Ore and Erd ˝os type conditions for long cycles in balanced bipartite graphs

We conjecture Ore and Erd˝os type criteria for a balanced bipartite graph of order 2 n to contain a long cycle C 2 n − 2 k , where 0 ≤ k < n/ 2 . For k = 0 , these are the classical hamiltonicity criteria of Moon and Moser. The main two results of the paper assert that our conjectures hold for k = 1 as well.


Introduction
One of the classical problems of graph theory is the study of sufficient conditions for a graph to contain a Hamilton cycle.In this paper we are primarily interested in two types of such conditions.Namely, the ones that put constraints on degree sums of pairs of non-adjacent vertices, and those that combine bounds on the size of a graph with bounds on its minimal degree.The first approach is due to Ore (see Section 2 for notation): Theorem 1.1 (Ore, [12]).Let G be a graph of order n ≥ 3, in which for every pair of non-adjacent vertices x and y.Then G contains a Hamilton cycle.
It follows immediately from Ore's theorem that the minimal size of a graph of order n ≥ 3 that guarantees hamiltonicity is n−1 2 + 2. Erdős generalized this condition by adding a bound on the minimal degree of a graph: Theorem 1.2 (Erdős, [9]).Let G be a graph of order n ≥ 3 and minimal degree δ(G) ≥ r, where 1 ≤ r < n/2.Then G contains a Hamilton cycle, provided The above conditions can, of course, be significantly strengthened in case of a balanced bipartite graph.The following two theorems are bipartite counterparts of Ore and Erdős criteria, respectively.Theorem 1.3 (Moon and Moser, [11]).Let G be a bipartite graph of order 2n, with colour classes X and Y , where |X| = |Y | = n ≥ 2. Suppose that d G (x) + d G (y) ≥ n + 1 for every pair of non-adjacent vertices x ∈ X and y ∈ Y .Then G contains a Hamilton cycle.
Our goal is to generalize the above criteria to long cycles, that is, cycles of length 2n − 2k, where 0 ≤ k < n/2.We state the following two conjectures, that include Theorems 1.3 and 1.4 as special cases (k = 0).
The main two results of this paper, Theorems A and B (Section 3), assert that our conjectures hold true for k = 1.We believe the conjectures to be significantly harder in case k ≥ 2.
It should be mentioned here that analogous generalizations to long cycles of Ore's and Erdős's theorems have been studied in ordinary graphs.Woodall [14,Thm. 11] gives a complete list of Erdős type conditions for a graph of order n to contain a cycle of length n − k for any 0 ≤ k ≤ n−3 2 .The Ore type criterion is conjectured in [1], and follows from a result of Linial [10] in case k ≤ 1.
Example 1.7.Let G 2 = (X, Y ; E) be a balanced bipartite graph obtained from the disjoint union of H 1 = K ⌊n/2⌋,⌊n/2⌋ and H 2 = K ⌈n/2⌉,⌈n/2⌉ by adding a single edge joining a vertex of H 1 with a vertex of H 2 .Then d G2 (x) + d G2 (y) ≥ n for every pair of non-adjacent vertices x ∈ X and y ∈ Y , nonetheless G 2 contains no cycle of length 2n − 2. In fact, G 2 contains no long cycle whatsoever.
The next section contains the inventory of basic definitions and results used throughout the paper.In Section 3 we state our main results, Theorems A and B, and their consequences.In particular, by combining Theorems A and B, we obtain a complete Erdős type characterisation of balanced bipartite graphs that do not contain cycles of length 2n − 2 (Theorem 3.6).The last two sections are devoted to proofs of the two main results.

Notation and tools
All graphs considered are undirected, have no loops and no multiple edges.Given a graph G, we denote by G the size (i.e., number of edges) of G, and by V (G) the vertex set of G.A bipartite graph is often denoted by G = (X, Y ; E), where X and Y are the two colour classes of G, and and N G (L) is the set of neighbours of all the vertices in L. Given distinct vertices x and y of G, an x − y path is a path in G with endvertices x and y.We denote by C l a cycle of length l, and by K n,n a complete balanced bipartite graph of order 2n.Finally, recall that a graph is called 2-connected if the removal of any single vertex does not disconnect G.
In this section we have gathered results used in the proofs of Theorems A and B. First of all, we recall two hamiltonicity criteria obtained by Moon and Moser [11].
Theorem 2.1 (Moon and Moser, [11]).Let G be a balanced bipartite graph of order 2n ≥ 4, with . Then G contains a Hamilton cycle.
Theorem 2.2 (Moon and Moser, [11]).Let G = (X, Y ; E) be a balanced bipartite graph of order 2n, , the sets S m and T m are of cardinalities less than m, then G is hamiltonian.
We shall need the following strengthening of Theorem 1.4.
Theorem 2.3 (Wojda and Woźniak, [13]).Let G(n, r) denote a bipartite graph with colour classes A bipartite graph of order 2n is called bipancyclic if it contains cycles of lengths 2k for all 2 ≤ k ≤ n.
for every pair of non-adjacent vertices x ∈ X and y ∈ Y , then G is bipancyclic.
We will also need to know the cycle structure of an n/2-regular hamiltonian bipartite graph G of order 2n.Notice that then G = n 2 /2, so the above theorem does not apply.We then have: Theorem 2.6 (J.Adamus, [2]).Let G be an n/2-regular hamiltonian bipartite graph of order 2n.Then G contains a cycle C of length 2n − 2.Moreover, if C can be chosen to omit a pair of adjacent vertices, then G is bipancyclic.
Given a balanced bipartite graph G = (X, Y ; E), one defines a k-biclosure BCl k (G) of G as the graph obtained from G by succesively joining pairs of non-adjacent vertices x ∈ X and y ∈ Y , with degree sum of at least k, until no such pair remains.Closely related to this construction is the notion of k-bistability: A property P defined on all balanced bipartite graphs of order 2n is called k-bistable when, whenever G + xy has the property P and d G (x) + d G (y) ≥ k, then G itself has the property P.

Long cycles in balanced bipartite graphs
Suppose we want to know whether a balanced bipartite graph G = (X, Y ; E) has the property of containing a long cycle C 2n−2k for some 0 ≤ k < n/2.Given Theorem 1.3 of Moon and Moser, a natural question arises: Can one impose such a property by decreasing the bound on the degree sum of nonadjacent vertices by k?We believe the answer to this question be positive (Conjecture A).As shown in Example 1.6, any lower bound on the degree sum of non-adjacent vertices x ∈ X and y ∈ Y which ensures C 2n−2k ⊂ G is at least n − k + 1.On the other hand, decreasing the bound below n + 1 imposes additional assumptions on the graph.Interestingly enough, without the 2-connectedness constraint the graph could contain no long cycles at all (see Example 1.7).The following result gives a positive answer to the above question in case k = 1.
≥ n for every pair of non-adjacent vertices x ∈ X and y ∈ Y .Then G contains an even cycle of length at least 2n − 2.
We postpone the proof of the theorem to Section 4. Right now we will show that Theorem A implies Conjecture A for k = 1.Proof: Let G = (X, Y ; E) be a balanced bipartite graph of order 2n that satisfies the assumptions of Conjecture A. By Theorem A above, G contains an even cycle of length at least 2n − 2, so without loss of generality one may assume that G is hamiltonian.
Let x ∈ X, say, be a vertex of minimal degree δ(G) in G. Then Y contains precisely n − δ(G) vertices non-adjacent to x, each of degree at least n − δ(G) (as Counting the edges incident with Y , we get and thus G contains C 2n−2 , by Theorem 2.5.If, in turn, δ(G) = n/2, then the result follows from Theorem 2.6.
Let us now turn to Erdős type criteria.In [3], the second author conjectured the following sufficient condition for a balanced bipartite graph to contain a long cycle C 2n−2k (proved in [3] under considerably stronger assumptions).Conjecture 3.2 (L.Adamus, [3]).Let G be a balanced bipartite graph of order 2n, where Notice that both assumptions of the conjecture are weakest possible, as shown by the following two examples.Interestingly, a similar graph was recently shown in [6] to be a counterexample to Győri's conjecture on C 2l -free bipartite graphs.
In light of Example 3.3 above, we ask: By how much can we decrease the lower bound on the size of a given graph G ensuring the existence of a cycle of length 2m − 2k, knowing that the minimal degree of G is greater than 1?We address this question in Conjecture B. Certain special cases of Conjecture B are known true: k = 0 is Theorem 1.4, k = r = 1 is done in [3].The following theorem (proved in Section 5 below) shows that the conjecture also holds for k = 1 and arbitrary r.
Theorem B. Let G = (X, Y ; E) be a balanced bipartite graph of order 2n and minimal degree δ(G) ≥ r ≥ 1, where n ≥ 4 and n ≥ 2r + 1.Let Then G contains a cycle of length 2n − 2, provided G ≥ g(n, r).
Notice that Theorems 2.1 and 1.4 can be put together as follows: Proof of Theorem 3.6: (1) Since n ≤ 2r − 1 iff r ≥ (n + 1)/2, then the degree sum is greater than or equal to n + 1 for every pair of vertices in G (in particular, for non-adjacent ones).By Theorem 2.4, G is then bipancyclic.
(2) The bound on the size of G together with δ(G) ≥ r = n/2 force 2-connectedness.Also, the degree sum is at least 2r = n for every pair of vertices in G. Hence, by Corollary 3.1, G contains C 2n−2 .
(3) This is Theorem B. We shall show first that G contains a Hamilton path.Suppose not.Let x ∈ X, y ∈ Y be non-adjacent vertices and let P be an x − y path in G of length 2n − 3; say,

Proof of Theorem A
where G[V (P )] denotes the subgraph of G induced by the vertex set of P .This shows that at least one of the vertices u 1 and v n−1 has a neighbour among the remaining vertices u n , v n of G−P ; say, v n−1 u n ∈ E. Notice that then u n v n / ∈ E, for otherwise u 1 . . .v n−1 u n v n u 1 would be a Hamilton path.Similarly, u 1 v n / ∈ E. Hence, in particular, I P contains indices of all the neighbours of u 1 in G, so Let now x ∈ X and y ∈ Y be a pair of non-adjacent vertices such that G contains a Hamilton x − y path P ; say, so that, for every 1 ≤ i ≤ n, Denote by x 1 , . . ., x d those of the vertices u 1 , . . ., u n that are adjacent to y, ordered according to the orientation of P (from x to y).Let y 1 , . . ., y d be the vertices of Y that lie on P next to the respective x 1 , . . ., x d ; then y d = y.
Observe that if Suppose now that, for every v ∈ V , N G (v) ⊂ U .Then, for all u ∈ X \ U and v ∈ V , u and v are non-adjacent, hence Thus G contains a complete bipartite graph K d,d spanned on the vertices of U and V , and a complete bipartite Now, G being 2-connected, it must contain two independent edges u i1 v j1 and u i2 v j2 for some i 1 , i 2 ≥ n − d + 1 and j 1 , j 2 ≤ n − d.One immediately verifies that such a graph contains a cycle of length 2n − 2, again contradicting the choice of G.
We can therefore conclude that there exists a vertex v j , with This contradiction completes the proof of the theorem.

Proof of Theorem B
Throughout this section we will frequently refer to the exceptional graph G(n, r) of Theorem 2.3.Recall that by G(n, r) we denote a balanced bipartite graph of order 2n, with colour classes Let, as before, g(n, r) = n(n − 1 − r) + r(1 + r) + 1.We shall first show the following lemma.Proof: Suppose G contains no cycle of length 2n − 2. Then G − {x, y} contains no such cycle either, and as δ(G − {x, y}) ≥ r, Theorem 2.3 implies that On the other hand, Hence d G (x) + d G (y) = n, the vertices x and y are non-adjacent, G − {x, y} equals G(n − 1, r), and r ≤ (n − 1)/2.Without loss of generality, we may assume that x belongs to the colour class of In the first case, x must have at least two neighbours in S or else at least one neighbour in both S and R. One easily verifies that then G contains a cycle of length 2n − 2, omitting y and a single vertex of P ; a contradiction.
If, in turn, d G (x) = r, then d G (y) = n − r and y must have neighbours in both P and Q, since r ≤ (n − 1)/2 < n/2.Consequently, G contains a cycle of length 2n − 2, omitting x and a vertex of S, which again contradicts the choice of G.
We are now in position to prove Theorem B.
For a proof by contradiction, consider a graph G satisfying the assumptions of Theorem B, that does not contain a cycle of length 2n − 2. Observe first that G > n 2 /2.Indeed, the difference g(n, r) − n 2 /2 is always positive.Hence, by Theorem 2.5, G is not hamiltonian.Consequently, Theorem 2.2 implies that there exists a positive integer m ≤ n/2 such that at least one of the sets S m = {x ∈ X : d G (x) ≤ m}, T m = {y ∈ Y : d G (y) ≤ m} has cardinality greater than or equal to m.
Let l be the least such m.Without loss of generality, we may assume that l is realized in X; i.e., |{x ∈ X : Then, by minimality of l, we have l = min{i : The rest of the proof proceeds in two cases, depending on l being equal to or greater than r.
Case 1: l = r.We will first show that all the vertices of Y have degrees greater than r.Suppose to the contrary that there exists y 1 ∈ Y with d G (y 1 ) = r.Then and δ(G − {x 1 , y 1 }) ≥ r − 1.On the other hand, by Theorem 2.3, Hence d G (x 1 )+d G (y 1 ) = 2r so that x 1 y 1 / ∈ E and G−{x 1 , y 1 } equals G(n−1, r−1).By comparison of degrees, one readily verifies that x 1 belongs to that colour class of G that contains P ∪Q of G(n−1, r−1); in fact, L = {x 1 } ∪ P .Consider the sets R and S of the other colour class of G(n − 1, r − 1).As |N G (x 1 )| = r > |R| and x 1 y 1 / ∈ E, it follows that either x 1 has neighbours in both R and S or else it has at least two neighbours in S. In any case, as in the proof of Lemma 5.1, one easily finds a cycle of length 2n − 2 in G, omitting y 1 and a vertex of P ; a contradiction.Thus d G (y) ≥ r + 1 for every y ∈ Y .
Next observe that every vertex of Y has a neighbour in L. Suppose otherwise, and let y 1 ∈ Y be such that N G (y 1 ) ⊂ X \ L. Notice that all vertices of X \ L have degrees greater than r, for otherwise g(n, r) ≤ G ≤ (r + 1)r + (n − r − 1)n = g(n, r) − 1.Consequently, by removing y 1 and a vertex of L, say x 1 , we do not decrease the minimal degree in the remainder of G. But, as y) ≥ n + 1 for every pair of non-adjacent vertices x ∈ X and y ∈ Y .Therefore, joining all the non-adjacent vertices of X and Y in G with degree sum of at least n + 1 yields a complete bipartite graph K n,n .As G was obtained from G also by joining certain non-adjacent vertices of X and Y with degree sum of at least n + 1, this shows that the (n + 1)-biclosure of G equals K n,n .Thus, by Theorem 2.7, G contains a Hamilton cycle, which, as we observed at the begining of this proof, is impossible.
Suppose first that d G (x) + d G (y) ≥ n + 2 for every pair of non-adjacent x ∈ X \ L and y ∈ Y .Let G ′ be the bipartite graph obtained from G by joining all the non-adjacent vertices of X \ L and Y .We claim that every y ∈ Y has a neighbour in L (in G ′ ).Suppose otherwise, and let y 1 ∈ Y be such that Let now G (3) be the graph obtained from G (2) by joining x l−2 with all the non-adjacent vertices of Y .In general, let G (m) (m ≥ 3) be obtained from G (m−1) by joining x l−m+1 with all the non-adjacent vertices of Y .Then G (l) = K n,n , and G (m) is obtained from G (m−1) by joining only pairs of vertices with degree sum of at least n + 1.Thus G (l) = BCl n+1 (G), so that the (n + 1)-biclosure of G is a complete bipartite graph.Now Theorem 2.7 implies that G contains a Hamilton cycle, which again leads to contradiction.
To complete the proof, it remains to consider the case when there is a pair of non-adjacent x 0 ∈ X \ L and y 0 ∈ Y with d G (x 0 ) + d G (y 0 ) ≤ n + 1.This however can only happen when n = 2r + 2 or n = 2r + 3.For let us suppose that n ≥ 2r + 4, and put f (l) = l 2 + (n − l − 1)(n − 1) + n + 2. We show G < f (l) and f (l) ≤ g(n, r), and thus obtain a contradiction with the assumption G ≥ g(n, r).If G contains a pair of non-adjacent vertices x ∈ X \ L and y ∈ Y with d G (x) + d G (y) ≤ n + 1, then As the derivative of f equals f ′ (l) = −n + 2l + 1, it follows that f (l) is decreasing for l ≤ (n − 1)/2, and hence maximal at l = r + 1.One immediately verifies that f (r + 1) ≤ g(n, r) for n ≥ 2r + 4. If, on the other hand, l > (n − 1)/2, then l = n/2 (since l ≤ n/2), and it is again immediate to check that f (n/2) ≤ g(n, r) for n ≥ 2r + 4.

Remark 1 . 5 . 5 .
Both the degree sum condition of Conjecture A and the bound on the size of Conjecture B are sharp, as can be seen in Example 1.6 below.It is also necessary to assume 2-connectedness in Conjecture A (Example 1.7).Finally, a quick look at C 6 and C 8 shows that Conjecture A would fail for n < Example 1.6.Let G 1 be a balanced bipartite graph, with colour classes X and Y , |X| = |Y | = n, where X = A ∪ B, Y = C ∪ D, |A| = k + r, |B| = n − k − r, |C| = r, and |D| = n − r.Moreover, assume that N G1 (x) = C for all x ∈ A, and N G1 (x) = Y for all x ∈ B. Then d G1 (x) + d G1 (y) = n − k for every pair x ∈ A and y ∈ D, and, in general, d G1 (x) + d G1 (y) ≥ n − k for every pair of x ∈ X and y for all x ∈ P , and N G(n,r) (x) = Y for all x ∈ Q.Let G be a balanced bipartite graph of order 2n ≥ 4, minimal degree δ(G) ≥ r ≥ 1, and size G ≥ n(n − r) + r 2 .Then G contains a Hamilton cycle, else r ≤ n/2 and G is isomorphic to G(n, r).

As 2 -
connectedness of a graph G implies δ(G) ≥ 2, the assertion of the theorem holds true for n ≤ 3, by Theorem 2.1.Suppose then there exists n ≥ 4 for which the assertion fails.Let G = (X, Y ; E) be a maximal 2-connected balanced bipartite graph of order 2n, in which d G (x) + d G (y) ≥ n for all non-adjacent x ∈ X, y ∈ Y , without a cycle of length at least 2n − 2. By maximality of G, G + xy contains a cycle of length at least 2n − 2, and hence G contains an x − y path of length 2n − 3 or 2n − 1 for every pair of non-adjacent x ∈ X, y ∈ Y .

Lemma 5 . 1 .
Let G = (X, Y ; E) be a balanced bipartite graph of order 2n and minimal degree δ(G) ≥ r ≥ 1, where n ≥ 4 and n ≥ 2r + 1.Let G ≥ g(n, r), and assume there exists a pair of vertices x ∈ X and y ∈ Y such that d G (x) + d G (y) ≤ n and δ(G − {x, y}) ≥ r.Then G contains a cycle of length 2n − 2.
and by Lemma 5.1, G contains a cycle of lenth 2n − 2; a contradiction.Consider the graph G − L. Notice that G − L ≥ g(n, r) − r 2 = n 2 − n − nr + r + 1.Moreover, we claim that d G−L (x) + d G−L (y) ≥ n for every pair of non-adjacent x ∈ X \ L and y ∈ Y .For if d G−L (x) + d G−L (y) ≤ n − 1 for a pair of non-adjacent x ∈ X \ L and y ∈ Y , then, by the above inequality, (G − L) − {x, y} ≥ n 2 − 2n − nr + r + 2 > (n − r − 1)(n − 1), which contradicts (G − L) − {x, y} being a bipartite graph with colour classes of cardinality n − r − 1 and n − 1. Taking into account that every vertex in Y has a neighbour in L, we now obtain that d G (x) + d G (y) ≥ n + 1 for all non-adjacent y ∈ Y and x ∈ X \ L. Let G be the bipartite graph obtained from G by joining all the non-adjacent vertices of Y and X \ L. As |X \ L| = n − r and every y ∈ Y has a neighbour in L, we get that d e G (y) ≥ n − r + 1 for all y ∈ Y .Hence d e G (x) + d e G r, as all the vertices in X \ L have degrees of at least l ≥ r + 1, and d G ′ (y) ≥ n − l ≥ l ≥ r + 1 for all y ∈ Y .Then Lemma 5.1 implies that G ′ contains a cycle of length 2n − 2, and hence, by Theorem 2.8, so does G; a contradiction.Notice that G ′ was obtained from G by joining only pairs of vertices with degree sum of at least n + 2. Also, as every vertex y ∈ Y has a neighbour in L (in G ′ ), we haved G ′ (y) ≥ n − l + 1. Recall that d G ′ (x l ) = d G (x l ) = l and d G ′ (x l−1 ) = d G (x l−1 ) = l.Hence d G ′ (x l ) + d G ′ (y) ≥ n + 1 and d G ′ (x l−1 ) + d G ′ (y) ≥ n + 1 for all y ∈ Y.Let G(2) be the graph obtained from G ′ by joining x l and x l−1 with all the vertices of Y .Then d G (2) (y) ≥ n − l + 2 for all y ∈ Y , and as d G (2) (x l−2 ) = d G (x l−2 ) ≥ l − 1 (by minimality of l), we get that d G (2) (x l−2 ) + d G (2) (y) ≥ n + 1 for all y ∈ Y.