Abstract
The enhanced hypercube is a well-known variant of the hypercube and can be constructed from a hypercube by adding an edge to every pair of vertices with complementary addresses. Let denote the set of faulty vertices in an n-dimensional enhanced hypercube . In this paper, we conclude that if , then every fault-free edge of lies on a fault-free cycle of every even length from 4 to , and if and k have the different parity, then every fault-free edge of lies on a fault-free cycle of every possible odd length from to , where .
1. Introduction
The well-known hypercue has several excellent properties, such as recursive structure, regularity, symmetry, small diameter, low degree, and logarithmic diameter [1]. One variant of the hypercube that has been the focus of a great deal of research is the enhanced hypercube [2,3], which can be obtained from the well-known n-dimensional hypercube by adding each edge from the vertex to the vertex . The n-dimensional enhanced hypercube is proposed to improve the efficiency of the hypercube structure , as it possesses many attractive properties that are superior to that of the hypercube [4,5,6,7,8,9,10,11]. Moreover, the folded hypercube is the special case of the enhanced hypercube when [12,13,14,15,16,17,18,19,20,21].
In computer network topology design, one of the central issues in evaluating a network is to study the network embedding problem. The embedding of one guest graph into another host graph is a one-to-one mapping m from the vertex set of to the vertex set [1]. Recently, the multiprocessor system is becoming prevalent and significant. Using the fault-tolerant embedding properties to evaluate the reliability of a parallel computing system is a significant issue. Therefore, many research fields and topics focus on the reliability analysis problems regarding the fault-tolerant embedding of distributed networks [19].
The concept of ISTs was first introduced by Itai and Rodeh [22]. At a later time, a large number of researchers were attracted by the problems regarding the reliability of parallel and distributed networks. The construction of ISTs is obtained to receive high levels of fault-tolerant properties and security. To pursue the above goals, one is to design an efficient construction or investigate the fault-tolerant embedding properties. Note that the class of enhanced hypercube is a general case of the folded hypercube. The enhanced hypercubes have attracted much attention, e.g., the diagnosability, embedding, and others. Fault-tolerant cycle embedding with respect to vertex is related to investigating the property of a more cost-effective constructure.
Problems regarding the fault-tolerant embedding for hypercubes and folded hypercubes have been studied in [14,15,23,24]. Let and be the sets of faulty vertices and faulty edges, respectively. Tsai [25] proved that every fault-free edge of lies on a fault-free cycle of every even length from 4 to inclusive, where . Furthermore, Hsieh and Shen [26] extended the above result to show that every fault-free edge of lies on a fault-free cycle of every even length from 4 to , where and . Xu et al. [21] showed that every fault-free edge of lies on a fault-free cycle of every even length from 4 to and also lies on a fault-free cycle of every odd length from to if n is even, where . Cheng et al. [13] proved that every fault-free edge of lies on a fault-free cycle of every even length from 4 to if , and also lies on a fault-free cycle of every odd length from to if n is even and , where . After that, Kuo and Stewart [16] further proved that every fault-free edge of lies on a fault-free cycle of every even length from 4 to if , and also lies on a fault-free cycle of every odd length from to if is even, where . Due to the above motivations, in this paper, we consider the faulty enhanced hypercube with , where denotes the set of faulty vertices of , proving that every fault-free edge of lies on a fault-free cycle of every even length from 4 to if , and every fault-free edge of lies on a fault-free cycle of every possible odd length from to if and k have different parity.
The remainder of this paper is organized as follows. In Section 2, we introduce some basic definitions and lemmas used in our discussion. We give the main results related to even cycles and odd cycles embedding in the faulty enhanced hypercube in Section 3 and Section 4 respectively. Finally, we conclude this paper in Section 5.
2. Preliminaries
For the graph theoretical terminology and notations not mentioned here, see [27]. A graph is an ordered pair in which V is a finite set and E is a subset of is an unordered pair of . We call V as the vertex set and E as the edge set. For a set of edges or vertices S in G, the graph is a subgraph of G by deleting all elements in S from G. Two vertices u and v are adjacent if . A path, represented as , is a sequence of distinct vertices in which any two consecutive vertices are adjacent. We call and the end-vertices of the path . A path forms a cycle if and . The length of a path P (respectively, a cycle C) is denoted by (respectively, ). Let and be the sets of faulty vertices and faulty edges in G, where , . A vertex v is fault-free if . An edge is fault-free if the two end-vertices and the edge between them are not faulty, i.e., . A path (respectively, a cycle) is fault-free if it contains no faulty edges.
The n-dimensional hypercube, denoted by , is a graph with vertices which are labeled as binary strings of length n from to . Two vertices u and v in are linked by an edge if and only if u and v differ exactly on one bit position. For convenience, we define the vertex and the vertex , where is the complement of , i.e., for some and . In other words, u and have the different binary strings exactly on the ith position. We call the edge as an ith dimension edge which is along dimension i. Let be the set of ith dimensional edges. Clearly, . For a given integer i, partition along dimension i into two -dimensional cubes, then (respectively, ) denotes the subgraph of induced by (respectively, ), where , . Obviously, we have and being isomorphic to .
A graph G is bipartite if the vertex set V can be divided into two disjoint partite subsets and such that each edge in G connects one end-vertex in and another in . A bipartite graph is hyper-Hamiltonian laceable if for any vertex , , there exists a Hamiltonian path of between any two vertices in .
The distance between u and v denoted by is the length of the shortest path between u and v in G. The Hamming distance between two vertices and in is denoted by , where and . The Hamming weight of the vertex , denoted by , is the number of i’s such that . We can use to check the parity of the vertex u, i.e., u is an even vertex (respectively, an odd vertex) if is even (respectively, is odd). Note that is a bipartite graph with two disjoint partite subsets is odd} and is even}. Clearly, , .
Definition 1.
Ref. [2] Enhanced hypercube is an undirected simple graph. Its vertex set is . Its edge set is ; for clarity, , , and y satisfies one of the following two conditions: , or .
One can observe that the enhanced hypercube is obtained from the well known hypercube by adding the edges in the set , which is called the set of complementary edges, denoted by . As mentioned above, and . We can define the edge set of as . Note that is -regular, vertex-transitive, but not edge-transitive [3]. For three-dimensional enhanced hypercubes and , see Figure 1.
Figure 1.
Illustrations of and .
Definition 2.
Ref. [6] An i-partition on , where , is a partition of along dimension i into two -dimensional cubes.
For , , can be partitioned into two -dimensional hypercubes, we call (respectively, ) as the subgraph of induced by (respectively, ). And all the edges in are between and , i.e., .
For , , can be partitioned into two -dimensional enhanced hypercubes, we call (respectively, ) as the subgraph of induced by (respectively, ). And we have .
Lemma 1.
Ref. [8] For any positive integers , an automorphism σ of is denoted as . Moreover, if , it follows that is isomorphic to (represented as , ). Particularly, if , . However, if , it implies that is a disconnected graph.
Let and denote the sets of faulty vertices and faulty edges, respectively.
Lemma 2.
Ref. [25] Let be with , every fault-free edge of lies on a fault-free cycle, whose length is of every even length from 4 to inclusive.
Lemma 3.
Ref. [28] Assume that u and v are any two distinct vertices in . Thus, there exists a path of length l joining u and v with and .
Lemma 4.
Ref. [29] Let be with and . Assume that u and v are any two distinct fault-free vertices in . Thus, contains a fault-free path joining u and v, whose length l satisfies that and .
Lemma 5.
Ref. [30] Let X and Y be any two partite subsets of . Assume x and u are two distinct vertices of X, and y and v are two distinct vertices of Y. Then there exist two vertex-disjoint paths and such that connects x and y, connects u and v. Moreover, and spanning , i.e., .
Lemma 6.
Ref. [31] Let be with . Then is hyper-Hamiltonian-laceable.
3. Even Cycles Embedding in with
Theorem 1.
Let be with , . Then every fault-free edge of lies on a fault-free cycle of every even length from 4 to .
Proof.
Applying Definition 1, it follows that , and . Let be an arbitrary fault-free edge. In , since , two 4-cycles, and , contain all the edges of ; i.e., every edge of lies on a fault-free cycle of length 4. Now we consider the case . We have the following two subcases according to the distributions of the fault-free edge e.
Case 1: .
Lemma 2 ensures that the theorem holds.
Case 2: .
For any arbitrary edge , applying Lemma 1, we know that there exists an automorphism such that . Since , and is isomorphic to , it implies that . Note that e is a fault-free edge, so it implies that . Lemma 2 indicates that lies on a fault-free cycle of every even length from 4 to .
In summary, all cases have been concerned, so the proof is completed. □
4. Odd Cycles Embedding in with
In this section, let denote the set of faulty vertices in . We will prove that every fault-free edge of lies on a fault-free cycle of every possible odd length from to if and have different parity. We first give three useful lemmas as follows.
Lemma 7.
Partition along dimension n into two -dimensional hypercubes and by Definition 2. Let be a jth dimensional edge in such that and , where . Then, for , , we have and ; for , , we have and .
Proof.
Assume is an arbitrary jth dimensional edge in . Note that is partitioned along dimension n. Obviously, . We check two possible situations regarding dimensions of the selected edge .
- (i)
- For , . We can denote the two vertices in by the binary strings as and . By the definition and connection of the binary strings of two vertices, we have ... and . Accordingly, by the calculating of the numbers of the different bits in the binary bits of the two selected vertices, we can conclude that . As an immediate result, we have .
- (ii)
- For , . We can denote the two vertices in by the binary strings as and . By the definition and connection of the binary strings of two vertices, we have and . Similarly as above, it follows that and .
In summary, for any , there exist distinct vertices v’s in , such that or (respectively, or ). □
Lemma 8.
Partition along dimension n into two -dimensional hypercubes and by Definition 2. Assume is an ith dimensional edge in , where . Then we can select distinct w’s in such that , and the edges and are both the jth dimension edges in and ; i.e., and , where , and .
Proof.
Let be an arbitrary ith dimensional edge in . We distinguish two possible situations regarding dimensions of the selected edge as follows:
When , , we can select w’s in such that are both the jth dimensional edges, , . We distinguish the following two subcases:
- For , we can select distinct w’s in such that and are both the jth dimensional edges in , i.e., , and . For clarity, let and be two vertices in . Thus is an ith dimensional edge. We can select a vertex , and . It implies that and . It follows that the vertices can be denoted as , , , and . As an immediate result, we have and .
- For , we can select distinct w’s in such that and are both the jth dimensional edges in , i.e., . For clarity, let and be two vertices in . Thus, is an ith dimensional edge. We can select a vertex , . It implies that and . Subsequently, we have and .
When , , we can select w’s in such that are both the jth dimensional edges, , . We distinguish the following two subcases:
- For , we can select distinct w’s in such that and are both the jth dimensional edges in , i.e., . For clarity, let and be two vertices in . Thus is an ith dimensional edge. We can select a vertex , . It implies that and . It is easy to see that and .
- For , we can select distinct w’s in such that and are both the jth dimensional edges in , i.e., , and . For clarity, let and be two vertices in . Thus, is an ith dimensional edge. We can select a vertex in , . Accordingly, we have and . As a consequence, we have and .
In summary, for any ith dimension edge in , there exist distinct w’s in such that , and , simultaneously. □
Lemma 9.
Let be with , where and k have the different parity. Let be an ith dimensional fault-free edge in . If (respectively, ), then the edge lies on a fault-free cycle of every possible odd length l with (respectively, ).
Proof.
The proof of this lemma is in Appendix A. □
Theorem 2.
Let be with , where and k have the different parity. For an ith dimensional edge , if , then the edge lies on a fault-free cycle of every possible odd length l with in ; if , then the edge lies on a fault-free cycle of every possible odd length l with in .
Proof.
The proof of this theorem is by induction on n. It is trivial to check the theorem holds for and . Assume the theorem holds for , where m and k have the different parity. We now would like to show the theorem holds for every , where m and k have the different parity. Recall that Lemma 9 proved the theorem holds for . In the following, we consider . Let , and , be two arbitrary distinct faulty vertices in . Thus, there exists an integer i, , such that . Applying Definition 2, if we partition along dimension i, where , then we can obtain two -dimensional cubes, and each of the cubes contains at least one faulty vertex. Let be an arbitrary fault-free edge in . We distinguish the following subcases according to the partition of (see Table 1).
Table 1.
Cases in Theorem 2 for the desired cycle containing the edge e.
Case 1: , . Without loss of generality, we can assume . Definition 2 ensures that is partitioned into two -dimensional enhanced hypercubes, denoted as and . Denote , , , and . By the partition of , it follows that and . We have two subcases according to the distributions of the fault-free edge e.
- First, . By the symmetric structure of and , and the distribution of faulty vertices, without loss of generality, we can assume that . On one hand, in , , by induction hypothesis, the edge e lies on a fault-free cycle of every possible odd length from to in . Let be a cycle of length in containing the edge e. Let denote that . Note that we can select an edge such that , and . (If not, it implies that . Thus, we have for , a contradiction. Specially, for , Lemma 9 implies that contains a cycle of length 7 when and . Thus, it is easy to select the desired edge on the cycle.) For clarity, we set , and . On the other hand, we can construct the desired cycle as , whose length is , i.e., .
- Now, . , and . Recall that , Theorem 1 implies that the fault-free edge lies on a fault-free cycle of every even length from 4 to in . Assume that and . Thus, . Note that u have n neighbors in , i.e., , , where , and , . We can observe that there exist n cycles of length four containing the edge in common, i.e., , . Recall that . Thus, there exists at least one fault-free pair , such that the cycle of length 4 is fault-free. Assume forms such a fault-free cycle of length 4 containing the edge . Obviously, . On one hand, by induction hypothesis, lies on a fault-free cycle of every odd length from to . For clarity, . Therefore the desired cycle of every odd length from to can be constructed as . On the other hand, we can construct the desired cycle of every odd length from to . Let be a fault-free cycle of length in , which contains the edge . Denote . Applying Theorem 1, lies on a cycle of every even length from 4 to . For clarity, , and . Subsequently, merging the two paths and as well as the two fault-free edges and , the desired cycle can be constructed as , and the cycle is of every odd length from to .
Case 2: , . Without loss of generality, we can select . Definition 2 ensures that can be partitioned into two -dimensional hypercubes, denoted as and . Denote , , and . It follows that and . We have three subcases according to the distributions of the fault-free edge e.
- First, . Without loss of generality, we can assume that .
- (i)
- and . Lemma 8 ensures that there exist distinct vertices ’s such that , and or . Assuming the vertex u as a faulty vertex temporarily, since , we obtain for . Lemma 4 implies that contains a fault-free path joining v and . Merging the path and the edges , and (respectively, , and ), there exist cycles ’s containing the edge , i.e., , or . Obviously, the cycles ’s have the common vertices in the path . We can observe that there are at most cycles ’s with faulty vertices in . Since , we have . It implies there are at least cycles ’s with fault-free vertices in . Since , there exist at least fault-free cycles ’s in . Without loss of generality, we can assume that is such a fault-free cycle and . Lemma 8 indicates that . Obviously, . In , and , applying Lemma 4, there exists a fault-free path of every odd length joining and , where . As an immediate result, forms the desired cycle of every possible odd length in . Since and , we can obtain (see Figure 2a). (In Figure 2, Figure 3, Figure 4 and Figure 5, we use white vertices and black vertices to distinguish the different parity of the vertices, and we use gray vertices to denote the vertices with unknown parity.)
Figure 2. Illustrations of (a) Case 2, (b) Case 2, and (c) Case 2 in the proof of Theorem 2.
Figure 3. Illustrations of (a) Subcase 1.1, (b) Subcase 1.2, and (c) Subcase 2.1 in the proof of Lemma 9.
Figure 4. Illustrations of (a) Subcase 2.2, (b) Subcase 3.1, and (c) Subcase 3.2.1 in the proof of Lemma 9.
Figure 5. Illustrations of (a) Subcase 3.2.1, (b) Subcase 3.2.2, and (c) Subcase 4.2.1 in the proof of Lemma 9. - (ii)
- and . Specially, Lemmas 3 and 7 imply that we can construct the cycles and , whose length is or . Since , it follows that there exists at least one of the above fault-free cycle of odd length or containing the edge . Generally, we construct the desired cycle of every odd length from to . Since , applying Lemma 2, lies on a fault-free cycle of length in , where . Assume that is an edge on , . For clarity, , . Since , we have or . It implies that we can assume and are both fault-free vertices in . Note that or . Lemma 4 ensures that contains a fault-free path of every even length , where . Accordingly, merging the two paths and as well as the two edges and , we can construct the desired cycle as , whose length is .
- Next, . Obviously, . Applying Lemma 7, we can select distinct vertices adjacent to u in such that or , and for . Since , Lemma 4 implies that contains a path joining and . Subsequently, there exist cycles ’s denoted as , where . Note that . It implies that there exists at least one fault-free cycle . Assume is such a fault-free cycle. For clarity, , it follows that . Recall that , Lemma 4 ensures that contains a fault-free path of every odd length joining u and v, where . Accordingly, the desired cycle containing the edge e can be constructed as , whose length is . As a result, (see Figure 2b).
- Finally, . Note that is the set of complementary edges between and . It follows that . Lemma 7 implies that contains distinct vertices ’s, which are adjacent to u and satisfy or , and for . As mentioned above, there exist cycles ’s denoted as , where . Since , there exists at least one fault-free cycle . Assume is such a fault-free cycle in . Lemma 4 indicates that there exists a fault-free path of every even length in , and there also exists a fault-free path of every odd length in , where and . It is easy to see that forms the desired odd cycle, whose length is . It follows that (see Figure 2c).
In summary, all cases have been concerned, so the proof is completed. □
5. Concluding Remarks
Let be the set of faulty vertices in . In this paper, we consider the faulty enhanced hypercube with faulty vertices. For a fault-free edge of , we show that it lies on a fault-free cycle of every even length from 4 to , where ; moreover, it lies on a fault-free cycle of every possible odd length from to in , where and k have different parity.
Funding
This research is supported by the Ningbo Natural Science Foundation (No. 2023J186), the General Scientific Research Foundation of Zhejiang Province Education Department (No. Y202351606), and the Ph.D. Research Startup Foundation of Ningbo University of Technology (No. 2090011540015).
Data Availability Statement
Data are contained within the article.
Conflicts of Interest
The author declares no conflicts of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript; or in the decision to publish the results.
Appendix A. The proof of Lemma 9
Proof.
Since and k have the different parity, we first need to check the lemma holds for . Since is vertex-transitive [3], we can assume the vertex 000 is faulty. For , we can find the cycles of length containing all the ith dimensional fault-free edges of , i.e., , , , . For , we can find the cycles of length containing all the ith dimensional fault-free edges of , i.e., , , , . For , we can find the cycles of length 7 containing all the ith dimensional fault-free edges of , i.e., , . It implies that for each ith dimensional edge in , if , it lies on a cycle of length 5 and 7; if , it lies on a cycle of length and 7.
Now we consider the cases that . Definition 2 ensures can be partitioned along dimension n into two -dimensional hypercubes, denoted as and . Let f be the faulty vertex in , i.e., . Recall that is vertex-transitive [3]. Without loss of generality, we may assume that . Choose an ith dimensional arbitrary fault-free edge in , . Let l be the length of the desired fault-free cycle. We have four subcases according to the distributions of the fault-free edge e (see Table A1).
Table A1.
Cases in Lemma 9 for the desired cycles containing the edge e.
Table A1.
Cases in Lemma 9 for the desired cycles containing the edge e.
| Case | The Distribution of e | The Length l of the Desired Cycle |
|---|---|---|
| Case 1 | ||
| Case 2 | ||
| Case 3 | ||
| Case 4 |
Case 1: . Obviously, . We distinguish two subcases according to the length l.
- Subcase 1.1:
- (respectively, ), for the ith dimensional edge , where (respectively, ).
- Since , it follows that . When and , we have . Thus, forms the desired cycle of length . Lemma 2 implies that lies on a fault-free cycle in of every even length from 4 to . Select in . For clarity, , where satisfies . Since , we have . Thus, forms the desired cycle of length , i.e., . Let be a cycle of length . We can select three distinct edges , and which satisfies that and , where and . Since , we have , . Thus, forms the desired cycle of length .
- Now we consider the case . Specially, Lemmas 3 and 7 ensure that contains a fault-free path of length with (respectively, ) when is an ith dimensional edge, where (respectively, ). Then forms the desired cycle containing the edge of length with (respectively, ) when is the ith dimensional edge, where and (respectively, ).
- Generally, applying Lemma 2, contains a fault-free cycle of every even length from 4 to containing the edge . Select an edge in . For clarity, , where satisfies . Lemma 3 indicates that contains a fault-free path of even length with . Consequently, forms the desired fault-free cycle of length , i.e., (see Figure 3a).
- Subcase 1.2:
- .Applying Lemma 2, the edge lies on a fault-free cycle of length . Note that and . It implies that there must exist a fault-free vertex . Note that w and f have different parity. Since for , we can select an edge on the cycle , such that and (or we can select an edge on the cycle , such that and ). Without loss of generality, we can assume the first situation holds. For clarity, and . Without loss of generality, we can assume is even. Since is odd, it follows that is odd, and is odd. Lemma 5 indicates that there exist two vertex-disjoint paths and spanning , that is, . For clarity, . So the desired cycle containing the edge e can be constructed as , whose length is in (see Figure 3b).
Case 2: . Obviously, . We distinguish two subcases according to the length l.
- Subcase 2.1:
- . (respectively, ), for the ith dimensional edge, where (respectively, ).
- First, we consider the case . Lemma 7 indicates that . Since , with out loss of generality, we can assume . Obviously, we have (respectively, ) when (respectively, ). For clarity, (respectively, ) forms the desired cycle of length (respectively, ) when (respectively, ). Lemma 2 indicates that lies on a fault-free cycle of length , where . We can select an edge on such that is an jth dimensional edge, where . Note that and (respectively, ) when (respectively, ). Thus, (respectively, ) forms the desired cycle, whose length is , i.e., (respectively, , i.e., ).
- Now we construct the desired cycle of every odd length from to . Let be a cycle of length and contains the edge . If every edge on the cycle are the ith dimensional edge, , then we can replace the edge by the path . Thus, the desired cycle is of length . If there exists an edge is an ith dimensional edge, , then we have and is even. Lemma 2 implies that lies on a cycle of length . Thus, forms the desired cycle of length , i.e., .
- Finally, we consider , since n and k have different parity, we have . Note that and is fault-free. Let be an ith dimensional edge in , where (respectively, ). Lemma 8 implies that there exist distinct vertices ’s in such that , is a fault-free jth dimensional edge in , where . Note that (respectively, ) when (respectively, ). Since , for the ith dimensional edge , (respectively, ), we can choose such a vertex satisfies and . It follows thatthe edge is a jth dimensional edge in , where . Obviously, Lemma 3 ensures that contains a fault-free path joining v and w, whose length is (respectively, ) when (respectively, ). Specially, by Lemma 8, if the fault-free path is of length (respectively, ), then it does not contain the vertex u. Assuming u as a faulty vertex temporarily, we can conclude that . Lemma 4 implies that there exists a fault-free path of every possible odd length joining v and w in , where (respectively, ). Consequently, (respectively, ). Since , by Lemma 4, there exists a fault-free path of every odd length joining and in , where . Note that is a fault-free edge in . Thus, . As a result, for the ith dimensional edge , (respectively, ), forms the desired cycle of every possible odd length in , i.e., (respectively, ) (see Figure 3c).
- Subcase 2.2:
- .Lemma 2 ensures that there exists a cycle containing e of length in . Since for , we can select an edge such that , . For clarity, , . Note that is even, is even and is odd. Without loss of generality, we can assume is odd and is odd. Lemma 6 indicates that contains a fault-free Hamiltonian path joining and with length . As a consequence, forms the desired cycle containing the edge e of odd length in (see Figure 4a).
Case 3: . Obviously, . Assume that and . Obviously, . We distinguish two subcases according to the length l.
- Subcase 3.1:
- .Note that is a nth dimensional edge between and . Specially, if , we can find that . Lemma 3 indicates that there exists a fault-free path of length in . Thus, forms the desired cycle of length . Lemma 7 ensures that there exist distinct vertices ’s adjacent to u in for , such that or . Generally, since , we can select a fault-free vertex such that is a jth dimensional edge in , where and . Lemma 7 indicates that . Lemma 4 indicates that contains a fault-free path of every odd length from 3 to . For convenience, we denote , . In , Lemma 3 ensures that contains a path of every even length joining and , where . Accordingly, the desired cycle can be constructed as with every possible odd length in , (see Figure 4b).
- Subcase 3.2:
- .By Lemma 2, contains a fault-free cycle of length . Note that and . It implies that there must exist a fault-free vertex . According to the distribution of the node , we consider the following subcases:
- Subcase 3.2.1
- , i.e., .Since the number of vertices that adjacent to the vertex u in is 2, there must exist such a vertex adjacent to the vertex u such that . Hence, the cycle can be represented as . Therefore, . Considering the relationship between and , we distinguish the following subcases:
- First, we consider the case that . One can observe that we may assume is odd and is even. It implies that is odd and is odd. Applying Lemma 5, there exist two vertex-disjoint paths and spanning , whose length totally is . For clarity, . Consequently, forms the desired cycle with length . Since , , it follows that (see Figure 4c).
- Now, we consider the case that . Note that is even. Applying Lemma 6, contains a fault-free Hamiltonian path joining and with length . As a result, forms the desired cycle containing the edge e, whose length is (see Figure 5a).
- Subcase 3.2.2:
- , i.e., .Recall that for . It follows that we can select an edge such that . For convenience, we may assume that is even and is odd. It implies that is odd and is odd. Lemma 5 indicates that contain two vertex-disjoint paths and with total length , that is, . For clarity, . As an immediate result, forms the desired cycle of length l with (see Figure 5b).
Case 4: . Obviously, . Note that is the set of complementary edges between and . Without loss of generality, we can assume and . Obviously, . We distinguish two subcases according to the length l.
- Subcase 4.1:
- .Specially, when , one can observe that . Lemma 3 indicates that contains a fault-free path of length . Thus, forms the desired cycle of length . In , Lemma 7 ensures that there exists distinct vertices ’s adjacent to the vertex u in such that or . Generally, since , we can select a fault-free vertex such that is a jth dimensional edge in when and . Lemma 7 indicates that . Applying Lemma 4, contains a fault-free cycle of every odd length from 3 to joining u and w. For clarity, , . In , Lemma 3 indicates that there exists a fault-free path of every even length joining and , where . Consequently, forms the desired cycle of length l with in . Since , , it follows that .
- Subcase 4.2:
- .Applying Lemma 2, contains a fault-free cycle of length . Note that . It implies that there must exist a fault-free vertex . Considering the distribution of the vertex , we distinguish the following subcases:
- Subcase 4.2.1:
- , i.e., .Since the number of vertices that adjacent to the vertex u in is 2, there must exist such a vertex adjacent to the vertex u such that . For clarity, , . According to the relationship between and , we distinguish the following subcases:
- First, we consider the case that . Note that we can assume is odd and is odd. Lemma 5 ensures that there exist two vertex-disjoint paths and spanning , that is, . Accordingly, forms the desired cycle of length in (see Figure 5c).
- Now, we consider the case that . Note that is even, is odd and . Lemma 6 indicates that contains a fault-free Hamiltonian path joining and with length . As a result, forms the desired cycle with length is .
- Subcase 4.2.2:
- , i.e., .This proof is similar to that in Case 3.2.2.
In summary, all cases have been concerned, so the proof is completed. □
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