Graphs and digraphs with all 2 factors isomorphic

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1 Graphs and digraphs with all 2 factors isomorphic M. Abreu, Dipartimento di Matematica, Università della Basilicata, I Potenza, Italy. abreu@math.fau.edu R.E.L. Aldred, University of Otago, P.O. Box 56, Dunedin, New Zealand. raldred@maths.otago.ac.nz M. Funk, Dipartimento di Matematica, Università della Basilicata, I Potenza, Italy. funk@unibas.it Bill Jackson, School of Mathematical Sciences, Queen Mary College, London E1 4NS, England. b.jackson@qmul.ac.uk D. Labbate, Dipartimento di Matematica, Politecnico di Bari, I Bari, Italy. labbate@poliba.it J. Sheehan, Department of Mathematical Sciences, King s College, Old Aberdeen AB24 3UE, Scotland. j.sheehan@maths.abdn.ac.uk This research was supported by the EPSRC. This research was supported by the Italian Ministry MIUR. This research was carried out within the activity of G.N.S.A.G.A. of the Italian C.N.R. and supported by the Italian Ministry MURST. 1

2 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 2 Abstract We show that a digraph which contains a directed 2-factor and has minimum in-degree and out-degree at least four has two non-isomorphic directed 2-factors. As a corollary we deduce that every graph which contains a 2- factor and has minimum degree at least eight has two non-isomorphic 2- factors. In addition we construct: an infinite family of strongly connected 3-diregular digraphs with the property that all their directed 2-factors are isomorphic, an infinite family of 2-connected 4-regular graphs with the property that all their 2-factors are isomorphic, and an infinite family of cyclically 6-edge-connected cubic graphs with the property that all their 2-factors are hamiltonian cycles. 1 Introduction All graphs considered are assumed to be simple (i.e. without loops or multiple edges) unless stated otherwise. We will use the term multigraph when we allow the possibility of multiple edges. By a digraph we mean a directed graph without loops or parallel arcs. A digraph D is k diregular if each vertex has in degree and out degree equal to k. A directed 2-factor of a digraph D is a 1 diregular spanning subdigraph of D. The family DU(k) is the set of all k diregular digraphs D such that all directed 2-factors of D are isomorphic. The subfamily HDU(k) is the set of those digraphs in DU(k) such that all their directed 2-factors are hamiltonian. The family U(k) is the set of all (connected) k regular graphs G such that G has a 2 factor and all 2 factors of G are isomorphic. We use BU(k) to denote the set of graphs in U(k) which are also bipartite, HU(k) is the set of graphs in U(k) which are also hamiltonian, and HBU(k) are those graphs in U(k) which are also hamiltonian and bipartite. In [1] we proved the following Theorem 1.1 BU(k) = /0 for k 4. An infinite family of graphs in HBU(3) is given in [3]. It is conjectured in [3] that all 3 connected graphs in HBU(3) belong to this family and, in [1], that all 3-connected graphs in BU(3) belong to HBU(3). Diwan [2] has shown that there are no planar graphs in HU(3). In section 3, we show that a digraph which contains a directed 2-factor and has minimum in-degree and out-degree at least four has two non-isomorphic 2- factors. As a corollary we deduce that DU(k) = /0 for all k 4. We also construct an infinite family of graphs in HDU(3). In section 4, we prove that every graph which contains a 2-factor and has minimum degree at least eight has two non-isomorphic 2-factors. As a corollary we

3 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 3 deduce that U(k) = /0 for k 8. We also construct infinite families of graphs in U(4) and cyclically-6-edge-connected graphs in HU(3). The only 3-connected graph we know in U(4) is K 5 and it is possible that this is the only 3 connected graph in U(4). 2 Preliminaries An r factor of a graph G is an r regular spanning subgraph of G. An r factorization of G is a partition of the edge set of G into r factors. A directed r-factor of a digraph D is an r/2 diregular spanning subdigraph of D. A directed r factorization of a digraph D is a partition of the edge set of D into directed r factors. A set S of edges of a graph G is a cyclic cutset if G S has two components each of which contains a cycle. The cyclic edge connectivity of a graph G, λ c (G), is the size of the smallest cyclic cutset of G. We say that a graph G is cyclically m edge connected if λ c (G) m. (We consider graphs without cyclic cutsets to be cyclically m edge connected for all m 1.) Let G be a bipartite graph with bipartition (X,Y ) such that X = Y, and A be its bipartite adjacency matrix. In general 0 det(a) per(a). We say that G is det extremal if det(a) = per(a). Let X = {x 1,x 2,...,x n } and Y = {y 1,y 2,...,y n } be the bipartition of G. For F a 1 factor of G define the sign of F, sgn(f), to be the sign of the permutation of {1,2,...,n} corresponding to F. Thus G is det extremal if and only if all 1 factors of G have the same sign. The following result is a special case of [4, Lemma 8.3.1]. Lemma 2.1 Let F 1,F 2 be 1 factors in a bipartite graph G and t be the number of circuits in F 1 F 2 of length congruent to zero modulo four. Then sgn(f 1 )sgn(f 2 ) = ( 1) t. The following is a special case of a result by Thomassen [5]. Lemma 2.2 Let G be a det extremal bipartite graph. Then G has a vertex of degree at most three. 3 The Class DU(k) For v a vertex of a digraph D, let d + (v) and d (v) denote the out degree and in degree of v. Theorem 3.1 Let D be a digraph with d + (v) 4 and d (v) 4 for all v V (D). Let X be a directed 2-factor of D. Then D has a directed 2-factor Y with Y = X.

4 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 4 Proof. Suppose all directed 2-factors of D are isomorphic to X. Construct the associated bipartite graph G for the digraph D in the following way. For each vertex u V (D), make two copies u and u in V (G). Each directed edge (u,v) E(D) becomes the undirected edge (u,v ) in E(G). Additionally, there are the edges (u,u ) E(G), for all u V (D). Note that G has minimum degree at least five and that the edges (u,u ) form a 1 factor, F 0, of G. Let F be any 1 factor of G disjoint from F 0. Then F F 0 is a 2 factor in G in which each cycle has alternatingly edges of F and edges of F 0. This 2-factor gives rise to a directed 2-factor in D When we contract each edge of F 0, which by assumption must be isomorphic to X. Hence F F 0 = X where X is a 2 factor of G in which each cycle corresponds exactly to one cycle of X but with twice the length. This implies that for any 1 factor F of G, disjoint from F 0, the number of cycles in F F 0 of length congruent to 0 modulo 4 is equal to the number of even cycles in X. We will call this constant number t. Using Lemma 2.1, we get that for any 1 factor F of G, disjoint from F 0, sgnf sgnf 0 = ( 1) t. Since t is constant, we may conclude that all 1-factors of G, disjoint from F 0 have the same sign. Hence, G F 0 is det-extremal. This contradicts Lemma 2.2 since G F 0 has minimum degree at least four. Corollary 3.2 DU(k) = /0 for k 4. Constructions A directed n cycle is a directed cycle of length n. We denote by (u 1,...,u n ) the directed n cycle given by the directed edges (u i,u i+1 ) for i = 1,...,n 1, and (u n,u 1 ). Let Γ be a finite group and S be a subset of Γ \ {id}. The Cayley digraph Cay(Γ,S) of Γ with respect to S is the digraph having vertex set Γ and edge set {(a,b) a,b Γ,ba 1 S}. It is well known that Cayley digraphs are vertex transitive. Let D 7 be the Cayley digraph Cay(Z 7,S), with S = {1,2,4}. The elements of S give rise in D 7 to the following edge disjoint directed hamiltonian cycles (1,2,3,4,5,6,7); (1,3,5,7,2,4,6); (1,5,2,6,3,7,4) which form a directed 2-factorization of D 7. Note that if we delete the fixed 1-factor F 0 from the bipartite graph associated to D 7, we obtain the Heawood graph. Proposition 3.3 D 7 HDU(3). Proof. The digraph D 7 has a hamiltonian directed 2-factor by construction. If D 7 has any non hamiltonian directed 2-factor, it must consist of a directed 3 cycle and a directed 4 cycle. Note that each vertex of D 7 belongs to exactly six directed 3 cycles.

5 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 5 Let (a,b,c) be any directed 3 cycle in D 7. We will show that there exists a vertex d V (D 7 ) {a,b,c} such that either {(a,d),(b,d),(c,d)} E(D 7 ) or {(d,a),(d,b),(d,c)} E(D 7 ). Since D 7 is vertex transitive, it is enough to check this for the six directed 3 cycles at the vertex 1, which are (1,2,4), (1,2,6), (1,3,4), (1,3,7), (1,5,6), (1,5,7) and the corresponding vertex d is 7, 3, 5, 6, 4, 2, respectively. Thus there cannot be a directed 4 cycle among the vertices of V (D 7 ) {a,b,c}, since all the in edges (or all the out edges) of one of its vertices have been already used. Let D 1 and D 2 be digraphs in HDU(3) and let u V (D 1 ) and v V (D 2 ). Suppose that (u 1,u), (u 2,u), (u 3,u), (u,u 4 ), (u,u 5 ), (u,u 6 ) are the directed edges incident with u and (v,v 1 ), (v,v 2 ), (v,v 3 ), (v 4,v), (v 5,v), (v 6,v) are those incident with v. If we delete the vertices u and v, we add a new vertex w and the directed edges (u 1,v 1 ), (u 2,v 2 ), (u 3,v 3 ), (w,u 4 ), (w,u 5 ), (w,u 6 ), (v 4,w), (v 5,w), (v 6,w), then we obtain a 3 diregular digraph D = D 1 D 2. We define D 1 and D 2 to be the subdigraphs of D corresponding to D 1 {u} and D 2 {v}. Remark 3.4 Let x D 1 and y D 2. (i) Any (x,y) path in D passes through at least one edge in (u 1,v 1 ), (u 2,v 2 ), (u 3,v 3 ); (ii) Any (y,x) path in D passes through w, hence it passes through exactly one directed edge in (w,u 4 ), (w,u 5 ), (w,u 6 ) and one in (v 4,w), (v 5,w), (v 6,w); (iii) Any directed cycle in D through w passes through exactly one edge (u 1,v 1 ), (u 2,v 2 ), (u 3,v 3 ). Proposition 3.5 If D 1,D 2 HDU(3) then D = D 1 D 2 HDU(3). Proof. Let H 1 and H 2 be hamiltonian directed 2-factors of D 1 and D 2, respectively. Up to a relabelling of the vertices adjacent to u and to v, the directed edges of H 1 passing through u are (u 1,u) and (u,u 4 ) and those of H 2 passing through v are (v,v 1 ) and (v 4,v). Then, H = (H 1 {u}) (H 2 {v}) {(u 1,v 1 ),(v 4,w),(w,u 4 )} is a hamiltonian directed 2-factor in D. Suppose that D contains a disconnected directed 2-factor F and let C be the component of F containing w. Relabelling if necessary, we may suppose that C passes through (u 1,v 1 ), (v 4,w) and (w,u 4 ). Each remaining component of F lies completely in either D 1 or in D 2. For i = 1,2, let X i be the set of components of F that lie completely in D i. By symmetry we may suppose that X 2 /0. Let P 2 be the directed (v 1,v 4 ) path in C. Then, X 2 (P 2 {(v,v 1 ),(v 4,v)} is a disconnected directed 2-factor of D 2, a contradiction since D 2 HDU(3). Note: Proposition 3.5 gives rise to an infinite family of digraphs in HDU(3) starting from D 7. A digraph D is strongly m connected if for each pair of distinct vertices x,y there are at least m internally disjoint paths from x to y.

6 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 6 Problem 3.6 Is D 7 the only strongly 2 connected digraph in HDU(3)? For t 2, let M t be the Cayley digraph Cay(Z 2t+1,S), with S = {1,2}. The digraph M t has girth t +1 and it can easily be seen that M t is strongly 2 connected. Proposition 3.7 M t HDU(2). Proof. Suppose that M t has a disconnected directed 2-factor F. Since M t has 2t + 1 vertices, there is at least one component of F of length less than t + 1, a contradiction to the fact that M t has girth t + 1. The digraphs M t have an odd number of vertices for t 2. The 6 vertex digraph having as directed edges the directed cycles (1,2,3,4,5,6) and (1,3,5,2,6,4) is an example of a strongly 2 connected digraph in HDU(2) with an even number of vertices. 4 The Class U(k) Theorem 4.1 Let G be a graph with d(v) 8 for all v V (G). Let X be a 2 factor of G. Then G has a 2 factor Y with Y = X. Proof. Let G 1 = G E(X) and U be the set of vertices of odd degree in G 1. Let M be a matching between the vertices of U and G 2 be the multigraph obtained by adding the edges of M to G 1. Note that we do not require M to be contained in or disjoint from E(G 1 ) and hence G 2 may contain multiple edges. Each vertex of G 2 has even degree, and hence each component of G 2 has an Euler tour. Thus we can construct a digraph D 2 by orienting the edges of G 2 in such a way that d D + 2 (v) = dd 2 (v) for all v V (D 2 ). Let D 1 be the digraph obtained from D 2 by deleting the arcs corresponding to edges in M. Then d D + 1 (v) 3 and dd 1 (v) 3 for all v V (D 1 ). Let X 1 be a 1 diregular digraph obtained by directing the edges of X and D be the digraph obtained from D 1 by adding the arcs of X 1. Then d D + (v) 4 and dd (v) 4 for all v V (D). By Theorem 3.1, D has a directed 2-factor Y 1 with Y 1 = X1. Clearly, Y 1 corresponds to a 2-factor Y of G with Y = X. Theorem 4.1 immediately gives: Corollary 4.2 U(k) = /0 for k 8. Problem 4.3 What is the largest value of k such that U(k) /0? Conjecture 4.4 HU(4) = {K 5 }.

7 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 7 Constructions (1) We construct an infinite family of 2-connected graphs in U(4). A loyal edge of a graph G is an edge e such that there exists a 2 factor that contains e and a 2 factor that does not, and has the additional property that e belongs to only one length of a cycle in every 2 factor of G that contains it. Definition 4.5 Let G U(4) such that G has at least one loyal edge, e. Let G i, i = 1,...,4, be four isomorphic copies of G and let e i = x i y i E(G i ) be the loyal edges corresponding to e. We construct a 4 regular graph G, called a 4 seed graft of G as follows. and V (G ) = ( 4 i=1v (G i )) {u,v} E(G ) = ( 4 i=1(e(g i ) {e i }) ( 4 i=1{x i u,y i v}). We call the new vertices u,v clips and we refer to G as a seed for G. Proposition 4.6 Let G U(4) and e be a loyal edge of G. Let G be the 4 seed graft of G. Then, (i) G is 4 regular and of connectivity 2; (ii) G U(4) HU(4), (iii) Each edge of G adjacent to a clip is loyal. Proof. (i) is straightforward. (ii) Let F i and H i be two distinct 2 factors of G i such that e i E(F i ) and e i / E(H i ). Suppose that C i is the component of F i containing e i. Let F be a 2 factor of G and C be the cycle in F containing u. Then v V(C) and C = (C i {e i }) (C j {e j }) {x i u,x j u,y i v,y j v}, for some i j. Thus F = (F i C i ) (F j C j ) C t {i, j} H t for some i j. Hence all 2 factors of G are isomorphic and disconnected. (iii) Each edge adjacent to a clip is either x i u or y i v, for some 1 i 4. Let F i j = (F i C i ) (F j C j ) C t {i, j} H t for all 1 i < j 4. Then F i j is a 2 factor of G containing x i u and y i v. For s i and t i, F st is a 2 factor of G containing neither x i u nor y i v. Furthermore, any 2 factor of G containing x i u or y i v is of the form F i j for some j i by (ii), and by the loyalty of e i, we have that x i u and y i v belong to only one length of a cycle in G. Note: Proposition 4.6 gives rise to an infinite family of 2-connected graphs in U(4) starting from K 5, since K 5 HU(4) and each edge of K 5 is loyal. (2) An infinite family of 3 connected graphs in HBU(3) is given in [3]. All but two of these graphs have cyclic edge-connectivity three and we conjecture in [3] that K 3,3 and the Heawood graph are the only cyclically 4-edge-connected graphs

8 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 8 in HBU(3). (3) We construct an infinite family of cyclically 6 connected graphs in HU(3). Definition 4.7 Let t 3 be an odd integer. The graph A(t) has vertex set and edge set V (A(t)) = {h i,u i,v i,w i : i = 1,2,...,t} E(A(t)) = {h i u i,h i v i,h i w i,u i u i+1,v i v i+1,w i w i+1 : i = 1,2,...,t} where the subscript addition is modulo t. It can be checked that Lemma 4.8 A(t) is cyclically 6 connected for t 7, and is cyclically t connected for t = 3,5. To assist in our discussion of the properties of the graphs defined above we introduce the following terminology. Definition 4.9 Let C t be a cycle of length t. For i = 1,2,...,t, we call the subgraph IC i of A(t) induced by the vertices {h i,u i,v i,w i }, the i th interchange of A(t). The vertex h i V (IC i ) and the edges {h i u i, h i v i, h i w i } E(IC i ) are called respectively a hub and the spokes of the interchange. The set of edges edges {u i u i+1,v i v i+1,w i w i+1 } linking IC i to IC i+1 are referred as the i th link L i of A(t). The edge u i u i+1 L i is called the u channel of the link. The subgraphs of A(t) induced by the vertices {u i : i = 1,2,...,t}, {v i : i = 1,2,...,t}, and {w i : i = 1,2,...,t}, are each isomorphic to C t and are called the base cycles of A(t). Recall that in a cubic graph G, a 1 factor, F, determines a corresponding 2 factor; namely G F. In studying 2 factors in cubic graphs it is frequently more convenient to consider the structure of the corresponding 1 factor. Lemma 4.10 Let t 3 be an odd integer. If F is a 1 factor of A(t), then each of the t links of A(t) contain precisely one edge from F. Proof. Clearly, the 1 factor F must contain precisely one spoke from each interchange to cover the hubs. All other edges in F must pair vertices within the base cycles. Consequently, no link can contain three edges from F.

9 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 9 Suppose, by way of a contradiction, that the i th link L i does not contain precisely one edge from F. Then, L i contains either no edge from F or two edges from F. We shall assume first that the L i contains no edge from F. Thus, if h i x i is a spoke in F, y i y i 1,z i z i 1 F forcing h i 1 x i 1 F where {x,y,z} = {u,v,w}. Consequently, L i 1 contains precisely two edges from F and L i 2 contains no edge from F. Continuing in this way, we see that the links of A(t) must alternately contain precisely zero and precisely two edges from F. Since t is odd, this is not possible. It is clear that if we had started with precisely two edges from F in L i, the same contradiction would be reached. In the light of Lemma 4.10, we see that a 1 factor may be completely specified by the ordered t-tuple (a 1,a 2,...,a t ) where a i {u,v,w} for each i = 1,2,...,t and indicates which edge in the i th link L i belongs to F. Together these edges leave a unique spoke in each interchange to cover its hub. Note that a i a i+1,i = 1,2,...,t. To read off the corresponding 2 factor, simply start at a vertex in a base cycle at the first interchange. If the corresponding channel to the next interchange is not barred by F, proceed along this channel to the next interchange. If the channel is barred, proceed to the hub (this spoke cannot be in F) and then along the remaining unbarred spoke and continue along the now unbarred channel before you. Continue until reaching a vertex already encountered, so completing a cycle. If at this stage all vertices in the first interchange have been encountered, stop as all vertices in A(t) have been covered. Otherwise, choose an unused vertex in the first interchange and repeat the process to produce the next cycle in your 2 factor. Example 4.11 The 5 tuple (u,v,w,v,w) in A(5) corresponds to the 1 factor F = {u 1 u 2,v 2 v 3,w 3 w 4,v 4 v 5,w 5 w 1,h 1 v 1,h 2 w 2,h 3 u 3,h 4 u 4,h 5 u 5 }. From the 1 factor F, we obtain the 2 factor which is a hamiltonian cycle. (u 1 h 1 w 1 w 2 w 3 h 3 v 3 v 4 h 4 w 4 w 5 h 5 v 5 v 1 v 2 h 2 u 2 u 3 u 4 u 5 u 1 ), Lemma 4.12 Let F be a 1 factor of A(t) corresponding to the t tuple (a 1, a 2,...,a t ) with a i {u,v,w} for each i = 1,2,...,t. Suppose that, for some j : 1 j t, a j = a j+2. Then, A(t) F is a hamiltonian cycle in A(t) if and only if A(t 2) F is a hamiltonian cycle in A(t 2), where F is the 1 factor in A(t 2) corresponding to the (t 2) tuple (a 1,a 2,...,a j 1,a j,a j+3,a j+4,...,a k ). Proof. Without loss of generality, we may assume that F is based upon the the t tuple (u,v,u,a 4,...,a t ). Thus the 2 factor A(t) F passes between the first and second interchanges along the v and w channels. We shall follow these paths

10 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 10 through the second and third interchanges. First the v channel. This path segment is as follows: v 1 v 2 h 2 u 2 u 3 h 3 v 3 v 4. Following the w channel we have w 1 w 2 w 3 w 4. Thus the path segment through the v channel leaving the first interchange traverses the second and third interchanges before entering the fourth interchange along the v channel. The path segment through the w channel leaving the first interchange traverses the second and third interchanges before entering the fourth interchange along the w channel. The two path segments do not intersect and together cover all vertices in the second and third interchanges. Clearly, if A(t) F is a hamiltonian cycle, replacing each of the indicated segments of the hamiltonian cycle by the edges v 1 v 4 and w 1 w 4 respectively would give a hamiltonian cycle in A(t 2) F. The converse is clear. Lemma 4.12 suggests that the hamiltonian nature of 2 factors in A(t) will be closely related to the hamiltonian nature of 2 factors corresponding to t-tuples that cycle through u,v and w. To make the discussion of such 2 factors easier, we make the following definition. Definition 4.13 A 1 factor F, corresponding to the t-tuple (a 1,a 2,...,a t ), is called a cyclic 1 factor in A(t) if for each i = 1,2,...,t we have {a i,a i+1, a i+2 } = {u,v,w}. Lemma 4.14 Let t 3 be an odd integer and let F be a cyclic 1 factor in A(3t). Then A(3t) F is hamiltonian if and only if A(3t 6) F is hamiltonian, where F is a cyclic 1 factor of A(3t 6). Proof. The proof of Lemma 4.14 is very similar to that of Lemma We may assume that F corresponds to the 3t tuple (u,v,w,u,v,w,u,v,w,..., u,v,w). Following the path leaving the first interchange through the v channel we obtain the path segment v 1 v 2 h 2 u 2 u 3 u 4 h 4 w 4 w 5 w 6 h 6 v 6 v 7 v 8 joining us to the eighth interchange. Starting in the w channel as we leave the first interchange we follow the path segment w 1 w 2 w 3 h 3 v 3 v 4 v 5 h 5 u 5 u 6 u 7 h 7 w 7 w 8 joining us to the eighth interchange. Thus, if A(3t) F is a hamiltonian cycle, replacing the above segments by the edges v 1 v 8 and w 1 w 8 respectively yields a hamiltonian cycle A(3t 6) corresponding to the cyclic 1 factor F based on the (3t 6)-tuple (u,v,w,..., u,v,w). The converse is clear. Theorem 4.15 A(t) HU(3) for t odd and t 3. Proof. First consider the case when t = 3. Let e be any edge in a link, say u i u i+1. It can be easily checked that there are exactly two 1 factors containing u i u i+1, i.e. F 1 := {u i u i+1,h i v i,h i+1 w i+1,v i+1 v i+2,h i+2 u i+2,w i+2 w i } F 2 := {u i u i+1,h i w i,h i+1 v i+1,w i+1 w i+2,h i+2 u i+2,v i+2 v i }

11 M. Abreu, R.Aldred, M.Funk, B.Jackson, D.Labbate, J.Sheehan 11 To each of these 1 factors corresponds a 2 factor that turns out to be hamiltonian A(3) F 1 = H 1 = (h i u i u i+2 u i+1 h i+1 v i+1 v i v i+2 h i+2 w i+2 w i+1 w i ) A(3) F 2 = H 2 = (h i u i u i+2 u i+1 h i+1 w i+1 w i w i+2 h i+2 v i+2 v i+1 v i ). Since the choice of the link is arbitrary then A(3) HU(3). Let t 5 be the smallest odd integer such that A(t) / HU(3) and let F be a 1 factor in A(t) whose corresponding 2 factor is not hamiltonian. From Lemma 4.12, the t tuple(a 1,a 2,a 3,...,a t ) corresponding to F contains no pair a j = a j+2, 1 j t. Thus, for each j = 1,2,...,t: {a j,a j+1,a j+2 } = {u,v,w} and F is a cyclic 1 factor. Hence, t = 3t for some odd integer t 3. From our choice of t, A(t 6) HU(3) and thus, in particular, for F a cyclic 1 factor of A(t 6) the corresponding 2 factor A(t 6) F is hamiltonian. Hence, from Lemma 4.14, A(t) F is hamiltonian, contradicting our choices of t and F. The only cyclically 7 edge-connected graphs in HU(3) which we know of are K 4 and K 3,3. (Indeed, the only cyclically 4 edge-connected graphs in HU(3) with n 2 mod 4 vertices which we know of are K 3,3 and the Heawood graph.) Conjecture 4.16 There exists an integer m such that the only cyclically m connected graphs in HU(3) are K 4 and K 3,3. References [1] R. Aldred, M. Funk, B. Jackson, D. Labbate and J. Sheehan, Regular bipartite graphs with all 2 factor isomorphic, submitted. [2] A.A. Diwan. Disconnected 2 factors in planar cubic bridgeless graphs, J. Combin. Th. Ser. B, 84, (2002), [3] M. Funk, B. Jackson, D. Labbate and J. Sheehan, 2 factor hamiltonian graphs, J. of Combin. Th. Ser. B, 87, (2003), no.1, [4] L. Lovász and M.D. Plummer, Matching Theory., Ann. Discrete Math., 29, North Holland, Amsterdam, [5] C. Thomassen, Sign nonsingular matrices and even cycles in directed graphs, Linear Algebra Appl., 75 (1986),