MATCHINGS EXTEND TO HAMILTONIAN CYCLES IN 5-CUBE 1

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1 Discussiones Mathematicae Graph Theory xx (xxxx) 5 doi:.75/dmgt.2 MATCHINGS EXTEND TO HAMILTONIAN CYCLES IN 5-CUBE Fan Wang 2 School of Sciences Nanchang University Nanchang, Jiangxi, P.R. China wangfan62@6.com and Weisheng Zhao Institute for Interdisciplinary Research Jianghan University Wuhan, Hubei 456, P.R. China weishengzhao@aliyun.com Abstract Ruskey and Savage asked the following question: Does every matching in a hypercube n for n 2 extend to a Hamiltonian cycle of n? Fink confirmed that every perfect matching can be extended to a Hamiltonian cycle of n, thus solved Kreweras conjecture. Also, Fink pointed out that every matching can be extended to a Hamiltonian cycle of n for n {2,,4}. In this paper, we prove that every matching in 5 can be extended to a Hamiltonian cycle of 5. Keywords: hypercube, Hamiltonian cycle, matching. 2 Mathematics Subject Classification: 5C8, 5C45. This work is supported by NSFC (grant nos and 269) and the science and technology project of Jiangxi Provincial Department of Education(grant No. 26BAB2). 2 Corresponding author. Download Date 2//7 7:27 PM

2 2 F. Wang and W. Zhao. Introduction Let [n] denote the set {,...,n}. The n-dimensional hypercube n is a graph whose vertex set consists of all binary strings of length n, i.e., V( n ) = {u : u = u u n and u i {,} for every i [n]}, with two vertices being adjacent whenever the corresponding strings differ in just one position. The hypercube n is one of the most popular and efficient interconnection networks. It is well known that n is Hamiltonian for every n 2. This statement dates back to 872 [9]. Since then, the research on Hamiltonian cycles in hypercubes satisfying certain additional properties has received considerable attention [2,, 4, 6, 2]. A set of edges in a graph G is called a matching if no two edges have an endpoint in common. A matching is perfect if it covers all vertices of G. A cycle in a graph G is a Hamiltonian cycle if every vertex in G appears exactly once in the cycle. Ruskey and Savage [] asked the following question: Does every matching in n for n 2 extend to a Hamiltonian cycle of n? Kreweras [] conjectured that every perfect matching of n for n 2 can be extended to a Hamiltonian cycle of n. Fink [5, 7] confirmed the conjecture to be true. Let K( n ) be the complete graph on the vertices of the hypercube n. Theorem. [5, 7]. For every perfect matching M of K( n ), there exists a perfect matching F of n, n 2, such that M F forms a Hamiltonian cycle of K( n ). Also, Fink [5] pointed out that the following conclusion holds. Lemma.2 [5]. Every matching in n can be extended to a Hamiltonian cycle of n for n {2,,4}. Gregor [8] strengthened Fink s result and obtained that given a partition of the hypercube into subcubes of nonzero dimensions, every perfect matching of the hypercube can be extended on these subcubes to a Hamiltonian cycle if and only if the perfect matching interconnects these subcubes. The present authors [4] proved that every matching of at most n edges in n can be extended to a Hamiltonian cycle of n for n 4. In this paper, we consider Ruskey and Savage s question and obtain the following result. Theorem.. Every matching in 5 can be extended to a Hamiltonian cycle of 5. It is worth mentioning that Ruskey and Savage s question has been recently done for n = 5 independently by a computer search [5]. In spite of this, a direct proof is still necessary, as it may serve in a possible solution of the general question. Download Date 2//7 7:27 PM

3 Matchings Extend to Hamiltonian Cycles in 5-Cube 2. Preliminaries and Lemmas Terminology and notation used in this paper but undefined below can be found in []. The vertex set and edge set of a graph G are denoted by V(G) and E(G), respectively. For a set F E(G), let G F denote the resulting graph after removing all edges in F from G. Let H and H be two subgraphs of G. We use H +H to denote the graph with the vertex set V(H) V(H ) and edge set E(H) E(H ). For F E(G), we use H+F to denote the graph with the vertex set V(H) V(F) and edge set E(H) F, where V(F) denotes the set of vertices incident with F. Thedistancebetweentwoverticesuandv isthenumberofedgesinashortest path joining u and v in G, denoted by d G (u,v), with the subscripts being omitted when the context is clear. Let j [n]. An edge in n is called an j-edge if its endpoints differ in the jth position. The set of all j-edges in n is denoted by E j. Thus, E( n ) = n i= E i. Let n,j and n,j, with the superscripts j being omitted when the context is clear, be the (n )-dimensional subcubes of n induced by the vertex sets {u V( n ) : u j = } and {u V( n ) : u j = }, respectively. Thus, n E j = n + n. We say that n splits into two (n )-dimensional subcubes n and n at position j; see Figure for example. E 4 Figure. 4 splits into two -dimensional subcubes and at. The parity p(u) of a vertex u in n is defined by p(u) = n i= ui (mod 2). Then there are 2 n vertices with parit and 2 n vertices with parity in n. Vertices with parit and are called black vertices and white vertices, respectively. Observe that n is bipartite and vertices of each parity form bipartite sets of n. Thus, p(u) p(v) if and only if d(u,v) is odd. A u,v-path is a path with endpoints u and v, denoted by P uv when we specify a particular such path. We say that a spanning subgraph of G whose components are k disjoint paths is a spanning k-path of G. A spanning -path thus is simply a spanning or Hamiltonian path. We say that a path P (respectively, a cycle C) passes through a set M of edges if M E(P) (respectively, M E(C)). Download Date 2//7 7:27 PM

4 4 F. Wang and W. Zhao Lemma 2. []. Let u,v,x,y be pairwise distinct vertices in with p(u) = p(v) p(x) = p(y) and d(u,x) = d(v,y) =. If M is a matching in {u,v}, then there exists a spanning 2-path P ux +P vy in passing through M. Lemma 2.2 []. For n {,4}, let u,v V( n ) be such that p(u) p(v). If M be a matching in n u, then there exists a Hamiltonian path in n joining u and v passing through M.. Proof of Theorem. Let M be a matching in 5. If M is a perfect matching, then the theorem holds by Theorem.. So in the following, we only need to consider the case that M is not perfect. Since 5 has 2 5 vertices, we have M 5. Choose a position j [5] such that M E j is as small as possible. Then M E j. Without loss of generality, we may assume j = 5. Split 5 into 4 and 4 at. Then 5 E 5 = Let α {,}. Observe that every vertex u α V( α 4 ) has in α 4 a unique neighbor, denoted by u α. Let M α = M E( α 4 ). We distinguish four cases to consider. Case. M E 5 =. We say that a vertex u is covered by M if u V(M). Otherwise, we say that u is uncovered by M. Since M is not perfect in 5, there exists avertex uncovered by M. By symmetry we may assume that the uncovered vertex lies in 4, and denote it by. In other words, V( 4 )\V(M). First apply Lemma.2 to find a hamiltonian cycle C in 4 passing through M. Let v be a neighbor of u on C such that u v / M. Since M is a matching, this is always possible. Since p( ) p( ) and M is a matching in 4, by Lemma 2.2 there exists a Hamiltonian path P u in 4 passing through M. Hence P u + C + { u, v } u v is a Hamiltonian cycle in 5 passing through M, see Figure 2. P u u v v C 4 4 Figure 2. Illustration for Case. Case 2. M E 5 =. Let M E 5 = { u }, where u α V( α 4 ). Let v α be a Download Date 2//7 7:27 PM

5 Matchings Extend to Hamiltonian Cycles in 5-Cube 5 neighbor of u α in α 4 for α {,}. Then p() p( ) and p(u ) p(v ). Since u M E 5, we have u α / V(M α ) for every α {,}. In other words, M α is a matching in α 4 u α. By Lemma 2.2 there exist Hamiltonian paths P uαv α in α 4 passing through M α for every α {,}. Hence P u +P u v +{ u, v } is a Hamiltonian cycle in 5 passing through M. Case. M E 5 = 2. Let M E 5 = { u, v }, where u α,v α V( α 4 ). If p( ) p( ), then p(u ) p(v ), the proof is similar to Case 2. So in the following we may assume p( ) = p( ). Now p(u ) = p(v ). In α 4, since there are already matched two vertices with the same color, we have M α 6 for every α {,}. Thus, i [4] M E i = M + M 2 and M 4. Choose a position k [4] such that M E k is as small as possible. Then M E k. Without loss of generality, we may assume k = 4. Let α {,}. Split α 4 into α and α at. For clarity, we write α and α as αl and αr, respectively, see Figure. Then α 4 E 4 = αl + αr. Let M αδ = M α E( αδ ) for every δ {L,R}. Note that every vertex s αl V( αl ) has in αr a unique neighbor, denoted by s αr, and every vertex t αr V( αr ) has in αl a unique neighbor, denoted by t αl. By symmetry, we may assume M E 4 M E 4. Since M E 4 = M E 4 + M E 4, we have M E 4. Since V( 4 ), without loss of generality we may assume V( L ). Now u V( L ). We distinguish two cases to consider. L L R R 4 4 Figure. 5 splits into four -dimensional subcubes L, R, L and R. Subcase.. V( R ). Now v V( R ). Subcase... M E 4 =. Apply Lemma.2 to find a Hamiltonian cycle C in 4 passing through M. Since u has only one neighbor in R, we may choose Download Date 2//7 7:27 PM

6 6 F. Wang and W. Zhao a neighbor x of u on C such that x V( L ). Similarly, we may choose a neighbor y of v on C such that y V( R ). Since {u, v } M, we have {u x,v y } M =. Since p( ) p( ) and M L is a matching in L, by Lemma 2.2 there exists a Hamiltonian path P u in L passing through M L. Similarly, there exists a Hamiltonian path P v in R passing through M R. Hence P u +P v +C +{ u, x, v, y } {u x,v y } is a Hamiltonian cycle in 5 passing through M, see Figure 4. L R 4 x x v y y u 4 L R Figure 4. Illustration for Subcase... Subcase..2. M E 4 =. Now M E 4 2. Let M E 4 = {s L s R }, where s δ V( δ ). Since p() = p( ) and p(s L ) p(s R ), without loss of generality, we may assume p( ) = p( ) = p(s L ) p(s R ). First, weclaimthatthereexistsahamiltoniancyclec in 4 passingthrough M such that the two neighbors of v on C both belong to V( R ). If M E 4 = 2, then M E 4 =. So M E i = for every i [4]. Since M α 6foreveryα {,}and M + M = i [4] M E i, wehave M α = 6 for every α {,}. Let M E 4 = {a L a R,b L b R }, where a δ,b δ V( δ ). Then p(a δ ) p(b δ ) for every δ {L,R}. (Otherwise, if p(a δ ) = p(b δ ), then M δ 2. Moreover, either p(u ) = p(a L ) = p(b L ) or p(v ) = p(a R ) = p(b R ), so M L or M R. Thus, M 2++2 = 5, a contradiction). If M E 4 =, let M E 4 = {a L a R }, where a δ V( δ ). Since a R has three neighbors in R, we may choose a neighbor b R of a R in R such that b R v. Now p(a δ ) p(b δ ) for every δ {L,R}. For the above two cases, since M δ is a matching in δ a δ, by Lemma 2.2 there exist Hamiltonian paths P aδ b δ in δ passing through M δ for every δ {L,R}. Let C = P al b L +P ar b R +{a L a R,b L b R }. In the former case, since { v,a L a R,b L b R } M, we have v / {a R,b R }. In the latter case, since { v,a L a R } M, we have v a R, and therefore, v / {a R,b R }. Hence C is a Hamiltonian cycle in 4 passing through M such that the two neighbors of v on C both belong to V( R ), see Figure 5(). Download Date 2//7 7:27 PM

7 Matchings Extend to Hamiltonian Cycles in 5-Cube 7 Next, choose a neighbor x of u on C such that x V( L ) and choose a neighbor y of v on C such that s R. Since M R is a matching in R, by Lemma 2.2 there exists a Hamiltonian path P v in R passing through M R, see Figure 5(2). Since s R / {, }, we may choose a neighbor t R of s R on P v such that t L. Now,,s L,t L are pairwise distinct vertices in L, and p() = p(s L ) p( ) = p(t L ), and d(, ) = d(s L,t L ) =. Since M L is a matching in L {,s L }, by Lemma 2. there exists a spanning 2-path P u +P sl t L in L passing through M L. Hence P u + P sl t L +P v +C +{ u, x, v, y,s L s R,t L t R } {u x,v y,s R t R } is a Hamiltonian cycle in 5 passing through M, see Figure 5(2). L R s L s R u u v L L u u x x s L t L C s tr R v v C y y R R () (2) L R Figure 5. Illustration for Subcase..2. Subcase.2. V( L ). Now v V( L ). Let x be the unique vertex in L satisfying d(x,v ) =. Then d(x,u ) =. Since M is a matching in 4 u, by Lemma 2.2 there exists a Hamiltonian path P u x in 4 passing through M. Since v has only one neighbor in R, we may choose a neighbor y of v on P u x such that y V( L ). Since d(x,v ) =, we have y x. Then u,x,v,y arepairwisedistinctvertices, andp(u ) = p(v ) p(x ) = p(y ), and d(u,x ) = d(v,y ) =, and the same properties also hold for the corresponding vertices,,,. If we can find a spanning 2-path P u +P v in 4 passing through M, then P u + P v + P u x + { u, x, v, y } v y is a Hamiltonian cycle in 5 passing through M. So in the following, we only need to show that the desired spanning 2-path P u + P v exists. We distinguish several cases to consider. Subcase.2.. M E 4 =. Since M L is a matching in L {, }, by Lemma 2. there exists a spanning 2-path P u +P v in L passing through M L. Let M E 4 = {s L s R }, where s δ V( δ ). Without loss of generality assumes L V(P v ). Chooseaneighbort L ofs L onp v. Sinces L s R M, we have s L t L / M. Since M R is a matching in R s R, by Lemma 2.2 there Download Date 2//7 7:27 PM

8 8 F. Wang and W. Zhao exists a Hamiltonian path P sr t R in R passing through M R. Let P u = P u and P v = P v +P sr t R +{s L s R,t L t R } s L t L. Then P u +P v is the desired spanning 2-path in 4, see Figure 6. u u L s L t L x x v y y L R s R t R 4 4 R Figure 6. Illustration for Subcase.2.. Subcase.2.2. M E 4 =. It suffices to consider the case that M L is maximal in L {, } and M R is maximal in R. In L, since p() = p( ), we have, are different in two positions, so there is one possibility of {, } up to isomorphism. Since d(, ) =, the vertex is fixed by. Sinced(, ) =, therearetwochoicesof uptoisomorphism. Thus, thereare two possibilities of {,,, } up to isomorphism, see Figure 7(a)(b). When {,,, } is the case (a), since M L is a maximal matching in L {, }, there are three possibilities of M L up to isomorphism, see Figure 7() (). When {,,, } is the case (b), there are seven possibilities of M L, see Figure 7(4) (). In R, there are three non-isomorphic maximal matchings, denoted by P,P 2 and P, see Figure 8. v ( a ) ( b) () (2) () (4) (5) (6) (7) (8) (9) () Figure 7. All possibilities of {,,,,M L } up to isomorphism. Download Date 2//7 7:27 PM

9 Matchings Extend to Hamiltonian Cycles in 5-Cube 9 Before the proof, we point out that if M R is isomorphic to the matching P or P 2, then there exists a Hamiltonian cycle in R passing through M R {e} for any e / M R, see Figure 9. ( ) P ( P2 ) P Figure 8. Three non-isomorphic maximal matchings in R. ( ) ( P - ) ( P - 2) ( P - ) ( P2 - ) ( P2-2) Figure 9. Hamiltonian cycles passing through M R {e} for any e / M R in R when M R is isomorphic to P or P 2. First, suppose that M R is isomorphic to P. Since M L is a matching in L {, }, by Lemma 2. there exists a spanning 2-path P u + P v in L passing through M L. Since E(P u +P v ) = 6 > M L + M R, there exists an edge s L t L E(P u +P v )\M L such that s R t R / M R. Choose a Hamiltonian cycle C R in R passing through M R {s R t R }. Hence P u + P v +C R +{s L s R,t L t R } {s L t L,s R t R } is the desired spanning 2-path in 4. (Note that the construction is similar to Subcase.2., so the readers may refer to the construction in Figure 6.) Next, suppose that M R is isomorphic to P 2. We say that a set S of edges crossesapositioniifs E i. If{,,,,M L }isisomorphictooneofthe cases (2) () in Figure 7, then we may choose a spanning 2-path P u +P v in L such that the set E(P u + P v ) \ M L crosses at least two positions, see Figure (2) (). Since all the edges in M R lie in the same position, there exists an edge s L t L E(P u + P v ) \ M L such that s R t R / M R. If {,,,,M L } is isomorphic to the case () in Figure 7, then we may choose two different spanning 2-paths P u + P v in L such that the two sets E(P u +P v )\M L cross two different positions, see Figure ( ), ( 2), and therefore, at least one of them is different from the position in which M R lies. Thus, we may choose a suitable spanning 2-path P u + P v such that there exists an edge s L t L E(P u + P v ) \ M L and s R t R / M R. The remaining construction is similar to the above case. Last, suppose that M R is isomorphic to P. Without loss of generality, we may assume M R (E 2 E ). Download Date 2//7 7:27 PM

10 F. Wang and W. Zhao If {,,,,M L } is isomorphic to the case (5) or (8) in Figure 7, we may choose a spanning 2-path P u + P v in L passing through M L such that (E(P u +P v )\M L ) E, see Figure. Let s L t L (E(P u +P v )\ M L ) E. Then s R t R E. One can verify that there exists a Hamiltonian cycle C R in R passing through M R {s R t R }. Hence P u +P v +C R + {s L s R,t L t R } {s L t L,s R t R } is the desired spanning 2-path in 4. ( - ) ( - 2) (2) () (4) (5) (6) (7) (8) (9) () Figure. Spanning 2-paths P u +P v in L with the possible edges s L t L lined by \\. (5 - ) (5-2) (8) Figure. The possible spanning 2-paths P u +P v such that (E(P u +P v )\M L ) E. If {,,,,M L } is isomorphic to one of the cases (), (6), (7) or () in Figure 7, then choose a spanning 2-path P u +P v in L passing through M L, see Figure 2(), (6), (7), (). If (E(P u +P v )\M L ) E, then the proof is similar to the above case. If (E(P u +P v )\M L ) E =, then the set E(P u +P v )\M L crosses the positions 2 and, and therefore, M R has two choices for every case, see Figure 2. Then we can find a spanning 2-path P u +P v in 4 passing through M, see Figure 2. If {,,,,M L } is isomorphic to one of the cases (), (2), (4) or (9) in Figure 7, we observe that there exist two vertices in V( L ) at distance, denoted by s L,t L, such that there is a spanning 2-path P u +P v in L + s L t L passing through M L {s L t L }, see Figure. Next, we can verify that there exists a Hamiltonian path P sr t R in R passing through M R. Hence P u +P v +P sr t R +{s L s R,t L t R } s L t L is the desired spanning 2-path in 4. Download Date 2//7 7:27 PM

11 Matchings Extend to Hamiltonian Cycles in 5-Cube Case 4. M E 5 =. Let M E 5 = { u, v,w w }, where u α,v α,w α V( α 4 ). Now M E i = for every i [5] and M = 5. Hence there are two vertices of {u α,v α,w α } in one partite set and one vertex in the other partite set. Otherwise, if p(u α ) = p(v α ) = p(w α ), then M α 5, and therefore, M, a contradiction. Without loss of generality, we may assume p(u α ) = p(v α ) p(w α ). v () (6) v v ( - ) ( - 2) (6 - ) (6-2) y y v v (7) () (7 - ) (7-2) ( - ) ( - 2) Figure 2. Spanning 2-paths P +P in 4 passing through M. sl sl sl sl () tl tl (2) tl (4) (9) tl Figure. Spanning 2-paths P u +P v in L +s L t L passing through M L {s L t L }. Split α 4 into two -cubes αl and αr at some position k such that u α V( αl ) and v α V( αr ). Without loss of generality, we may assume k = 4. Since p(u α ) = p(v α ) p(w α ), by symmetry we may assume w α V( αl ). Since M E 4 + M E 4 = M E 4 =, by symmetry we may assume M E 4. Let M αδ = M α E( αδ ) for every δ {L,R}. Subcase 4.. M E 4 =. Since p(u ) p(w ) and M is a matching in 4 u, by Lemma 2.2 there exists a Hamiltonian path P u w in 4 passing Download Date 2//7 7:27 PM

12 2 F. Wang and W. Zhao through M. Since v has only one neighbor in L, we may choose a neighbor y of v on P u w such that y V( R ). Now V( R ) and p() = p( ) p(w ) = p( ). Since M L is a matching in L and M R is a matching in R, by Lemma 2.2 there exist Hamiltonian paths P u w in L and P v in R passing through M L and M R, respectively. Hence P u w +P u w +P v + { u,w w, v, y } v y is a Hamiltonian cycle in 5 passing through M, see Figure 4. L R w w v y y u L R Figure 4. Illustration for Subcase 4.. Subcase 4.2. M E 4 =. Now M E 4 = 2. Let M E 4 = {s L s R } and M E 4 = {a L a R,b L b R }, where s δ V( δ ) and a δ,b δ V( δ ). Since M = 5, 5 has exactly two vertices uncovered by M, one in L and the other in R. Thus, p(a L) p(b L ), and p( ) p(s R ), and M L is a perfect matching in L {u,w,a L,b L }. Since p(u ) p(w ) and p(a L ) p(b L ), without loss of generality, we may assume p(u ) = p(b L ) p(w ) = p(a L ). Thus, p(v ) = p(a R ) p(b R ). L u u L s L sr tl tr w w v v bl b R a L a R R y R Figure 5. Illustration for Subcase 4.2. Since p( ) p(w ) and M L is a matching in L, by Lemma 2.2 there exists a Hamiltonian path P u w in L passing through M L, see Figure 5. Since s L / {,w }, we may choose a neighbor t L of s L on P u w such that t R. Since p(s R ) p(t R ) and M R is a matching in R s R, Download Date 2//7 7:27 PM

13 Matchings Extend to Hamiltonian Cycles in 5-Cube by Lemma 2.2 there exists a Hamiltonian path P sr t R in R passing through M R, see Figure 5. Since / {s R,t R }, we may choose a neighbor of on P sr t R such that y b R. Now v,y,a R,b R are pairwise distinct vertices, and p(v ) = p(a R ) p(y ) = p(b R ), and d(v,y ) =, and M R is a matching in R {v,a R }. u R M L R w ( ) a L R M L R b L a L ( w ) L Figure 6. The spanning 2-path P ua L +P wb L (or P uw +P alb L ) in L. If d(a R,b R ) =, then by Lemma 2. there is a spanning 2-path P v y + P ar b R in R passing through M R, see Figure 7(). Since M L is a perfect matching in L {u,w,a L,b L }, we have M L {u w,a L b L } is a perfect matching in K( L ). By Theorem., there exists a perfect matching R in L ). Hence M L R forms a spanning 2-path in L. Note that each path of the spanning 2- pathisan(r,m L )-alternatingpathbeginningwithanedgeinrandendingwith an edge in R. So the number of vertices in each path is even. Since 5 is a bipartite graph, the two endpoints of each path have different parities. Hence one path joins the vertices u and a L, and the other path joins the vertices w and b L, see Figure 6 for example. Denote the spanning 2-path by P u a L +P w b L. Note that s L t L / M and / M. Hence P u w +P sr t R +P u a L +P w b L +P v y + P ar b R + { u,w w, v, y,a L a R,b L b R,s L s R,t L t R } {,s L t L } is a Hamiltonian cycle in 5 passing through M, see Figure 7(). such that M L {u w,a L b L } R forms a Hamiltonian cycle in K( L L u u L L u u L s L sr tl t R w w v v bl b R a L a R s L sr tl t R w w v v bl b R a L a R R y R R () (2) Figure 7. Illustration for Subcase 4.2. y R Download Date 2//7 7:27 PM

14 4 F. Wang and W. Zhao If d(a R,b R ) =, then d(v,b R ) = d(a R,y ) =. Since M R is a matching in R {v,a R }, by Lemma 2. there is a spanning 2-path P v b R + P ar y in R passing through M R, see Figure 7(2). Since M L {u a L,w b L } is a perfect matching in K( L ), similar to the above case, there is a spanning 2-path P u w +P al b L in L passing through M L. Hence P u w +P sr t R +P v b R + P y a R +P u w +P al b L +{ u,w w, v, y,a L a R,b L b R,s L s R,t L t R } {,s L t L } is a Hamiltonian cycle in 5 passing through M, see Figure 7(2). The proof of Theorem. is complete. Acknowledgements The authors would like to express their gratitude to the anonymous referees for their kind suggestions on the original manuscript. References [] J.A. Bondy and U.S.R. Murty, Graph Theory with Applications (North-Holland, NewYork-Amsterdam-Oxford, 982). [2] R. Caha and V. Koubek, Spanning multi-paths in hypercubes, Discrete Math. 7 (27) doi:.6/j.disc [] D. Dimitrov, T. Dvořák, P. Gregor and R. Škrekovski, Gray codes avoiding matchings, Discrete Math. Theoret. Comput. Sci. (29) [4] T. Dvořák, Hamiltonian cycles with prescribed edges in hypercubes, SIAM J. Discrete Math. 9 (25) doi:.7/s [5] J. Fink, Perfect matchings extend to Hamilton cycles in hypercubes, J. Combin. Theory Ser. B 97 (27) doi:.6/j.jctb [6] J. Fink, Connectivity of matching graph of hypercube, SIAM J. Discrete Math. 2 (29) 9. doi:.7/ [7] J. Fink, Matching graphs of hypercubes and complete bipartite graphs, European J. Combin. (29) doi:.6/j.ejc [8] P. Gregor, Perfect matchings extending on subcubes to Hamiltonian cycles of hypercubes, Discrete Math. 9 (29) 7 7. doi:.6/j.disc [9] L. Gros, Théorie du Baguenodier (Aimé Vingtrinier, Lyon, 872). [] G. Kreweras, Matchings and Hamiltonian cycles on hypercubes, Bull. Inst. Combin. Appl. 6 (996) Download Date 2//7 7:27 PM

15 Matchings Extend to Hamiltonian Cycles in 5-Cube 5 [] F. Ruskey and C. Savage, Hamilton cycles that extend transposition matchings in Cayley graphs of S n, SIAM J. Discrete Math. 6 (99) doi:.7/462 [2] J. Vandenbussche and D. West, Matching extendability in hypercubes, SIAM J. Discrete Math. 2 (29) doi:.7/ [] F. Wang and H.P. Zhang, Two types of matchings extend to Hamiltonian cycles in hypercubes, Ars Combin. 8 (25) [4] F. Wang and H.P. Zhang, Small matchings extend to Hamiltonian cycles in hypercubes, Graphs Combin. 2 (26) doi:.7/s [5] E. Zulkoski, V. Ganesh and K. Czarnecki, MathCheck: A Math Assistant via a Combination of Computer Algebra Systems and SAT Solvers, in: A.P. Felty and A. Middeldorp, Proc. CADE-25 (Ed(s)), (LNCS 995, 25) doi:.7/ Received 27 September 26 Revised 4 November 26 Accepted 4 November 26 Download Date 2//7 7:27 PM