(n2)-fault-tolerant edge-pancyclicity of möbius cubes MQn

Huifeng Zhang1,2, Jun Zhu1, Xirong Xu1, Peng Zhang3
1Zhejiang Lab, Hangzhou,311100,China
2School of Computer Science and Technology Dalian University of Technology, Dalian, 116024, China
3Department of Computer Science Zhongshan College of Dalian Medical University, Dalian, 116085, China

Abstract

The n-dimensional Möbius cube MQn is an important variant of the hypercube Qn, which possesses some properties superior to the hypercube. This paper investigates the fault-tolerant edge-pancyclicity of MQn, and shows that if MQn (n5) contains at most n2 faulty vertices and/or edges then, for any fault-free edge uv in MQni(i=0,1) and any integer with 7i2nfv, there is a fault-free cycle of length containing the edge uv, where fv is the number of faulty vertices. The result is optimal in some senses.

Keywords: Combinatorics, Möbius cube, Edge-pancyclicity, Fault-tolerant, Interconnection network

1. Introduction

It is well known that interconnection networks are of interest in parallel and distributed computing systems because they determine the performance of the systems on a large scale. As topological structures, interconnection networks can be represented by a graph G=(V,E), where V is the vertex-set of G and E is the edge-set of G. |V(G)| and |E(G)| denote the numbers of vertices and edges of G, respectively. A path denoted by (v1,v2,,vk) is a sequence of vertices where two consecutive vertices are adjacent in G. A cycle is a path (v1,v2,,vk) where v1=vk. A graph G is pancyclic if, for every girth|V(G)|, G has a cycle of length . A graph G is edge-pancyclic if, for any edge e of G and every girth|V(G)|, G has a cycle of length containing e. A graph G is vertex-pancyclic if, for any vertex u of G and every girth|V(G)|, G has a cycle of length containing u. Obviously, if a graph is edge-pancyclic, it is also vertex-pancyclic. Edge-pancyclic and vertex-pancyclic on various interconnection networks were studied, including hypercubes, crossed cubes, twisted cubes, locally twisted cubes, augmented cubes, star graphs, and others.

Edge and/or vertex failures are inevitable when a large parallel and distributed computer system is running. Thus, the fault-tolerant capacity of network is an important issue for parallel and distributed computing. A graph G is k-fault-tolerant pancyclic(resp., vertex-pancyclic, edge-pancyclic) if GF is still pancyclic(resp., vertex-pancyclic, edge-pancyclic) for any FE(G)V(G) with |F|k. Pancyclicity and fault-tolerant pancyclicity have been widely studied for many well-known networks, see Xu and Ma [14] for a detail survey on these topics.

The Möbius cube has many properties superior to the hypercube. Though both the Möbius cubes and the ordinary hypercube have the same number of vertices and the same vertex degree, the diameter of the Möbius cube is approximately half that of the ordinary hypercube. Due to nearly half the diameter and better graph embedding capability as compared with its hypercube counterpart of the same size, the Mö¡§bius cubes have been proposed as promising candidates for interconnection topology, and have received considerable attention [1,2,3,6,7,8,11,13,5,4].

With regard to the fault-tolerant Hamiltonicity of Möbius cubes, Huang et al. [7] showed that an n-dimensional Möbius cube is Hamiltonian in the presence of up to n2 node and edge faults. As concerns the pancyclicity and fault-tolerant pancyclicity of Möbius cubes, Fan [3] proved that Möbius cubes are four-pancyclic. Hsieh and Chen [6] found that an n-dimensional Möbius cube with up to n2 edge faults is four-pancyclic.

This paper investigates the fault-tolerant edge-pancyclicity of MQn, and shows that if MQn(n5) contains at most n2 faulty vertices and/or edges then, for any fault-free edge uv in MQni(i=0,1) and any integer with 7i2nfv, there is a fault-free cycle of length containing the edge uv, where fv is the number of faulty vertices.

The remainder of this paper is organized as follows. In Section 2, we recall the definition of MQn, and introduce some properties of MQn to be used in our proofs. In Section 3, we give the proof of our result. Finally, we give some concluding remarks in Section 4.

2. Definitions and properties

The n-dimensional Möbius cube, denoted by MQn, is such an undirected graph, its vertex set is the same as the vertex set of Qn, the vertex X=x1x2xn connects to n other vertices Yi, (1in), where each Yi satisfies one of the following equations: Yi={x1x2xi1x¯ixi+1xn when xi1=0,x1x2xi1x¯ix¯i+1x¯n when xi1=1.

From the above definition, X connects to Yi by complementing the bit xi if xi1=0 or by complementing all bits of xi,,xn if xi1=1. The connection between X and Y1 is undefined, so we can assume that x0 is either equal to 0 or equal to 1, which gives us slightly different network topologies. If we assume x0=0, we call the network a ¡®0-Möbius cube¡¯; and if we assume x0=1, we call the network a ¡®1-Möbius cube¡¯, denoted by MQn0 and MQn1, respectively. The graphs shown in Fig.1 are MQ40 and MQ41, respectivily.

According to the above definition, it is not difficult to see that MQn0 (respectively, MQn1) can be recursively constructed from MQn10 and MQn11 by adding 2n1 edges. MQn0 is constructed by connecting all pairs of vertices that differ only in the 1-th bit, and MQn1 is constructed by connecting all pairs of vertices that differ in the 1-th through the n-th bits. The superscript i of notation MQni, i=0,1, can be omitted if there is no ambiguity arise.

For convenience, we say that MQn10 and MQn11 are two sub-Möbius cubes of MQn, where MQn10 (respectively, MQn11) is an (n1)-dimensional 0-Möbius cube (respectively, 1-Möbius cube) which includes all vertices 0x2xn (respectively, 1x2xn), xi{0,1}. More simply, let L=MQn10 and R=MQn11. In addition, we define the set of crossing edges of MQn to be EC={xyE(MQn)|xV(MQn10)yV(MQn11)}. For any edge xyEC, vertices x and y are crossing vertices of each other. Indeed, there are 2n1 crossing edges and 2n1 pairs of crossing vertices in MQn.

The Möbius cube MQn was first proposed by Cull and Larson [1]. Like Qn, MQn is an n-regular n-connected graph with 2n vertices and n2n1 edges. Moreover, MQn has a diameter of (n+2)2 for MQn10 and (n+1)2 for MQn11. However, for n4, MQn is neither vertex-transitive nor edge-transitive.

Cull and Larson[1] first proved the existence of hamiltonian cycles in MQn by proving that in MQn0 or MQn1, there are 2nk disjoint cycles of length 2k for any k2.

Figure 1 (a) MQ40, (b) MQ41

We need the following two definitions.

Definition 2.1. For any edge e=(x1x2xn,y1y2yn)E(MQn), let λ(e) be the smallest positive integer i{1,2,,n} such that xiyi, then e is called a λ-dimensional edge.

According to Definition 2.1, if we use uL to denote a vertex in L, then uR always denotes its unique neighbor in R, that is, uLuR is an 1-dimensional edge. Let e=uv be a i1-dimensional edge, then denote v=ui1. Let ui1,,ij1,ij=(ui1,,ij1)ij for j2. Let Pt1(u0)=(u0,u1,ujut1) be a path of length =t1, then uj=uj1ij for 1jt1. We use (i1,i2,,it1) to denote Pt1(u0) for short. If u0ut1E, then we use (i1,i2,,it1) to denote cycle Ct(u0) of length =t containing edge u0ut1 for short.

For example, for C7(000001)=(000001,001001,011001,010110,110110,101001,100001) in MQn0, we use (3,2,3,1,2,3) to denote the cycle C7(000001).

Definition 2.2.

If |F|=n2 and there exists a vertex w with NMQnF(w)={w1,w2}, then w is called a weak 2-degree vertex and {w1,w2} is called a w-weak vertex pair(or a weak vertex pair, for short).

Since there is no triangle in MQn, we can obtain the Proposition 2.3 as follows.

Proposition 2.3.

If xyE(MQn), then (x,y) is not a weak vertex-pair in MQnF with |F|n2.

Lemma 2.4.

( Xu et al. [9]) If for any vertex uLL(uRR), vL(vR) is a neighbor of uL(uR) in L(R) then, the distance between uR(uL) and vR(vL) is 1 or 2.

Lemma 2.5.

( Xu et al. [11]) MQn is edge-pancyclic for n2.

Lemma 2.6.

( Xu et al. [10]) For any two different vertices x and y with distance d in MQn(n3), there exists an xy-path of every length from d to 2n1 except for d+1.

Let F be a set of faulty elements in MQn. We need the following lemmas:

Lemma 2.7. If any edge uLuREC in MQn0F for |F|n2, there exists a fault-free 4-cycle or 5-cycle containing the edge uLuR.

Proof. Let uL=0x2xn, we show the lemma according to the following two cases.

Case 1. x2=0.

We can find (n2) disjoint 4-cycles and a 5-cycle except the common edge uLuR as follows. {(i+2,1,i+2),         1in2,(2,1,3,2),      i=n1.

Since |F|n2, there exists a fault-free 4-cycle or 5-cycle containing the edge uLuR. The lemma holds.

Case 2. x2=1.

There exist at most n2 disjoint 4-cycles as Ci4(uL)=(i,1,i)=(uL,uLi,uLi,1,uR) for 3in except the common edge uLuR.

If one of 4-cycle Ci4(uL)(3in) is fault-free, the lemma holds.

If each 4-cycles Ci4(uL)(3in) contains a faulty vertices. Consider 4-cycles (3,1,3)=(uL,uL3,uL3,1,uR).

  1. If uL3,1F, then uL3F. We can find a fault free 5-cycle (3,2,1,2)=(uL,uL3,uL3,2,uL3,2,1,uR) containing the edge uLuR.

  2. If uL3F, then uL3,1F. We can find a fault-free 5-cycle (2,1,2,3)=(uL,uL2,uL2,1,uL3,1,uR) containing the edge uLuR.

The proof is completed. ◻

Lemma 2.8. If any edge uLuREC in MQn1F for |F|n2, there exists a fault-free 5-cycle containing the edge uLuR.

Proof.  Let uL=x1x2xixi+1xn. There exist (n2) disjoint 5-cycles and a 4-cycle except the common edge uLuR as follows. {(i+1,i+2,1,i+1),  1in2xi+1=0,(i+1,1,i+2,i+1),  1in2xi+1=1,(n,1,n),           i=n1.

Since |F|n2, there exists a fault-free 4-cycle or 5-cycle containing the edge uLuR◻

By Lemma 2.7 and 2.8, we can obtain the following result.

Corollary 2.9.

If any edge uLuREC in MQnF for |F|n2, there exists a fault-free 4-cycle or 5-cycle containing the edge uLuR.

Lemma 2.10.

(Xu et al. [9]) If |F|n3 and n3 then, for any fault-free edge e in MQn and any integer with 62nfv, there is a fault-free cycle of length containing the edge e in MQn.

Lemma 2.11.

(Xu et al. [12]) If FV(MQn)E(MQn) and |F|n2, then for any two distinct fault-free vertices u and v, there exists a fault-free path Puv of every length with 2n112nfv1α, where α=0 if vertices u and v is not a weak vertex-pair in MQnF and α=1 if vertices u and v form a weak vertex-pair in MQnF(n5).

Figure 2 Four disjoint 6-cycles in L except the common edge (001011,001100)
Lemma 2.12.

If FV(MQn1) with |F|n2(n6), then for any edge uLvLL, there exists a fault-free 6-cycle containing the edge uLvL in MQn1F.

Proof.  By Lemma 2.10, the lemma holds for |F|n3. We only need to consider the case of |F|=n2.

Let uL=x1x2xjxn. We can assume that |FR||FL|. Otherwise, if |FL||FR|, then |FL|n22n4(n6). By Lemma 2.10, there exists a fault-free 6-cycle containing the edge uLvL in LFL, and so in MQn1F.

If |FL|n4, by Lemma 2.10 , there exists a fault-free 6-cycle containing the edge uLvL.

If |FL|n3, then |FR|1. We can prove the result according to the relationship between the vL and uL as follows.

Case 1. vL=uLj(j=2).

There exist n1 disjoint 6-cycles as Ci6(uL)(1in1) except the common edge uLvL as follows. Ci6(uL)={(1,3,1,2,3),i=1,(i+1,2,1,3,1),i=2,(i+1,5,2,5,4),i=3,(i+1,n,2,n,i+1),4in2x4=0,(i+1,4,2,4,i+1),4in2x4=1,(i+1,4,2,4,n),i=n1.

Since |F|n2, there exists a fault-free 6-cycle containing the edge uLvL.

Case 2. vL=uLj(3jn).

Case 2.1. uR,vRF.

For 3jn2, there exist two disjoint uRvR-paths as Pi3(uR) of length =3 in RFR as follows. Pi3(uR)={(i+j2,j,j1),i=1,(i+j2,j+1,j+2),i=2xj1=xj+1,(i+j2,j+2,j+1),i=2xj1xj+1.

For jn1, there exist two disjoint uRvR-paths as Pi3(uR) of length =3 in RFR as follows. Pi3(uR)={(2,j,2),i=1,(j1,j,j1),i=2.

Since |FR|1, there exists a fault-free uRvR-path P3(uR) of length =3 in RFR. Then C=uLuR+P3(uR)+vRvL+uLvL is a fault-free 6-cycle containing the edge uLvL.

Case 2.2. |{uR,vR}F|=1. Without loss of generality, assume that uRF. Let F=F{uR}, then |F|=n3.

For j=3, there exist n2 6-cycles as Ci6(uL)(1in2) containing the common edge uLvL in MQn1F as follows. Ci6(uL)={(i+1,1,3,1,2),i=1,i<j1.(i+2,3,1,4,1),i=2,n3x2=0i<j1.(i+2,n,3,n,i+2),3in3x2=0,n3x20i<j1.(i+2,3,1,i+2,1),3in3x20,(i+2,3,1,n,1),i=n2.

For 4jn2, there exist n2 6-cycles as Ci6(uL)(1in2) containing the common edge uLvL in MQn1F as follows: Ci6(uL)={(i+1,j,j+1,2,j+1),i=1,(i+1,n,j,n,i+1),2ij3xi=0,(i+1,j1,j,j1,i+1),2ij3xi0,(i+1,j+1,j+2,j,j1),i=j2xj2=xjxj2=xj+1,(i+1,j+2,j+1,j,j1),i=j2xj2=xjxj2xj+1,(i+1,j,j+2,j+1,j1),i=j2xj2xj((xj1=xj+1xj2=0)(xj1xj+1xj2=1)),(i+1,j,j+1,j+2,j1),i=j2xj2xj((xj1xj+1xj20)(xj1=xj+1xj21)),(i+2,j,1,j+1,1),i=j1,(i+2,2,j,2,i+2),jin2xj1=0,(i+2,j,1,i+2,1),jin2xj10.

For j=n1, there exist n2 6-cycles as Ci6(uL)(1in2) containing the common edge uLvL in MQn1F as follows. Ci6(uL)={(i+1,n,j,n,i+1),in4xi=0,(i+1,j1,j,j1,i+1),in4xi0,(i+1,n,1,n2,1),i=n3xn2=0,(i+1,1,n1,1,n2),i=n3xn20,(i+2,n1,1,n,1),i=n2.

For j=n, there exist n2 6-cycles as Ci6(uL)=(i+1,i,n,i,i+1)(1in2) containing the common edge uLvL in MQn1F.

Note that n2 6-cycles are disjoint in L and contain no uR in MQn1F. Since |F|=n3, we have that there exists a fault-free 6-cycles in MQn1F, and so in MQn1F.

Take edge uLvL=(001011,001100) in MQ61 for an example. We have uR=110100,vR=110011. Assume 110100F, then 110011F. There exist four disjoint 6-cycles in L except a common edge uLvL=(001011,001100) as follows(see Figure 2). {(001011,011011,011100,011111,001111,001100),(001011,000011,000010,000000,000100,001100),(001011,001001,001110,110001,110011,001100),(001011,001010,001101,110010,110011,001100).

 ◻

Lemma 2.13. If FR with |F|=n2(n6), then for any edge uRvRR, there exists a fault-free 6-cycle containing the edge uRvR in MQn1F.

Proof. Let uL,vL be the adjacent vertices of uR,vR respectively in L and uR=x1x2xixn. Since FR, FL=0. There exists a fault-free uLvL-path P of length =3 in L as follows. P={(2,4,3),vR=uR2x3=0,(2,3,4),vR=uR2x30,(j1,j,j1),vR=uRj(3jn1),(2,n,2),vR=uRn.

Then C=vRvL+P+uLuR+uRvR is a fault-free 6-cycle containing the edge uRvR in MQn1F◻

 

Lemma 2.14. For any edge uLuREC in MQniF(i=0,1)(n6) with |F|n2, there exists a fault-free cycle of length =7i,7,8 containing the edge uLuR in MQniF(i=0,1).

Proof.  Let uL=x1x2x3xixn. We prove the lemma according to edge uLuR in MQn0 or MQn1 as follows.

Case 1. uLuRE(MQn0).

Case 1.1. =7.

For 1in1, we can find n1 disjoint 7-cycles as Ci7(uL) except the common edge uLuR as follows. C17(uL)={(2,3,n1,1,n1,2),x3=0x2=0,(2,3,2,1,4,3),x3=0x20,(2,1,2,5,4,3),x30x2=0x4=0,(2,1,5,2,4,3),x30x2=0x40,(2,n,1,n,2,3),x30x20, C27(uL)={(3,2,3,1,2,3),x3=0x2=0,(3,2,3,1,3,2),x3=0x20,(3,4,2,1,5,2),x30x2=0x4=0,(3,5,2,1,4,2),x30x2=0x40,(3,4,2,1,4,2),x30x20x4=0,(3,5,2,1,5,2),x30x20x40,  Ci7(uL)={(i+1,2,3,1,2,i+1),3in1x2=0,(i+1,3,2,1,2,i+1),3in1x20.

Since |F|n2, there exists a fault-free 7-cycle containing the edge uLuR.

Case 1.2. =8.

For 1in1, we can find n1 disjoint 8-cycles as Ci8(uL) except the common edge uLuR as follows. C18(uL)={(2,1,n1,3,n1,2),x2=0,(2,3,2,1,2,3),x20,

Ci8(uL)=(i+1,2,n,2,1,n),2in2,

Cn18(uL)={(n,2,1,n1,3,n1),x2=0,(n,2,1,3,4,2),x20x3=0,(n,2,1,4,3,2),x20x30.

Since |F|n2, there exists a fault-free 8-cycle containing the edge uLuR.

Take edge (010010,110010) in MQ60 for an example, there exist 5 disjoint 8-cycles except the common edge (010010,110010) as follows(see Figure 3). {(010010,000010,001010,011010,111010,100101,101101,110010),(010010,011101,001101,001100,011100,111100,111101,110010),(010010,010110,000110,000111,010111,110111,110110,110010),(010010,010000,000000,000001,010001,110001,110000,110010),(010010,010011,000011,100011,101011,101100,110011,110010).

Case 2. uLuRE(MQn1).

Case 2.1. =6.

For 1in1, we can find n1 disjoint 6-cycles as Ci6(uL) except the common edge uLuR as follows. Ci6(uL)={(i+1,1,i+3,i+2,i+1),1in4xi=xi+2xi+1=0,(i+1,i+2,i+3,1,i+1),1in4xi=xi+2xi+10,(i+1,i+3,1,i+2,i+1),1in4xixi+2xi+1=0,(i+1,i+2,1,i+3,i+1),1in4xixi+2xi+10, Cn36(uL)={(n2,n1,n2,1,n1),xn2=0,(n2,n1,1,n2,n),xn20, Cn26(uL)={(n1,n2,1,n2,n),xn2=0,(n1,1,n2,n1,n2),xn20, Cn16(uL)={(n,n2,1,n1,n2),xn2=0,(n,n2,1,n2,n1),xn20.

Since |F|n2, there exists a fault-free 6-cycle containing the edge uLuR.

Case 2.2. =7.

For 1in1, we can find n1 disjoint 7-cycles as Ci7(uL) except the common edge uLuR as follows.

Ci7(uL)={(i+1,i+3,i+2,1,i+3,i+1),1in4xi+1=0,(i+1,i+3,1,i+2,i+3,i+1),1in4xi+10, Cn37(uL)={(n2,1,n,n2,n,n1),xn2=0,(n2,1,n,n1,n2,n),xn20, Cn27(uL)={(n1,n2,n3,1,n3,n),xn2=0xn3=0,(n1,n,n3,1,n3,n2),xn2=0xn30xn4=0,(n1,n3,n2,1,n3,n),xn2=0xn30xn40,(n1,2,n2,2,1,n2),xn20, Cn17(uL)={(n,n3,1,n3,n,n2),xn2=0xn3=0,(n,n2,n3,1,n3,n),xn2=0xn30xn4=0,(n,n2,n1,1,n,n2),xn2=0xn30xn40,(n,2,n1,2,1,n1),xn20.

Since |F|n2, there exists a fault-free 7-cycle containing the edge uLuR.

Case 2.3. =8.

For 1in1, we can find n1 disjoint 8-cycles as Ci8(uL) except the common edge uLuR as follows. Ci8(uL)={(i+1,n,1,n,i+3,i+2),1in4xi=xi+2xi+1=0,(i+1,i+2,i+3,n,1,n),1in4xi=xi+2xi+10,(i+1,i+3,n,1,n,i+2),1in4xixi+2xi+1=0,(i+1,i+2,n,1,n,i+3),1in4xixi+2xi+10, Cn38(uL)={(n2,1,n1,1,n,1),xn2=0,(n2,1,n,1,n1,1),xn20, Cn28(uL)={(n1,2,n,3,1,2),x2=0,(n1,2,n,1,3,2),x20, Cn18(uL)={(n,n2,1,2,n2,2),x2=0xn2=0,(n,n1,n2,n3,1,n3),x2=0xn20xn3=0,(n,n1,n3,1,n3,n2),x2=0xn20xn30,(n,2,3,2,1,3),x20x3=0,(n,2,4,3,1,2),x20x30.

Since |F|n2, there exists a fault-free 8-cycle containing the edge uLuR.

Take edge (010010,101101) in MQ61 for an example, there exist 5 disjoint 8-cycles except the common edge (010010,101101) as follows (see Figure 4): {(010010,000010,001010,001101,001100,110011,110010,101101),(010010,011101,011110,011111,100000,100001,100101,101101),(010010,010110,101001,101011,010100,010101,101010,101101),(010010,010000,000000,000001,111110,110001,101110,101101),(010010,010011,000011,001011,011011,100100,101100,101101).

 ◻

Figure 4 5 disjoint 8-cycles except the common edge (010010, 101101) in MQ61

3. Fault-tolerant edge-pancyclicity of MQn

In this section, we investigate the fault-tolerant edge-Pancyclicity of MQn and show that MQn is (n2)-fault-tolerant edge-Pancyclic.

Let F be a set of faulty elements in MQn, FL=FL, FR=FR, FC=FEC, Fv=FV(MQn), Fe=FE(MQn) FLv=FLV(L) and FRv=FRV(R), fv=|Fv|, fe=|Fe|.

Theorem 3.1. If fv+fen2 and n5 then, for any fault-free edge e in MQni(i=0,1) and any integer with 7i2nfv, there is a fault-free cycle of length containing the edge e.

In this section, we will give the proof of Theorem 3.1. The theorem follows from Lemma 2.10 if |F|n3. We only need to consider |F|=n2. Start with the following lemma.

Lemma 3.2. If Theorem 3.1 holds for any subset FV(MQn) with |F|=n2, then Theorem 3.1 also holds for any subset FV(MQn)E(MQn) with |F|=n2.

Proof. The lemma holds for fe=0 by hypothesis of Lemma 3.2. Assume that the lemma holds for any t with 0<tn3 and fe=t. We prove the lemma holds for fe=t+1. Let xy be an edge in F.

Case 1. xF or yF.

Without loss of generality, assume xF. Let F=Fxy, then |F|=n3. By Lemma 2.10, this result has been proved.

Case 2. x,yF.

Let F=F+{x}{xy}, then |F|=n2, and F contains at most t edges and fv+1 vertices. By induction hypothesis, for every integer with 7i2nfv1, there is a fault-free -cycle containing the edge e in MQniF(i=0,1). By Proposition 2.1, for any e=uvE(MQnF), (u,v) is not a weak vertex-pair in MQnF. For =2nfv, by Lemma 2.11, there is a fault-free uv-hamiltonian path in MQnF, i.e., there is a fault-free -cycle containing the edge e in MQnF.

The proof of lemma is completed. ◻

Proof of Theorem 3.1. By Lemma 3.2, we only need to prove the theorem with FV(MQn). The proof proceeds by induction on n5. The result holds for n=5 by developing computer program using depth first searching technique combining with backtracking and branch and bound algorithm.

Assume that the theorem holds for any k with 6k<n. Then we must show the theorem holds for n.

Let e=uv be a fault-free edge in MQn. By Proposition 2.3, (u,v) is not a weak vertex-pair in MQnF. Let 2n12nfv and =+1, where 2n112nfv1 then, by Lemma 2.11, there exists a fault-free uv-path P of length in MQn. Then P+uv is a fault-free cycle of length containing the edge e in MQn. Thus, we only need to consider with 7i2n11 in MQni(i=0,1)(n6).

Case 1. |FR||FL|.

Case 1.1. eE(LFL). Let e=uLvL.

By Lemma 2.12, there exists a fault-free 6-cycle containing the edge e in MQn1F. Thus, we only need to consider the length of 72n11.

Case 1.1.1. |FL|n3. Then |FR|n4.

Case 1.1.1.1. 72n1|FLv|.

By induction hypothesis, there is a fault-free cycle of length containing the edge e in L, so in MQn.

Case 1.1.1.2. 2n1|FLv|+12n11.

Write =1+1+2, where 2n2|FLv|+112n21 and 2=2n21. Since 2n2|FLv|+12n2|FL|+12n2n+4>7 for n6 and |FL|n3, by induction hypothesis, there is a fault-free cycle C of length 1 containing the edge uLvL in L. Note that a cycle of length 1 contains a matching M with |M|=12. Consider the following inequality. 12|FC||FR||{e}|2n2|FLv|+12|FC||FR|12n3|F|1 2n3n+1.

Let f(x)=2x3x+1. Since f(x)=2x3ln210 for x6, f(x) is an increasing function, which implies that 12|FC||FR|f(6)=2636+1=3. It follows that, there is such an edge, say xLyL, in M that xLyLuLvL and |{xR,yR,xLxR,yLyR}F|=0. By Lemma 2.1, there is a fault-free xRyR-path P of length 2 in R. Then CxLyL+yLyR+P+xRxL is a fault-free cycle of length (=1+1+2) containing e (see Figure 5 (a)).

Case 1.1.2. |FL|=n2. In this case, |FR|=0.

Let uR and vR be neighbors of uL and vL in R, respectively.

Let =+3, where 42n14. By Lemma 2.4, duRvR2. Since |FR|=|FC|=0, by Lemma 2.6, there is a fault-free uRvR-path P of length in R, and so P+vRvL+vLuL+uLuR is a fault-free cycle of length containing the edge uLvL (see Figure 5 (b)).

Figure 5 The illustrations of Case 1.1

Case 1.2. eE(RFR). Let e=uRvR.

Case 1.2.1. |FL|n3. Then |FR|n4.

Case 1.2.1.1. 7i2n1|FRv| in MQni(i=0,1).

By induction hypothesis, there is a fault-free cycle of length containing the edge e in R, so in MQni(i=0,1).

Case 1.2.1.2. 2n1|FRv|+12n11.

The proof is similar to Case 1.1.1.2.

Case 1.2.2. |FL|=n2. In this case, |FR|=0.

Since |FR|=0, by Lemma 2.5, there exists a fault-free cycle of length with 7i2n11 containing the edge e in R, and so in MQniF(i=0,1).

Case 1.3. e(ECFC). Let e=uLuR.

By Lemma 2.14, there exists a fault-free cycle of length =7i,7,8 containing the edge uLuR in MQniF(i=0,1). Thus, we only need to consider the length 92n11.

Case 1.3.1. |FL|n3. Then |FR|n4.

By Corollary 2.1., there exists a fault-free 4-cycle C (or 5-cycle) containing the edge uLuR.

Case 1.3.1.1. C=C4=(uL,sL,sR,uR).

For 92n1|FRv|1. Let =+2, where 72n1|FRv|3. By induction hypothesis, there is a fault-free cycle C of length containing the edge uRsR in R. Then C=CuRsR+sRsL+sLuL+uLuR is a fault-free cycle of length containing the edge e=uLuR.

For 2n1|FRv|2n11. Let =1+2, where 1=2n2 and 2n2|FRv|22n21. Since 2n2>7 and 2n2|FRv|>7 for n6, by induction hypothesis, there exists a fault-free cycle C1 of length 1 containing the edge uLsL in L and a fault-free cycle C2 of length 2 containing the edge uRsR in R. Then C=C1uLsL+sLsR+C2sRuR+uRuL is a fault-free cycle of length containing the edge e=uLuR in MQn (see Figure 6(a)).

Case 1.3.1.2. C=C5=(uL,wL,wR,x,uR).

Since |FR|n4. Let FR1=FR{wR}, then |FR1|=|FR|1n3.

For 92n1|FRv|1. Let =+3, where 62n1|FRv|4. By induction hypothesis, there is a fault-free cycle C of length containing the edge uRx in RFR1. Then C=CuRx+xwR+wRwL+wLuL+uLuR is a fault-free cycle of length containing the edge e=uLuR.

For 2n1|FRv|2n11. Let =1+2+2, where 1=2n21 and 2n2|FRv|122n22. Since 2n2|FRv|1>7 for n6, by induction hypothesis, there exists a fault-free cycle C1 of length 2 containing the edge xuR in RFR1 and, by Lemma 2.11, there exists a fault-free uLwL-path P of length 1 in L. Then C=C1uRx+xwR+wRwL+P+uLuR is a fault-free cycle of length containing the edge e=uLuR in MQn (see 6(b)).

Figure 6 The illustration of Case 1.3.1.1. and Case 1.3.1.2.

Case 1.3.1.3. C=C5=(uL,x,wL,wR,uR).

For 92n1|FRv|1. Let =+3, where 62n1|FRv|4. By induction hypothesis, there is a fault-free cycle C of length containing the edge uRwR in R. Then C=CuRwR+wRwL+wLx+xuL+uLuR is a fault-free cycle of length containing the edge e=uLuR.

For 2n1|FRv|2n11. Let =1+2+1, where 1=2n2 and 2n2|FRv|122n22. Since 2n2|FRv|1>7 for n6, by induction hypothesis, there exists a fault-free cycle C1 of length 2 containing the edge uRwR in R and, by Lemma 2.11, there exists a fault-free uLwL-path P of length 1 in L. Then C=C1uRwR+wRwL+P+uLuR is a fault-free cycle of length containing the edge uLuR in MQn (see Figure 7(a)).

Case 1.3.2. |FL|=n2. In this case, |FR|=0.

Since |FL|=n2, there exists a fault-free neighbor vL of uL in L. Let vR be neighbors of vL in R.

Let =+3, where 62n14. By Lemma 2.4, duRvR2. By Lemma 2.6, there is a fault-free uRvR-path P of length in R. So P+uRuL+uLvL+vLvR is a fault-free cycle of length containing the edge uLuR.

Case 2. |FL||FR|.

Case 2.1. eE(RFR). Let e=uRvR.

Case 2.1.1. |FR|n3. Then |FL|n4.

For 7i2n1|FRv| in MQni(i=0,1). By induction hypothesis, there is a fault-free cycle of length containing the edge e in R, so in MQni(i=0,1).

For 2n1|FRv|+12n11. The proof is similar to Case 1.1.1.2.

Case 2.1.2. |FR|=n2. In this case, |FL|=0.

By Lemma 2.13, there exists a fault-free 6-cycle containing the edge e in MQn1F. Thus, we only need to consider the length of 72n11.

Let uL and vL be neighbors of uR and vR in L, respectively.

Let =+3, where 42n14. By Lemma 2.4, duLvL2. Since |FL|=|FC|=0, by Lemma 2.6, there is a fault-free uLvL-path P of length in L, and so P+vLvR+vRuR+uRuL is a fault-free cycle of length containing the edge uRvR.

Case 2.2. eE(LFL). Let e=uLvL.

Case 2.2.1. |FR|n3. Then |FL|n4.

For 7i2n1|FLv| in MQniF(i=0,1). Since |FL|n4, by Lemma 2.10, there exists a fault-free cycle of length containing the edge e in MQniF(i=0,1).

For 2n1|FLv|+12n11. The proof is similar to Case 1.1.1.2.

Case 2.2.2. |FR|=n2. In this case, |FL|=0.

Since |FL|=0, by Lemma 2.5, there exists a fault-free cycle of length with 7i2n11 containing the edge e in L, and so in MQni(i=0,1).

Case 2.3. e(ECFC). Let e=uLuR.

By Lemma 2.14, there exists a fault-free cycle of length =7i,7,8 containing the edge uLuR in MQniF(i=0,1). Thus, we only need to consider the length 92n11.

Case 2.3.1. |FR|n3. Then |FL|n4.

By Corollary 2.1., there exists a fault-free 4-cycle C (or 5-cycle) containing the edge uLuR.

Case 2.3.1.1. C=C4=(uL,sL,sR,uR).

The proof is similar to Case 1.3.1.1.

Case 2.3.1.2. C=C5=(uL,wL,wR,x,uR).

For 92n1|FLv|1. Let =+3, where 62n1|FLv|4. Since |FL|n4, by Lemma 2.10, there is a fault-free cycle C of length containing the edge uLwL in LFL. Then C=CuLwL+wLwR+wRx+xuR+uRuL is a fault-free cycle of length containing the edge e=uLuR.

For 2n1|FLv|2n11. Let =1+2+1, where 2n2|FLv|112n22 and 2=2n2. Since 2n2|FLv|1>7 for n6, by induction hypothesis, there exists a fault-free cycle C1 of length 1 containing the edge uLwL in LFL and, by Lemma 2.11, there exists a fault-free uRwR-path P of length 2 in R. Then C=C1uLwL+wLwR+P+uRuL is a fault-free cycle of length containing the edge e=uLuR in MQn (see Figure 7(b)).

Figure 7 The illustration of Case 1.3.1.3. and Case 2.3.1.2.

Case 2.3.1.3. C=C5=(uL,x,wL,wR,uR).

The proof is similar to Case 1.3.1.3.

Case 2.3.2. |FR|=n2. In this case, |FL|=0.

Since |FR|=n2, there exists a fault-free neighbor vR of uR in R. Let vL be the neighbor of vR in L. Let =+3, where 62n14. By Lemma 2.4, duLvL2. By Lemma 2.6, there is a fault-free uLvL-path P of length in L. So P+uLuR+uRvR+vRvL is a fault-free cycle of length containing the edge uLuR.

The proof of the theorem is completed. ◻

4. Conclusion and remarks

As one of the most fundamental networks for parallel and distributed computation, cycles are suitable for developing simple algorithms with low communication cost. Edge and/or vertex failures are inevitable when a large parallel computer system is put in use. Therefore, the fault-tolerant capacity of a network is a critical issue in parallel computing. The fault-tolerant edge-pancyclicity of an interconnection network is a measure of its capability of implementing ring-structured parallel algorithms in a communication-efficient fashion in the presence of faults.

In view of the fact that the hypercube network Qn contains only even cycles, MQn is superior to Qn in fault-tolerant pancyclicity. This shows that, when the MQn is used to model the topological structure of a large-scale parallel processing system, our result implies that the system has larger capability of implementing ring-structured parallel algorithms in a communication-efficient fashion in the hybrid presence of edge and vertex failures than one of the hypercube network.

We make some remarks on the optimality of our result in the following sense.

  1. For =4, in MQ50, taking e=(11011,11100), there are only two cycles: (11011,11010,11101,11100), (11011,01011,01100,11100) of length 4 containing the edge e (we calculate it by computer).

    If F={01011,11101}, then there exists no fault-free cycle of length 4 containing the edge e in MQ50F. In MQ51, taking e=(11101,11110), there exists only one cycle (11101,11100,11111,11110) of length 4 containing the edge e. If F={11100} or F={11111}, then there exists no fault-free cycle of length 4 containing the edge e in MQ51F.

  2. For =5, in MQ5, taking e=(11110,11111), the cycles of length 5 containing the edge e are as Table 1 and Table 2(we calculate it by computer):

    Table 1 Cycles of length 5 containing the edge e=(11110,11111) in MQ50
    (11110 11101 11010 11000 11111) (11110 11101 10010 10000 11111)
    (11110 11001 11011 11100 11111) (11110 10001 10011 11100 11111)
    Table 2 Cycles of length 5 containing the edge e=(11110,11111) in MQ50
    (11110 11101 11010 11000 11111) (11110 11101 10010 10000 11111)
    (11110 11101 00010 00000 11111) (11110 11001 11011 11100 11111)
    (11110 10001 10011 11100 11111) (11110 00001 00011 11100 11111)

    If F={11100,11101}, then there exists no fault-free cycle of length 5 containing the edge e in MQ5F.

  3. For =6, in MQn0, taking e=(01011,01100), the cycles of length 6 containing the edge e are as Table 3(we calculate it by computer):

    Table 3 Cycles of length 6 containing the vertex e=(01011,01100) in MQ50
    (01011 01010 01000 00000 00100 01100) (01011 01010 01101 01110 01111 01100)
    (01011 01010 01101 00101 00100 01100) (01011 01010 01101 11101 11100 01100)
    (01011 01010 00010 00000 00100 01100) (01011 01010 11010 11011 11100 01100)
    (01011 01010 11010 11101 11100 01100) (01011 01010 11010 11101 01101 01100)
    (01011 01001 01000 01010 01101 01100) (01011 01001 01000 00000 00100 01100)
    (01011 01001 00001 00000 00100 01100) (01011 01001 00001 00101 00100 01100)
    (01011 01001 00001 00101 01101 01100) (01011 01001 11001 11011 11100 01100)
    (01011 00011 00010 00000 00100 01100) (01011 00011 00010 01010 01101 01100)
    (01011 00011 00001 00000 00100 01100) (01011 00011 00001 00101 00100 01100)
    (01011 00011 00001 00101 01101 01100) (01011 11011 11010 11101 11100 01100)
    (01011 11011 11010 11101 01101 01100) (01011 11011 11010 01010 01101 01100)
    (01011 11011 11100 11101 01101 01100) (01011 11011 11100 11111 01111 01100)

If F={00100,01101,11100}, then there exists no fault-free cycle of length 6 containing the edge e in MQ50F.

Acknowledgements

The research is supported by National Natural Science Foundation of China under Grant No.U22A2005, No. U21A20491, No. U1936109, No.U1908214, No. KLSA201906 and Key Research and Development Program of Zhejiang (2024SSYS0001).

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