Hamiltonian Cycles With All Small Even Chords

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1 Hamiltonian Cycles With All Small Even Chords Guantao Chen, Katsuhiro Ota, Akira Saito, and Yi Zhao Abstract. Let G be a graph of order n 3. An even squared hamiltonian cycle (ESHC) of G is a hamiltonian cycle C = v 1 v... v nv 1 of G with chords v i v i+3 for all 1 i n (where v n+j = v j for j 1). When n is even, an ESHC contains all bipartite -regular graphs of order n. We prove that there is a positive integer N such that for every graph G of even order n N, if the minimum degree δ(g) n +9, then G contains an ESHC. We show that the condition of n being even cannot be dropped and the constant 9 can not be replaced by 1. Our results can be easily extended to even kth powered hamiltonian cycles for all k. 1. Introduction In this paper, we will only consider simple graphs finite graphs without loops or multiple edges. The notations and definitions not defined here can be found in Diestel [7]. Let G = (V, E) be a graph with vertex set V and edge set E. For a vertex v V and a subset S V, let Γ(v, S) denote the set of neighbors of v in S, and deg(v, S) = Γ(v, S). Given another set U V, define Γ(U, S) = u U Γ(u, H) as the set of common neighbors in S of all the vertices in U, and let deg(u, S) = Γ(U, S). When U = {v 1,...,v k }, we simply write Γ(U, S) and deg(v, S) as Γ(v 1...v k, S) and deg(v 1...v k, S), respectively. When S = V, we only write Γ(U) and deg(u). A graph G is called hamiltonian if it contains a spanning cycle. The Hamiltonian Problem, determining whether a graph has a hamiltonian cycle, has long been one of few fundamental problems in graph theory. In this paper we fix G to be a graph of order n 3. Dirac [8] proved that if minimum degree δ(g) n/ then G is hamiltonian. Ore [3] extended Dirac s result by replacing the minimum degree condition with that deg(u)+deg(v) n for all nonadjacent vertices u and v. Many results have been obtained on generalizing these two classic results (see Gould [13] for a recent survey in this area). A -regular subgraph (-factor) of G consists of disjoint cycles of G. Aigner and Brandt [] proved that if minimum degree δ(g) n 1 3 then G contains all -factors as subgraphs (Alon and Fischer [3] proved this for sufficiently large n). If to-be-embedded - factors have at most k odd components, then by a conjecture of El-Zahar [9], the minimum degree condition can be reduced to δ(g) (n + k)/ (Abbasi [1] announced a proof of El-Zahar s conjecture for large n). Another generalization to Aigner and Brandt s result is to find one specific subgraph of G which contains all -factors. A squared hamiltoinian cycle of G is a hamiltonian cycle v 1 v... v n v 1 together with edges v i v i+ for all 1 i n. Note that we always assume that v n+i = v i for i 1. It is easy to see that a squared hamiltonian A. Saito s research is partially supported by Japan Society for the Promotion of Science, Grant-in-Aid for Scientific Research (C), , 004, and The Research Grant of Nihon University, College of Humanities and Sciences. Y. Zhao s research is partially supported by NSA H and H

2 GUANTAO CHEN, KATSUHIRO OTA, AKIRA SAITO, AND YI ZHAO cycle contains all -factors of G. Pósa (see [10]) conjectured that every graph G of order n 3 with δ(g) 3n contains a squared hamiltonian cycle. Fan and Kiestead [1] proved this conjecture approximately; Komlés, Sárközy, and Szemerédi [16] proved the conjecture for sufficiently large n. More generally, the k-th powered hamiltonian cycle is a hamiltonian cycle v 1 v...v n v 1 with chords v i v i+j for all 1 i n and 1 j k. Komlés, Sárközy, and Szemerédi [18, 19] proved a conjecture of Seymour for sufficiently large n: every n-vertex graph G with δ(g) (k 1)n/k contains a kth powered hamiltonian cycle. Böttcher, Schacht and Taraz [5] recently proved a conjecture of Bollobás and Komlós [15], which asymptotically includes all the results mentioned above. Given an integer b, a graph H is said to have bandwidth at most b, if there exists a labeling of the vertices by v 1, v,..., v n, such that j i b whenever v i v j E(H). It is shown in [5] that for any ε > 0 and integers r,, there exists β > 0 with the following property. Let G and H be n-vertex graphs for sufficiently large n. If δ(g) ((r 1)/r + ε)n and H is r-chromatic 1 with maximum degree and bandwidth at most βn, then G contains a copy of H. We are interested in the situation when the error term εn in the conjecture of Bollobás and Komlós can be reduced to a constant. According to the El-Zahar Conjecture, every n-vertex graph G with minimum degree δ(g) n/ contain all -factors with even components. Given a graph G, we define an Even Squared Hamiltonian Cycle (ESHC) as a hamiltonian cycle C = v 1 v...v n v 1 of G with chords v i v i+3 for all 1 i n. When n 7, an ESHC is 4-regular with chromatic number χ = for even n and χ = 3 for odd n. It is not hard to check that an n-vertex ESHC contains all bipartite graphs of order n with maximum degree (e.g., by using the fact that every ESHC of even order contains a ladder graph defined below). Below is our main result. Theorem 1.1. There exists N > 0 such that for all even integers n N, if G is a graph of order n with δ(g) n + 9, then G contains an ESHC. We show that the constant 9 in Theorem 1.1 can not be replaced by 1. Proposition 1.. Suppose that n 10. Let G be the union of two copies of K n sharing 4 vertices. Then δ(g) = n + 1 but G contain no ESHC. We also show that the condition of n being even is necessary for Theorem 1.1 even if we replace 9 by n/8 1/. Proposition 1.3. There are infinitely many odd n and graphs G of order n such that δ(g) n + n/8 1/ and G contains no ESHC. More generally, an Even kth powered Hamiltonian Cycle (EkHC) of a graph G is a hamiltonian cycle v 1 v...v n v 1 with edges v i v i+j 1 for all 1 i n and 1 j k. Then an E1HC is simply a hamiltonian cycle while an EHC is an ESHC. Using the same proof techniques for Theorem 1.1, we can derive the following result, whose proof is omitted. Theorem 1.4. For any positive integer k, there exist a constant s = s(k) and a positive integer N such that if G is a graph of even order n N and δ(g) n+s then G contains an EkHC. + 1 The kth powered hamiltonian cycle of order n has chromatic number k + 1 or k + depending on the value of n. The authors of [5] make their result applicable even when H is r + 1-chromatic but one of its color classes is fairly small, e.g., the kth powered hamiltonian cycle.

3 HAMILTONIAN CYCLES WITH ALL SMALL EVEN CHORDS 3 One may view the even kth cycle of order n = N as the following bipartite graph. Let B k (N) be the bipartite graph (X Y, E) with X = {x 1,...,x N } and Y = {y 1,...,y N } such that x i y j E if and only if i j (mod N) { k + 1,..., 1, 0, 1,...,k}. In particular, B (N), or ESHC, contains the ladder graph defined by Czygrinow and Kierstead [6], which has the same vertex sets X and Y but x i is adjacent to y j if and only if i j (mod N) { 1, 0, 1}. Note that the the ladder graph contains all -factors with even components. The structure of the paper is as follows. We prove two (easy) Propositions 1. and 1.3 in the next section. Following the approach of [16] on squared hamiltonian cycles, we prove Theorem 1.1 by the regularity method. In Section 3 we state the Regularity Lemma and the Blow-up Lemma. In Section 4 we prove Theorem 1.1 by proving the non-extremal case and two extremal cases separately. It seems harder to handle the extremal cases here than in [16]; this is also the reason why we need a large constant 9 in Theorem 1.1. The last section gathers open problems with a remark.. Proofs of Propositions 1. and 1.3 Proof of Proposition 1.. Given a graph G, a pair (A, B) of vertex subsets is called a separator of G if V (G) = A B, both A B and B A are non-empty and E(A B, B A) =. It is easy to see that Proposition 1. follows from the following more general statement. Suppose that G is a graph with an EHSC. If (A, B) is a separator of G with A B 3 and B A 3, then A B 6. Let H be an ESHC of G with supporting cycle C. Assign C an orientation. An segment P 1 [x, z] of C is called an A B-path if x A B and z B A. We claim that if P 1 [x 1, z 1 ] is an A B-path such that (A B) V (P 1 ) 1 and (B A) V (P 1 ) 1, then C V (P 1 ) contains another A B-path P. In fact, C[z 1, x 1 ] = C E(P 1 ) must contain two internal vertices x A B and z B A. Then P := C[x, z ] is an A B-path. We can further assume that V (P ) {x, z } A B by letting P be a minimal A B-path of C V (P 1 ). Now let P 1 [x 1, z 1 ] be an A B-path such that V (P 1 ) {x 1, z 1 } A B. Since A B 3, B A 3, by the claim above, C V (P 1 ) contains another A B-path P [x, z ] such that V (P ) {x, z } A B. If e(p i ) 4 for i = 1,, then A B e(p 1 )+e(p ) 6 and we are done. On the other hand, we know that e(p i ) {1, 3} for i = 1, because x i z i E(G), which follows from e(a B, B A) =. Without loss of generality, assume that e(p 1 ) =, or P 1 = x 1 y 1 z 1. By following the orientation of C, let x 1 be the predecessor of x 1 and z 1 + be the successor of z 1. Since x 1 z 1, x 1 z 1 + E(G), we have x 1, z+ 1 A B. Since P 1 and P are vertex disjoint A B-paths, V (P ) {x 1, y 1, z 1 + } =. If P contains at least three internal vertices, then A B 6 and we are done. So we may assume that P = x y z, and consequently {x, y, z + } A B. If z+ 1 x and z+ x 1, then all x 1, y 1, z 1 +, x, y, z + are distinct, and consequently A B 6. Otherwise, without loss of generality, assume that z 1 + = x. Then P 1 = P 1z 1 + P is an A B-path with two vertices in each of A B and B A. Since A B 3 and B A 3, by the earlier claim, there is an A B-path P 3 [x 3, z 3 ] vertex-disjoint from P 1 such that V (P 3) {x 3, z 3 } A B. If e(p 3 ) 4, then A B 6 because P 1, P 3 are disjoint and P 1 contains three vertices y 1, z 1 +, y from A B. Otherwise P 3 = x 3 y 3 z 3 for some y 3 A B, then {x 3, y 3, z 3 + } A B. Since six vertices y 1, z 1 +, y, x 3, y 3, z 3 + are contained in A B, we have A B 6.

4 4 GUANTAO CHEN, KATSUHIRO OTA, AKIRA SAITO, AND YI ZHAO x 1 A x 1 y 1 z 1 z + 1 B Figure 1. A segment connecting A and B Proof of Proposition 1.3. Let q be an odd prime power, by using projective planes, one can construct (e.g. [11]) C 4 -free graphs H of order h = q + q + 1 with δ(h) q. Let G := H + K h q, i.e. a graph obtained from H by adding h q vertices such that each new vertex is adjacent to all vertices of H. Let X := V (G) V (H) and n := V (G) = h q. Then n is odd and δ(g) h = n + q n n because n = q + q + < (q + 1). To see that G does not have any ESHC, consider an arbitrary hamiltonian cycle C of G (if it exists). Since X is an independent vertex set and n is odd, we conclude that C X is the union of vertex-disjoint paths such that e(c X) is odd. In particular, one path P[x, y] of C X has odd length. If V (P) 4, then xx +++ / E(G), where x +++ = ((x + ) + ) +, since H contains no C 4. Otherwise V (P) =, which in turn shows that x, y + X. Since X is independent, x y + / E(G). In all cases G does not have an ESHC based on C. 3. The Regularity Lemma and Blow-up Lemma As in [16, 19], the Regularity Lemma of Szemerédi [4] and Blow-up Lemma of Komlós, Sárközy and Szemerédi [17] are main tools in our proof of Theorem 1.1. For any two disjoint vertex-sets A and B of a graph G, the density of A and B is the ratio d(a, B) := e(a, B)/( A B ), where e(a, B) is the number of edges with one endvertex in A and the other in B. Let ε and δ be two positive real numbers. The pair (A, B) is called ε-regular if for every X A and Y B satisfying X > ε A, Y > ε B, we have d(x, Y ) d(a, B) < ε. Moreover, the pair (A, B) is called (ε, δ)-super-regular if (A, B) is ε-regular and deg B (a) > δ B for all a A and deg A (b) > δ A for all b B. Lemma 3.1 (Regularity Lemma Degree Form). For every ε > 0 there is an M = M(ε) such that, for any graph G = (V, E) and any real number d [0, 1], there is a partition of the vertex set V into l + 1 clusters V 0, V 1,...,V l, and there is a subgraph G of G with the following properties: l M, V i ε V for 0 i l, V 1 = V = = V l, deg G (v) > deg G (v) (d + ε) V for all v V, G [V i ] = (i.e. V i is an independent set in G ), for all i, each pair (V i, V j ), 1 i < j l, is ε-regular with d(v i, V j ) = 0 or d(v i, V j ) d in G.

5 HAMILTONIAN CYCLES WITH ALL SMALL EVEN CHORDS 5 The Blow-up Lemma allows us to regard a super-regular pair as a complete bipartite graph when embedding a graph with bounded degree. We need a bipartite version of this lemma which also restricts the mappings of a small number of vertices. Lemma 3. (Blow-up Lemma - bipartite version). For every δ,, c > 0, there exists an ε = ε(δ,, c) > 0 and α = α(δ,, c) > 0 such that the following holds. Let (X, Y ) be an (ε, δ)-super-regular pair with X = Y = N. If a bipartite graph H with (H) can be embedded in K N,N by a function φ, then H can be embedded in (X, Y ). Moreover, in each φ 1 (X) and φ 1 (Y ), fix at most αn special vertices z, each of which is equipped with a subset S z of X or Y of size at least cn. The embedding of H into (X, Y ) exists even if we restrict the image of z to be S z for all special vertices z. 4. Proof of Theorem 1.1 A partition V 1...V k of a set V is called balanced if V i V j 1 for all i j. In this paper we often consider a balanced bipartition of V of order n for an even n, in which both partition sets are of size n/. Let us first define two extremal cases. Extremal Case 1: There exists a balanced partition of V into V 1 and V such that the density d(v 1, V ) 1 α for some small α (to be specified later). Extremal Case : There exists a balanced partition of V into V 1 and V such that the density d(v 1, V ) α for some small α (to be specified later). The following three theorems deal with the non-extremal case and two extremal cases separately. Theorem 4.1. For every α > 0, there exist β > 0 and a positive integer n 0 such that the following holds for every even integer n n 0. For every graph G of order n with δ(g) ( 1 β)n, either G contains an ESHC or G is in one of the extremal cases with parameter α. Theorem 4.. Suppose that 0 < α 1 and n is a sufficiently large even integer. Let G be a graph on n vertices with δ(g) n + 3. If G is in Extremal Case 1 with parameter α, then G contains an ESHC. Theorem 4.3. Suppose that 0 < α 1 and n is a sufficiently large even integer. Let G be a graph on n vertices with δ(g) n + 9. If G is in Extremal Case with parameter α, then G contains an ESHC. It is easy to see that Theorem 1.1 follows from Theorems 4.1, 4. and 4.3. For this purpose we can even use a weaker version of Theorem 4.1 with β = 0, but the current Theorem 4.1 may have other applications. If G is in Extremal Case with parameter α, then there exists x V such that deg(x) < (1 + α)n/ and in turn δ(g) < (1 + α)n/. Theorems 4.1 and 4. together implies the following remark, which is a special case of the theorem of Böttcher, Schacht and Taraz [5]. Remark 4.4. For any α > 0, there exists a positive integer n 0 such that every graph G of even order n n 0 with δ(g) ( 1 + α)n contains an ESHC. This remark will be used in the proof of Theorem Non-extremal Case. In this section we prove Theorem 1.1.

6 6 GUANTAO CHEN, KATSUHIRO OTA, AKIRA SAITO, AND YI ZHAO Proof. We fix the following sequence of parameters (1) ε d β α and specify their dependence as the proof proceeds. Actually we let α be the minimum of the two parameters defined in the extremal cases. Then we choose d β such that they are much small than α. Finally we choose ε = 1 d ε(d, 4, 4 ) following the definition of ε in the Blow-up lemma. Choose n to be sufficiently large. In the proof we omit ceiling and floor functions if they are not crucial. Let G be a graph of order n such that δ(g) ( 1 β)n and G is not in either of the extremal cases. Applying the Regularity Lemma (Lemma 3.1) to G with parameters ε and d, we obtain a partition of V (G) into l + 1 clusters V 0, V 1,...,V l for some l M = M(ε), and a subgraph G of G with all described properties in Lemma 3.1. In particular, for all v V, deg G (v) > deg G (v) (d + ε)n ( 1 β ε d)n (1 β)n, provided that ε+d β. On the other hand, e(g ) e(g) (d+ε) n e(g) dn by using ε < d. We further assume that l = k is even; otherwise we eliminate the last cluster V l by removing all the vertices in this cluster to V 0. As a result, V 0 εn and () (1 ε)n ln = kn n. For each pair i and j with 1 i j l, we write V i V j if d(v i, V j ) d. As in other applications of the Regularity Lemma, we consider the reduced graph G r, whose vertex set is {1,...,l}, and two vertices i and j are adjacent if and only if V i V j. From δ(g ) > ( 1 β)n, we easily derive that δ(g r) ( 1 β)l as follows. Fix a cluster V i and a vertex v V i. Let m be the number of clusters that adjacent to V i. Since in G the vertex v has no neighbor in any cluster that is not adjacent to V i, we have ( 1 β)n deg G (v) mn. With (), this gives that m (1 β)n/n (1 β)l. The rest of the proof consists of the following five steps. Step 1: Show that G r contains a hamiltonian cycle X 1 Y 1...X k Y k. Step : For each i, initiate a connecting ESP (even squared path) P i between Y i 1 and X i (where Y 0 = Y k ) with two vertices from each Y i 1 and X i. Step 3: For each i, move at most εn vertices from X i Y i to V 0 such that the graph on the remaining vertices of X i Y i has the minimum degree at least (d ε)n. Step 4: Extend P 1,...,P k to include all the vertices in V 0 such that X i j V (P j ) = Y i j V (P j ) (1 d /)N for all i. Step 5: Apply the Blow-up Lemma to each (X i, Y i ) and obtain an ESP consisting of all the remaining vertices of X i Y i. Concatenating these ESP s with P 1,...,P k, we obtain the desired ESHC of G. We now give details to each step. The assumption that G is not in either of the extremal cases leads to the following claim, which will be used in Step 1 and Step 4. Claim 4.5. (1) G r contains no independent set U 1 of size at least ( 1 8β)l. () G r contains no two disjoint subsets U 1, U of size at least ( 1 6β)l such that e Gr (U 1, U ) = 0.

7 HAMILTONIAN CYCLES WITH ALL SMALL EVEN CHORDS 7 Proof. (1). Suppose instead, that G r contains an independent set U 1 of size ( 1 8β)l. We will show that G is in the Extremal Case 1 with parameter α. Let A = i U 1 V i and B = V (G) A. By (), ( 1 9β)n (1 8β)Nl = U 1 N = A < ( 1 β)n. For each x A, since deg G (x, A) deg G (x, A) + (d + ε)n < βn, then deg G (x, B) > ( 1 β)n βn (1 β)n. Hence e G(A, B) ( 1 9β)n(1 β)n > ( β)n. Now move at most 9βn vertices from B to A such that A and B are of size n/. We still have ( 1 e G (A, B) > 4 11 ) β n 9βn n ( ) 14 ( n ) = 10β n = (1 40β). By specializing 40β α in (1), we see that G is in the Extremal Case 1 with parameter α. (). Suppose instead, that G r contains two disjoint subsets U 1, U of size ( 1 6β)l such that e Gr (U 1, U ) = 0. We will show that G is in the Extremal Case with parameter α. Let A = i U 1 V i and B = i U V i. Since e Gr (U 1, U ) = 0, we have e G (A, B) = 0. Since e(g) e(g ) + dn, we have e G (A, B) e G (A, B) + dn = dn. Note that A = U 1 N = ( 1 6β)lN > (1 7β)n. Similarly, B > (1 7β)n. By adding at most 7βn vertices to each of A and B, we obtain two subsets of size n/ and still name them as A and B, respectively. Then, e(a, B) dn + (7βn)(n/) = 8βn, which in turn shows the density d(a, B) = e(a, B)/( n ) 3β. Since α > 3β, we obtain that G is in the Extremal Case with parameter α. Step 1. To show that G r is hamiltonian, we need the following theorem of Nash-Williams. Theorem 4.6 (Nash-Williams[]). Let G be a -connected graph of order n. If minimum degree δ(g) max{(n + )/3, α(g)}, then G contains a hamiltonian cycle. We first show that G r is βl-connected. Suppose, to the contrary, let S be a cut of G r such that S < βl and let U 1 and U be two components of G r S. Since δ(g r ) ( 1 β)l, we have U i ( 1 3β)l for i = 1,. Since e(u 1, U ) = 0, we obtain a contradiction to Claim 4.5 (). Since n = Nl + V i (l + )εn, we have l 1/ε 3/β, provided that β 5ε. Then βl 3, and G r is 3-connected. By Claim 4.5 (1), the independence number of G r is at most ( 1 8β)l, which is less than ( 1 β)l, or the minimum degree of G r. By Theorem 4.6, G r is hamiltonian. Following the order of a hamiltonian cycle of G r, we denote all the clusters of G except for V 0 by X 1, Y 1,...,X k, Y k (recall that l = k is even). We call X i the partner of Y i and vice versa. Step. For each i = 1,...k, we initiate an ESP P i of G connecting X i and Y i 1 (with Y 0 = Y k ) as follows. Given an ε-regular pair (X, Y ) of clusters and a subset Y Y, we call a vertex x X typical to Y if deg(x, Y ) (d ε) Y. By the regularity of (X, Y ), at most εn vertices of X are not typical to Y whenever Y > εn. Fix 1 i k. First let a i X i be a vertex typical to both Y i 1 and Y i and let b i X i be a vertex typical to both Γ(a i, Y i 1 ) and Γ(a i, Y i ). Since both pairs (Y i 1, X i ) and (X i, Y i ) are ε-regularity with density at least d, all but εn vertices of X can be chosen as a i. Since Γ(a i, Y i 1 ), Γ(a i, Y i ) (d ε)n > εn, all but at most εn + 1 vertices of X can be chosen as b i (note that b i a i ). Recall that Γ(a i b i, Y i 1 ) = Γ(a i, Y i 1 ) Γ(b i, Y i 1 ). The way we select a i and b i guarantee that Γ(a i b i, Y i 1 ) (d ε) N εn +.

8 8 GUANTAO CHEN, KATSUHIRO OTA, AKIRA SAITO, AND YI ZHAO Now let c i 1, d i 1 Γ(a i, b i, Y i 1 ) be two (distinct) vertices of Y i 1 such that c i 1 is typical to both X i 1 and X i, and d i 1 is typical to both Γ(c i 1, X i 1 ) and Γ(c i 1, X i ). All but at most εn vertices of Γ(a i b i, Y i 1 ) can be chosen as c i 1 and d i 1. In summary P i = c i 1 a i d i 1 b i is an ESP with c i 1, d i 1 Y i 1, a i, b i X i such that (3) Step 3. For each i 1, let deg(c i 1 d i 1, X i 1 ) (d ε) N, deg(a i, Y i 1 ) (d ε)n, deg(a i b i, Y i ) (d ε) N, deg(d i 1, X i ) (d ε)n. X i := {x X i, deg(x, Y i ) (d ε)n} and Y i := {y Y i, deg(y, X i ) (d ε)n}. Since (X i, Y i ) is ε-regular, we have X i, Y i (1 ε)n. If X i Y i, say X i > Y i, then we pick an arbitrary subset of X i of size Y i and still name it X i. As a result, we have X i = Y i. Let V 0 := V k 0 i=1 (X i X i ) (Y i Y i ). From X i X i + Y i Y i εn, we derive that V 0 εn + (εn)k = 3εn by using Nk n from (). In addition, the minimum degree δ(g[x i, Y i ]) (d ε)n εn. Step 4. Consider a vertex x V (G) and an original cluster A (X i or Y i for some i), we say that x is adjacent to A, denoted by x A, if deg(x, A) (d ε)n. Given two vertices u, w, we define a u, w-chain of length t as distinct clusters A 1, B 1,...,A t, B t such that u A 1 B 1...A t B t w and each A j and B j are partners, in other words, {A j, B j } = {X ij, Y ij } for some i j. Claim 4.7. Let L be the list of at most εn pairs of vertices of G. For each {u, w} L, we can find u, w-chains of length at most four such that every cluster is used in at most d N/0 chains. Proof. Suppose that we have found chains of length at most four for the first m < εn pairs such that no cluster is contained in more than d N/0 chains. Let Ω be the set of all clusters that are used in exactly d N/0 chains. Since each chain use at most four clusters, we have d kn N Ω 4m 8εn 8ε 0 1 ε, where the last inequality follows from (). Therefore Ω 30ε k 30ε l βl provided (1 ε)d d that 1 ε 1 and 30ε d β. Now consider a pair {u, w} L. Our goal is to find a u, w-chain of length at most four by using clusters not in Ω. Let U be the set of all clusters adjacent to u but not in Ω, and W be the set of all clusters adjacent to w but not in Ω. Let P(U) and P(W) be the set of the partners of clusters in U and W, respectively. The definition of chains implies that a cluster A Ω if and only if its partner P(A) Ω. Therefore (P(U) P(W)) Ω =. We claim that P(U) = U ( 1 3β)l. To see it, we first observe that any vertex v V is adjacent to at least ( 1 β)l clusters. For instead, ( ) ( ) ( β n deg G (v) β ln + dnl + 3εn < 3 ) β n, a contradiction, provided that β d + 3ε. Since Ω βl, we thus have U (1 3β)l. Similarly P(W) = W ( 1 3β)l. If E Gr (P(U), P(W)), then there exist two adjacent clusters B 1 P(U), A P(W). If A 1, B are partners of each other, then u A B 1 w gives a u, w-chain of length two.

9 HAMILTONIAN CYCLES WITH ALL SMALL EVEN CHORDS 9 Otherwise assume that A 1 = P(B 1 ) and B = P(A ). Then u A 1 B 1 A B w gives a u, w-chain of length four. Note that all A i, B i Ω. Otherwise assume that (4) E Gr (P(U), P(W)) =. If P(U) P(W) =, then (4) contradicts with Claim 4.5 () because P(U) ( 1 3β)l and P(W) ( 1 3β)l,. Otherwise assume that A P(U) P(W). Then by (4), A is not adjacent to any cluster in P(U) P(W). Since deg Gr (A) ( 1 β)l, we derive that P(U) P(W) ( 1 + β)l. Since P(U) (1 3β)l and P(W) (1 3β)l, then P(U) P(W) ( 1 8β)l. By (4), P(U) P(W) is an independent set in G r, which contradicts with Claim 4.5 (1). We arbitrarily partition V 0 into at most εn pairs (note that V 0 is even because X i = Y i for all i). Applying Claim 4.7, we find chains of length at most four for each pair such that every cluster is used in at most d N/0 chains. For each i let m i denote the number of chains that contains (X i, Y i ). Claim 4.8. We can extend connecting ESP s to include all the vertices in V 0 such that the following holds for all i. The resulting ESP s P i = u 1 v 1...u t v t satisfies u 1, u Y i 1, v t 1, v t X i and deg(u (5) 1 u, X i 1 ) (d ε) N, deg(v 1, Y i 1 ) (d ε)n, deg(v t v t 1, Y i ) (d ε) N, deg(u t, X i ) (d ε)n. The sets Xi = X i jv (P j ) and Yi = Y i jv (P j ) satisfy (6) Xi = Yi (1 ε)n 7m i. Proof. We prove by induction on m := V 0 /. When m = 0, by (3), the initial P i = c i 1 a i d i 1 b i satisfies (5). The initial Xi = X i {a i, b i } and Yi := Y i {c i, d i }) satisfy (6). Suppose that m 1 and we have extended the connecting ESP s to include m 1 pairs from V 0 such that (5) and (6) hold for all i. Let {x, y} be the last pair from V 0. We first consider the case when the x, y-chain has length two. Without loss of generality, assume that x Y i X i y for some i. Let P i = u 1 v 1...u t v t be the current connecting ESP between Y i 1 and X i and Yi := Y i jv (P j ) and Xi := X i jv (P j ). To include x, we extend P i to P i = P iy 1 x 1 y x y 3 x 3 y 4 x with four vertices y 1, y, y 3, y 4 Yi and three vertices x 1, x, x 3 Xi such that in addition y 1 Γ(v t 1 v t ), y Γ(v t ), y 3, y 4 Γ(x), x 1 Γ(u t ), (7) y 4 is typical to X i, and x 3 is typical to Γ(x, Y i ). This is possible by using Lemma 3. (actually we only need the regularity between X i and Y i ; but applying the Blow-up Lemma makes our proof shorter). To see it, first note that (6) implies that Xi, Y i > (1 d /)N because m i d N/0 by Claim 4.7. Then, by (5), Γ(v t 1 v t, Yi ) deg(v t 1 v t, Y i ) d N (d ε) N d N > d 4 N. Similarly we can show that Γ(v t, Yi ), Γ(u t, Xi ) (d d )N. The definition of x Y i also guarantees that Γ(x, Yi ) (d d )N. Finally (7) only forbids additional εn vertices when choosing y 4 and x 3. Therefore we can apply Lemma 3. to find such an P i. By (7), we have deg(y 4, X i ) (d ε)n and deg(x 3 x, Y i ) (d ε) N. Consequently P i satisfies (5).

10 10 GUANTAO CHEN, KATSUHIRO OTA, AKIRA SAITO, AND YI ZHAO Since x behaves like a vertex of X i in P i, we call such a procedure inserting x into X i (by extending P i ). Since y X i, we can similarly insert y to Y i by extending P i+1 to P i+1 = yx 4y 5 x 5 y 6 x 6 y 7 x 7 P i+1 with x j Xi and y j Yi. Since each X i and Yi lose seven vertices totally, (6) holds. Now consider the case when the x, y-chain has length four. Assume that x A 1 B 1 A B y. We first insert x to B 1, then pick any (available) vertex in B 1 that is typical to A and insert it to B, and finally insert y to A. As a result, A 1, B 1 each loses four vertices to some connecting paths while A, B each loses seven vertices to some connecting paths. Thus (6) holds. Step 5. Fix 1 i k. Suppose that at present P i = u 1 v 1...u t v t and P i+1 = w 1 z 1...w s z s for some integers s, t, and Xi = X i jv (P j ) and Yi = Y i jv (P j ). By Claim 4.8, P i and P i+1 satisfy (5) and Xi, Y i (1 d /)N. Since (X i, Y i ) is regular, the Slicing Lemma of [14] says that (Xi, Y i ) is (ε, d/)-super-regular (note that (d ε)n d N/ > dn/). We now apply the Blow-up Lemma to each G [Xi, Y i ] to obtain an spanning ESP y 1 x 1 y... x Ni 1, y Ni, x Ni, where N i = Xi = Y i such that y 1 Γ(v t v t 1, Yi ), x 1 Γ(u t, Xi ), y Γ(v t, Yi ), x Ni 1 Γ(w 1, Xi ), y Ni Γ(z 1, Yi ), x N i Γ(w 1 w, Xi ). u t 1 w 1 w u t x Ni y Ni x Ni 1 y x 1 y 1 v t 1 v t z 1 z Y i 1 X i Y i X i+1 Figure. An ESP covering X i and Y i The restrictive mapping of y 1, x 1, y, x Ni 1, y Ni, x Ni is possible because by (5), all the targeting sets are of size at least (d ε) N d N/ > d N/4. We now complete the proof of the non-extremal case. 4.. Extremal Case 1. In this subsection we prove Theorem 4.. We start with a lemma which gives a balanced spanning bipartite subgraph that we will use throughout the section. Lemma 4.9. Suppose that 0 α (1/9) 3. Let G = (V, E) be a graph on n vertices with δ(g) n + 3 and a balanced partition V 1 V such that d(v 1, V ) 1 α. Then G contains a balanced spanning bipartite subgraph G with parts U 1, U such that

11 HAMILTONIAN CYCLES WITH ALL SMALL EVEN CHORDS 11 There is a set W of at most α /3 n vertices such that we can find vertex disjoint 4-stars (stars with four edges) in G with the vertices of W as their centers. deg G (x) (1 α 1/3 α /3 )n/ for all x W. Proof. For simplicity, let α 1 = α 1/3 and α = α /3. For each i = 1,, we define { V i = x V : deg(x, V 3 i ) (1 α 1 ) n }. Since d(v 1, V ) 1 α, we have V i V i α n/ and consequently V i (1 α )n/ for i = 1,. For any x V i, (8) deg(x, V 3 i) > (1 α 1 ) n α n. Let V 0 = V V 1 V. Then V 0 α n. For each v V 0 and i = 1,, we have deg(v, V i ) (1 α 1 ) n, which implies that deg(v, V n i) α 1 and (9) deg(v, V i ) (α 1 α ) n. We now separate cases based on V 1 and V. Case 1: V 1, V n/. In this case we partition V 0 into W 1 W such that W i = n/ V i for i = 1,. For each vertex w W i, we greedily find four neighbors from V 3 i such that the neighbors for all the vertices of W i are distinct. This is possible for i = 1, because of (9) and (α 1 α ) n 4α n 4 V 0 provided that α 1 9α or α 1 1/9. Define U i = V i W i for i = 1,. Then U 1 = U = n/. With W = V 0, the second assertion of Lemma 4.9 follows from (8). Case : one of V 1, V, say, V 1 is greater than n/. Let V 0 1 be the set of vertices v V 1 such that deg(v, V 1 ) α 1n/. First assume that V1 0 V 1 n/. Then we form a set W with V 1 n/ vertices of V 0 1 and all the vertices of V 0. Let U 1 = V 1 W and U = V W. Then U 1 = U = n/. Since (1 α )n/ V n/, we have W α n/. Then V 1 U 1 α n/, by (9) and the definition of V1 0, we have (10) deg(v, U 1 ) deg(v, V 1) α n/ (α 1 α )n/ α n/. for all v W. With α 1 6α, we have (α 1 α )n/ 4α n/ 4 W. Therefore we can greedily find four neighbors for each vertex v W such that the neighbors for all the vertices of W are distinct. The second assertion of Lemma 4.9 follows from (8) and V 1 U 1 α n/. V 0 Otherwise assume that V 0 1 < V 1 n/. In this case let U 1 = V 1 V 0 1 and U = V 1 V 0. Then U 1 = n +t 1 for some positive integer t 1 α n/. Since deg(v, V 1 ) < α 1n/ for every v U 1, the induced graph G[U 1 ] has the maximum degree α 1 n/. By the minimum degree condition δ(g) n/ + 3, G[U 1 ] has the minimum degree at least ( n ) ( n ) δ(g) U + 3 t 1 t We now need the following simple fact. Fact In a graph G 1 of order n 1 with the maximum degree (G 1 ) and the minimum degree δ(g 1 ) t, the number of disjoint 4-stars is at least (t 3)n 1 5( +t 3).

12 1 GUANTAO CHEN, KATSUHIRO OTA, AKIRA SAITO, AND YI ZHAO To see it, suppose G 1 has a largest family of disjoint 4-stars on M of size m. Then (t 3)(n 1 5m) e(m, V (G) M) 5m and the fact follows. Applying Fact 4.10, there are at least n (t ) U 1 5( + t ) t 1 U 1 n 5(α 1 + t 1) t 1 5(α 1 + α ) n t 1 vertex disjoint 4-starts in G[U 1 ]. Pick t 1 such 4-stars and move their centers to U. As a result, U 1 = U = n/. Let W 0 = V1 0 V 0 and W be the union of W 0 with the new vertices of U. Since V 1 U 1 = V 1 n/ α n/, we have (10) for all v W 0. Since W = n/ V α n/, we can find disjoint 4-stars in G[U 1, W 0 ] with all the vertices of W 0 as centers such that these 4-stars are also disjoint from the existing 4-stars. In addition, the second assertion of Lemma 4.9 holds as before. Proposition 4.11 below shows that if G is a graph or a bipartite graph with large minimum degree and it contains not many vertex disjoint 4-stars, then we can find an ESC or ESP containing all the vertices in these stars. We need its part () for this subsection, and part (1) for Extremal Case. Proposition Fix 0 < ε 1 1/5. (1) Let G be a graph of order N with a subset W of size t ε 1 N/8. Suppose that G contains t vertex-disjoint 4-stars with the vertices of W as centers. For every vertex x / W, we have deg(x) (1 ε 1 )N. Then G contains an ESC C of length 8t which contains all the vertices of W such that any two nearest vertices of W are separated by exactly seven vertices not in W. () Let G be a bipartite graph on two parts U 1, U of size N. Let W be a vertex subset of size t ε 1 N/5. Suppose that G contains t vertex-disjoint 4-stars with the vertices of W as centers. For every vertex x / W, we have deg(x) (1 ε 1 )N. Then G contains an ESP of length 8t + 4 which contains all the vertices of W and whose first and last three vertices are not from W. Proof. (1). Suppose that W = {x 1,...,x t }, and denote the four leaves of x i by a i, b i, c i, d i. For each x i, we greedily choose three new vertices u i, v i, w i (that are not contained in any existing star) such that u i Γ(c i 1 d i 1 a i b i ), v i Γ(d i 1 a i b i c i ), w i Γ(c i d i a i+1 b i+1 ), in which the indices are modulo t. This is possible because each a i, b i, c i, d i, 1 i t, has at least (1 ε 1 )N neighbors and any four of them have at least (1 4ε 1 )N ε 1 N 8t common neighbors (where 8t is the total number of vertices used at the end of this greedy algorithm). We thus obtain an ESC u i a i v i b i x i c i w i d i : i = 1,...,t, in which the vertices of W are distributed evenly. (). Partition W = W 1 W with W 1 = U 1 W and W = U W. For each W i, we follow the procedure in (1) to find two vertex disjoint ESC s C 1 and C of length 8 W 1 and 8 W in which the vertices of W are distributed evenly. The calculation is similar except that any four vertices in U 1 W (or U W) have at least (1 4ε 1 )N ε 1 N > 4t common neighbors in U (or U 1 ), where 4t is the total number of the vertices used in one partition set. We next break C 1 into P 1 = x 1 x x 3...u 3 u u 1 and break C into P 1 = v 1 v v 3...y 3 y y 1 such that x i, u i, v i, y i / W for i = 1,, 3 and u 1, u 3, v U 1, u, v 1, v 3 U. Assume that

13 HAMILTONIAN CYCLES WITH ALL SMALL EVEN CHORDS 13 t otherwise we are done. Choose four new vertices not in W (in this order) z 1 Γ(u 1 u 3 ), z 3 Γ(u 1 v ) and z Γ(u z 1 z 3 v 1 ), z 4 Γ(z 1 z 3 v 1 v 3 ). This is possible because the number of common neighbors of any four vertices not in W is at least (1 4ε 1 )N 5t 4t +, where 4t + is the total number of the vertices used in one partition set. As a result, P 1 z 1 z z 3 z 4 P is an ESP which contains all the vertices of W and whose first and last three vertices are not from W. We finally observe that a bipartite graph with very large minimum degree is superregular. Proposition 4.1. Given 0 < ρ < 1, let G be a bipartite graph on X Y such that (11) δ(x, Y ) (1 ρ) Y, δ(y, X) (1 ρ) X. Then G is ( ρ, 1 ρ)-super-regular. Proof. It suffices to show that G is ρ-regular. Consider subsets A X, B Y with A = ε 1 X and B = ε Y for some ε 1, ε > ρ. By (11), we have δ(a, Y ) Y ρ Y and consequently δ(a, B) B ρ Y = (ε ρ) Y. The density between A and B satisfies d(a, B) δ(a, B) A A B (ε ρ) Y B = ε ρ ε > 1 ρ ρ = 1 ρ. Since 1 ρ < d(a, B) 1 and in particular, 1 ρ < d(x, Y ) 1, we have d(a, B) d(x, Y ) < ρ. We are ready to prove Theorem 4. now. Proof of Theorem 4.. Let 0 α 1, in particular α (1/9) 3. Write α 1 = α 1/3 and α = α /3. Let G = (V, E) be a graph oh n vertices with δ(g) n + 3. Suppose G is in the extremal case 1 with parameter α. We first apply Lemma 4.9 to G and obtain a bipartite subgraph G with two partition sets U 1, U of size n/ which contains at most α /3 n vertex disjoint 4-stars. Denote by W the set of the centers of the 4-stars. We also have (1) deg G (x) (1 α 1 α )n/ for all x W. Since α 1 + α 1/5 and α n α 1+α n 5, we may apply Proposition 4.11 () to G with ε 1 = α 1 + α and N = n/. We thus obtain an ESP P 0 = x 1 x x 3...y 3 y y 1 of length 8 W + 4 which contains all the vertices of W such that x i, y i V 1 V for 1 i 3. In order to find an ESHC of G, it suffices to find an ESP P = u 1 u u 3...v 3 v v 1 on V (G) V (P 0 ) such that x 3 x x 1 P y 1 y y 3 = x 3 x x 1 u 1 u u 3...v 3 v v 1 y 1 y y 3 is also an ESP. Let U i = U i V (P 0 ) for i = 1,, and n = U 1 = U. Then n = n/ (4 V 0 + ) n/ (4α n + ). By (1), the bipartite subgraph G[U 1, U ] has its minimum degree at least (1 α 1 α ) n (4α n + ) (1 3α 1 ) n (1 3α 1)n. by using α 1 9α. Similarly the degree from x i or y i, i = 1,, 3, to U 1 or U is at least (1 3α 1 )n. By Proposition 4.1, (U 1, U ) is ( 3α 1, 3 )-super-regular (using α 1 1/9 again).

14 14 GUANTAO CHEN, KATSUHIRO OTA, AKIRA SAITO, AND YI ZHAO Since 3α 1 1, we can apply the Blow-up Lemma to obtain an ESHP u 1 u u 3... v 3 v v 1 of G[U 1, U ] such that u 1 Γ(x 1, x 3 ), u Γ(x ), u 3 Γ(x 1 ), v 1 Γ(y 1, y 3 ), v Γ(y ), v 3 Γ(y 1 ). Since Γ(x i ), Γ(y i ) (1 3α 1 )n for i = 1,, 3 and Γ(x 1, x 3 ), Γ(y 1, y 3 ) (1 6α 1 )n, the restrictive mapping of u 1, u, u 3 and v 1, v, v 3 is possible Extremal Case. In this subsection we prove Theorem 4.3. In Extremal Case 1, we used the Blow-up Lemma to find an ESHP with certain properties in a bipartite graph with very large minimum degree. In this subsection we first prove such a lemma for arbitrary graphs. Lemma Let k 3 and n 1 be sufficiently large. Let G be a graph of order k+n Suppose that G contains an ESP P 0 = u 1...u k. Let X = V (G) V (P 0 ). Suppose that x 1 x x 3 and y 1 y y 3 are two paths of X and let X be the set of the remaining vertices of X (then X = n 1 ). If deg(x, X ) 7 8 n for all vertices x X and deg(u j, X) 7 8 X + 1 for j = 1,, 3, k, k 1, k, then G contains an ESHP that starts with x 3 x x 1, finishes with y 1 y y 3 and contains P 0 as an internal path. Proof. Our proof consists of three steps. Step 1: we find an ESC on X. Let G 1 = G[X ]. Since δ(g 1 ) 7 8 n 1 and n 1 is sufficiently large, by Remark 4.4, G 1 contains an ESHC. Step : we find an ESP on X such that it starts with x 3 x x 1 and finishes with y 1 y y 3. Let v 1,...,v n1 be the ESC given by Step 1. We will form an ESP x 3 x x 1 v i...v 1 v n1 v n1 1...v i+1 y 1 y y 3. for some 1 i n 1. It suffices to have the following adjacency (13) x 3 v i, x v i 1, x 1 v i, x 1 v i, y 1 v i+1, y 1 v i+3, y v i+, y 3 v i+1, in which we assume that v j = v j+n1 for all integers j. Since deg(x, X ) 7n 1 /8 + 1 for any vertex x x 1, x, x 3, y 1, y, y 3, the number of 1 i n 1 satisfying (13) is at least n 1 8(n 1 /8 1) = 8. Thus (13) hods for some 1 i n 1. Step 3: we find an ESHP of G that starts with x 3 x x 1, finishes with y 1 y y 3 and contains P 0 as an internal path. Let n = X = n Denote the ESP found in Step by v 1,...v n, where v 1 = x 3, v = x, v 3 = x 1, v n = y 1, v n 1 = y, v n = y 3. Our goal is to find an index 3 i n 3 such that v 1...v i P 0 v i+1...v n is an ESP. Since P 0 = u 1...u k, it suffices to have the following adjacency (14) u 1 v i, u v i 1, u 3 v i, u 1 v i, u k v i+1, u k v i+3, u k 1 v i+, u k v i+1, for some 3 i n 3. Since deg(u j, X) 7n /8 + 1 for j = 1,, 3, k, k 1, k, the number of 1 i n satisfying (13) is at least n 8(n /8 1) = 8. Thus (13) hods for some 3 i n 3.

15 HAMILTONIAN CYCLES WITH ALL SMALL EVEN CHORDS 15 Proof of Theorem 4.3. We start with defining two new sets, which are variants of V 1 and V. Let α 1 = α 1/3 and α = α /3. We define { V i n } = x V : deg(x, V 3 i ) < α 1. for i = 1,. Since δ(g) > n/, we have deg(x, V i ) > (1 α 1 )n/ for every x V i. Since d(v 1, V ) α, we have V i V i α n/ and V i (1 α )n/ for i = 1,. Consequently, (15) deg(x, V i n ) > deg(x, V i ) α 1 (1 α 1 α ) n for all x V i Let V 0 = V V 1 V. Then V 0 α n and deg(x, V i ) (α 1 α )n/ for all x V 0. Our proof consists of the following two steps which together provide an ESHC of G. Step 1. Find two disjoint ESP s x 1...x p and y 1...y p of length 6 p 14 such that x 1, x, x 3, y 1, y, y 3 V 1 and x p, x p 1, x p, y p, y p 1, y p V. Step. Find two ESP s P 1 and P consisting of all the remaining vertices in V 1 and V V 0, respectively, such that x 3 x x 1 P 1 y 1 y y 3, and x p x p 1 x p P y p y p 1 y p are also ESP s. While Step follows from Proposition 4.11 and Lemma 4.13 easily, Step 1 is much harder (at least from the point of our view) it is where we need the large constant 9 in the min-degree condition. Below we present Step first Step : Find two ESP s covering the remaining vertices. Let P 1 = x 1...x p and P = y 1...y p be the two ESP s of length 6 p 14 provided by Step 1. Let S = V (P 1 ) V (P ). Let U 1 = V 1 S and U 1 = U 1 {x 1, x, x 3, y 1, y, y 3 }. Fix x U 1, by (15), we have (16) deg(x, U 1) (1 α 1 α )n/ p (1 α 1 α )n/ + 1, where the second inequality follows from α n/ 9 p + 1. Using U 1 (1 + α )n/ and α 1 3α, we derive that deg(x, U 1 ) (1 α 1) U Using α , we have deg(x, U 1 ) 7 8 U We then apply Lemma 4.13 to G[U 1] with P 0 = and obtain an ESP x 3 x x 1...y 1 y y 3 on U 1. On the other hand, let U = V V 0 S and U = U {x p, x p 1, x p, y p, y p 1, y p }. Partition U into V = V S and V 0 = V 0 S. We have V 0 α n and for each x V 0, deg(x, V ) (α 1 α ) n p α 1 n 4 V 0, by using α n/ p and α 1 16α. We then greedily find V 0 disjoint 4-stars with the vertices of V 0 as centers and vertices in V as leaves. Let N = (1 + α )n/. Then U V V 0 = n V 1 N. For any vertex x V, the arguments above for G[U 1] give that (17) deg(x, V ) (1 α 1 )N + 1 We then apply Proposition 4.11 (1) to G[U ] and obtain an ESC C 0 of length k = 8 V 0 8α n such that it contains all the vertices of V in a way that every two nearest vertices of V 0 0 are separated by exactly seven vertices of V. We then break C 0 into an ESP P 0 = u 1...u k such that u 1, u, u 3, u k, u k 1, u k V. Let X = U V (P 0) and X = X {x p, x p 1, x p, y p, y p 1, y p }. For any vertex x V, by (17) and using 8α n α 1 n α 1N, deg(x, X ) (1 α 1 )N + 1 8α n (1 3α 1 )N + 1.

16 16 GUANTAO CHEN, KATSUHIRO OTA, AKIRA SAITO, AND YI ZHAO Since X V, u 1, u, u 3, u k, u k 1, u k V and X < X N, by letting α 1 1 4, the degree conditions in Lemma 4.13 hold. We then apply Lemma 4.13 to G[U ] and obtain the desired ESHP x p x p 1 x p...p 0...y p y p 1 y p Step 1: Connect V 1 and V. Given two disjoint sets A and B, an ESP on A B is called an (A, B)-connector if its first three vertices are from A, and the last three vertices are from B. Our goal is to find two disjoint (V 1, V )-connectors of length at most 14. The simplest connector is an ESP x 1...x 6 with x 1, x, x 3 V 1 and x 4, x 5, x 6 V. Unfortunately such a simple connector may not exist if e(v 1, V ) is very small but e(v i, V 0) is relatively large. Let us sketch our proof. We first separate the vertices of V 0 with large degree to both V 1 and V : V 0 = {x V 0 : deg(x, V 1), deg(x, V ) n/6 α n} The reason why we choose n/6 can be seen from (0), in which we use n/ = 3(n/6). If V 0 > 165, then we can find two disjoint copies of T,3, from the 3-partite subgraph G[V 1, V 0, V ], where T,3, is the union of two copies K,3 sharing the three vertices in one partition set. Each copy of T,3, can be easily extended to an (V 1, V )-connector. If V 0 165, then V 0 will not be used any more. We add the vertices of V 0 V 0 into V 1 or V forming two new (disjoint) sets U 1, U such that any three vertices in U i, i = 1,, have many common neighbors. What remains is to find two disjoint (U 1, U )-connectors by using the minimum degree condition δ(g) (n+ V 0 )/+9. One way to construct such a connector is to find two adjacent vertices x U 1, y U such that there is 4-vertex path between Γ(x, U 1 ) and Γ(y, U ). If this can not be done, then we find six vertices x 1, x, x 3 U 1 and x 4, x 5, x 6 U such that x 1 x 4 and x 3 x 6 are edges and x, x 3, x 4, x 5 form a copy of K, in G[U 1, U ]. After finding one connector, we remove all or some of its vertices and repeat the procedure above. Note that ignoring V 0 of size at most 165 is the major reason for the large constant 9 in δ(g); for example, when V 0 =, then δ(g) n/ + 9 suffices. We need the following propositions on the existence of T,3, and K,, whose proofs are standard counting arguments. Given two disjoint vertex sets A, B in a graph H, we write K s,t (A, B) if H[A, B] contains a copy of K s,t with s vertices from A and t vertices from B. Similarly T,3, (A, B, C) means that there are subsets X A, Y B and Z C such that H[X, Y ] = H[Z, Y ] = K,3. We denote by δ(a, B) the minimum degree deg(a, B) over all a A. Proposition (1) Give any integer t > 0, there exist ε 0 > 0 and n 0 such that the following holds for any ε ε 0 and n n 0. Let A, B be two disjoint vertex set in a graph such that δ(b, A) n/6 3εn, A (1 + ε)n/ and B > 9(t 1). Then K,t (A, B). () There exist ε 0 > 0 and n 0 such that the following holds for any ε ε 0 and n n 0. Let A, B, C be three disjoint vertex set in a graph such that δ(b, A), δ(b, C) n/6 3εn, A, C (1 + ε)n/ and B > 16. Then T,3, (A, B, C). Proof. (1). If the graph contains no K,t with vertices in A and t vertices in B, then ( ) ( ) deg(x, A) A (18) (t 1). x B

17 HAMILTONIAN CYCLES WITH ALL SMALL EVEN CHORDS 17 Since δ(b, A) n/6 3εn and A (1 + ε)n/, ( ) ( ) n/6 3εn (1 + ε)n/ B (t 1). As n and ε 0, we obtain B 9(t 1), contradiction. (). Since B > 9(19 1), we can apply (1) with t = 19 to (A, B) and obtain a copy of K,19 on X A of size and Y B of size 19. Then since Y = 19 > 9(3 1), we can apply (1) again with t = 3 to (C, Y ) and obtain a copy of K,3 on Z C of size and Y Y of size 3. The next proposition easily follows from a classical result of Kővári, Sós, and Turán [0]. For completeness, we include its proof below. Proposition Let H = (A B, E) be a bipartite graph such that A = n, B = m. Then H contains a copy of K, if either of the following holds. (1) deg(x) m for all x A and n > m + m, () e := E max{3n, m /}. Proof. Suppose instead, H contains no copy of K,. Then (18) hold with t =. (1). We have ( ) m n ( ) ( ) deg(x) m, x A m(m 1) which implies that n m( m 1) = m( m + 1), a contradiction. (). By convexity, we have n ( ) ( e/n m ), which implies e(e/n 1) < m. Since e m /, we have (e/n 1) < or e < 3n, a contradiction. Now we start our proof. First assume that V 0 > 165. Since δ(v 0, V i ) n/6 α n and V i (1 + α )n/, we can apply Proposition 4.14 () to the 3-partite subgraph on V 1 V 0 V and find a copy of T,3, on X V 1, Y V 0 and Z V such that X = Z =, Y = 3. Let V = V 1 X, V 0 = V 0 Y, and V = V Z. Then V 0 > 16 and δ(v 0, V i ) n/6 α n. We apply Proposition 4.14 () again to 3-partite subgraph on V 1 V 0 V and find another copy of T,3,. We next extend each copy of T,3, to a (V 1, V )-connector of length 11 as follows. Assume X = {x 3, x 5 }, Y = {x 4, x 6, x 8 }, and Z = {x 7, x 9 }. Then x 3, x 4,...,x 8, x 9 is an ESP but it is not a (V 1, V )-connector because x 4 V 1 and x 8 V. We extend this ESP by adding two vertices from V 1 in the beginning and two vertices from V at the end. Since x 3, x 5 V 1, by (15), we can find a vertex x Γ(x 3 x 5, V 1 ). Since deg(x, V 1 ) > V 1 α 1n/ and deg(x 4, V 1 ) > n/6 α n, we can find a vertex x 1 Γ(x x 4, V 1 ), which is different from x 3, x 5. Therefore x 1 x... x 9 is an ESP. Similarly we find x 10, x 11 V such that x 1x...,x 10 x 11 is an ESP, which is a (V 1, V )-connector. Now assume that c 0 := V We will not use the vertices of V 0 any more. Since V 0 α n, all vertices x V 0 satisfy deg(x, V 1 V ) > n/ α n. If x V 0 V 0, then exactly one of deg(x, V 1 ) and deg(x, V ) is less than n/6 α n. We thus partition V 0 V 0 into W 1 and W such that W i = {x V 0 V 0 : deg(x, V 3 i ) < n/6 α n}. For i = 1,, we have δ(w i, V i ) δ(w i, V 1 V ) n 6 + α n n α n n 6 + α n = n 3 + α n.

18 18 GUANTAO CHEN, KATSUHIRO OTA, AKIRA SAITO, AND YI ZHAO Let U i = V i W i for i = 1,. The above bound for δ(w i, V i ) and (15) together imply that (19) δ(u i, V i ) n 3 + α n. Since V i (1 + α )n/, for any three vertices x 1, x, x 3 U i, we have (0) deg(x 1 x x 3, V i ) 3( n 3 + α n) V i α n. Without loss of generality, assume that U 1 U. Since U 1 (1 α )n/, we have U (1 + α )n/. It suffices to find two disjoint (U 1, U )-connectors u 1...u q and v 1...v q for some q 8. In fact, if any of u 1, u, u 3 is not from V 1 (note that u 1, u, u 3 U 1 by the definition of (U 1, U )-connectors), then we find at most three new vertices x 1, x, x 3 V 1 such that x 1 x x 3 u 1 u u 3 is an ESP. For example, assume that u 1 V 1. Then we first find x 3 Γ(u 1 u 3, V 1 ), then x Γ(x 3 u, V 1 ), and finally x 1 Γ(x u 1, V 1 ) by applying (0) three times. Similar we add at most three new vertices x q+1, x q+, x q+3 from V such that u q u q 1 u q x q+1 x q+ x q+3 is an ESP. The resulting ESP x 1 x x 3 u 1...u q x q+1 x q+ x q+3 is a (V 1, V )-connector of length at most 14. We obtain the other connector similarly. The following technical lemma is the main step in our proof, we postpone its proof to the end. Lemma Let ε 1 and n be sufficiently large. Suppose that G is a graph of order n with a vertex partition U 0 U 1 U such that U 0 = c 0 n; U 1 U (1 + ε)n/; U 1 contains a subset V 1 such that δ(v 1, V 1 ) (1 ε)n/; δ(u, U ) n/3. If δ(g) n+c 0 + 6, then G[U 1, U ] contain either of the following 6-vertex subgraphs. H 1 : Two vertices x U 1, x 5 U are adjacent; two vertices x 1, x 3 Γ(x, V 1 ) and two vertices x 4, x 6 Γ(x 5, U ) form a path x 1 x 4 x 3 x 6. H : Four vertices x, x 3 U 1 and x 4, x 5 U form a copy of K, ; a vertex x 1 Γ(x 4, U 1 ), and a vertex x 6 Γ(x 3, U ). We observe that subgraphs H 1 and H given by Lemma 4.16 can be easily converted to (U 1, U )-connectors. In the case of H 1, x 1 x x 3 x 4 x 5 x 6 is an ESP and thus a (U 1, U )- connector. In the case of H, by (0), x 1, x, x 3 have a common neighbor x 7 V 1, and x 4, x 5, x 6 have a common neighbor x 8 V. The x 1x 7 x x 4 x 3 x 5 x 8 x 6 is an ESP and thus a (U 1, U )-connector. x x 5 x 1 x 4 x 1 x 4 x x 5 x 3 x 6 x 3 x 6 H 1 H Figure 3. Convert H 1 and H to ESP s