Linearly many faults in arrangement graphs

Linearly Many Faults in Arrangement Graphs Eddie Cheng and László Lipták Department of Mathematics and Statistics, Oakland University, Rochester, Michigan 48309 Allen Yuan Harvard University, Cambridge, Massachusetts 02138 The star graph proposed by Akers et al. (Proc Int Conf Parallel Process, University Park, PA, 1987, pp. 393–400) has many advantages over the n -cube. However, it suffers from having large gaps in the possible number of vertices. The arrangement graph was proposed by Day and Tripathi (Inf Process Lett 42 (1992), 235–241) to address this issue. Since it is a generalization of the star graph, it retains many of the nice properties of the star graph. In fact, it also generalizes the alternating group graph (Jwo et al., Networks 23 (1993), 315–326). There are many different measures of structural integrity of interconnection networks. In this article, we prove results of the following type for the arrangement graph: If h (r , n , k ) vertices are deleted from the arrangement graph An ,k , the resulting graph will either be connected or have a large component and small components having at most r − 1 vertices in total. Our result is tight for r ≤ 3, and it is asymptotically tight for r ≥ 4. Moreover, we also determine the cyclic vertex-connectivity of the arrangement graph. © 2012 Wiley Periodicals, Inc. NETWORKS, Vol. 000(00), 000–000 2012 Keywords: connectivity; fault tolerance; arrangement graphs 1. INTRODUCTION Parallel computing allows for high speed operation, but poses the problem of finding a fault-resistant architecture for static interconnection networks. These networks are represented by graphs whose vertices correspond to individual computers and whose edges correspond to links between the processors. The earliest networks were structured as hypercubes. Star graphs were proposed later by Akers et al. [1] and had the advantage of lower cost. These graphs are governed by a parameter n; the n-dimensional hypercube has 2n vertices while the n-dimensional star graph has n! vertices. Received November 2011; accepted May 2012 Correspondence to: L. Liptak; e-mail: liptak@oakland.edu DOI 10.1002/net.21476 Published online in Wiley Online Library (wileyonlinelibrary.com). © 2012 Wiley Periodicals, Inc. NETWORKS—2012—DOI 10.1002/net However, the star graph has the disadvantage of having a factorial number of vertices, which creates large gaps in the number of possible vertices; for example, the smallest star graph with at least 6000 vertices has 40,320 vertices. Several alternatives to the star graph have been proposed, such as alternating group graphs [27] and split stars [6] but they have the same scaling problem. The alternative that we shall consider in this article is the arrangement graph, proposed by Day and Tripathi [15], which is a generalization of the star graph and the alternating group graph. Many properties of interconnection networks have been studied, such as topological properties, broadcasting issues, fault-tolerant Hamiltonian properties [21, 23–25], and fault-tolerant routing. Other ways of evaluating the strength of networks were developed and examined in [3, 4, 6, 8, 11, 19, 20, 33]. In particular, fault tolerance is of prime significance in interconnection networks, because the failure of computers in poorly structured networks can cause networks to become disconnected. It is of course desirable that the graph remain connected after vertex deletion. However, deleting as many vertices as the smallest degree of a vertex can disconnect the graph. Because the graph may not remain connected, it is desirable that when vertices are deleted, the remaining graph has a large component and small components with a small number of vertices. Then a large part of the network will still be connected, allowing it to continue to function. The first natural question to ask is what happens when the number of faults is approximately twice the smallest degree; this was examined in [5, 7, 9, 10, 34]. Subsequently, it was explored what happens when a linear number of vertices are deleted in hypercubes [35], in Cayley graphs generated by transpositions, including star graphs [9], and Cayley graphs generated by 2-trees, including alternating group graphs [12]. In this article, we examine the fault resistance of the arrangement graph An,k when a small number of vertices are deleted and generalize it to the case of deleting a linear number of vertices. The main result is a bound in terms of the positive integers n, k, and r on the greatest number of vertices that can be deleted from an arrangement graph such that the remaining graph must still be either connected or have a large component and some small components with at most r −1 vertices in total. In section 2, we introduce the necessary definitions, in section 3, we examine what happens when the number of deleted vertices is at most three times the common degree, and in section 4 we examine the case of deleting linearly many vertices. 2. DEFINITIONS AND PRELIMINARIES A graph G = (V , E) with vertex set V and edge set E is r-regular if the degree of every vertex of G is r. If W ⊂ V is a set of vertices of G, then the graph obtained by deleting the vertices of W from G will be denoted by G − W . A noncomplete graph G is r-connected if deleting any set of fewer than r vertices results in a connected graph. A complete graph with r + 1 vertices is k-connected for k ≤ r. An rregular graph is maximally connected if it is r-connected. The arrangement graph, denoted by An,k , is defined for positive integers n and k such that n > k ≥ 1. The vertex set of the graph is all the permutations of k elements of the set {1, 2, . . . , n}. Two vertices corresponding to the permutations [a1 , a2 , . . . , ak ] and [b1 , b2 , . . . , bk ] are adjacent if and only if there exists exactly one integer 1 ≤ i ≤ k such that ai = bi . Figure 1 shows A4,2 . (For convenience, we write the (n, k)-permutation [i, j] as ij in this figure, for example [1, 4] as 14.) There have been much research on this class of interconnection networks including embeddings, Hamiltonicity and surface area [2, 13, 14, 16–18, 22, 26, 28, 31]. We mentioned earlier that An,k generalizes the star graphs and the alternating group graphs. Indeed, An,n−1 is isomorphic to the n-dimensional star graph (n! vertices with regularity n−1) and An,n−2 is isomorphic to the n-dimensional alternating group graph (n!/2 vertices with regularity 2(n − 2)). The graph An,1 is the complete graph Kn , so we will always assume k ≥ 2. In fact, in terms of applicability as interconnection networks, not all arrangement graphs are useful. Certainly An,1 is not useful as it is a complete graph. The basic requirement is to have many vertices with respect to the regularity of the graph. Because An,k has n(n − 1) · · · (n − k + 1) vertices and its regularity is k(n − k), we use n as the guiding parameter and consider only those networks that have exponentially many or even factorially many vertices with respect to n. Let Hi be the set of vertices representing permutations whose kth element is i for 1 ≤ i ≤ n, and let T denote a set of vertices to be deleted. Define T ∩ Hi = Ti and |Ti | = ti for 1 ≤ i ≤ n. This notation will be used throughout the article. It can easily be seen that An,k is k(n − k)-regular because for any vertex, all neighbors differ in one of the k positions in the permutation, and for each position there are n − k other choices for the number in that position. Let us first note some other preliminary facts about An,k , which are easy to check. 1. Hi is isomorphic to An−1,k−1 when n > k ≥ 2. This is because removing i from all the permutations in Hi results in permutations of k − 1 elements from 2 NETWORKS—2012—DOI 10.1002/net FIG. 1. A4,2 . {1, 2, . . . , n} − {i}. This fact is highly useful in the inductive proofs of the article, as we can often use the induction hypothesis on Hi . n! 2. An,k has (n−k)! vertices, which is the number of permutations of k elements from an n-element set. It follows that Hi has (n−1)! (n−k)! vertices for all 1 ≤ i ≤ n. 3. For any j vertices in Hi , there are exactly j(n − k) distinct vertices outside Hi that are adjacent to at least one of the j vertices. This follows from the fact that each vertex in Hi has n − k neighbors outside Hi and that no two vertices share a common neighbor outside Hi . 4. For each pair Hi and Hj with i  = j, there are exactly (n−2)! (n−k−1)! independent edges (that is, edges such that no two are incident to a common vertex) between them. Note that every edge between Hi and Hj must be between vertices whose permutations differ in their kth element. Thus the number of edges between Hi and Hj is just the number of permutations of k − 1 elements from {1, 2, . . . , n} − {i, j}. 3. BEYOND CONNECTEDNESS Day and Tripathi [15] proved the following theorem about the arrangement graphs: Theorem 3.1. An,k is maximally connected, that is, its connectivity is k(n − k). (One way to see the validity of Theorem 3.1 is to note that An,k is edge transitive and apply a result of Watkins [32].) This result determines that the graph stays connected when the number of vertices deleted is less than the degree, k(n−k). So one may wonder whether deleting an optimal disconnecting set (of size k(n−k)) can leave a resulting graph with many components. Ideally, the resulting graph will have exactly two components, one of which is a singleton. A graph is tightly super-connected if this is true for every optimal disconnecting set. Indeed, for An,k this is true even if we delete approximately 2k(n − k) vertices except for small cases (for example, A4,2 is not tightly super-connected as we can split it into two 4-cycles by deleting four vertices). We will now present this result. Because An,k is governed by two parameters, there is a class of initial cases, which are the boundary cases when k = 2. So we first consider An,2 before looking at the general arrangement graph. Proposition 3.2. Let n ≥ 5, and let T be a subset of vertices of An,2 such that |T | ≤ 3n − 8. Then An,2 − T is either connected or has a large component and a singleton (isolated vertex). Proof. Define T ∩ Hi = Ti and |Ti | = ti for 1 ≤ i ≤ n as before. Note that the subgraphs formed by the vertices in Hi are (n − 1)-cliques. Suppose for the sake of contradiction that An,2 − T is not connected and is not a large component and a singleton. Then we can partition the vertices into two sets A and B such that |A − T |, |B − T | ≥ 2, and A − T and B − T have no edges between them. If Hi ⊆ T , then put Hi into A. Note that each Hi can only have vertices in A or in B, but not both, thus we can associate each Hi with either A or B. Let a be the number of these (n − 1)-cliques associated with A, and let b be the number associated with B; we have a + b = n. Consider a vertex v ∈ A labeled by [x, y]. If v ∈ / T , then each neighbor of v in B must be deleted, and there are at least b − 1 of these, since at most one of the Hi ’s associated with B does not have a neighbor of [x, y]. In contrast, each of these neighbors in B is adjacent to at most a vertices in A. Let m be the number of vertices in A − T . The number of vertices in A that are deleted is a(n − 1) − m, and the number of vertices in B that are deleted is at least m( b−1 a ). Thus, at least a(n − 1) − m + m( b−1 a ) = a(n − 1) + n−2a−1 m( b−a−1 ) = a(n − 1) + m( ) vertices are deleted. a a Without loss of generality we can assume that a ≤ b, so 2a ≤ n. We consider two cases. Case 1. a = 2n . Since m ≤ a(n − 1) = n2 (n − 1), we delete at least n2 (n − 1)+m(− n2 ) ≥ 2n (n −1)−(n −1) = 21 (n −1)(n −2) vertices. When n ≥ 8, we have 21 (n − 1)(n − 2) > 3n − 8, which is not possible. Since n ≥ 5 and n is even, the only case remaining is n = 6. Here 21 (n − 1)(n − 2) = 3n − 8 = 10, thus there must be equality in every inequality we used. In particular, we have m = a(n − 1) = 15, so T ∩ A = ∅, and we must delete all neighbors of three of the Hi ’s in A6,2 . It is easy to see that this requires deleting every vertex in B, that is, 15 vertices, which is more than allowed. Case 2. 2a + 1 ≤ n. is non-negative. Because k ≥ 2, the number Then n−2a−1 a of vertices deleted is a(n − 1) + k( n−2a−1 ) ≥ a(n − 1) + a n−2a−1 2 2( a ) = (a + a )(n − 1) − 4 ≥ 3(n − 1) − 4 > 3n − 8, a contradiction. Because we reach a contradiction in both cases, An,2 −T is either connected or has a large component and a singleton. ■ The result in Proposition 3.2 is sharp because we can isolate two adjacent vertices in An,2 by deleting all their neighbors, altogether 3n − 7 vertices. We are now ready to handle An,k for every k. The star graph An,n−1 has already been given an extensive treatment in [7, 9, 10], so we will consider only when n − 1 > k. We also need the following for the base case: Lemma 3.3. Let T be a set of vertices in A5,3 such that |T | ≤ 8. Then A5,3 − T is either connected or has a large component and a singleton (isolated vertex). Proof. Similar properties for Cayley graph generated by 2-trees were studied in [12]. This class of graphs and the class of arrangement graphs have the alternating group graph An,n−2 in their intersection. It was proved in [12] that if up to 4n − 12 vertices are deleted from An,n−2 , the resulting graph is either connected or has a large component and a singleton. Because 4n − 12 = 8 for n = 5, the result follows. ■ We note that it is possible to prove Lemma 3.3 directly via an ad-hoc method as it is only one (small) graph. Theorem 3.4. Let n and k be positive integers such that n − 1 > k ≥ 2, n ≥ 5. Let T be a subset of the vertices of An,k such that |T | ≤ (2k − 1)(n − k) − 2. Then An,k − T is either connected or has a large component and a singleton. Proof. Define Hi , Ti , ti for i = 1, . . . , n as before. We proceed by induction on k. The case k = 2 follows from Proposition 3.2. We also need the base case when n = 5 and k = 3, which is Lemma 3.3. Now suppose that the statement holds for all pairs (n, k) such that k < k ′ for some k ′ ≥ 3 and such that 2 ≤ k < n − 1 and n ≥ 5, and consider deleting at most (2k ′ − 1)(n − k ′ ) − 2 vertices from An,k ′ where n ≥ 6. We have three cases based on the location of the deleted vertices: Case 1. ti ≤ (k ′ − 1)(n − k ′ ) − 1 for all i. Then every Hi − Ti is connected by Theorem 3.1. Because (n−2)! every pair Hi and Hj have (n−k ′ −1)! independent edges between them, at least this many vertices must be deleted from the graph in order for the graph to become disconnected (otherwise at least one edge would remain between (n−2)! any two Hi − Ti and Hj − Tj ). If k ′ > 3, then (n−k ′ −1)! ≥ ′ (n − 2)(n − 3)(n − 4) ≥ 3(n − 2)(n − 4) > (2k − 1)(n − k ′ ). Thus, we only need to consider the case when k ′ = 3, but then ti ≤ (k ′ − 1)(n − k ′ ) − 1 = 2n − 7, so we have ti + tj ≤ 4n − 14 < 4(n − 3) ≤ (n − 2)(n − 3) for any i = j. Thus there is an edge between Hi − Ti and Hj − Tj for any i  = j, hence An,k − T is connected. Case 2. ti ≤ (2k ′ − 3)(n − k ′ ) − 2 for all i. By the inductive hypothesis, each Hi − Ti is either connected or has a large component and a singleton. First we show that the large components of all the Hi − Ti are all part of one large component Y in An,k − T . As in Case 1, if k ′ > 3, it is easy to see that between any two Hi − Ti there are at least (n−2)! two edges, because (n−k ′ −1)! −2 ≥ (n−2)(n−3)(n−4)−2 ≥ NETWORKS—2012—DOI 10.1002/net 3 3(n−2)(n−4)−2 ≥ 3(n−2)(n−k ′ )−2 > (2k ′ −1)(n−k ′ ) (the last inequality follows from (3n − 6) − (2k ′ − 1) = n + 2(n − k ′ ) − 5 ≥ n − 3 ≥ 3 and n − k ′ ≥ 2). Thus in these cases all the large components are part of Y . If k ′ = 3, then a total of 5(n−3)−2 vertices are deleted, implying that at most four vertices are deleted from some Hi . Thus for any other j, there are at least two edges between Hi − Ti and Hj − Tj because ti + tj ≤ (2k ′ − 3)(n − k ′ ) − 2 + 4 = 3(n − 3) + 2 and (n−2)! (n−k ′ −1)! −2 = (n−2)(n−3)−2 ≥ 5(n−3)−2 > 3(n−3)+2 when n ≥ 7. The only remaining case is when n = 6 and k ′ = 3. Here, we have |T | = 13 and ti ≤ 7 for all i, and there are 12 independent edges between any two Hi ’s. By the pigeonhole principle, there exists a j such that tj ≤ 2. Then for all i = j, ti + tj ≤ 9 < 12 − 2, so there are at least two edges between Hi − Ti and Hj − Tj for all i = j, so all the large components are part of Y . Now if ti ≤ (k ′ − 1)(n − k ′ ) − 1 for all i, then An,k − T is connected by Case 1. If there are at least three i’s for which Hi ≥ (k ′ − 1)(n − k ′ ), then we deleted at least 3(k ′ − 1)(n − k ′ ) > (2k ′ − 1)(n − k ′ ) − 2 vertices, which is not possible. If there is only one i for which Hi ≥ (k ′ − 1)(n − k ′ ), then we are also done because at most one vertex is not part of Y . The only remaining case is if there are exactly two integers i and j such that ti , tj ≥ (k ′ − 1)(n − k ′ ), and there are two vertices u ∈ Hi and v ∈ Hj such that u and v are not part of Y . If u and v are adjacent, then it is easy to see that there are 2(k ′ −1)(n−k ′ )+(n−k ′ −1) > (2k ′ −1)(n−k ′ )−2 vertices adjacent to at least one of u and v. If u and v are not adjacent, then at most one vertex in Hi is adjacent to v and at most one vertex in Hj is adjacent to u. Furthermore, no vertices outside Hi and Hj are adjacent to both u and v. Since u and v have degree k ′ (n − k ′ ) and at most two common neighbors, at least 2k ′ (n − k ′ ) − 2 > (2k ′ − 1)(n − k ′ ) − 2 vertices are adjacent to at least one of u and v. Hence in each case at least one of u or v must still have a neighbor in Y , and thus it is part of Y . Case 3. ti > (2k ′ − 3)(n − k ′ ) − 2 for some i. Then fewer than 2(n − k ′ ) ≤ (k ′ − 1)(n − k ′ ) vertices are deleted outside of Hi , so Hj − Tj is connected for every j = i; in fact, they are part of one large component, say Y , in An,k − T by similar reasoning as in Case 1. Now suppose that at least two vertices in Hi − Ti are not part of Y . By property (3) in section 2, these vertices have at least 2(n − k ′ ) different neighbors outside Hi . However, since fewer than |T | − ti = (2k ′ − 1)(n − k ′ ) − 2 − (2k ′ − 3)(n − k ′ ) + 2 = 2(n − k ′ ) vertices are deleted outside Hi , one of these neighbors must belong to Y , a contradiction, finishing the proof. ■ proof, we need to consider the boundary cases when k = 2. Proposition 3.5 provides the result for An,2 . It turns out that our argument for An,k requires special treatment when k = 3. So we present the case k = 3 separately in Proposition 3.6. Finally, Theorem 3.7 gives the general result. We remark that Proposition 3.5 requires n ≥ 7 because the claim is not true for n = 6 as the number of vertices in A6,2 is too small. So for An,3 , we have to require n ≥ 8 as Proposition 3.5 needs to be applicable for the n copies of An,2 in An,3 . These results will add the requirement n − k ≥ 5 in the statement of Theorem 3.7. This requirement is necessary, because we will be applying induction and the base cases here are the pairs (n, k) when k = 3, but Proposition 3.6 applies only when n ≥ 8. The additional requirement n−k ≥ 5 may seem to devalue Theorem 3.7. However, the corresponding result is known for n − k = 1 (see [8], corresponding to the class of star graphs) and for n−k = 2 (see [12], corresponding to the class of alternating group graphs). Nevertheless, it does leave the cases n − k = 3 and n − k = 4 without a result. Fortunately, in the next section we will provide a general result that will be meaningful for these cases apart from small n. Proposition 3.5. Let n ≥ 7 and let T be a subset of the vertices of An,2 such that |T | ≤ 4n−12. Then An,2 −T is either connected, or has a large component and small components with at most two vertices in total, or |T | = 4n−12 and An,2 − T has a large component and a 4-cycle over the elements [a, b], [a, d], [c, d], [c, b] for some four distinct integers a, b, c, d. Proof. The proof will be similar to that of Proposition 3.2. Define Hi , Ti , ti as before, and suppose that An,2 −T is not connected, moreover, it is not simply a large component and small components with at most two vertices in total. We will show that the only other possible case is when An,2 − T has a large component and a 4-cycle. Define and construct A, B, a, b as before in Proposition 3.2, noting that now |A−T |, |B−T | ≥ ) vertices 3. Again, we have that at least a(n − 1) + k( n−2a−1 a are deleted (where k = |A − T | ≥ 3), and we may assume that 2a ≤ n. We consider two cases depending on a. Case 1. a = 2n . Then k ≤ a(n − 1) = n2 (n − 1), so we delete at least 2 n 1 n 2 (n − 1) + k(− n ) ≥ 2 (n − 1) − (n − 1) = 2 (n − 1)(n − 2) 1 vertices. When n ≥ 7 and n is even, we have 2 (n−1)(n−2) > 4n − 12, a contradiction. Case 2. 2a + 1 ≤ n. The bound in Theorem 3.4 is best possible, since two adjacent vertices have (2k − 1)(n − k) − 1 different neighbors, and deleting them creates a component of size 2. Now that we have described the number of vertices needed to be deleted to get at least two vertices disconnected from the rest of the graph, the natural task is to find the number of vertices needed to be deleted to get at least three vertices in the small components. Given that induction is the method of 4 NETWORKS—2012—DOI 10.1002/net Then the number n−2a−1 is non-negative. Because k ≥ 3, a the number of vertices deleted is a(n − 1) + k( n−2a−1 ) ≥ a n−2a−1 3 a(n − 1) + 3( a ) = (a + a )(n − 1) − 6. When a = 2, this is at least 4(n − 1) − 6, which is greater than 4n − 12. When a = 2, if k > 4, then a(n − 1) + k( n−2a−1 ) > a 4n − 12. Thus, we only need to consider the case when all neighbors of three or four vertices in two of the Hi ’s are deleted. It can be easily checked that a 4-cycle in the form [a, b], [a, d], [c, d], [c, b] has 4n − 12 neighbors and that all other possibilities for the vertices will have more than this many neighbors (in particular, the neighbor set of any three out of the four vertices of this 4-cycle has 4n − 11 vertices, showing that our conclusion is tight). The desired conclusion follows. ■ Proposition 3.6. Let n ≥ 8 and let T be a subset of the vertices of An,3 such that |T | ≤ 7n−25. Then An,3 −T is either connected or has a large component and small components with at most two vertices in total. Proof. Define Hi , Ti , ti as before. We consider cases depending on the ti ’s: Case 1. ti < 2n − 6 for all i. This implies by Theorem 3.1 that Hi − Ti is connected for all i. Furthermore, for any i = j, we have ti + tj < 4n − 13 < (n − 2)(n − 3) when n ≥ 7. Since there are (n − 2)(n − 3) independent edges between Hi and Hj , at least one edge remains between Hi − Ti and Hj − Tj . Thus An,3 − T is connected. Case 2. ti > 4n − 16 for some i. Then fewer than 3n − 9 vertices are deleted outside Hi . By the same argument as above, at least one edge remains between Hj − Tj and Hm − Tm for j, m = i, so there is a component Y in An,3 − T that contains the large component of Hj − Tj for all j = i. Suppose that a vertex in Hj − Tj is not in Y for some j = i. Then tj ≥ 2(n − 3) = 2n − 6 by Theorem 3.1. Consider any m = i, j. We have tj + tm ≤ 3n − 10, so tm < 2n − 6, hence Hm − Tm is connected. It follows that Y contains Hm − Tm for all m = j, i. The number of vertices deleted outside of Hi and Hj is at most |T |−ti −tj ≤ (7n−25)−(4n−15)−(2n−6) = n − 4. However, every vertex in Hi or Hj has at least n − 4 neighbors outside Hi and Hj (the vertex has a neighbor in n − 2 other Hm ’s, two of which may be in Hi or Hj ). Thus, the n − 4 deleted vertices must be the neighbors of any vertex in Hj − Tj but not in Y . Hence at most one vertex in Hj − Tj is not in Y , because no two vertices of Hj have the same set of neighbors outside Hj and Hi . By the same argument on Hi − Ti , at most one vertex of Hi − Ti is not in Y . Thus at most two vertices of An,3 − T are not in Y as desired. Otherwise Y contains every vertex not in (Hi − Ti ) ∪ T . Suppose that three vertices in Hi − Ti are not in Y . Then these vertices have 3(n − 3) = 3n − 9 different neighbors outside Hi , so at least one of these neighbors belongs to Y , because fewer than 3n − 9 vertices are deleted outside Hi , contradicting that these vertices are not in Y . Thus at most two vertices of An,3 − T are not in Y as desired. Case 3. ti ≤ 4n − 16 for all i, but there exists a j such that tj ≥ 2n − 6. By Proposition 3.5, for all i, Hi − Ti is either connected, or it has a large component and small components with at most four vertices in total. We first show that the large components are all part of one large component Y in An,3 − T . Because tj ≥ 2n − 6, the pigeonhole principle implies that there exists an m such that tm ≤ 4. Thus ti + tm ≤ 4n − 16 + 4 = 4n − 12 < (n − 2)(n − 3) − 3 for any i  = m and n ≥ 7. Because (n − 2)(n − 3) is the number of independent edges between Hi and Hm , at least three of these independent edges remain. If there are no edges between the large components of Hi − Ti and Hm − Tm , then all these independent edges are between their small components. Hence Hi − Ti and Hm − Tm each have at least three vertices in small components, thus they both have a large component and a 4-cycle. However, Proposition 3.5 implies that ti = tm = 4n − 16, yielding ti + tm = 8n − 32 ≥ 7n − 25 with equality if and only if n = 7. This is a contradiction unless n = 7 and all deleted vertices are in Hi and Hm . But then each vertex in the 4-cycles has a neighbor remaining outside of Hi and Hm , so An,3 − T remains connected. Hence there is a component Y in An,3 − T that contains the large component of Hi − Ti for all i. Proposition 3.5 implies that for all i, Hi − Ti either has a 4-cycle not in Y or has no more than two vertices not in Y . Furthermore, Proposition 3.2 and Theorem 3.1 yield ti > 3n − 11 if at least two vertices of Hi − Ti are not in Y , and ti ≥ 2n−6 if at least one vertex of Hi −Ti is not in Y . Clearly, if a vertex of Hi − Ti is not in Y , then it is not in the large component of Hi −Ti . Suppose that at least four of the Hi −Ti ’s are not connected. Then |T | ≥ 4(2n−6) = 8n−24 > 7n−25, a contradiction. Thus at most three of the Hi − Ti ’s are not connected. If Hi − Ti is connected for all i, then we are done because An,3 − T is also connected. Now we consider cases depending on how many Hi − Ti ’s have vertices not in Y . Case 3a. Exactly three of the Hi −Ti ’s have at least one vertex not in Y . Then at least 2n − 6 are deleted from each of these Hi ’s. If at least 3n − 10 vertices are deleted from some Hi − Ti , then |T | ≥ 7n − 22, which is a contradiction. Thus each of these three Hi − Ti ’s has at most one vertex not in Y . Let the three vertices be u = [a1 , a2 , a3 ], v = [b1 , b2 , b3 ], and w = [c1 , c2 , c3 ], where a3 , b3 , and c3 are pairwise distinct. Vertex u has 2(n − 3) neighbors in Ha3 , and similarly for v and w in respectively Hb3 and Hc3 . In addition, u has at least n − 5 neighbors outside Ha3 , Hb3 and Hc3 , thus |T | ≥ 6(n − 3) + (n − 5) = 7n − 23, which is a contradiction. Case 3b. Exactly two of the Hi − Ti ’s have at least one vertex not in Y . Suppose one of these, Hj − Tj , has a large component and a 4-cycle not in Y ; it follows that tj = 4n − 16. Then no other Hi − Ti can have a large component and a 4-cycle, hence at most two neighbors of the 4-cycle can remain in An,3 . The 4cycle must be in the form [a, b, j], [a, c, j], [d, c, j], [d, b, j] by Proposition 3.5. There are 4(n − 4) neighbors of this 4-cycle in Hj and 4(n − 3) neighbors outside Hj , altogether 8n − 28 neighbors. At most two of these neighbors can remain, so |T | ≥ 8n − 30 > 7n − 25, which is a contradiction. Hence none of the Hi − Ti ’s has a 4-cycle not in Y . It follows that NETWORKS—2012—DOI 10.1002/net 5 there exist integers j and m such that Hj −Tj and Hm −Tm both have either one or two vertices not in Y . If both have only one vertex not in Y , then at most two vertices are not in Y , so we are done. Thus, we may assume without loss of generality that Hj −Tj has two vertices not in Y . By Proposition 3.2 these two vertices have at least 3n − 10 neighbors in Hj , and each has at least n−4 neighbors outside Hj and Hm , which are all distinct. Finally, consider Hm . If exactly one vertex of Hm − Tm is not in Y , then it has 2(n − 3) neighbors in Hm . Otherwise, if two vertices of Hm − Tm are not in Y , then again by Proposition 3.2, there are at least 3n − 10 neighbors of these vertices in Hm − Tm . Hence |T | ≥ (3n − 10) + 2(n − 4) + 2(n − 3) = 7n − 24 > 7n − 25, which is a contradiction. Case 3c. Exactly one of the Hi − Ti ’s has at least one vertex not in Y . If Hi − Ti has a large component and a 4-cycle, then the 4-cycle has 8n − 28 neighbors as in Case 3b, and so An,3 − T is connected. Otherwise, Hi − Ti has a large component and at most two vertices not in Y , and we are done. We have considered all cases, so our proof is complete. ■ Proposition 3.6 is sharp, since three vertices of a 4-cycle in An,3 (as in Case 3b) have exactly 7n − 24 neighbors, whose deletion creates a component containing three vertices. We are now ready to prove the corresponding theorem for An,k : Theorem 3.7. Let n and k be positive integers such that n ≥ 8, k ≥ 2, n − k ≥ 5. Let T be a subset of the vertices of An,k such that |T | ≤ (3k−2)(n−k)−4. Then An,k −T is either connected or has a large component and small components with at most two vertices in total unless k = 2 and |T | = 4n−12, in which case An,k −T could have a large component and a 4-cycle. Proof. Define Hi , Ti , ti as before. We proceed by induction on k. The cases k = 2 and k = 3 follow, respectively, from Proposition 3.5 and Proposition 3.6. Now suppose that the statement holds for all pairs (n, k) such that k < k ′ for some k ′ ≥ 4, n ≥ 8, and n − k ≥ 5. Consider deleting at most (3k ′ − 2)(n − k ′ ) − 4 vertices from (n−2)! An,k ′ . There are (n−k ′ −1)! ≥ (n − 2)(n − 3)(n − 4) ≥ 4(n − ′ 2)(n − 3) ≥ 4k (n − k ′ ) ≥ (3k ′ − 2)(n − k ′ ) independent edges between any two of the Hi ’s. Thus after deleting T , at least four edges remain between Hi − Ti and Hj − Tj for every i = j. Hence there is a component Y in An,k − T that contains the large component of Hi − Ti for every i for which Hi − Ti has a large component and at most two vertices in total in small components. We consider two cases based on the ti ’s: Case 1. ti ≤ (3k ′ − 5)(n − k ′ ) − 4 for all i. By the inductive hypothesis, for every i, Hi − Ti is either connected or has a large component and small components with at most two vertices in total. Hence, all the large components are part of Y . If Hi − Ti is connected for all i, then we are done. If exactly one of the Hi − Ti ’s is disconnected, then we are also done because its small components have at 6 NETWORKS—2012—DOI 10.1002/net most two vertices in total. If at least four of the Hi − Ti ’s are disconnected, then at least (k ′ − 1)(n − k ′ ) vertices must be deleted from each of them, so |T | ≥ 4(k ′ − 1)(n − k ′ ) = (3k ′ + (k ′ − 4))(n − k ′ ) > (3k ′ − 2)(n − k ′ ) − 4, which is a contradiction. Thus we have two cases remaining: Case 1a. Exactly two of the Hi − Ti ’s have at least one vertex not in Y . Let these be Hi − Ti and Hj − Tj with i = j. If neither has two vertices outside Y , then we are done because there are at most two vertices outside Y in An,k − T . Thus without loss of generality, let Hi − Ti have two vertices outside of Y , say u and v. Then ti ≥ (2k ′ − 3)(n − k ′ ) − 1 by Theorem 3.4. Suppose that Hj − Tj has two vertices outside Y . Then ti + tj ≥ (4k ′ − 6)(n − k ′ ) − 2 ≥ (3k ′ − 2)(n − k ′ ) − 2 > (3k ′ − 2)(n − k ′ ) − 4, which is a contradiction. Thus Hj − Tj has exactly one vertex w outside Y . Vertex w has (k ′ − 1)(n − k ′ ) neighbors in Hj , and each of u and v has at least n − k ′ − 1 distinct neighbors outside Hi and Hj , hence |T | ≥ (2k ′ −3)(n−k ′ )−1+(k ′ −1)(n−k ′ )+2(n−k ′ −1) = (3k ′ − 2)(n − k ′ ) − 3 > (3k ′ − 2)(n − k ′ ) − 4 ≥ |T |, which is a contradiction. Case 1b. Exactly three of the Hi −Ti ’s have at least one vertex not in Y . If one of these, say Hi − Ti , has at least two vertices not in Y , then ti ≥ (2k ′ − 3)(n − k ′ ) − 1 by Theorem 3.4. Let Hj −Tj and Hm −Tm be disconnected as well. We have tj , tm ≥ (k ′ − 1)(n − k ′ ), thus ti + tj + tm ≥ (4k ′ − 5)(n − k ′ ) − 1 > (3k ′ − 2)(n − k ′ ) − 4, which is a contradiction. Thus Hi − Ti , Hj −Tj , and Hm −Tm must each have a large component and a singleton. Hence ti , tj , tm ≥ (k ′ −1)(n−k ′ ), and the singleton in Hi −Ti has at least (n−k ′ −2) neighbors outside Hi , Hj , and Hm . Hence |T | ≥ (3k ′ −2)(n−k ′ )−2 > (3k ′ −2)(n−k ′ )−4, which is a contradiction. Case 2. ti > (3k ′ − 5)(n − k ′ ) − 4 for some i. Then fewer than 3(n − k ′ ) vertices are deleted outside Hi . Since k ′ ≥ 4, this implies that Hj − Tj is connected for every j = i. Thus, Y contains all vertices of An,k −T except possibly some in Hi − Ti . If at least three vertices in Hi − Ti are not in Y , then these vertices have at least 3(n − k ′ ) neighbors in G − Hi . However, fewer than 3(n − k ′ ) vertices are deleted from G − Hi , contradicting that these three vertices are not in Y . Our induction is complete, and the theorem is proven. ■ Again, three vertices of a 4-cycle in An,k have exactly (3k− 2)(n−k)−3 neighbors, so deleting them creates a component containing three vertices, showing that Theorem 3.7 is sharp. 4. LINEARLY MANY FAULTS Theorems 3.1, 3.4, and 3.7 are analogous in that they determine tight bounds on the fault resilience of the arrangement graphs as the number of faulty vertices goes up to about three times the degree of each vertex. However, each theorem was significantly more complex to prove because of the structure of the graph when n and k are small. Thus, instead of a sharp general result, we settle for an asymptotic result. Let S be a set of r vertices. Suppose we want the small components to contain these r vertices. They have rk(n − k) neighbors counting multiplicities and the vertices in S. Let g(S) be the number of these doublecounted vertices. To find a tight formula, we need to find an S in which g(S) is largest. Let S ∗ (n, k) be an optimal solution for An,k . Then one has the result that if up to rk(n − k) − g(S ∗ (n, k)) − 1 vertices are deleted, then the resulting graph is either connected or has a large connected component and small components with at most r − 1 vertices in total. Since it may be too difficult to find the best g(S ∗ (n, k)), we approximate it by a function f (r, n, k) such that (rk(n − k) − g(S ∗ (n, k)) − 1)/(rk(n − k) − f (r, n, k) − 1) tends to 1 as k tends to infinity, thus giving an asymptotically tight result below. Before we prove this asymptotic result, we note that this bridges the gap mentioned earlier regarding Theorem 3.7, which is a tight result for r = 3 but requires n − k ≥ 5. We remarked earlier that by existing results, the gap exists only for n − k = 3 and n − k = 4. If we let r = 3 in Theorem 4.1, |T | is bounded by (3(k − 2) − 6)(n − k), a weaker bound than the tight bound given in Theorem 3.7. However, this bound works for n − k = 3 and n − k = 4 as it is meaningful for large n. For example, consider n − k = 4. The bound is (3(n − 6) − 6)4 = 12n − 96. It does not give a useful result for A6,2 but it has substance for large n. Theorem 4.1. Let n, k, r be positive integers such that n − 1 > k ≥ 2, r ≥ 1. If T is a subset of the vertices of An,k such 2 that |T | ≤ (r(k − 2) + 2 − r2 )(n − k), then An,k − T is either connected or has a large component and small components with at most r − 1 vertices in total. Proof. When r = 1, |T | ≤ (k − 21 )(n − k) < k(n − k), so An,k − T is connected by Theorem 3.1. When r = 2, |T | ≤ (2k − 4)(n − k) < (2k − 1)(n − k) − 2, thus by Theorem 3.4, when n − 1 > k ≥ 2 and n ≥ 5, An,k − T is either connected or has a large component and a singleton. When n < 5, we have k < 4, so (2k − 4)(n − k) < k(n − k), hence An,k − T is connected. Thus, we may assume r ≥ 3 for the rest of the proof. The claim follows easily from Theorem 3.1 when k ≤ 4, 2 because r(k − 2) + 2 − r2 < k when r ≥ 3. Thus we assume that k ≥ 5, and proceed by induction on r. Suppose that the statement holds for all triples of positive integers n, k, and r satisfying the conditions of the theorem and such that r < r ′ for some r ′ ≥ 3. It suffices to show that the statement holds for n, k, and r ′ . To prove this, we shall proceed by induction on k. Suppose that the statement holds for all k < k ′ for some k ′ ≥ 5. It remains to show that the theorem holds for n, k ′ , 2 and r ′ . Note that the function defined by g(x) = x(k ′ −2)− x2 is strictly decreasing for x > k ′ − 2. Thus if r ′ > k ′ − 2, then ′2 |T | ≤ (r ′ (k ′ − 2) + 2 − r2 )(n − k ′ ) ≤ ((r ′ − 1)(k ′ − 2) + ′ 2 ′ 2 − (r −1) 2 )(n − k ), and so the inductive hypothesis implies that An,k ′ − T is either connected or has a large component and small components with at most r ′ − 2 vertices in total. Thus we may assume that r ′ ≤ k ′ − 2. Define Hi , Ti , ti as before. Consider two cases depending on the distribution of the faults: r ′2 ′ 2 )(n − k ) for some i. Then fewer than r ′ (n − k ′ ) ≤ (k ′ − 2)(n − k ′ ) vertices are Case 1. ti > (r ′ (k ′ − 3) + 2 − deleted outside Hi . By Theorem 3.1, Hj − Tj is connected for all j = i, and by a similar argument we can easily show that they are all part of one component Y of An,k − T . Suppose that at least r ′ vertices in Hi − Ti are not in Y . These vertices have r ′ (n−k ′ ) neighbors outside Hi , so at least one of them is not in T . Thus, at least one of these r ′ vertices is in Y , which is a contradiction. Case 2. ti ≤ (r ′ (k ′ − 3) + 2 − r ′2 2 )(n − k ′ ) for all i. By the inductive hypothesis, each Hi − Ti is either connected or has a large component and small components with at most r ′ − 1 vertices in total. Consider Hi − Ti and Hj − Tj with i = j. The number of vertices deleted plus the number of vertices in the small components of Hi − Ti and Hj − Tj is ′2 at most (r ′ (k ′ − 2) + 2 − r2 )(n − k ′ ) + 2(r ′ − 1). However, (n−2)! (n−2)! there are (n−k ′ −1)! ≥ (n−6)! = (n − 2)(n − 3)(n − 4)(n − 5) independent edges between Hi and Hj . Since (n − 2)(n − 3)(n − 4)(n − 5) ≥ (n − 2)(n − 3)(n − 4)(n − k ′ ) ≥ (n − 3)2 (n − 4)(n − k ′ ) + (n − 3)(n − 4)(n − k ′ )
 r ′2 ′ ′ ≥ r (k − 2) + 2 − (n − k ′ ) + 2(r ′ − 1), 2 at least one edge must remain between Hi − Ti and Hj − Tj . Thus the large components of Hi − Ti for i = 1, . . . , n are all part of one component Y of An,k . Now we show that the total number of vertices in the small components is at most r ′ − 1. Suppose that there are at least r ′ vertices in the small components. 2 Define the function F(x) = (x(k ′ − 3) + 2 − x2 )(n − k ′ ). For any i such that Hi − Ti is disconnected, let qi denote the total number of vertices in the small components of Hi − Ti . Because at most r ′ − 1 vertices are in the small components in any Hi − Ti , we have qi ≤ r ′ − 1. The sum of the qi ’s is at least r ′ , thus at least two of the qi ’s are positive. Hence wecan find numbers p1 , . . . , pn such that pi ≤ qi for all i, ni=1 pi = r ′ , and at least two of the pi ’s are positive. Without loss of generality we may assume that p1 , p2 , . . . , pj are positive and all the other pi ’s are zero. Because at least pi vertices are in the small component of Hi − Ti , we have ti > F(pi ) for 1 ≤ i ≤ j. Hence the total number of deleted vertices is 
j n   r ′2 (n − k ′ ) ≥ ti ≥ (F(pi ) + 1) r ′ (k ′ − 2) + 2 − 2 i=1 i=1 NETWORKS—2012—DOI 10.1002/net 7 = j  i=1  p2 pi (k ′ − 3) + 2 − i 2  ′ (n − k )   j j   1 = pi (k ′ − 3) + 2j − p2i  (n − k ′ ) 2 i=1 i=1   j  1 = r ′ (k ′ − 3) + 2j − p2i  (n − k ′ ). 2 i=1 Note that since at least two of the pi ’s are positive and the j pi ’s sum to r ′ , we have i=1 p2i ≤ 1 + (r ′ − 1)2 , from which we obtain 
r ′2 ′ ′ (n − k ′ ) r (k − 2) + 2 − 2
 1 ≥ r ′ (k ′ − 3) + 2j − (1 + (r ′ − 1)2 ) (n − k ′ ). 2 Canceling n − k ′ and rearranging leads to 3 ≥ 2j. However, since j ≥ 2, this is a contradiction, and our induction is complete. ■ We note that Theorem 4.1 is indeed asymptotically tight 2 as (rk(n − k) − g(S ∗ (n, k)) − 1) ≥ (r(k − 2) + 2 − r2 )(n − k). Thus 1≤ (rk(n − k) − g(S ∗ (n, k)) − 1) r(k − 2) + 2 − ≤ r2 2 (n − k) (rk(n − k)) r(k − 2) + 2 − r2 2 . (n − k) 2 Since (rk(n − k))/(r(k − 2) + 2 − r2 )(n − k) tends to 1 as k tends to infinity, (rk(n − k) − g(S ∗ (n, k)) − 1)/(r(k − 2) + 2 2 − r2 )(n − k) also tends to 1 as k tends to infinity. 5. CYCLIC VERTEX-CONNECTIVITY In this section, we consider the concept of cyclic vertexconnectivity. In a graph G, a set of vertices T ⊆ V (G) is called a cyclic vertex-cut if G − T is disconnected and at least two components in G − T contain a cycle. The cyclic vertex-connectivity, denoted by κc (G), is the size of a smallest cyclic vertex-cut. We refer readers to [29, 30] for additional history behind cyclic vertex-connectivity. It is known that κc (An,n−1 ) = 6(n − 3). For other arrangement graphs we have the following: Theorem 5.1. For n − 5 ≥ k ≥ 3 and n ≥ 8, we have κc (An,k ) = (3k − 2)(n − k) − 2, while κc (An,2 ) = 4n − 12. Proof. The case k = 2 follows from Proposition 3.5. When k ≥ 3, Theorem 3.7 implies that if fewer than (3k−2)(n−k)−3 vertices are deleted, then if the graph is disconnected, the small components have at most two vertices 8 NETWORKS—2012—DOI 10.1002/net and cannot have a cycle. To get at least three vertices in the small components, at least (3k − 2)(n − k) − 3 vertices have to be deleted, and it can be seen from the proof of Theorem 3.7 that when this occurs, the small component is a 3-vertex path (part of a 4-cycle), which contains no cycles. Thus κc (An,k ) ≥ (3k−2)(n−k)−2. It is easy to see that the vertices of a 3-cycle (such as [1, 2, . . . , k−1, k], [1, 2, . . . , k−1, k+1], and [1, 2, . . . , k − 1, k + 2]) have exactly (3k − 2)(n − k) − 2 neighbors, whose deletion isolates this 3-cycle in a small component. The large component clearly contains a cycle, so the conclusion follows. ■ 6. CONCLUSIONS In this article, we gave a bound on how many vertices we can delete from an arrangement graph and still guarantee that the remaining graph is either connected or has a large component and small components with at most r −1 vertices in total. These results generalize those given in [5]. Finally, we applied these results to determine the cyclic vertex-connectivity of An,k . For r ≤ 3, Theorems 3.1, 3.4, and 3.7 give tight results. These results were then extended to the bound (r(k − 2) + 2 2 − r2 )(n − k) for r > 3, which is asymptotically tight. Our definition of asymptotically tight requires k to tend to infinity in addition to n. 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