5-Regular Subgraphs in Hypercubes

Y. M. Borse1, S. R. Shaikh1, J. B. Saraf1
1Department of Mathematics, Savitribai Phule Pune University, Ganeshkhind, Pune 411007, India

Abstract

One of the fundamental properties of the hypercube Qn is that it is bipancyclic as Qn has a cycle of length l for every even integer l with 4l2n. We consider the following problem of generalizing this property: For a given integer k with 3kn, determine all integers l for which there exists an l-vertex, k-regular subgraph of Qn that is both k-connected and bipancyclic. The solution to this problem is known for k=3 and k=4. In this paper, we solve the problem for k=5. We prove that Qn contains a 5-regular subgraph on l vertices that is both 5-connected and bipancyclic if and only if l{32,48} or l is an even integer satisfying 52l2n. For general k, we establish that every k-regular subgraph of Qn has 2k,2k+2k1 or at least 2k+2k1+2k3 vertices.

Keywords: Hypercube, k-connected, Regular, Bipancyclic, Subgraphs

1. Introduction

A graph G=(V,E) on n vertices is an n-vertex graph. A 2n-vertex graph is bipancyclic if it has a cycle of length l for all even integers l satisfying 4l2n. The Cartesian product of two graphs G and H is the graph G◻H with vertex set V(G)×V(H) in which two vertices (x,y) and (u,v) are adjacent if and only if either x=u and y is adjacent to v in H, or y=v and x is adjacent to u in G. Throughout the paper n denote a positive integer.

The n-dimensional hypercube Qn is the Cartesian product of n copies of the complete graph K2. It is an n-regular, n-connected, bipartite, and bipancyclic graph on 2n vertices with diameter n. Because of such rich properties, hypercubes are one of the most widely used interconnection network topologies [1]. The connectivity of a network is an important parameter to evaluate the reliability and fault tolerance of a network [2]. Bipancyclicity is a fundamental property of the hypercube networks as it allows the embedding of cycles of various lengths effectively into hypercubes. Cycle networks are used to design simple algorithms with low communication cost and it has applications in image processing and signal processing [1,3].

We consider the following problem of generalizing the property of bipancyclicity of hypercubes to the existence of l-vertex, k-regular subgraphs for various values of l that are also k-connected and bipancyclic. This will give subgraphs of Qn with less number of vertices which retain the important properties of Qn such as regularity, high connectivity, and bipancyclicity.

Problem 1. For a given integer k with 3kn, determine all integers l for which there exists an l-vertex, k-regular subgraph of Qn that is both k-connected and bipancyclic.

This problem is also related to the problem of embedding regular graphs into hypercubes. Cybenko et al. [4] proved that the problem of deciding whether or not a given graph is embeddable into a hypercube is NP-complete, in fact, the problem is NP-complete even for trees [5].

Since the hypercube Qn is a bipartite graph, every regular subgraph of it has even number of vertices. For k=3, Ramras [6] established that every even integer from 8 to 2n except 10 can be the number of vertices of a 3-regular subgraph of Qn. Borse and Shaikh [7] improved this result by showing that such a 3-regular subgraph can be bipancyclic also. They solved the above problem for k=3 and k=4 in [7] and [8], receptively. They established that, for k{3,4}, Qn has a k-regular, k-connected, bipancyclic subgraph on l vertices if and only if l=2k or l is an even integer with 2k+2k1l2n. The problem remains open for k5.

Besides hypercubes, the special case k=3 of the above problem is settled for the class of the Cartesian product of cycles in [9] and for the class of the Cartesian product of paths in [10]. Also, Borse et al. [11] proved the existence of a factorization of the Cartesian product of r cycles, each of length a power of 2, into isomorphic k-regular, k-connected and bipancyclic subgraphs with the number of vertices a power of 2, for 2k<2r. Moreover, the number of vertices of a smallest k-regular subgraph of an r-regular graph G is related to the conditional k-edge-connectivity of G [12].

In this paper, we settle Problem 1 for the case k=5. The following is the main result of the paper.

Theorem 1 (Main Theorem). For n5, there exists a 5-regular, 5-connected and bipancyclic subgraph of the hypercube Qn on l vertices if and only if l{32,48} or l is an even integer such that 52l2n.

For general k, Borse and Shaikh [8] obtained the following result about the non-existence of k-regular subgraphs of Qn on a certain number of vertices.

Theorem 2 ([8]). For a given integer k with 1kn, every subgraph of Qn with minimum degree at least k either is isomorphic to Qk or has at least 2k+2k1 vertices.

In this paper, we improve this theorem as follows.

Theorem 3. For a given integer k with 2kn, if H is a subgraph of Qn with minimum degree at least k, then one of the following holds:

  1. H is isomorphic to Qk.

  2. H is a spanning subgraph of the subgraph of Qn induced by V(C◻Qk2) for some cycle C of length six.

  3. H has at least 2k+2k1+2k3 vertices.

Thus, if 2kn and 1l<2k+2k1+2k3 with l{2k,2k+2k1}, then Qn does not have a k-regular subgraph on l vertices and hence no k-regular, l-vertex graph is embeddable into Qn.

We provide preliminary results in Section 2 and prove Theorem 3 in Section 3. The proof of Main Theorem 1 is divided into the next three sections. A construction of 5-regular subgraphs of Qn is given in Section 4. The connectivity and bipancyclicity properties of these subgraphs are dealt in Section 5 and Section 6, respectively.

2. Preliminaries

We can write Qn as Qnk◻Qk for 0kn, where Q0 is the complete graph K1. A k-cycle means a cycle of length k. We need the following lemmas.

Lemma 1 ([3]). Let Gi be an mi-regular and mi-connected graph for i=1,2. Then the graph G1◻G2 is (m1+m2)-regular and (m1+m2)-connected.

Lemma 2 ([13]). If P and Q are non-trivial paths and one of them has an even number of vertices, then the graph P◻Q is bipancyclic.

Hence C◻K2 is bipancyclic if C is a non-trivial path or a cycle of length at least three.

Lemma 3 ([8]). For n3, the hypercube Qn has a Hamiltonian cycle C with a chord e such that there is a 4-cycle in Qn containing e and three edges of C.

Lemma 4 ([8]). Let l be an even integer such that 6l2n2. Then there exists an l-cycle C in Qn containing six vertices x, y, z, u, v, w and there are two vertices g, h in V(Qn)V(C) such that

  1. g is adjacent to x, y, z;

  2. h is adjacent to u, v, w;

  3. xu, uy and yv are edges of C.

We obtain a similar result as follows.

Lemma 5. Let l be an even integer such that 8l2n2. Then there exists an l-cycle C in Qn and a vertex u in V(Qn)V(C) having four pairwise non-adjacent neighbours in C.

Proof. Suppose n=4. Clearly, C= <v1,v2,,v8,v1> is a required 8-cycle in Q4 as shown by bold lines in Figure 1. Replacing the edge v1v2 of C by a path of length 3 avoiding u gives a 10-cycle C and replacing the edge v3v4 of C by a path of length 3 or 5 that is edge-disjoint with C produces required 12-cycle and 14-cycle. Thus the result holds for n=4.

Figure 1. 8-Cycle C in Q4

Suppose n5. Write Qn as Qn2◻Q2. Then Qn is obtained from four copies Qn21, Qn22, Qn23, Qn24 of Qn2 such that Qn2i is joined to Qn2i+1 for i=1, 2, 3 by a matching between their corresponding vertices. Vertices of Qn21 are joined to the corresponding vertices of Qn24.

Figure 2. The l-cycle Z

Since n23, by Lemma 3, each Qn2i contains a Hamiltonian cycle Ci with a chord ei such that there is a 4-cycle in Qn2i containing ei and three edges from Ci. For simplicity let r=2n2. Label the set of vertices of Qn2i by {vpi | p=1,2,,r} so that Ci= <v1i,v2i,,vri,v1i> and e=v2ivr1i. We now construct a cycle Z of length l in Qn, as required, from the cycles C1,C2,C3,C4 and the chord e1 of C1. If 8l2n1+4, then l=2t+6, where 1t=l/232n21=r1. In this case, take Z as the cycle shown in Figure 2(a). If 2n1+6l2n4, then l=2m+2n1 with 3m=l/22n22n12n22=r2. In this case, choose Z to be the cycle shown in Figure 2(b). Finally, for l=2n2 take Z as the cycle given in Figure 2(c). In each case, Z is a cycle of length l in Qn, and further, the vertex v11 is not on Z but it has four pairwise non-adjacent neighbours v21,vr1,v12,v14 in Z. This completes the proof. ◻

3. Proof of Theorem 3

In this section, we prove Theorem 3. For a graph G, let G[S] denote the induced subgraph of G on a vertex subset SV(G). The minimum degree of G is denoted by δ(G). If G is isomorphic to a graph H, then we write GH. Since Qn=Qn1◻K2, we can split Qn into two copies Qn10 and Qn11 of Qn1. If H is a subgraph of Qn10, then there is a subgraph H of Qn11 isomorphic to H such that the vertex set of H is the set of neighbours of H in Qn11. We say that H is the subgraph of Qn11 corresponding to H.

We need the following result.

Lemma 6 ([8]). For a given integer k with 1k<n, if H is a subraph of Qn isomorphic to Qk, then every vertex in V(Qn)V(H) has at most one neighbour in H.

Proof of Theorem 3. By Theorem 2, (i) holds or |V(H)|2k+2k1. Suppose (i) does not hold. Then |V(H)|2k+2k1. We prove, by induction on k, that (ii) holds if |V(H)|=2k+2k1, otherwise (iii) holds. Suppose k=2. Then |V(H)|=6 or |V(H)|7. If |V(H)|=6, then it follows that H is a chordless 6-cycle or a 6-cycle with a chord and so (ii) holds. If |V(H)|7>22+221+223, then (iii) follows. Thus the result is true for k=2.

Suppose k3. Assume that the result holds for the subgraphs of Qn of minimum degree at least k1. Let e be an edge of H. Without loss of generality, we may assume that the end vertices of e differ in the first coordinate only. Write Qn=Qn10Qn11D, where D is the set of all edges in Qn whose end vertices differ in the first coordinate only. Then eD. Hence H intersects both Qn10 and Qn11. Let Hi=HQn1i for i=0,1. Then δ(Hi)k1. We may assume that |V(H0)||V(H1)|.

By induction hypothesis, H0Qk1 or H0 is a spanning subgraph of Qn[V(C◻Qk3)] for some 6-cycle C or |V(H0)|2k1+2k2+2k4. Hence |V(H0)|=2k1 or |V(H0)|=3(2k2) or |V(H0)|2k1+2k2+2k4. We consider these three cases separately.

Case (i). Suppose |V(H0)|2k1+2k2+2k4.

Consider |V(H)|=|V(H1)|+|V(H0)|2|V(H0)|=2k+2k1+2k3 as |V(H1)||V(H0)|. Therefore (iii) holds.

Case (ii). Suppose |V(H0)|=2k1.

In this case, H0 is isomorphic to Qk1. As δ(H)k, each vertex of H0 has a neighbour in H1. Let W1 be the subgraph of Qn11 corresponding to H0. Then W1Qk1 and V(W1)V(H1). Let W2 be the subgraph of H1 induced by V(H1)V(W1). Observe that no vertex of W2 has a neighbour in H0 and by Lemma 6, it has at most one neighbour in W1. Therefore δ(W2)k1. By Theorem 2, W2Qk1 or |V(W2)|2k1+2k2. In the later case, (iii) holds as |V(H))|=|V(H0)|+|V(W1)|+|V(W2)|2k1+2k1+2k1+2k2>2k+2k1+2k3. Suppose W2Qk1. Then the subgraph of Qn induced by the vertices of H0,W1,W2 is isomorphic to P3◻Qk1=(P3◻K2)◻Qk2, where P3 is a path on three vertices. Since P3◻K2 is a 6-cycle with a chord, (ii) holds.

Case (iii). H0 is a spanning subgraph of Qn[V(C◻Qk3)] for some 6-cycle C.

We consider two subcases depending on whether the cycle C has a chord or not.

Subcase (i). Suppose C is chordless.

Then H0 is (k1)-regular and in fact, H0C◻Qk3. Let W1 be the subgraph of Qn11 corresponding to H0. Then V(W1)V(H1). If V(W1)=V(H1), then W1=H1. As HH0◻K2(C◻Qk3)◻K2C◻Qk2, (ii) follows. Suppose V(H1)V(W1). Let W2 be the subgraph of H1 induced by V(H1)V(W1). As Qn is triangle-free, it follows from Lemma 6 that every vertex of W2 has at most three neighbours in W1. This shows that δ(W2)k3. By Theorem 2, |V(W2)|2k3. Thus |V(H)|=|V(H0)|+|V(W1)|+|V(W2)|3(2k2)+3(2k2)+2k3=2k+2k1+2k3. Therefore (iii) holds in this case.

Subcase (ii). Suppose C has a chord.

Then C is a spanning subgraph of P3◻K2 and hence H0 is a spanning subgraph of P3◻Qk2. Let 1,2,3 be the vertices of the path P3 in order and let Li the copy of Qk2 in H0 corresponding to the vertex i for i{1,2,3}. Let Ri be the subgraph of Qn11 corresponding to Li for i{1,2,3}. Then RiQk2. Since the degree of every member of V(L1)V(L3) is k1 in H0, we have V(R1)V(H1) and V(R3)V(H1). If V(H1)=V(R1)V(R2)V(R3), then (ii) holds (see Figure 3(a)). Suppose V(H1)V(R1)V(R2)V(R3) is non-empty and let W3 be the subgraph of H1 induced by this set. Then, by Lemma 6, δ(W3)k2. Therefore, by Theorem 2, W3Qk2 or |V(W3)|2k2+2k3. In the later case, (iii) holds.

Suppose W3Qk2. It follows from Lemma 6 that every vertex of W3 has a neighbour in Ri for i=1,3 but no neighbour in R2. Suppose V(R2)V(H1)=. Then V(H1)=V(R1)V(R3)V(W3) (see Figure 3(b)). It follows that H=Z◻Qk2, where Z is a 6-cycle whose six vertices correspond to L1,L2,L3,R3,W3,R1 in order, and so, (ii) holds. Suppose V(R2)V(H1). Then the graph R2H1 has minimum degree at least k3 and hence, it has at least 2k3 vertices by Theorem 2. Thus |V(H)|=|V(H0)|+|V(R1)|+|V(R3)|+|V(W3)|+|V(R2H1)|6(2k2)+2k3. Therefore (iii) holds. This completes the proof. ◻

Figure 3. The Subgraph H of Qn

Corollary 1. Every 5-regular subgraph of Qn has 32,48 or at least 52 vertices.

4. Construction of 5-regular Subgraphs of Qn

In this section, we give a construction of an l-vertex, 5-regular subgraph of the hypercube Qn. Suppose Qn has a 5-regular subgraph on l vertices. Then l is an even integer and by Corollary 1, we have l{32,48} or 52l2n. We prove that for every such l there exists a 5-regular subgraph of Qn with n6 that is both 5-connected and bipancyclic. The case when l is a multiple of 4 follows trivially from the following result.

Theorem 4 ([8]). If n4 and l is an even integer such that l=16 or 24l2n, then there exists a l-vertex, 4-regular subgraph of Qn that is both 4-connected and bipancyclic.

Lemma 7. If n6 and l is a multiple of 4 such that l{32,48} or 52l2n, then there exists a 5-regular, 5-connected and bipancyclic subgraph of Qn on l vertices.

Proof. We can write l=4m, where m=8 or m is an integer such that 12m2n2. Express Qn as Qn1◻K2. Since n14, Qn1 has a 4-regular, 4-connected and bipancyclic subgraph H on 2m vertices by Theorem 4. Therefore, by Lemma 1, the graph H◻K2 is a 5-regular and 5-connected subgraph of Qn on 4m=l vertices. As H is bipancyclic, it has a Hamiltonian cycle and so has a Hamiltonian path. Hence, by Lemma 2, H◻K2 is bipancyclic. ◻

Suppose l is an even integer such that 52l2n but not a multiple of 4. Then n6 and 54l2n2. We have the following four cases.

  1. (i) l=16k+2 with 4k2n41

  2. (ii) l=16k+6 with 3k2n41.

  3. (iii) l=16k+10 with 3k2n41.

  4. (iv) l=16k+14 with 3k2n41.

Case (i). l=16k+2 with 4k2n41.

In this case, n7 and 66l2n14. Write Qn=Qn3◻Q3. By Lemma 5, Qn3 contains a cycle C of length 2k and there is a vertex gV(Qn3)V(C) with four pairwise non-adjacent neighbours x,y,z,w in C. Let V(Q3)={a0,a1,,a7} so that a0,a1,a2,a3,a0 and a4,a5,a6,a7,a4 are 4-cycles and a0a7,a1a6,a2a5 and a3a4 are edges of Q3. Then C◻Q3 is a 5-regular subgraph of Qn with 16k vertices containing a copy Ci of C corresponding to the vertex ai of Q3. Let gi,xi,yi,zi,wi be the vertices of Qn3i corresponding to the vertices g,x,y,z,w, respectively.

Let L1 be the subgraph of Qn with V(L1)={xi,yi,zi,wi,gi:i=0,7} and E(L1)={gixi,giyi,gizi,giwi:i=0,7}{g0g7}. We define a graph H1 from C◻Q3 and L1 as follows. H1=(C◻Q3)L1{x0x7,y0y7,z0z7,w0w7}    (see Figure 4). Clearly, H1 is a 5-regular subgraph of Qn on 16k+2=l vertices.

Now, to construct 5-regular subgraphs in Cases (ii), (iii) and (iv), we choose a cycle C in Qn3 of length 2k by Lemma 4. Then there are six vertices x, y, z, u, v, w on C and two vertices g,h in V(Qn3)V(C) such that g is adjacent to x, y, z, and h is adjacent to u, v, w, and xu, uy, yv are edges of C. Let H=C◻Q3. Then H is a 5-regular subgraph of Qn on 16k vertices. As in Case (i), let Ci be a copy of C corresponding to the vertex ai of Q3 and let gi,hi,xi,yi,zi,ui,vi,wi be the vertices of Qn3i corresponding to g,h,x,y,z,u,v,w, respectively.

Figure 4. The Subgraph H1 on 16k+2 Vertices

Case (ii). Suppose l=16k+6 with 3k2n41.

Clearly, 54l2n10. Let L2 be the subgraph of Qn with vertex set {gi,xi,yi,zi:i=0,1,2,5,6,7} and edge set {gixi,giyi,gizi:i=0,1,2,5,6,7}{g0g1,g1g2,g2g5,g5g6,g6g7,g7g0}. Define a subgraph H2 of Qn from the graphs H and L2 as follows. H2=(HL2){x0x1,y0y1,z0z1,x2x5,y2y5,z2z5,x6x7,y6y7,z6z7}. The graph H2 is depicted in Figure 5. It is easy to see that H2 is a 5-regular subgraph of Qn on l vertices.

Figure 5. The Subgraph H2 on 16k+6 Vertices

Case (iii). l=16k+10 with 3k2n41.

Consider the three edge sets: F1={g0g1,g1g2,g2g5,g5g6,g6g7,g7g0}, F2={h2h3,h3h4,h4h5,h5h2}, and F3={x0x7,y0y7,z0z7,x1x2,y1y2,z1z2,x5x6,y5y6,z5z6,u2u3,v2v3,w2w3,u4u5,v4v5,w4w5}.

Let L3 be the subgraph of Qn with vertex set
{gi,xi,yi,zi:i=0,1,2,5,6,7}{hj,uj,vj,wj:j=2,3,4,5} and edge set {gixi,giyi,gizi:i=0,1,2,5,6,7}{hjuj,hjvj,hjwj:j=2,3,4,5}F1F2. We now define a subgraph H3 of Qn which is shown in Figure 6 as H3=(HL3)F3 Clearly, H3 is a 5-regular subgraph of Qn on 16k+10 vertices.

Figure 6. The subgraph H3 on 16k+10 vertices

Case (iv). l=16k+14 with 3k2n41.

We define the four edge sets M1,M2,M3,M4 as follows.

M1={g0g1,g1g2,g2g3,g3g4,g4g5,g5g6,g6g7,g7g0}; M2={h0h1,h1h2,h2h5,h5h6,h6h7,h7h0}; M3={u0u7,v0v7,w0w7,u1u6,v1v6,w1w6,u2u5,v2v5,w2w5} and M4={x0x1,y0y1,z0z1,x2x3,y2y3,z2z3,x4x5,y4y5,z4z5,x6x7,y6y7,z6z7}. Let L4 be the subgraph of Qn having vertex set {gi,xi,yi,zi:i=0,1,,7}{hj,uj,vj,wj:j=0,1,2,5,6,7} and edge set {gixi,giyi,gizi:i=0,1,,7}{hjuj,hjvj,hjwj:j=0,1,2,5,6,7}M1M2. We define a subgraph H4 of Qn as follows. H4=(HL4)(M3M4). The graph H4 is shown in Figure 7. Clearly, it is a 5-regular subgraph of Qn on 16k+14 vertices.

Figure 7. The subgraph H4 on 16k+14 vertices

Thus we have constructed 5-regular subgraphs H1, H2, H3, and H4 of the hypercube Qn on l vertices in each of the four cases. In the next two sections, we prove that these subgraphs are 5-connected and bipancyclic.

5. Connectivity

In this section, we prove that all the four subgraphs H1,H2,H3, and H4 of Qn that are constructed in Section 4 and shown in Figures 4, 5, 6, and 7, respectively are 5-connected.
If F is the matching in Hi consisting of edges having one end vertex in the lower side (L) and the other end vertex in the upper side (R) in the figure of Hi, then F is an edge-cut of Hi. The graph HiF has two components and further, the components are 4-connected and isomorphic to each other. We prove that these components together with the matching F give a 5-connected graph. We first prove the following observations for general graphs.

Lemma 8. Let G be a simple k-connected graph and let v1,v2,,vk be distinct vertices of G. Let G^ be a new graph obtained from G by adding a new vertex u and k edges uvi for i=1,2,,k. Then G^ is k-connected.

Proof. The graph G^ is shown in Figure 8(a). Suppose SV(G^) with |S|<k. If uS, then SV(G) and so GS is connected. Therefore G^S is connected as the vertex u has a neighbour in GS. Suppose uS. Then S{u} contains at most k2 vertices of G and hence G(S{u})=G^S is connected. Thus G^ is k-connected. ◻

Lemma 9. Let G be a simple k-connected graph with at least 2k vertices and an independent set {u1,u2,,ur}, where 1rk. Suppose G is a simple graph obtained from G by adding a new vertex u and k new edges each having one end vertex u including the r edges uu1,uu2,,uur. Let H be the graph obtained from the graph G◻K2 by deleting the matching consisting of r edges between the two copies of G corresponding to the vertices u1,u2,,ur. Then the graph H is (k+1)-connected.

Proof. The graph H is shown in Figure 8(b). The graph G◻K2 is obtained from G and a copy G of G by adding edges between their corresponding vertices. Let v and vi be the vertices of G corresponding to u and ui for 1ir, respectively and let M={u1v1,u2v2,,urvr}. Then H=(G◻K2)M. Since G has at least 2k+1 vertices, there are at least k+1 edges in H between G and G.

Suppose SV(H) with |S|k. It is sufficient to prove that HS is connected. Since G is k-connected, by Lemma 8, the graph G is k-connected. Suppose S intersects both V(G) and V(G). Then both GS and GS are connected and they are joined to each other by an edge. Hence HS is connected. Suppose SV(G). The degree of ui in G is at least k and so it is at least k+1 in G for any i{1,2,,r}. If GS has a component with vertex set a subset of the independent set {u1,u2,,ur}S, then that component has just one vertex and so |S|=k+1, a contradiction. Thus every component of GS has a neighbour in the connected graph G in HS. This shows that HS is connected. ◻

Figure 8. The Graphs of Lemma 8 and Lemma 9

Let H1, H2, H3, and H4 be the 5-regular subgraphs of Qn constructed in Section 4. We now prove that these graphs are 5-connected.

Lemma 10. The graph H1 is 5-connected.

Proof. The subgraphs of H1 induced by V(C0)V(C1)V(C2)V(C3) and by V(C4)V(C5)V(C6)V(C7) are isomorphic to C0◻Z for some 4-cycle Z. Hence, by Lemma 1, these two subgraphs are 4-connected. Now, by Lemma 9, H1 is 5-connected. ◻

Lemma 11. The graph H2 is 5-connected.

Proof. The subgraphs of H2 induced by V(C0)V(C3) and by V(C1)V(C2) are isomorphic to C0◻K2 and so, by Lemma 1, they are 3-connected. Hence, by Lemma 9, the upper half subgraph R of H2 that is induced by V(C0)V(C1)V(C2)V(C3){g0,g1} is 4-connected. If L=H2[V(C4)V(C5)V(C6)V(C7){g6,g7}], then L is isomorphic to R. Thus, by Lemma 9, H2 is 5-connected. ◻

Lemma 12. The graph H3 is 5-connected.

Proof. By Lemmas 1 and 9, the subgraphs H3[V(C0)V(C3)] and H3[V(C1)V(C2){g1,g2}] of H3 are 3-connected. By similar arguments of the proof of Lemma 11, we see that the upper half subgraph R of H3 induced by V(C0)V(C1)V(C2)V(C3){g1,g2,h2,h3} is 4-connected. If L=H3[V(C4)V(C5)V(C6)V(C7){g5,g6,h4,h5}], then L is isomorphic to R. Now, the result follows from Lemma 9◻

Lemma 13. The graph H4 is 5-connected.

Proof. Let V1=V(C0)V(C1){g0,g1} and let V2=V(C2)V(C3){g2,g3}. By Lemma 9, the subgraphs H4[V1] and H4[V2] of H4 are 3-connected. Therefore the graph H4[V1V2] is 4-connected. Hence the graph R=H4[V1V2{h0,h1,h2}] is also 4-connected. The subgraph L of H4 induced by V(H4)V(R) is isomorphic to R and so, it is 4-connected.

We now prove that H4 is 5-connected. Suppose SV(H4) with |S|4. It is sufficient to prove that H4S is connected. If S intersects both V(R) and V(L), then RS and LS are connected and they are joined to each other by an edge leaving H4S connected. Suppose SV(R). The set of vertices of R that do not have any neighbour in V(L) is A={h1,g1,g2,u0,v0,w0,u1,v1,w1,u2,v2,w2}. Assume that RS has a component D such that V(D)A. No member of A is isolated in RS as each has degree five in R. Hence δ(D)1. Observe that the subgraph of H4 induced by A is the forest as shown in Figure 9. Therefore D is a tree containing at least two pendant vertices. So |S|=4 and every pendant vertex of D is adjacent to all the four members of S. This gives K2,3 as a subgraph of H4 and so of Qn, a contradiction. Hence every component of RS has a neighbour in the connected graph L. Therefore H4S is connected. Similarly, H4S is connected if SV(L). ◻

Figure 9. Subgraph of H4 Induced by A

6. Bipancyclicity

In this section, we prove that the 5-regular graphs H1, H2, H3, and H4 constructed in Section 4 are all bipancyclic.

A ladder on n4 vertices has two edges at its two ends. If we identify one of these two edges with an edge of a k-cycle, then it follows that the resulting graph has cycles of every even length from k to k+n2. The following lemma is based on this fact.

Lemma 14. Let l,m,n be even integers and C be an m-cycle conatining an edge xy and L be a ladder on n vertices containing an end edge xy. Then the graph CL{xx,yy} has cycles of all even lengths from m+2 to m+n (see Figure 10).

Figure 10. The Graph CL{xx,yy}

Lemma 15. The graph H1 is bipancyclic.

Proof. Recall that H1 contains the eight copies C0,C1,,C7 of a 2k-cycle C. Let Eij be the set of edges of H1 with one end vertex in the cycle Ci and the other in the cycle Cj. Consider the subgraph W=C0C1C2C3C4C5C6C7E03E32E25E54E47E76E61E10 of H1. Then W is isomorphic to C◻Z, where Z is an 8-cycle. By Lemma 2, W is bipancyclic. Therefore W and so H1 contains a cycle of every even length from 4 to 16k=|V(W)|. Let Ci=<v1i,v2i,v2ki,v1i> for i=0,1,7 such that v1i is the label to the neighbor of gi. Then the cycle shown in Figure 11 is a Hamiltonian cycle of H1. Thus the graph H1 is bipancyclic. ◻

Figure 11. Hamiltonian Cycle in H1

Recall that we have written Qn as Qn3◻Q3 where the vertices of Q3 are labeled by a0,a1,,a7, so that a0,a1,a2,a3,a4,a5,a6,a7,a0 is a Hamiltonian cycle of Q3 with chords a0a3,a1a6,a2a5 and a4a7. Furthermore, C is a 2k-cycle in Qn3 and Ci is its copy in Qn corresponding to vertex ai of Q3 in the construction of the graphs H2,H3 and H4. Let Eij be the matching between cycles Ci and Cj in Qn corresponding to the edge aiaj of Q3. For each i, label the vertices of Ci by v1i,v2i,,v2ki so that Ci=<v1i,v2i,v3i,v4i,,vpi,,vqi,,v2ki,v1i> and v1i,v3i,vpi are the neighbours of the vertex gi, and v2i,v4i,vqi are the neighbours of the vertex hi for some p,q with 4<p,q2k and pq. Hence the neighbours xi,yi,zi of gi on Ci are relabeled as v1i,v3i,vpi, respectively. Similarly, the vertices ui,vi,wi are relabeled as v2i,v4i,vqi, respectively.

Lemma 16. The graph H2 is bipancyclic.

Proof. The graph H2 has 16k+6 vertices. Consider the subgraph W of H2, where W=C0C1C2C3C4C5C6C7E03E32E21E16E65E54E47E70. Then W2 is isomorphic to C0◻Z, where Z is an 8-cycle. By Lemma 2, it contains cycles of every even length from 4 to 16k. A cycle Z of length 16k+6 in H2 is shown in Figure 12. Furthermore, Z=(Z{g1,g6}){v11v16,g0g7} is a cycle of length 16k+4 while Z=(Z{g0,g7}){v10v17} is a cycle of length 16k+2 in H2. Thus H2 is bipancyclic. ◻

Figure 12. (16k+6)-cycle in H2
Figure 13. l-cycles in H3 for l{16k+6,16k+10}

.5cm

Lemma 17. The graph H3 is bipancyclic.

Proof. The graph H3 is on 16k+10 vertices. Consider the subgraph W of H3 on 12k vertices defined as W=C0C1C3C4C6C7E03E34E47E67E16E01. Then W is isomorphic to C0◻Z, where Z is a 6-cycle. Hence W contain cycles of all even lengths from 4 to 12k. Therefore these cycles are also contained in H3. Let L be the ladder on 4k vertices defined as L=(C2v12v22)(C5v15v25){vi2vi5:i=1,2,,2k}. Note that W has a Hamiltonian cycle Z1=<v21,v31,,v2k1,v11,v10,v2k0,v2k10,,v20,v23,,v2k3,v13,v14,v2k4,,v24,v27,,v2k7,v17,v16,v2k6,,v26,v21>. Then the subgraph Z1L{v22v21,v25v26} of H3 has 16k vertices. By Lemma 14, it has cycles of every even length from 12k+2 to 16k. Let Z2 and Z3 be the cycles of length 16k+6 and 16k+10 as shown in Figure 13(a) and 13(b), respectively. Then Z2=(Z2{g0,g1}){v10v11} is a cycle of length (16k+4) whereas (Z2{g6,g7}){v16v17} is a (16k+2)-cycle in H3. Finally, (Z3{h3,h4}){v23v24,h2h5} is a cycle of length 16k+8 in H3. Thus H3 is bipancyclic. ◻

Lemma 18. The graph H4 is bipancyclic.

Proof. Recall that H4 has 16k+14 vertices. We prove that H4 contain cycles of every even length from 4 to 16k+14. The subgraph of H4, W=C0C3C4C7E03E34E47 is bipancyclic as it is isomorphic to C0◻P4, where P4 is a path on four vertices. This implies that H4 contain cycles of every even length from 4 to 8k. Consider the Hamiltonian cycle Z1 of W, where
Z1=<v20,v30,,v2k0,v10,v13,v2k3,v2k13,,vk+23,vk+24,vk+34,,v2k4,v14,v17,v2k7,v2k17,,v27,v24,v34,,vk+14,vk+13,vk3,,v23,v20>. We get two paths P and Q each of length 4k from C1C6 and C2C5, respectively as follows. P=<v21,v31,,v2k1,v11,v16,v2k6,v2k16,,v26>, Q=<v22,v32,,v2k2,v12,v15,v2k5,v2k15,,v25>. Then L1=PQ({vi1vi2,vi5vi6:i=1,2,,2k}) is a ladder on 8k vertices. Hence, by Lemma 14, Z1L1{v20v21,v22v23} is a subgraph of H4 on 16k vertices contains cycle of length p for every even integer p with 8k+2p16k.

Let Z2 and Z3 be the cycles in H4 of length 16k+8 and 16k+14 as shown in Figures 14(a) and 14(b), respectively. If Z2=(Z2{g5,g6}){v15v16}; Z2=(Z2{g3,g4}){v13,v14} and Z2=(Z2{g1,g2}){v11,v12}, then Z2, Z2 and Z2 are cycles of lengths 16k+6, 16k+4 and 16k+2, respectively. Finally, Z3=(Z3{h5,h6}){v25v26} and Z3=(Z3{h1,h2}){v21v22} are cycles of lengths 16k+12 and 16k+10, respectively. Hence H4 is bipancyclic. ◻

Figure 14. l-cycles in H4 for l{16k+8,16k+14}

Remark 1. The problem of the existence of a k-regular subgraph of the hypercube Qn of a given order is completely solved for k=3, 4, 5. To solve the problem for the general value of k one needs to identify the even integers in between 2k and 2n that cannot be the order of any k-regular subgraph of Qn. By Theorem 3, we have two intervals (2k, 2k+2k1) and (2k+2k1, 2k+2k1+2k3) with the property that no even integer belonging to any of these intervals is the order of a k-regular subgraph of Qn. There is no further gap for k{3,4,5}. However, there seems a further gap for k6. The next interval of the gaps can be found by characterizing the k-regular subgraph of Qn on 2k+2k1+2k3 vertices. Thus, Problem 1 remains open for k6.

Acknowledgment

The authors are thankful to the referee for fruitful suggestions to improve the presentation of the paper. The first author is financially supported by DST-SERB, Government of India through the project MTR/2018/000447.

Conflict of Interest

The authors declare no conflict of interest

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