the graph, which is impossible as AQn0−1 has connectivity 2n − 3 since n ≥ 5.) Since we have three different almost perfect matchings in AQn0−1 − F0 , each missing a different vertex, we may indeed assume there is an almost perfect matching missing w in AQn0−1 − F0 such that |{w c , w x , (w, w c ), (w, w x )} ∩ F | ≤ 1, so we are done.  4. Conclusion In this paper, we studied the strong matching preclusion problem introduced in [25]. Given hypercubes are bipartite and hence not resilient under the strong matching preclusion measure as shown in [25], it is natural to consider non-bipartite variants of hypercubes. The class of augmented cubes is a natural choice due to its many attractive properties. We showed that these interconnection networks are indeed resilient under this measure. Acknowledgments We would like to thank the three anonymous referees for a number of helpful comments and suggestions. This was a project started at the 2011 Oakland University Summer Mathematics Institute under the supervision of EC. A preliminary version of this project received a United States Army Award and a United States Air Force Certificate of Achievement at the 2012 Detroit Science Fair. E. Cheng et al. / Theoretical Computer Science 491 (2013) 71–77 77 References [1] R.C. Brigham, F. Harary, E.C. Violin, J. Yellen, Perfect-matching preclusion, Congressus Numerantium 174 (2005) 185–192. [2] P. Bonneville, E. Cheng, J. Renzi, Strong matching preclusion for the alternating group graphs, Journal of Interconnection Networks 12 (2011) 277–298. [3] N.-W. Chang, S.-Y. Hsieh, Conditional diagnosability of augmented cubes under the PMC Model, IEEE Transactions on Dependable and Secure Computing 9 (2012) 46–60. [4] E. Cheng, P. Hu, R. Jia, L. Lipták, Matching preclusion and conditional matching preclusion for bipartite interconnection networks I: sufficient conditions, Networks 59 (2012) 349–356. [5] E. Cheng, P Hu, R. Jia, L. Lipták, Matching preclusion and conditional matching preclusion for bipartite interconnection networks I: Cayley graphs generated by transposition trees and hyper-stars, Networks (2012) 357–364. [6] E. Cheng, R. Jia, D. Lu, Matching preclusion and conditional matching preclusion for augmented cubes, Journal of Interconnection Networks 11 (2010) 35–60. [7] E. Cheng, M.J. Lipman, L. Lipták, D. Sherman, Conditional matching preclusion for the arrangement graph, Theoretical Computer Science 412 (2011) 6279–6289. [8] E. Cheng, L. Lesniak, M.J. Lipman, L. Lipták, Conditional matching preclusion sets, Information Sciences 179 (2009) 1092–1101. [9] E. Cheng, M.J. Lipman, L. Lipták, Matching preclusion and conditional matching preclusion for regular interconnection networks, Discrete Applied Mathematics 160 (2012) 1936–1954. [10] E. Cheng, L. Lipták, Matching preclusion for some interconnection networks, Networks 50 (2007) 173–180. [11] S.A. Choudum, V. Sunitha, Augmented Cubes, Networks 40 (2) (2002) 302–310. [12] A.A. Hagberg, D.A. Schult, P.J. Swart, Exploring network structure, dynamics, and function using NetworkX. in: Proceedings of the 7th Python in Science Conference, SciPy2008, 2008, pp. 11–15. [13] S.-Y. Hsieh, Y.-R. Cian, Conditional edge-fault Hamiltonicity of augmented cubes, Information Sciences 180 (2010) 2596–2617. [14] S.-Y. Hsieh, J.-Y. Shiu, Cycle embedding of augmented cubes, Applied Mathematics and Computation 191 (2007) 314–319. [15] H.-C. Hsu, L.-C. Chiang, J.J.M. Tan, Lih-Hsing Hsu, Fault Hamiltonicity of augmented cubes, Parallel Computing 31 (2005) 131–145. [16] L.-H. Hsu, C.-K. Lin, Graph Theory and Interconnection Networks, CRC Press, 2009. [17] W.-S Hong, S.-Y. Hsieh, Strong diagnosability and conditional diagnosability of augmented cubes under the comparison diagnosis model, IEEE Transactions on Reliability 61 (2012) 140–148. [18] X. Hu, H. Lou, The (conditional) matching preclusion for burnt pancake graphs, Discrete Applied Mathematics 161 (2013) 1481–1489. [19] T.-K. Li, J.J.M. Tan, L.-H. Hsu, Hyper Hamiltonian laceability on edge fault star graph, Information Science 165 (2004) 59–71. [20] H. Lü, X. Li, H. Zhang, Matching preclusion for balanced hypercubes, Theoretical Computer Science 465 (2012) 10–20. [21] M. Ma, G. Liu, J.M. Xu, Panconnectivity and edge-fault-tolerant pancyclicity of augmented cubes, Parallel Computing 33 (2007) 36–42. [22] M. Ma, G. Liu, J.-M. Xu, The super connectivity of augmented cubes, Information Processing Letters 106 (2008) 59–63. [23] J.-H. Park, Matching preclusion problem in restricted HL-graphs and recursive circulant g (2m , 4), Journal of KIISE 35 (2008) 60–65. [24] J.-H. Park, S.H. Son, Conditional matching preclusion for hypercube-like interconnection networks, Theoretical Computer Science 410 (2009) 2632–2640. [25] J.-H. Park, I. Ihm, Strong matching preclusion, Theoretical Computer Science 412 (2011) 6409–6419. [26] S. Wang, R. Wang, S. Lin, J. Li, Matching preclusion for k-ary n-cubes, Discrete Applied Mathematics 158 (2010) 2066–2070.