Disjoint Hamiltonian Cycles in Bipartite Graphs

Disjoint Hamiltonian Cycles in Bipartite Graphs Michael Ferrara 1, Ronald Gould 1, Gerard Tansey 1 Thor Whalen Abstract Let G = (X, Y ) be a bipartite graph and define σ (G) = min{d(x) + d(y) : xy / E(G), x X, y Y }. Moon and Moser [6] showed that if G is a bipartite graph on n vertices such that σ (G) n + 1 then G is hamiltonian, extending a classical result of Ore [7]. Here we prove that if G is a bipartite graph on n vertices such that σ (G) n + k 1 then G contains k edge-disjoint hamiltonain cycles. This extends the result of Moon and Moser and a result of R. Faudree, et al. [3] Keywords: Graph, Degree Sum, Bipartite, Disjoint Hamiltonain Cycles. 1 Introduction and Terminology For any graph G, let V (G) and E(G) V (G) V (G) denote the sets of vertices and edges of G respectively. An edge between two vertices x and y in V (G) shall be denoted xy. Furthermore, let δ(g) denote the minimum degree of a vertex in G. A useful reference for any undefined terms is [1]. If G is bipartite with bipartition (X, Y ) we will write G = (X, Y ). If X = Y then we will say that G is balanced. A proper pair in G is a pair of non-adjacent vertices (x, y) with x in X and y in Y. A cycle in G is a sequence of distinct vertices v 1, v 1,..., v t such that v i v i+1 (1 i t 1) and v 1 v t are in E(G). We shall denote a cycle on t vertices by C t. A hamiltonian cycle in a graph G is a cycle that spans V (G) and, if such a cycle exists, G is said to be hamiltonian. Hamiltonian graphs and their properties have been widely studied. A good reference for recent developments and open problems is [5]. In general, we are interested in degree conditions that assure hamiltonian cycles in a graph. Perhaps the most classical condition of this type was put forth by Dirac []. Theorem 1.1 (Dirac 195). If G be a graph on n 3 vertices such that δ(g) n, then G is hamiltonian. 1 Department of Mathematics and Computer Science, Emory University, Atlanta, GA Metron Inc., Reston, VA 1

For an arbitrary graph G, define σ (G) to be the minimum degree sum of non-adjacent vertices in G. Of greater interest for our work here is the famed Ore condition for hamiltonicity which uses this parameter. Theorem 1. (Ore 1960). If G is a graph of order n 3 such that σ (G) n, then G is hamiltonian. The reader should note that Dirac s theorem is a corollary of Ore s theorem. In a bipartite graph G, we are interested instead in the parameter σ (G), defined to be the minimum degree sum of a proper pair. Moon and Moser [6] extended Ore s theorem to bipartite graphs as follows. Theorem 1.3 (Moon, Moser 1960). If G = (X, Y ) is a balanced bipartite graph on n vertices such that σ (G) n + 1, then G is hamiltonian. Faudree, Rousseau and Schelp [3] were able to give degree-sum conditions that assured the existence of many disjoint hamiltonian cycles in an arbitrary graph. Theorem 1.4 (Faudree, Rousseau, Schelp 1984). If G is a graph on n vertices such that σ (G) n + k then for n sufficiently large, G contains k edge-disjoint hamiltonian cycles. In this paper we will extend the previous two results by proving the following. Theorem 1.5. If G = (X, Y ) is a balanced bipartite graph of order n, with n 18k such σ (G) n + k 1, then G has k edge-disjoint hamiltonian cycles. Veneering Numbers and k-extendibility To prove our main theorem, we need some results on path-systems in bipartite graphs. Our strategy is to develop k systems of edge-disjoint paths and show that they can be extended to k edge-disjoint hamiltonian cycles. The following definitions and theorems can be found in [4]. Let W k (G) be the family of all k-sets {(w 1, z 1 ),..., (w k, z k )} of pairs of vertices of G where w 1,..., w k, z 1,..., z k are all distinct. Let S k (G) denote the collection of edge-disjoint path systems in G that have exactly k paths. If W W k (G) lists the end-points of a path system P in S k (G), we say that P is a W -linkage. A graph G is said to be k-linked if there is a W -linkage for every W W k (G). A graph G is said to be k-extendible if any W -linkage of maximal order is spanning. In order to tailor the idea of extendible path systems to bipartite graphs, we introduce the idea of a veneering path system. Definition 1. A path system P veneers a bipartite graph G if it covers all the vertices of one of the partite sets.

Given a W W k (G), we denote by W 1 the set of bipartite pairs of W, by W X those pairs of W that are in X, and by W Y those that are in Y. Also, with a slight abuse of notion, we will let W X (resp. W Y ) be the set of vertices of X (resp. Y ) that are used in the pairs of W. Definition. Let G be a bipartite graph and W W k (G). The veneering number ϑ X Y (W ) of W is defined to be ϑ X Y (W ) = (X Y ) ( W X W Y ), = (X Y ) W X W Y. For a given path system P, let (P) denote the set of pairs of endpoints of paths in P and let P denote P (P). We define the veneering number of such a P to be the veneering number of P. The veneering number of a given set of endpoints is of interest, because it represents the minimum possible number of vertices left uncovered by a path system with those endpoints. We can now reformulate the notion of a k-extendible graph. If P 1 and P are two path-systems of G, we write P 1 P when every path of P 1 is contained in a path of P. The following fact will prove most useful. Proposition.1. Let G = (X Y, E) be a bipartite graph and P 1, P S(G) be such that P 1 P. Let G 1 = (X 1 Y 1, E 1 ) = G P 1, G = (X Y, E ) = G P, then ϑ X 1 Y 1 (P 1 ) = ϑ X Y (P ). We are now ready to give our definition of a k-extendible bipartite graph. Definition 3. Let G be a bipartite graph. Then G is said to be k-extendible if for any path-system P in S k (G) there exists some veneering path system P in S k (G) that preserves the endpoints of P. We will utilize the following in the proof of our main theorem. Theorem.. If k and G = (X, Y ) is a bipartite graph of order n such that X, Y > 3k and σ n+3k (G), then G is k-extendible. It is important to note that a maximal path system with veneering number zero is spanning. Thus, if a graph G that meets the σ bound for k-extendibility has some path system P in S k (G) such that ϑ(p) = 0, then G must have a spanning path system. We give two more results from [4] that will be very useful. 3

Theorem.3. If G = (X Y, E) is a balanced bipartite graph of order n with σ (G) n + k then for any set W in W k (G) comprised entirely of proper pairs of G, there exists a system of k edge-disjoint paths whose endpoints are exactly the pairs in W. Theorem.4. If G = (X Y, E) is a balanced bipartite graph of order n such that for any x X and any y Y, d(x) + d(y) n +, then for any pair (x, y) of vertices of G, there exists a Hamilton path between x and y. The degree sum condition is the best possible. 3 Main Theorem Before we begin the proof of Theorem 1.5, we need to establish a number of facts. Suppose the theorem is not true, and let G be a counterexample of order n with a maximum number of edges. The maximality of G implies that for any proper pair (x, y), G+xy contains k edge-disjoint Hamilton cycles, one of these containing the edge xy. Thus, to a proper pair (x, y) we will associate k 1 edge-disjoint Hamilton cycles H 1,..., H k 1 and an (x, y)-hamilton path P = (x = z 1, z,, z n = y). Let H denote the union of subgraphs H 1,..., H k 1, and L = L(x, y) denote the subgraph obtained from G by removing the edges of H. Before we go on proving our theorem we will state a few facts about L. Throughout these proofs, we must keep in mind that n 18k, (1) and for any vertex w of G, we have d L (w) = d G (w) (k 1), () so the degree sum condition on any proper pair (x, y) of G d G (x) + d G (y) n + k 1. (3) This yields the following: Fact 1. For any proper pair (x, y) of G, we have d L (x) + d L (y) n k + 3. (4) Fact. If there is a proper pair (x, y) of G, with v d G (x) + d G (y) n + 4k 3, or equivalently d L (x) + d L (y) n + 1, then L contains a Hamilton cycle. 4

If there were a proper pair (x, y) of G such that d G (x) + d G (y) n + 4k 3, then by (), d L (x) + d L (y) n + 1, hence if we consider the (x, y)-path P in L, we see that there must be a vertex z V (P ) such that z N(y) and z + N(x). Then xz + [z +, y] P yz [x, z] is a hamiltonian cycle in L. Note that the existence of P shows that L is connected In fact, L must be -connected. Lemma 3.1. If L is not -connected, then there are k edge-disjoint hamiltonian cycles in G. Proof: Suppose w is a cut-vertex of L; we assume, without loss of generality, that w X. Since L admits a h1amiltonian path, L w can only have two components, one of them being balanced. Let B be the subgraph of G induced by the balanced component of L w and A = G B. Note that w A, and E L (A X w, B) = E L (A Y, B) =. Let a = A X = A Y and b = B X = B Y. Claim 1. a, b > n k. Proof: Assume a n k. Then a(k ) + a < ak n, implying a(k ) < n a = b, so d H (A Y, B X ) < B X = b. Thus there is a vertex u B X such that E H (u, A Y ) =, so E G (u, A Y ) =. Take any v A Y. We have uv / E(G), so d(u) + d(v) A X + d H (v, B X ) + B Y + d H (u, A Y ) a + (k 1) + b < n + k 1, which contradicts the condition of our theorem. Claim 1 The two following claims give lower bounds on the degrees of the vertices in L. Claim. For any z A w, d L (z) A k+3 and for any z B, d L(z) B k+3. Proof: Assume z B Y (the cases z B X, z A X, z A Y are similar). By Claim 1 and the fact that n 18k, we have A X w = a 1 > n k 1 > (k 1), so there is a z A X w such that zz / E(H), thus zz / E(G), so that d L (z) + d L (z ) n k + 3. Then since d L (z ) a, we get d L (z) n k + 3 a = b k + 3. Claim Claim 3. d L (w) n k k + 3. Proof: If w is adjacent, in G, to all the vertices of A Y, then the theorem is obviously true. If not, there is a v A Y with wv / E(G), so that d L (w) + d L (v) n k + 3. Since d L (v) a = n b < n n k, we get d L (w) n k + 3 d L (v) > n k + 3 (n n k ) = n k + 3. k 5

Claim 3 Finally: Claim 4. E G (A X, B Y ), E G (A Y, B X ) k 1. Proof: If G[(A X, B Y )] is complete, the result is obvious. If not, there is a pair of nonadjacent vertices u A X and v B Y, so d(u) + d(v) n + k 1. Yet d(u, A Y ) a and d(v, B X ) b, so d(u, B Y ) + d(v, A X ) n + k 1 a b = k 1. The proof is identical for (A Y, B X ). Claim 4 By Claim, Claim 3, and the fact that n 18k we have, for any pair of vertices (u, v) V (A X ) V (A Y ) d A (u) + d A (v) A k + 3 + n k k + 3 > A + k. Thus, A, and by a similar computation B, satisfies the conditions of Theorem.4. Hence take k pairs (e i, e i ) of edges such that the e i are distinct edges of E G (A X, B Y ) and the e i are distinct edges of E G (A Y, B X ). These edges exist by Claim 4. Let u i A X and v i B Y be the end vertices of e i, and u i A X and v i B Y be the end vertices of e i. Since A and B both satisfy the conditions of Theorem.4, there are k edge-disjoint hamiltonian paths U 1,, U k in A such that u i and u i are the end-vertices of U i, and there are k edge-disjoint hamiltonian paths V 1,, V k in B such that v i and v i are the end-vertices of V i. Together with the e i and e i edges we get k edges disjoint hamiltonian cycles in G, which contradicts the assumption. Hence the lemma is proven. Now we show that the -connectedness of L assures the existence of a large cycle. Lemma 3.. If L is -connected, then it contains a cycle of order at least n 4k + 4. Proof: Let x 1 X and y n Y be two non-adjacent vertices of G and let P = x 1 y 1 x n y n be a hamiltonian path in L. This path induces an obvious ordering of the vertices; namely, we say that z < z if one encounters z before z when traversing P from x 1 to y n. We say that a vertex z is the minimum (maximum) vertex with respect to a given property if z < z (z < z) for all other vertices z satisfying that property. By the -connectivity of L, N(x 1 ) and N(y n ) are non-empty. Let x be the minimum vertex of N(y n ) and y be the maximum vertex of N(x 1 ). Note that if x < y, then the path along C between x and y has length at least 4k 1, since otherwise the xy-path must have length at most 4k 3, and C = x 1 y [y, y n ] y n x [x, x 1 ] would form a cycle of length at least n 4k + 4. But then d L (y n ) n 1 d L (x 1 ) k, 6

which gives that d L (x 1 )+d L (y n ) n k 1, a contradiction of the degree-sum condition. So y < x. We construct a cycle C using the following algorithm: Set z 0 = x 1 and z 3 = y Set i = 0 While z i+3 / [x +, y n ] do Set i = i + 1 Let z i and z i+3 be vertices of P such that z i < z i+1 < z i+3, z i z i+3 E(L), and z i+3 is maximum (note that the existence of this edge is ensured by the -connectivity of L) End While Set l = i + Set z 1 = x 1 and replace z 3 with the minimum vertex of N(x 1 ) [z +, y] Set z l = y n and z l be the maximum vertex of N(y n ) [x, z l 1 ] Set C = z 1 z 3 z l z l l i=1 z iz i+3 l i=1 [z i 1, z i ] Note that since y < x, the while loop is performed at least once, and l 3. We now show that C n 4k + 4. Let F = {x i y i (x i / N(y n )) (y i / N(x 1 ))}. Note that there cannot be any i with x i N(y n ) and y i N(x 1 ), since then [x 1, x i ] [y i, y n ] x 1 y i x i y n would be a hamiltonian cycle. This implies that F + {x i y i : y i N(x 1 )} + {x i y i : x i N(y n )} = F + d L (x 1 ) + d L (y n ) = n By Fact 1 we get that F k 3 (5) We show that all but perhaps vertices of V (G C ) are in V (F ). All the vertices of [z +, z 3 ], except z 3, are in V (F ) by the minimality of z 3. Similarly, by the maximality of z l, the only vertex that is in V (F ) [z + l, z l 1 ] is z+ l. Finally, by the maximality of y and the minimality of x we get V ( l 1 i= [z+ i, z i+1 ]) V ([y+, x ]) V (F ), hence (L V (F )) V (L C ). By (5) we get L C F + 4k 4, hence C n 4k + 4. 7

3.1 Path Systems In order to prove an important technical lemma, we must first establish some facts about extending paths and path systems. Lemma 3.3. Let G = (X Y, E) be a bipartite graph, and let P be a path system of G. Let X be a subset of ( P) X, and let Y be a subset of Y ( P) Y. Suppose that X = s + t, where s is the number of vertices in X arising from paths of P consisting of a single vertex. If δ(x, Y ) > t + s then there exists another path system, P, of G such that P P and ( P ) X =. Proof:We will first show that s may be assumed to be 0. If s > 0, let P 1, P,..., P s be the trivial paths contained in X. Now, for every i [s], replace P i = {x i } with a path P i on three vertices such that the endvertices of P i are new vertices added to X and the middle vertex of P i is a new vertex added to Y Y. In addition, let the endvertices of P i be adjacent to the neighbors of x i. Let P 1 be the new path system, and let X 1, consisting of X and the vertices added to X, be the new set of endvertices we wish to eliminate. The new system P 1 now contains no trivial paths, and X 1 = t + s. Thus, if our lemma were true for systems with no trivial paths, then the condition δ(x, Y ) > t + s = t + s ensures the existence of a path system P 1 such that P 1 P 1 and ( P 1 ) X 1 =. By replacing every P i by P i within the appropriate paths of P 1, we obtain the desired path system of G. So assume that X = {x 1,..., x t }. Note that the result clearly holds if t = 1, so assume that t. We exhibit an algorithm that produces a sequence of path systems P = P(0) P(1) P(t) = P where P(i) is obtained from P(i 1) by adding an edge from x i to Y, and P is the desired path system. Since this algorithm will attach an edge to every vertex of X, P will have no endvertices in X. Given a vertex x i X, let P i P(i 1) be the path having x i as an endvertex, and let w i be the other endvertex of P i. Let Z i be the set of internal vertices of the paths of P(i 1) in Y. The algorithm consists of the following steps: 8

1) Set P(0) = P ) For i = 1 to t do 3) Let y be any vertex of N(x i, Y ) w i Z i 4) Set P(i) to be P(i 1) with the path P i replaced with P i x i y 5) end of ) loop Clearly this algorithm will terminate if N(x i, Y ) w i Z i is never empty when the vertex y must be chosen in step 3), and if the algorithm terminates, we will have the desired path system. Note that since no path in P(0) had internal vertices in Y, the only way that a vertex a Y can be an internal vertex of a path of P(i) is if the algorithm selected a in step 3) on two passes through the algorithm. Thus which implies that Z i max{0, i } t, N(x i, Y ) w i Z i d(x, Y ) 1 Z i > t 1 t = 0. The following corollary is obtained from Lemma 3.3 by induction on k: Corollary 3.4. Let G = (X Y, E) be a bipartite graph, let P 1,..., P k be k edge-disjoint path systems, and let Y Y k i=1 int(p i) Y. For all i [k] let X i ( P i ) X and X i = s i + t i, where s i is the number of vertices of X i arising from paths of P i consisting of a single vertex. If for all i [k], δ(x i, Y ) > t i + s i + (k 1) then there exist k edge-disjoint path systems P 1,..., P k such that for all i [k], P i P i and ( P i ) X i =. 3. The Degree-Product Lemma Interestingly, the proof of Theorem 1.5 relies on a result pertaining to degree products as opposed to degree sums. We feel it would be interesting to investigate similar results. Lemma 3.5. If G has no proper pair (u, v) such that d L (u)d L (v) 1k(n 1k) then G has k edge-disjoint hamiltonian cycles. Proof: Suppose G has no such vertices. Let A be the subgraph of G generated by the vertices of degree less than 16k, and B the subgraph generated by the vertices of degree greater or equal to 16k. By (3) and (1) no bipartite pairs (u, v) of A are proper. Further, 9

no bipartite pairs (u, v) of B can be proper or else, by (), (1), Fact 1 and convexity, we would have d L (u)d L (v) (14k + )(n 16k + 1) 1k(n 1k). Thus A and B induce complete bipartite graphs. Assume without loss of generality, that A X A Y, and set λ = A X A Y = B Y B X. We can assume λ < 4k 3 since otherwise we could find a proper non-adjacent pair (x, y) V (B X ) V (A Y ) with d G (x) + d G (y) B Y + λ + A X + λ = n + λ n + 4k 3, and Fact would imply a Hamilton cycle in L, hence k edge-disjoint Hamilton cycles in G. Claim 5. We have δ(a X, B Y ) λ + k 1 and δ(a Y, B X ) k 1 λ Let x A X such that d(x, B Y ) = δ(a X, B Y ). By (3), every vertex y B Y N(x, B Y ) must verify so d G (y) n + k 1 d G (x) = n + k 1 A Y d(x, B Y ) = B Y + k 1 δ(a X, B Y ), d G (y, A X ) B Y + k 1 δ(a X, B Y ) B X = λ + k 1 δ(a X, B Y ) This implies that d G (A X x, B Y N(x, B Y )) ( B Y δ(a X, B Y ))(λ + k 1 δ(a X, B Y )) (6) yet, since the vertices of A X can be adjacent to no more than λ + 4k 1 vertices of B Y (by Fact ), we see that d G (A X x, B Y N(x, B Y )) ( A X 1)(λ + 4k 1). (7) Thus if λ + k 1 δ(a X, B Y ) > 0, (6) and (7) imply B Y ( A X 1)(λ + 4k 1) λ + k 1 δ(a X, B Y ) + δ(a X, B Y ) (16k)(8k 4) + k which contradicts the fact that n 18k, hence δ(a X, B Y ) λ + k 1. The proof of δ(a Y, B X ) k 1 λ is similar. Claim5 We distinguish two cases, according to the size of A X : 10

Case 1: Suppose 1 A Y k 1. Then Claim 5 and the completeness of A imply δ(a Y ) A X + k 1 λ = A Y + k 1 > A Y + (k 1). Now, we apply Corollary 3.4 with P i = X i = A Y for all i, and let Y = X. This implies, in the language of the corollary, that δ(x i, Y ) = δ(a Y ). Thus, we find that there are k edge-disjoint systems P 1,..., P k whose paths have all order three and whose endvertices are all in X. Further, since A is a complete bipartite graph, we may choose these path-systems so that they cover a subset A X of min( A X, A Y ) vertices of A X. That is to say, if A X A Y, A X = A X, so these systems each cover A entirely, and if A X > A Y, we require that they each cover the same proper subset A X of A X having order A Y. For all i [k] we let P i = P i when A X A Y, and P i = P i (A X A X ) when A X > A Y. In either case, we now have k edge-disjoint path systems which cover A. Again we wish to apply Corollary 3.4 to the P i with X i = ( P i ) A X, to extend to a family of k edge-disjoint systems P 1,..., P k such that every path in each of these systems has both endvertices in B. We may do so since if A X A Y then all t i = A X vertices of X i come from non-trivial paths, and if A X > A Y then t i = A Y vertices of X i also come from non-trivial paths, and s i = A X A Y of them come from paths consisting of exactly one vertex, so by Claim 5, d(a X, B Y ) λ + k 1 > t i + s i + (k 1). Consider some matching M 1 that contains exactly one edge from each non-empty path in P 1. Clearly, ϑx Y (M 1) = 0, and therefore by Proposition.1 we have that ϑ( (P 1)) = 0 (8) in G P 1. Thus, as (P 1 ) B, and B induces a complete bipartite graph, we can link the endpoints of the paths in P 1 to form a Hamiltonian cycle in G. Suppose then that we have extended P 1,..., P t 1 (t k) to the disjoint Hamiltonian cycles H 1,..., H t 1. As above, Proposition.1 implies that ϑ( (P t)) = 0 (9) in G P t. Assume that P t has exactly j paths, and let {x 1, y 1 },..., {x j, y j } denote the pairs of endpoints of these paths. Additionally, let the set W = {{y 1, x }, {y, x 3 },..., {y j, x 1 }}. As B induces a complete bipartite graph with each partite set having size at least n 11

A Y λ n 6k, it is simple to see that there is a W -linkage in G t := G P t t 1 i=1 E(H i). Note that there are at most j A Y < k paths in P t, so if we are able to show that G t is k-extendible we will be done. By Corollary., it suffices to show that σ (G t ) > V (G t) + 6k n k n k. (10) In G, the minimum degree of a vertex in the subgraph induced by B is n ( A Y + λ) n 6k. In removing the edges from the t 1 other hamiltonian cycles, each vertex loses t < k adjacencies. Thus, it is clear that σ (G t) certainly exceeds n k, completing this case. Case : Suppose A Y k. Let A X be a subset of A Y vertices of A X. As A is a complete bipartite graph, there are k edge-disjoint hamiltonian cycles in (A X A Y ) G, and we let x 1 y 1,, x k y k be independent edges of (A X A Y ) G such that x i y i is an edge of the i th hamiltonian cycle. Using Claim 5 we get that δ(a X, B Y ) k 1 and δ(a Y, B X ) k 1 λ so Let B = G A X A Y. We have δ(a Y, B X) A X A X + δ(a Y, B X ) k 1. σ (B ) δ(a X A X, B Y ) + B X B X + λ + k 1 = B + k 1 One may then use the edges of E(A X, B Y ) and E(A Y, X A X ) along with Theorem.3 to find k edge-disjoint hamiltonian cycles in G. Before we proceed to prove the main theorem, we give one final technical lemma. Lemma 3.6. Let G be a graph containing a hamiltonian cycle C and let S and R be nonempty disjoint subsets of V (G). If R + S E(R, S) then there are four distinct vertices c 1, c, c 3, c 4, encountered in that order on C, such that either (a) c 1, c 3 R, c, c 4 S, c 1 c E(G), and c 3 c 4 E(G), or (b) c 1, c 4 S, c, c 3 R, c 1 c 3 E(G), and c c 4 E(G). (c) c 1, c 4 R, c, c 3 S, c 1 c 3 E(G), and c c 4 E(G). Proof: First, note that if R = {r R : d(r, S) = 0} and S = {s S : d(s, R) = 0}, then R + S R + S E(R, S) = E(R, S ) 1

so we may assume that every vertex of R is adjacent to at least one vertex of S, and vica versa. Further, observe that the inequality in the statement of the lemma cannot hold if R = 1 or S = 1, so R, S. Let R = {u 1,..., u r }, where the labels of the vertices of R are determined by a chosen orientation of C. Let C and C be the two [u 1, u r ]-paths of C, C is the path containing the u i for 1 i r. In order to avoid (a), (b), and (c), N(R) S must be entirely contained in C and for every 1 i < j r, N(u i, S) N(u j, S) must be empty if i j +, and can have at most one element if i = j + 1. Thus, to avoid (a), (b) and (c), S must contain at least E(R, S) R + 1 E(R, S) R + 1 vertices. 3.3 Proof of Theorem 1.5 Proof: Let C be a cycle of L of maximal order which minimizes d L (T, C), where T = L C. By Lemma 3. t = T k (11) Let u T X and v T Y such that d L (u, C) + d L (v, C) is maximal. Let α = d L (u, C) and β = d L (v, C). We assume, without loss of generality, that α β. We may assume that α k + 4. (1) Indeed, by Fact 1, every vertex of V (Y ) N(u) has degree greater or equal to n k + 3 t α. If α k+3, this would yield that there are at least n t (k+3) (k 1) n 6k vertices that have degree at least n k + 3 t (k + 3) n 6k, implying that G is hamiltonian by Lemma 3.5. Note that or else C could be extended. α + β n t + 1 n k + 3 We must have N L (u, C) + N L (v, C) 1 and N L (u, C) N L (v, C) + 1. Let R = N L (v, C) + N G (u, C). Then R d L (v, C) d H (u, C) N L (u, C) N L (v, C) + β (k 1) 1 = β k + 1 (13) For every r R ru / E(G), so by Fact 1, d L (r) + d L (u) = d L (r, T ) + d L (r, C) + d L (u) n k + 3, 13

hence d L (r, C) n k + 3 d L (u, C) d L (u, T ) d L (r, T ) (14) n k + 3 α t t Together with the fact that r R d L(r, T ) t 1 (since otherwise, we could extend C), we get d L (R, C) = r R d L (r, C) r R ( n k + 3 dl (u, C) d L (u, T ) d L (r, T ) ) (15) = R (n k + 3) R (α + t) r R d L (r, T ) (16) R (n k + 3 α t) t + 1. (17) Let S = N L (u, C). We have d L (R, S) d L (R, C) C X S R (n k + 3 α t) t + 1 (n t) + S = R (n k + 3 α t) + S + 1 n Lemma 3.6 with R = R and S = S + gives S ( d L (R, S) R + 1 ) 0 S ( ( R (n k + 3 α t) + S + 1 n) R + 1 ) 0 n R (n k + α t) 0 (18) By (13) and (1), we have R α k + 1 3, so (18) yields n 3(n k + α t) 0 (19) 3α n k + 9 (0) α 3 n k 3t + 3. (1) 3 Yet, as α β, t k 1, and n 18k 46k), this would imply α + β 4 3 n 4 k 6(k 1) + 6 > n + k 3 contradicting (3.3). T heorem 1.5 14

References [1] G. Chartrand, L. Lesniak, Graphs and Digraphs 3 rd Edition, CRC Press, 1996. [] G.A. Dirac, Some Theorems on Abstract Graphs, Proc. London Math. Soc., (195), 69-81. [3] R. Faudree, C. Rousseau and R. Schelp, Edge-Disjoint Hamiltonian Cycles, Graph Theory with Applications to Algorithms and COmputer Science, 1984, 31-49. [4] R. Gould and T. Whalen, Connectivity, Linkage and Hamilton Path-Systems in Bipartite Graphs.(preprint) [5] R. Gould, Advances on the Hamiltonian Problem, A Survey, Graphs and Combinatorics, 19 (003), 7-5. [6] J. Moon and L. Moser, On Hamiltonian Bipartite Graphs, Israel J. Math, 1 (1963), 163-65. MR 8 # 4540 [7] O. Ore, A Note on Hamilton Circuits., Amer. Math. Monthly, 67 (1960), 55. 15