Hamiltonian-Laceability in Star Graphs with Conditional Edge Faults

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(1)Hamiltonian-Laceability in Star Graphs with Conditional Edge Faults. a. Ping-Ying Tsai a,*, Jung-Sheng Fu b, Gen-Huey Chen a Department of Computer Science and Information Engineering, National Taiwan University, Taipei, Taiwan, ROC b Department of Electronics Engineering,National United University, Miaoli, Taiwan, ROC * E-mail address: bytsai0808@gmail.com. ABSTRACT The star graph Sn possess many nice topological properties. It is one of the most versatile and efficient interconnection networks (networks for short) so far discovered for parallel computation. Edge fault tolerance is an important issue for a network since the edges in the network may fail sometimes. It is known that Sn is a bipartite graph. In this paper, we show that for any Sn (n ≥ 4) with ≤ 2n − 7 edge faults in which each node is incident to at least two healthy edges is strongly Hamiltonian laceable. It is also shown that our result is optimal.. 1: INTRODUCTIONS The star graph [1], which is a Cayley graph, has been recognized as an attractive to the hypercube. It possesses many nice topological properties, such as recursiveness, symmetry, maximal fault tolerance, sublogarithmic degree and diameter, and strong resilience, which are all desirable when we are building an interconnection topology for a parallel and distributed system. The star graph can embed rings [16], grids [10], trees [3], and hypercubes [15]. Many efficient communication algorithms for shortest-path routing [16], multiple-path routing [4], broadcasting [14], and scattering [5] were proposed. Linear arrays and rings, which are two of the most fundamental networks for parallel and distributed computation, are suitable for designing simple algorithms with low communication costs. Numerous efficient algorithms designed on linear arrays and rings for solving various algebraic problems and graph problems can be found in [2], [11]. Linear arrays and rings can also be used as control/data flow structures for distributed computation in arbitrary networks. These applications motivate the embedding of paths and cycles in networks. Suppose that W is an interconnection network. A path (or cycle) in W is called a Hamiltonian path (or Hamiltonian cycle) if it contains every node of W exactly once. W is called Hamiltonian if there is a Hamiltonian cycle in W. A Hamiltonian network can embed a longest cycle with dilation 1, congestion 1, load 1, and expansion 1. Since processor or link faults may develop in real. world networks, it is important to consider faulty networks. The problems of diameter, routing, gossiping, multicasting, broadcasting, and embedding have been solved on various faulty networks. This study considers the embedding problem on a faulty star graph. Previously related work can be found in [7], [8], [9], [13]. Let Sn denote an n-dimensional star graph. In this study, we show that for any Sn (n ≥ 4) with ≤ 2n − 7 edge faults in which each node is incident to at least two healthy edges, we can obtain a fault-free Hamiltonian path between two arbitrary nodes in different partite sets and a fault-free path of length n! − 2 between two arbitrary nodes in the same partite set. In the next section, necessary definitions and notations are first introduced. Then, the embedding is shown in Section 3. Finally, this paper concludes with some remarks in Section 4.. 2: PRELIMINARIES For convenience of discussing, a network topology is represented by a simple undirected graph, which is loopless and without multiple edges. For the graph definition and notation, we follow [17]. A graph G is a triple consisting of a vertex set V(G), an edge set E(G), and a relation that associates with each edge two vertices called its endpoints. When vertices u and v are the endpoints of an edge, they are adjacent and are neighbors. Since a graph is simple undirected, we use (u, v) to denote the edge that connects vertex u and v. For a node u, N(u) denotes the neighborhood of u, which is the set {v | (u, v) ∈ E}. A path is a simple graph whose vertices can be ordered so that two vertices are adjacent if and only if they are consecutive in the list. A path is denoted by a sequence of adjacent vertices <v0, v1, … , vk>, in which v0, v1, … , vk are distinct except that possibly v0 = vk. The length of the path is k. For ease of description, it is abbreviated to a v0-vk path, and we may use P or <v0, P, vk> to denote the path. The degree of vertex v in G, denoted by dG(v) = |N(v)|. The minimum degree of G is denoted by δ(G) = min {dG(v)| v ∈ V(G)}. We use <n> to denote the set {1, 2, … , n}. An independent set in a graph is a set of pairwise nonadjacent vertices. A graph G = (V0 ∪ V1, E) is bipartite if V0 ∩ V1 = ∅ and E ⊆ {(u, v)| u ∈ V0 and v ∈ V1}. Star graphs are bipartite graphs. In [18], Wong introduced the concept of Hamiltonian laceability for the class of bipartite graphs. A bipartite. - 144 -.

(2) graph G = (V0 ∪ V1, E) with |V0| = |V1| is Hamiltonian laceable if there is a Hamiltonian path between every vertex of V0 and every vertex of V1, where V0 and V1 are the two partite sets of G. We note that any path between two vertices of the same partite set has length at most |V0| + |V1| − 2. It is meaningful to extend the concept of Hamiltonian laceability so that a longest path between every two vertices of the same partite set is required. As [7], we say that a Hamiltonian laceable graph G = (V0 ∪ V1, E) is strongly Hamiltonian laceable if G, additionally, has the property that there is a path of length |V0| + |V1| − 2 between every two vertices of the same partite set. In other words, there is a longest path between every two vertices of a strongly Hamiltonian laceable graph. Lewinter and Widulski [12] introduced extended concept of hyper Hamiltonian laceability. G is hyper Hamiltonian laceable if it is Hamiltonian laceable and, for any vertex v ∈ Vi, there is a Hamiltonian path in G − v between any two vertices in V1−i, i ∈ {0, 1}. Throughout this paper, the paired terms network and graph, node and vertex, and link and edge are used interchangeably. The following is a formal definition of star graph. Definition 1. A n-dimensional star graph is denoted by Sn, which is a graph with V(Sn) = {a1a2 … an | a1a2 … an is a permutation of 1, 2, …, n}, and E(Sn) = {(a1a2 … an, aia2 … ai−1a1ai+1… an) | a1a2 … an ∈V(Sn) and 2 ≤ i ≤ n}. 1234. 4231 2134. 3214. 3241. 3124. 2431 2341. 2314. 3421 1324. 4321. 3412. 2413. 4312. 1423 4213. 1432 1342. 4132 3142. 4123. 1243 2143. Fig.1. The structure of S4 Sn has n! nodes in which each node is one permutation of 1, 2, …, n. Two nodes are adjacent if and only if they can be obtained from each other by swapping the leftmost number with one of the other n − 1 numbers. Apparently, Sn is regular of degree n − 1. The structures of S1, S2, and S3 are a node, an edge, and a cycle with a length of six, respectively. The structure of S4 is shown in Fig. 1. For convenience, we say that the edge between nodes a1a2 … an and aia2 … ai−1a1ai+1… an is along the dimension i, where 2 ≤ i ≤ n. If v ∈ V(Sn), we use e(v) to denote the set of incident edges of v in Sn and ei(v) denote the incident edge of v along dimension i. Ei(Sn) denote the set of edges along dimension i in Sn. An Sn is a recursive structure that contains many smaller stars, which are referred to as embedded Srs of Sn, where 1 ≤ r ≤ n. We use. S kbk +1 ...bn to denote the induced subgraph of Sn. that Sn consists of n disjoint embedded Sn−1s connected with the edges along dimension i. In this situation, we say that we have partitioned Sn over dimension i. There are (n − 2)! edges between S ni −1 and S nj−1 for 1 ≤ i ≠ j ≤ n. Conveniently, we use Ei,j(Sn) to denote the set of these edges. The following lemmas, which are results in [13], will also be used very often. Lemma 1. [13] Sn is (n − 3)-edge fault tolerant strongly Hamiltonian laceable, where n ≥ 4. This means that in Sn with at most n − 3 faulty edges, we can find a Hamiltonian path between every two distinct nodes of different partite sets and a path of length n! − 2 between every two distinct nodes of the same partite set. Lemma 2. [13] The star graph Sn is hyper Hamiltonian laceable for n ≥ 4. Meanwhile, nodes and edges that are not faulty are referred to as healthy nodes/edges.. 3: MAIN RESULT In this section, we would show that for any Sn (n ≥ 4) with ≤ 2n − 7 edge faults in which each node is incident to at least two healthy edges, a fault-free path of length n! − 1(or n! − 2) between two arbitrary nodes in different (or the same) partite sets could be obtained. The basic idea uses an inductive proof. First, Sn+1 will be partitioned into n + 1 disjoint Sns such that every node is incident with at least two healthy edges in each Sn. Then, we can combine the paths of each Sn into a new path in the Sn+1. Hence, we should have a method to partition star graph. The following lemma in [6] will be used in our proof. Lemma 3. [6] Let F ⊂ E(Sn+1), |F| ≤ 2n − 5 and δ (Sn+1 − F) = 2, where n ≥ 3. We can partition Sn+1 over some dimension i ∈ <n+1> − {1} such that for all q ∈ <n+1>, δ (<*i−1q*n+1−i>n − F) = 2 and |Ei(Sn+1) ∩ F| ≥ 1. We also need the following lemmas to prove our main theorem. Lemma 4. Let F ⊂ E(Sn+1), |F| ≤ 2n − 5, q ∈ <n+1> and δ (Sn+1 − F) = 2, where n ≥ 3. Suppose that there is an s-t path P with length n! − 1 or n! − 2 in S nq − F. Then there exists an edge (v, u) of P such that en+1(v), en+1(u) ∈ En+1(Sn+1) − F. Proof. Since the length of P is n! − 1or n! − 2, we have at least n! − 2 choices. If none of the edges of P meet the requirements of such an edge (v, u), then |En+1(Sn+1) ∩ F| ≥ ⎡( n!− 2 ) / 2 ⎤ . (Because an edge in En+1(Sn+1) eliminates two edges of P.) Moreover, ⎡( n!− 2 ) / 2 ⎤ > 2n − 5 for n ≥ 3. This contracts |F| ≤ 2n − 5. Therefore, we can always find such (v, u) in P. □ Lemma 5. Given a fixed n, let F ⊆ E(Sn), A ⊆ <n>, and let SA denote the subgraph of Sn induced by V ( U S nk−1 ) . k∈A. For all distinct integers i, j∈A, if |Ei,j(Sn) ∩ F| < (n − 2)! /2,. with vertex set {a1a2 … an | ak+1… an = bk+1… bn}. Note. - 145 -.

(3) and S ni −1 −F, S nj−1 −F are strongly Hamiltonian laceable, then SA−F is strongly Hamiltonian laceable. Proof. For all distinct integers i, j∈ A and |Ei,j(Sn) ∩ F| < (n − 2)! / 2, there are at least two edges (ui, vj), (ui', vj') ∈ Ei,j(Sn) − F such that ui, ui' ∈ V( S ni −1 ), vj, vj' ∈ V( S nj−1 ),. subpaths of P1. An s-t path Q in SA−F is constructed as follows (see Fig. 3): < s, P, v1, u2, P2, v2, ... , um, Pm, vm, u1, P', t >. s. Path. and ui and ui'(vj and vj', respectively) are in different partite sets. Suppose A = {k1, k2, … , km}, where m ≤ n and each k is distinct. Let s and t be arbitrary two nodes in SA−F. We want to construct a longest s-t path in SA−F. Two cases are considered: Case 1. s ∈ V( S na−1 ) and t ∈ V( S nb−1 ), where a, b ∈ A. Link u2. for all ki ∈ A, there is a ui-vi Hamiltonian path Pi in S nk−i 1 − F for all i ∈ <m−1>, and a um-t path Pm with a length (n − 1)! − 1 ((n − 1)! − 2, respectively) in S nk−m1 − F if s and t are in different partite set (in the same partite set, respectively). An s-t path Q in SA−F is constructed as follows (see Fig. 2): < s, P1, v1, u2, P2, v2, ... , um, Pm, t >. Path Link. v1 s. v2. u2. S nk1−1. t. um. S nk−21. Snk−m1. Fig.2. s and t are in different substars of Lemma 5. The length of Q is (m − 1) × ((n − 1)! − 1) + (m − 1) + ((n − 1)! − 1) = m × (n − 1)! − 1 if s and t are in different partite set, and the length of Q is (m − 1) × ((n − 1)! − 1) + (m − 1) + ((n − 1)! − 2) = m × (n − 1)! − 2 if s and t are in the same partite set. So SA−F is strongly Hamiltonian laceable. Case 2. s, t ∈ V( S nq−1 ) for some q ∈ A. Without loss of generality, let k1 = q. Since S nk1−1 − F is strongly Hamiltonian laceable, there is an s-t path P1 with a length (n − 1)! − 1 ((n − 1)! − 2, respectively) in S nk1−1 − F if s and t are in different partite set (in the same partite set, respectively). In addition, by Lemma 4, there exists an edge (v1, u1) of P1 such that (v1, u2) , (vm, u1) ∈ En(Sn) − F, where u2 ∈ V( S nc−1 ) and vm ∈ V( S nd−1 ), where c, d ∈ A − {k1}. Without loss of generality, let k2 = c and km = d. Similar to Case 1, we can find (v2, u3), (v3, u4), … , (vm−1, um) ∈ En(Sn) − F, where ui, vi ∈ V( S nk−i 1 ), ui and vi are in different partite sets, for all i ∈ <m> − {1}. Since S nk−i 1 − F is strongly Hamiltonian laceable for all ki ∈ A, there is a ui-vi Hamiltonian path Pi in S nk−i 1 − F for all i ∈ <m> − {1}. Additionally, let P and P' denote s-v1 and u1-t. v1. S nk−2 1 v. and a ≠ b. Without loss of generality, let k1 = a and km = b. Let u1 = s. From above discussion we can find edges (v1, u2), (v2, u3), (v3, u4), … , (vm−1, um) ∈ En(Sn) − F, where ui, vi ∈ V( S nk−i 1 ), ui and vi are in different partite sets, for all i ∈ <m>. Since S nk−i 1 − F is strongly Hamiltonian laceable. t. 2. S nk1−1. u1. vm. S nk−m1. um. v m −1. u3 v3. u m −1. S nk−m1−1 Fig.3. s and t are in the same substar of Lemma 5. The length of Q is (m − 1) × ((n − 1)! − 1) + m + ((n − 1)! − 1) − 1 = m × (n − 1)! − 1 if s and t are in different partite set, and the length of Q is (m − 1) × ((n − 1)! − 1) + m + ((n − 1)! − 2) − 1 = m × (n − 1)! − 2 if s and t are in the same partite set. So SA−F is strongly Hamiltonian laceable. □ Lemma 6. Let s, t ∈ V(Sn). For any edge e ≠ (s, t) in Sn(n ≥ 4), there exists a s-t path including e with a length of n! − 1 (or n! − 2) in Sn if s and t are in different partite sets (or in the same partite set). Proof. We can use a computer program to check that the result is true for S4, and then proceed by induction on n. As per our induction hypothesis, assume that the result holds for Sn for some n ≥ 4. Now consider Sn+1. Let e = (a, b). Since a and b are adjacent, they differ only in the 1st position and the ith position for some i ∈ <n+1> − {1}. We can partition Sn+1 over some dimension j ∈ <n+1> − {1, i}, hence make a and b in the same Sn. Without loss of generality, let j = n + 1, that is, each embedded Sn is S nr S nk−3 1. for all r ∈ <n+1>. Suppose e ∈ E( S nk ) for some k ∈ <n+1>. Five cases are considered: Case 1. s ∈ V( S nk ), t ∈ V( S nq ) for some q ∈ <n+1> − {k}. Let k = i1, q = in+1. Let (v1, u2) ∈ En+1(Sn+1) such that v1 ∈ V( S ni1 ) − {a, b}, u2 ∈ V( S ni2 ), v1 and s are in different partite sets, where i2 ∈ <n+1> − {i1, in+1}. With the induction hypothesis, there is a s-v1 Hamiltonian path P1 including e for S ni1 . Let A = <n+1> − {i1}. Since each Sn is strongly Hamiltonian laceable, by Lemma 5, SA is strongly Hamiltonian laceable. Hence there is an u2-t path P with a length of n × n! − 1 (or n × n! − 2) in SA if s and t are in different partite sets (or in the same partite set). An s-t path including e in Sn+1 is constructed as follows (see Fig. 4): < s, P1, v1, u2, P, t >. The length of the s-t path is (n × n! − 1) + 1 + (n! − 1) = (n + 1)! − 1 if s and t are in different partite sets, and the. - 146 -.

(4) length of the s-t path is (n × n! − 2) + 1 + (n! − 1) = (n + 1)! − 2 if s and t are in the same partite set. Path. v1. u2. s. S ni1. i2}. Let (v2, u3) ∈ En+1(Sn+1) such that v2 ∈ V( S ni2 ) − {a, b},. Link. v2. t. v2 and u2 are in different partite sets, u3 ∈ V( S ni3 ), where. u n +1. S nin+1 Fig.4. s and t are in different substars. Case 2. t ∈ V( S nk ), s ∈ V( S nq ) for some q ∈ <n+1> − {k}. The construction of an s-t path including e in Sn+1 is similar to that of Case 1. Case 3. s ∈ V( S np ), t ∈ V( S nq ), where p, q ∈ <n+1> − S ni2. {k} and p ≠ q. Let i1 = p, in+1 = q, i2 = k. Let (v1, u2), (v2, u3) ∈ En+1(Sn+1) such that v1 ∈ V( S ni1 ) , u2, v2 ∈ V( S ni2 ) − {a, 3. 1. 2. Moreover, with Lemma 4, we can find an edge (v1, u1) of P1 such that (v1, u2), (vn+1, u1) ∈ En+1(Sn+1), where u2 ∈ V( S ni2 ) − {a, b}, vn+1 ∈ V( S nin +1 ), and in+1 ∈ <n+1> − {i1,. i3 ∈ <n+1> − {i1, i2, in+1}. With the induction hypothesis, there is a u2-v2 Hamiltonian path P2 including e for S ni2 . Let A = <n+1> − {i1, i2}. Since u3 and vn+1 are in different partite sets, by Lemma 5, there is an u3-vn+1 Hamiltonian path PA in SA. Additionally, let P and P' denote s-v1 and u1-t subpaths of P1. An s-t path including e in Sn+1 is constructed as follows (see Fig. 5): < s, P, v1, u2, P2, v2, u3, PA, vn+1, u1, P', t >. s. t Path. 2. b}, u ∈ V( S ) ,v and s, v and u are in different partite i3 n. sets, where i3 ∈ <n+1> − {i1, i2, in+1}. By Lemma 1, there is a s-v1 Hamiltonian path P1 for S ni1 . With the induction. Link. strongly Hamiltonian laceable, by Lemma 5, SA is strongly Hamiltonian laceable. Hence there is an u3-t path P with a length of (n − 1) × n! − 1 (or (n − 1) × n! − 2) in SA if s and t are in different partite sets (or in the same partite set). An s-t path including e in Sn+1 is constructed as follows (see Fig. 4): < s, P1, v1, u2, P2, v2, u3, P, t >. The length of the s-t path is ((n − 1) × n! − 1) + 2 + 2 × (n! − 1) = (n + 1)! − 1 if s and t are in different partite sets, and the length of the s-t path is ((n − 1) × n! − 2) + 2 + 2 × (n! − 1) = (n + 1)! − 2 if s and t are in the same partite set. Case 4. s, t ∈ V( S nk ). Let k = i1. With the induction hypothesis, there is a s-t path P1 including e in S ni1 with a length n! − 1 (or n! − 2) when s and t are in different partite sets (or in the same partite set). Moreover, with Lemma 4, we can find an edge (v1, u1) ≠ e of P1 such that (v1, u2), (vn+1, u1) ∈ En+1(Sn+1), where u2 ∈ V( S ni2 ), vn+1 ∈. V( S nin +1 ), and i2, in+1 ∈ <n+1> − {i1}. Let A = <n+1> − {i1}. Since u2 and vn+1 are in different partite sets, by Lemma 5, there is an u2-vn+1 Hamiltonian path PA in SA. Additionally, let P and P' denote s-v1 and u1-t subpaths of P1. An s-t path including e in Sn+1 is constructed as follows (see Fig. 5): < s, P, v1, u2, PA, vn+1, u1, P', t >. The length of the s-t path is (n! − 1) − 1 + n × (n! − 1) + (n + 1) = (n + 1)! − 1 if s and t are in different partite sets, and the length of the s-t path is (n! − 2) − 1 + n × (n! − 1) + (n + 1) = (n + 1)! − 2 if s and t are in the same partite set. Case 5. s, t ∈ V( S nq ) for some q ∈ <n+1> − {k}. Let i1 = q, and i2 = k. By Lemma 1, there is an s-t path P1 in S. i1 n. with a length n! − 1 (or n! − 2) when s and t are in different partite sets (or in the same partite set).. v. u2. hypothesis, there is a u2-v2 Hamiltonian path P2 including e for S ni2 . Let A = <n+1> − {i1, i2}. Since each Sn is. S ni2. 1. Sni1. u. 1. v n +1. S nin+1. u n +1. v2. vn. u3 v3. S ni3. un. S nin. Fig.5. s and t are in the same substar. The length of the s-t path is (n! − 1) − 1 + n × (n! − 1) + (n + 1) = (n + 1)! − 1 if s and t are in different partite sets, and the length of the s-t path is (n! − 2) − 1 + n × (n! − 1) + (n + 1) = (n + 1)! − 2 if s and t are in the same partite set □ Lemma 7. Let (s, t), (u, v) denote two distinct edges in Sn(n ≥ 3). Then there exists a Hamiltonian cycle in Sn including (s, t) and (u, v). Proof. The result is trivial for n = 3. When n ≥ 4, since s and t are in different partite sets, by Lemma 6, there exists a Hamiltonian path P from s to t including (u, v). □ Then P and (s, t) form a Hamiltonian cycle in Sn. q Lemma 8. Let u ∈ V( S n −1 ) for some q ∈ <n>. Then |en(N(u)) ∩ Ei,q(Sn)| = 1 for all i ∈ <n> − {q}. Proof. Suppose u = a1a2 … an, where {a1, a2, … , an} = <n> and an = q. Let v ∈ N(u). This means v = aja2 … aj−1a1aj+1… an for some 2 ≤ j ≤ n, hence en(v) incident to a S n −j 1 . For w ≠ v ∈ N(u), w = aka2 … ak−1a1ak+1… an for some 2 ≤ k ≤ n, then k ≠ j. Since |N(u)| = n − 1 and |N(u) ∩ V( S nq−1 )| = n − 2, if v ∈ N(u) − V( S nq−1 ), then en(v) = en(u) = (u, v) ∈ E a1 , q ( S n ) . Else en(v) ∈ E a j , q ( S n ) , where 2 ≤ j ≤. n − 1. So we have |en(N(u)) ∩ Ei,q(Sn)| = 1 for all i ∈ <n> □ − {q}. In the following, we start our main proof.. - 147 -.

(5) Theorem 1. Let F ⊂ E(Sn), n ≥ 4, where |F| ≤ 2n − 7 and δ (Sn − F) = 2. Then, Sn − F is strongly Hamiltonian laceable. Proof. We proceed by induction on n. Since n − 3 = 2n − 7 when n = 4, according to Lemma 1, the theorem holds for n = 4. As per our induction hypothesis, assume that the result holds for Sn for some n ≥ 4. Consider Sn+1 with |F| ≤ 2n − 5 and δ (Sn+1 − F) = 2. By Lemma 3, we can partition Sn+1 over some dimension i ∈ <n+1> − {1} such that for all q ∈ <n+1>, δ (<*i−1q*n+1−i>n − F) = 2 and |Ei(Sn+1) ∩ F| ≥ 1. Without loss of generality, let i = n + 1, that is, each embedded Sn is S nq for all q ∈ <n+1>. Since |En+1(Sn+1) ∩ F| ≥ 1, we have |E( S nq ) ∩ F| ≤ 2n − 6, for all. q ∈ <n+1>. Let s and t be arbitrary two nodes in Sn+1−F. We want to construct a longest s-t path in Sn+1−F. Two cases are considered: Case 1. |F ∩ E( S nq )| ≤ 2n − 7, for all q ∈ <n+1>. We have. ∑| E i≠ j. i, j. ( S n +1 ) ∩ F | ≤ 2n − 5 ≤. ((n + 1) − 2)! for all i, j 2. ∈ <n+1> and i ≠ j, where n ≥ 4 since |F| ≤ 2n − 5. If there are two distinct integers i', j' ∈ <n+1> such that | E i ', j ' ( S n +1 ) ∩ F | = ((n+1)−2)! /2, then we have n = 4 and |Ei,j(Sn+1) ∩ F| = 0 < ((n+1)−2)! /2 for all i, j ∈ <n+1>, where i ≠ j and {i', j'} ≠{i, j}. Two cases are further considered: Case 1.1. n > 4. Let A = <n+1>. In this case, we have |Ei,j(Sn+1) ∩ F| < ((n+1)−2)! /2 for all distinct i, j∈A. With the induction hypothesis, S nq − F is strongly Hamiltonian laceable for all q ∈ A. By Lemma 5, Sn+1 − F is strongly Hamiltonian laceable. Case 1.2. n = 4. If |Ei,j(S5) ∩ F| < (n − 1)! / 2 for all distinct integers i, j ∈ <5>, then the discussion is the same as when n > 4. We assume that | Ei', j' (S5 ) ∩ F | = ((4+1)−2)! /2 = 3, for some distinct integers i', j' ∈ <5>. First, consider that s ∈ V( S 4a ) and t ∈ V( S 4b ), where a, b ∈ <5> and a ≠ b. Let i1 = a and i5 = b. We can always find r ∈ <4> such that i' = ir and j' ≠ ir+1. Then the construction of an s-t path in S5 is similar to Case 1 of Lemma 5. Now, consider that s, t ∈ V( S 4q ) for some q ∈ <5>. Let i1 = q. If i1 ∈ {i', j'}, we can let i3 ∈ {i', j'} and i3 ≠ i1. Then i,j | E i1 ,i3 ( S n +1 ) ∩ F |= 3 and |E (S5) ∩ F| = 0 for all i, j ∈ <5>, where i ≠ j and {i, j} ≠{i1, i3}. So the rest of the construction is similar to Case 2 of Lemma 5. If i1 ∉ {i', j'}, then no edge in E5(S5) ∩ F connected to S 4i1 . Let (v1, u1) be an edge of the s-t path of length 4! − 1 or 4! − 2 in S 4i1 , where (v1, u2), (v5, u1) ∈ E5(S5), u2 ∈ V( S4i ' ), v5 ∈. V( S 4j ' ). Let i2 = i' and i5 = j'. Then the rest of the construction is similar to Case 2 of Lemma 5. Case 2. |F ∩ E( S nk )| = 2n − 6 for some k ∈ <n+1>. Because |F| ≤ 2n − 5 and |En+1(Sn+1) ∩ F| = 1, we have |Ei,j(Sn+1) ∩ F| ≤ 1 < ((n+1)−2)! /2 for all distinct i, j ∈. <n+1>, and |F ∩ E( S nq )| = 0 for all q ∈ <n+1> − {k}, where n ≥ 4. Five cases are further considered: Case 2.1. s, t ∈ V( S nk ). In this case, let i1 = k and (v1,. u1) ∈ E( S ni1 ) ∩ F such that en+1(v1), en+1(u1) ∉ F, where s ≠ v1 or u1. Since | S ni1 ∩ (F− {(v1, u1)})| = 2n − 7, with the induction hypothesis, there is an s-t path P in S ni1 − (F− {(v1, u1)}) with length n! − 1 (n! − 2, repectively) if s and t are in different partite sets (in the same partite set, respectively). Assume that P contains (v1, u1) (otherwise, the discussion is easier). The construction of an s-t path of Sn+1 is similar to Case 2 of Lemma 5. Case 2.2. s ∈ V( S nk ) and t ∈ V( S nq ) for some q ∈ <n+1> − {k}. Let (v+, u+) ∈ E( S nk )∩ F such that en+1(v+),. en+1(u+) ∉ F, where s ≠ v+ or u+. Let i1 = k and in+1 = q. Two cases are further considered: Case 2.2.1. Neither v+ nor u+ connects to S nin +1 . Let v1 ∈ V( S ni1 ) − {s} such that v1 and t are in the same partite set, en+1(v1) ∉ F. And let (v+, u2), (v3, u+), (v1, u4) ∈ En+1(Sn+1) − F, where u2 ∈ V( S ni2 ), v3 ∈ V( S ni3 ) , u4 ∈. V( S ni4 ), and i2, i3, i4 ∈ <n+1> − {i1, in+1}. Note that | S nk ∩ (F − {(v+, u+)})| = 2n − 7. With the induction hypothesis, there is an s-v1 path P with a length of n! − 1 (or n! − 2) in S ni1 if s and t are in different partite sets (or in the same partite set). Assume that P contains (v+, u+) (otherwise, the discussion is easier). Let A = {i2, i3} and B = {i4, i5, ... , in+1}, where {i5, i6, ... , in} = <n+1> – {i1, i2, i3, i4, in+1}. Since u2 and v3, u4 and t are in different partite sets, by Lemma 5, there is an u2-v3 Hamiltonian path P1 in SA−F and an u4-t Hamiltonian path P2 in SB−F. An s-t path in Sn+1 is constructed as follows (see Fig. 6): < s, ... , v+, u2, P1, v3, u+, ... , v1, u4, P2, t >. B. Path. s. Link. v1. v4. u4. u+. v+. S. u2. v. 2. i1 n. Sni4. un. v3. u3. vn t. u n +1. S nin. S ni3. S nin+1 Fig.6. Illustration for case 2.2.1. The length of the s-t path is (n! − 1) − 1 + (n! − 1) × n + (n + 1) = (n + 1)! − 1 if s and t are in different partite sets, and the length of the s-t path is (n! − 2) − 1 + (n! − 1) × n + (n + 1) = (n + 1)! − 2 if s and t are in the same partite set. Case 2.2.2. One of v+ or u+ connects to S nin +1 . S ni2. Without loss of generality, we may assume u+ connects to S nin +1 . Two cases are further considered:. - 148 -.

(6) Case 2.2.2.1. u+ does not connect to t. Let (u+, v−), (v+, u ) ∈ En+1(Sn+1) − F, where u2 ∈ V( S ni2 ), v− ∈ V( S nin +1 ). Now, consider that w = t or (w, v2) ∈ F. Since | F ∩ E( S nin +1 ) | = 0 and |En+1(Sn+1) ∩ F| = 1, by Lemma 8, we. and i2 ∈ <n+1> − {i1, in+1}. We can find (u1, vn) ∈ En+1(Sn+1) − F, where u1 ≠ s ∈ V( S ni1 ), u1 and t are in the. can always find (v−, w') ∈ E( S nin +1 ) such that w' ≠ t and (w',. 2. same partite set, v ∈ V( S nin ), and in∈ <n+1> – {i1, i2, n. in+1}. Similar to case 2.2.1, assume there is an s-u1 path in S ni1 containing (v+, u+) with a length of n! − 1 (or n! − 2) if s and t are in different partite sets (or in the same partite set). By Lemma 8, we can find (v−, w) ∈ E( S nin +1 ) with (w,. v2) ∈ En+1(Sn+1), where v2 ∈ V( S ni2 ). First, consider the case w ≠ t and (w, v2) ∉ F. Let un+1 = w and (t, vn+1) ∈ E( S nin +1 ) such that (vn+1, u3) ∈ En+1(Sn+1) − F, where u3 ∈. V( S ni3 ), i3 ∈ <n+1> – {i1, i2, in, in+1} (since | F ∩ E( S nin +1 ) | = 0 and |En+1(Sn+1) ∩ F| = 1, with Lemma 8, we can find such vn+1.) Additionally, with Lemma 7, there exists a Hamiltonian cycle including (t, vn+1) and (v−, un+1). Let A = {i3, i4, ... , in}, where {i4, i5, ... , in−1} = <n+1> – {i1, i2, i3, in, in+1}. Since u2 and v2, u3 and vn are in different partite sets, by Lemma 1 and Lemma 5, there is an u2-v2 Hamiltonian path P1 in S ni2 and an u3-vn Hamiltonian path P2 in SA−F. An s-t path in Sn+1 is constructed as follows (see Fig. 7a): < s, ... , v+, u2, P1, v2, un+1, ... , vn+1, u3, P2, vn, u1, ... , u+, − v , ... , t >. (a). v+. S. i1 n. v3) ∈ En+1(Sn+1) − F, where v3 ∈ V( S ni3 ), i3 ∈ <n+1> – {i1, i2, in, in+1}. Let un+1 = w' and (t, vn+1) ∈ E( S nin +1 ) such that (vn+1, u4) ∈ En+1(Sn+1) − F, where u4 ∈ V( S ni4 ), i4 ∈ <n+1> – {i1, i2, i3, in, in+1} when n > 4 (the discussion is similar when n = 4). By Lemma 7, there exists a Hamiltonian cycle including (t, vn+1) and (v−, un+1). Let A = {i2, i3} and B = {i4, i5, ... , in}, where {i5, i6, ... , in−1} = <n+1> – {i1, i2, i3, i4, in, in+1}. Since u2 and v3, u4 and vn are in different partite sets, by Lemma 5, there is an u2-v3 Hamiltonian path P1 in SA−F and an u4-vn Hamiltonian path P2 in SB−F. An s-t path in Sn+1 is constructed as follows (see Fig. 7b): < s, ... , v+, u2, P1, v3, un+1, ... , vn+1, u4, P2, vn, u1, ... , u+, − v , ... , t >. The length of the s-t path is 2 × ((n! − 1) − 1) + (n! − 1) × (n − 2) + 5 + (n − 3) + (n! − 1) = (n + 1)! − 1 if s and t are in different partite set, and the length of the s-t path is 2 × ((n! − 1) − 1) + (n! − 1) × (n − 2) + 5 + (n − 3) + (n! − 2) = (n + 1)! − 2 if s and t are in the same partite set. B. s. u+. Link. t. vn. S ni2. v2. S v−. v2. in n. S ni2. t. v3. u. S. (b). v+ u. S ni2. v. u3. n +1. v. i n +1 n. S. n +1. S. i3 n. i1 n. s. u. un. u1. +. v. n. S nin u. S. 3. v−. v. v3. u. n +1. S nin +1. S nin+1. u3. S ni3. Fig.8. Illustration for case 2.2.2.2. Case 2.2.2.2. u+ connects to t. Let (v+, u2) ∈ En+1(Sn+1) − F, where u2 ∈ V( S ni2 ) and i2 ∈ <n+1> − {i1, in+1}. We can find (u1, vn) ∈ En+1(Sn+1) − F with u1 ≠ s ∈ V( S ni1 ), vn ∈ V( S nin ), u1 and t are in the same partite set, where in ∈. v. 2n − 7, with the induction hypothesis, there is an s-u1 path P in S nk −(F−{(v+, u+)}) with a length n! − 1 (or n! − 2) when s and t are in different partite sets (or in the same partite set). Assume that P contains (v+, u+) (otherwise, the discussion is easier). Let (v2, un+1), (vn+1, u3) ∈ En+1(Sn+1) − F, where v2 ∈ V( S ni2 ), u3 ∈ V( S ni3 ), un+1, vn+1 ∈ V( S nin +1 ), v2 and u2 are in different partite sets, un+1 and. t. i3 n. u n +1. v3. v n +1. <n+1> − {i1, i2, in+1}. Note that | S nk ∩ (F − {(v+, u+)})| =. 2. 2. in n. S ni1. u2. u1. S. u+. v+. un. u2. Path. un. vn. Link. Path. s. u1. n +1. u4. S ni4. Fig.7. Illustration for case 2.2.2.1.. 4. vn+1 are in the same partite set, and i3 ∈ <n+1> − {i1, i2, in, in+1}. By Lemma 1 and Lemma 2, there is an u2-v2 Hamiltonian path P1 for S ni2 and an un+1-vn+1 Hamiltonian path P2 for S nin +1 − {t}. Let A = {i3, i4, ... , in}, where {i4,. i5, ... , in−1} = <n+1> – {i1, i2, i3, in, in+1}. Since u3 and vn. - 149 -.

(7) are in different partite sets, by Lemma 5, there is an u3-vn Hamiltonian path P3 in SA−F. An s-t path in Sn+1 is constructed as follows (see Fig. 8): < s, ... ,v+, u2, P1, v2, un+1, P2, vn+1, u3, P3, vn, u1, ... , u+, t >. The length of the s-t path is (n! − 1) × (n − 1) + (n! − 2) + (n + 2) + ((n! − 1) − 1) = (n + 1)! − 1 if s and t are in different partite set, and the length of the s-t path is (n! − 1) × (n − 1) + (n! − 2) + (n + 2) + ((n! − 2) − 1) = (n + 1)! − 2 if s and t are in the same partite set. Case 2.3. t ∈ V( S nk ) and s ∈ V( S nq ) for some q ∈ <n+1> − {k}. Let i1 = q and in+1 = k. The construction of an s-t path in Sn+1 is similar to that of Case 2.2. Case 2.4. s, t ∈ V( S nq ) for some q ∈ <n+1> − {k}. Let. i1 = q and (v+, u+) ∈ E( S nk ) ∩ F such that en+1(v+), en+1(u+) ∉ F. If one of v+ or u+ connects to S ni1 , then let i2 = k, u2 = u+, v2 = v+. Otherwise, let i3 = k, u3 = u+, v3 = v+. In the following we only consider when k = i3. The case k = i2 is similar to the case k = i3. Without loss of generality, let (u+, v2), (v+, u4) ∈ En+1(Sn+1) − F, where v2 ∈ V( S ni2 ), u4 ∈. u2) ∈ En+1(Sn+1) − F with v1 ∈ V( S ni1 ), u2 ∈ V( S ni2 ), v2 and u2 are in different partite sets. Since | F ∩ E( S ni1 ) | = | F ∩ E( S ni2 ) | = 0, by Lemma 1, there is an s-v1 Hamiltonian path P1 and an u2-v2 Hamiltonian path P2. Since | S nk ∩ (F − {(v+, u+)})| = 2n − 7, u3 and v3 are in different partite sets, with the induction hypothesis, there is an u3-v3 Hamiltonian path P3 in S ni3 −(F−{(v+, u+)}). Let A = {i4,. i5, ... , in+1}, where {i5, i6, ... , in} = <n+1> – {i1, i2, i3, i4, in+1}. By Lemma 5, SA−F is strongly Hamiltonian laceable and we can find a corresponding u4-t path P in SA−F. An s-t path in Sn+1 is constructed as follows (see Fig. 4): < s, P1, v1, u2, P2, v2, u3, P3, v3, u4, P, t >. The length of the s-t path is (n! − 1) × n + n + (n! − 1) = (n + 1)! − 1 if s and t are in different partite set, and the length of the s-t path is (n! − 1) × n + n + (n! − 2) = (n + 1)! − 2 if s and t are in the same partite set. Path. t. S nin+1. an edge of S ni1 with (v1, u2), (u1, vn+1) ∈ En+1(Sn+1) − F,. v1. where u2 ∈ V( S ni2 ), vn+1 ∈ V( S nin +1 ), v1 and u+ are in. u2. different partite sets, and in+1 ∈ <n+1> − {i1, i2, i3, i4}. Since | F ∩ E( S ni1 ) | = 0, with Lemma 6, we can find an. S. i2 n. s-t path including (v1, u1) with a length n! − 1 (or n! − 2) when s and t are in different partite sets (or in the same partite set) in S ni1 . Since | S nk ∩ (F − {(v+, u+)})| = 2n − 7, u3 and v3 are in different partite sets, with the induction hypothesis, there is an u3-v3 Hamiltonian path P3 in S ni3 −(F−{(v+, u+)}). Let A = {i4, i5, …, in+1}, where {i5, 2. u n +1. Link. V( S ni4 ) and i2, i4 ∈ <n+1> − {i1, i3}. Let (v1, u1) ≠ (s, t) be. 2. v−. s. S. i1 n. u1 vn. v2. un. v n −1. u3. S ni3. S nin. v3. u. n −1. S nin −1. 4. i6, …, in} = <n+1> – {i1, i2, i3, i4, in+1}. Since u and v , u and vn+1 are in different partite sets, by Lemma 1 and Lemma 5, there is an u2-v2 Hamiltonian path P2 in S ni2. and an u4-vn+1 Hamiltonian path P in SA−F. An s-t path in Sn+1 is constructed as follows (see Fig. 5): < s, ... , v1, u2, P2, v2, u3, P3, v3, u4, P, vn+1, u1, ... , t >. The length of the s-t path is (n! − 1) × n + (n + 1) + (n! − 1) − 1 = (n + 1)! − 1 if s and t are in different partite set, and the length of the s-t path is (n! − 1) × n + (n + 1) + (n! − 2) − 1 = (n + 1)! − 2 if s and t are in the same partite set. Case 2.5. s ∈ V( S nq ) and t ∈ V( S nr ) for some q, r ∈ <n+1> − {k}, where q ≠ r. Let i1 = q and in+1 = r. Let (v+, u+) ∈ E( S nk ) ∩ F such that en+1(v+), en+1(u+) ∉ F. Three cases are further considered: Case 2.5.1. Neither v+ nor u+ connects to S ni1 or. S nin +1 . Let i3 = k, u3 = u+, v3 = v+. Without loss of generality, let u+ and s are in the same partite set. Let (v+, u4), (u+, v2) ∈ En+1(Sn+1) − F, where v2 ∈ V( S ni2 ), u4 ∈. V( S ni4 ), and i2, i4 ∈ <n+1> − {i1, i3, in+1}. We can find (v1,. Fig.9. Illustration for case 2.5.2. Case 2.5.2. Exactly one of v+ or u+ connect to S ni1 or. S nin +1 . Without loss of generality, we may assume u+ connects to S ni1 . Let i2 = k, u2 = u+, v2 = v+, (u+, v1), (v+,. u3) ∈ En+1(Sn+1) − F, where v1 ∈ V( S ni1 ), u3 ∈ V( S ni3 ) and i3 ∈ <n+1> − {i1, i2, in+1} (Note that it is possible that v1 = s.) Since | S nk ∩ (F − {(v+, u+)})| = 2n − 7, u2 and v2 are in different partite sets, with the induction hypothesis, there is an u2-v2 Hamiltonian path P2 in S ni2 −(F−{(v+, u+)}). We can find (v1, u1) ∈ E( S ni1 ) with (u1, vn) ∈ En+1(Sn+1) −. F, where u1 ≠ s ∈ V( S ni1 ), vn ∈ V( S nin ), in ∈ <n+1> − {i1, i2, i3, in+1}. Let (un+1, v−) ∈ En+1(Sn+1) − F with v− ≠ s ∈ V( S ni1 ), un+1 ∈ V( S nin +1 ), v− and t are in the same partite set. By Lemma 1 and Lemma 6, we can find an un+1-t Hamiltonian path Pn+1 in S nin +1 and an s-v− path including (v1, u1) with a length n! − 1 (or n! − 2) in S ni1 when s and t are in different partite sets (or in the same. - 150 -.

(8) partite set). Let A = {i3, i4, …, in}, where {i4, i5, …, in−1} = <n+1> – {i1, i2, i3, in, in+1}. Since u3 and vn are in different partite sets, by Lemma 5, there is an u3-vn Hamiltonian path P in SA−F. An s-t path in Sn+1 is constructed as follows (see Fig. 9): < s, ... ,v1, u2, P2, v2, u3, P, vn, u1, ... , v−, un+1, Pn+1, t >. The length of the s-t path is (n! − 1) × n + (n + 1) + (n! − 1) − 1 = (n + 1)! − 1 if s and t are in different partite sets, and the length of the s-t path is (n! − 1) × n + (n + 1) + (n! − 2) − 1 = (n + 1)! − 2 if s and t are in the same partite set. Case 2.5.3. Both v+ and u+ connect to S ni1 and S nin +1 . Without loss of generality, we may assume u+ connects to S ni1 and v+ connects to S nin +1 . Let i2 = k. Two cases are. Note that v1 and t are in the same partite set, hence v1, u1, u2, u3, u4 are all in the same partite set (see Fig. 10), thus, u4 and vn are in different partite sets. By Lemma 5, we can find a u4-vn Hamiltonian path P3 in SA−F. An s-t path in Sn+1 is constructed as follows: < s, ... , v−, u+, ... , v2, u3, P1, v3, u1, ... , v1, un+1, P2, vn+1, 4 u , P3, vn, u2, ... , v+, t >. The length of the s-t path is (n! − 1) × (n − 1) − 1 + (n! − 2) + (n + 3) + (n! − 1) − 1 = (n + 1)! − 1 if s and t are in different partite set, and the length of the s-t path is (n! − 1) × (n − 1) − 1 + (n! − 2) + (n + 3) + (n! − 2) − 1 = (n + 1)! − 2 if s and t are in the same partite set. v1. further considered: v. 1. Path. u n +1. s. u. S. u+. u. v. S nin+1. +. vn. un. S ni2. v3. S ni4. S nin. u2. v2. v4. u3. u2. S u3. +. u4. v+. v2. S ni3. v. v n +1. u−. S ni1. n +1. S nin+1. i1 n. v3. Link. t. v−. 1. Link. v−. u1. Path. u n +1 t. s. u4. v4. S ni3. i2 n. S ni4 Fig.11. Illustration for case 2.5.3.2. Case 2.5.3.2. u+ does not connect to s and v+ does not connect to t. Let (u+, v−), (v+, u−) ∈ En+1(Sn+1) − F, where v− ∈ V( S ni1 ) and u− ∈ V( S nin +1 ). Since | F ∩ E( S ni1 ) | = 0,. un. vn. S nin. Fig.10. Illustration for case 2.5.3.1. Case 2.5.3.1. At least one of u+ and v+ connects to s or t. Without loss of generality, assume v+ connects to t. Let (u+, v−) ∈ En+1(Sn+1) − F, where v− ∈ V( S ni1 ) (It is possible that v− = s). We can find (u1, v−) ∈ E( S ni1 ) with u1 ≠ s, (u1,. v3) ∈ En+1(Sn+1) − F, where v3 ∈ V( S ni3 ) and i3 ∈ <n+1> – {i1, i2, in+1}. Select v1 and t are in the same partite set such that (v1, un+1) ∈ En+1(Sn+1) − F, where v1 ≠ s ∈ V( S ni1 ) and. we can find (u1, v−), (s ,v1) ∈ E( S ni1 ) such that (u1, v3), (v1,. un+1) ∈ En+1(Sn+1) − F, where um, vm ∈ V( S nim ) for m ∈ {1, 3, n + 1}, u1 ≠ s , un+1 ≠ t, and i3 ∈ <n+1> – {i1, i2, in+1}. Then by Lemma 7, there exists a Hamiltonian cycle including (s, v1) and (u1, v−). Let (v2, u3), (u2, u4) ∈ En+1(Sn+1) − F, where um, vm ∈ V( S nim ) for m ∈ {2, 3, 4},. un+1 ∈ V( S nin +1 ). Since | F ∩ E( S ni1 ) | = 0, with Lemma 6,. u2 and v2, u3 and v3 are in different partite sets, and i4 ∈ <n+1> – {i1, i2, i3, in+1}. By Lemma 1, there is a u3-v3 Hamiltonian path P1 in S ni3 . Note that | S nk ∩ (F − {(v+,. there is a s-v1 path including (u1, v−) with a length of n! − 1 (or n! − 2) in Sn if s and t are in different partite sets (or in the same partite set). Let (v2, u3), (u2, vn), (vn+1, u4) ∈ En+1(Sn+1) − F, where um, vm ∈ V( S nim ) for m ∈ {2, 3, 4, n,. u+)})| = 2n − 7, with the induction hypothesis, there is an u2-v2 Hamiltonian path P in S ni2 −(F−{(v+, u+)}). Assume. n + 1}, u2 and v2, u3 and v3 are in different partite sets, vn+1 and un+1 are in the same partite set, and i4, in ∈ <n+1> – {i1, i2, i3, in+1} (if n = 4, then i4 = in). By Lemma 1 and Lemma 2, there is a u3-v3 Hamiltonian path P1 in S ni3. that P contains (v+, u+) (otherwise, the discussion is easier). Since | F ∩ E( S nin +1 ) | = 0, we can find (u−, vn+1) ∈. E( S nin +1 ) such that vn+1 ≠ t and (vn+1, vn) ∈ En+1(Sn+1) − F,. and a un+1-vn+1 Hamiltonian path P2 in S nin +1 − {t}. Note that | S nk ∩ (F − {(v+, u+)})| = 2n − 7, with the induction hypothesis, there is an u2-v2 Hamiltonian path P in S ni2 −(F−{(v+, u+)}). Assume that P contains (v+, u+) (otherwise, the discussion is easier). Let A = {i4, i5, ... , in}, where {i5, i6, ... , in−1} = <n+1> – {i1, i2, i3, i4, in, in+1}.. where vn ∈ V( S nin ) and in ∈ <n+1> – {i1, i2, i3, i4, in+1} (if. n = 4, then in = i4). Since un+1 and s are in the same partite set and | F ∩ E( S nin +1 ) | = 0, with Lemma 6, there is a un+1-t path including (u−, vn+1) with a length of n! − 1 (or n! − 2) in S nin +1 if s and t are in different partite sets (or in the same partite set). Let A = {i4, i5, ... , in}, where {i5, i6, ... , in−1} = <n+1> – {i1, i2, i3, i4, in, in+1}. Note that u+, u−, u1, u3, u2, vn are all in the same partite set (see Fig. 11). Thus, u4 and vn are in different partite sets. By Lemma 5,. - 151 -.

(9) we can find a u4-vn Hamiltonian path P2 in SA−F. An s-t path in Sn+1 is constructed as follows: < s, ... , v−, u+, ... , v2, u3, P1, v3, u1, ... , v1, un+1, ... , u−, + v , ... , u2, u4, P2, vn, vn+1, ... , t >. The length of the s-t path is 2 × ((n! − 1) − 1) + (n! − 1) × (n − 2) + (n + 3) + ((n! − 1) − 1) = (n + 1)! − 1 if s and t are in different partite set, and the length of the s-t path is 2 × ((n! − 1) − 1) + (n! − 1) × (n − 2) + (n + 3) + ((n! − 2) − 1) = (n + 1)! − 2 if s and t are in the same partite set. □ The result above is optimal. It means for any Sn (n ≥ 4) with ≥ 2n − 6 edge faults in which each node is incident to at least two healthy edges, we probably could not find a fault-free Hamiltonian path between two arbitrary nodes in different partite sets. The proof is in the following theorems. Theorem 2. Let F ⊂ E(Sn), |F| ≥ 2n − 6 and δ (Sn − F) = 2, where n ≥ 4. Then Sn − F might not be Hamiltonian laceable. Proof. When |F| = 2n − 6, consider the condition in Fig. 12, where dashed edges denote edges in F and real ones denote edges not in F. In Sn, each node has degree n − 1. Suppose (u, x), (x, y), (y, v) ∉ F and the other edges incident to x or y are in F, so |e(x) ∩ F| = n − 3 = |e(y) ∩ F|. Now let u be the starting point and v is the ending point. It is clear that u and v are in different partite sets. Since < u, x, y, v > is the only path from u to v in Sn − F pass □ though x or y, it can’t be Hamiltonian laceable.. REFERENCES [1]. [2] [3]. [4]. [5]. [6] [7] [8]. [9]. u. n - 3 faulty edges. …. .... x. y. v. Fig.12. Illustration for Theorem 2.. 4: CONCLUDING REMARKS Since processor or link faults may occur in real world networks, fault tolerance is an important research subject of the multi-process computer systems. 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