Cycles with many chords

How many edges in an n$$ n $$ ‐vertex graph will force the existence of a cycle with as many chords as it has vertices? Almost 30 years ago, Chen, Erdős and Staton considered this question and showed that any n$$ n $$ ‐vertex graph with 2n3/2$$ 2{n}^{3/2} $$ edges contains such a cycle. We significantly improve this old bound by showing that Ω(nlog8n)$$ \Omega \left(n\kern0.2em {\log}^8n\right) $$ edges are enough to guarantee the existence of such a cycle. Our proof exploits a delicate interplay between certain properties of random walks in almost regular expanders. We argue that while the probability that a random walk of certain length in an almost regular expander is self‐avoiding is very small, one can still guarantee that it spans many edges (and that it can be closed into a cycle) with large enough probability to ensure that these two events happen simultaneously.


Introduction
One of the classical problem frameworks in combinatorics deals with questions of the following type.How many edges does an n-vertex graph need to have to contain a subgraph with a certain prescribed structure?In many instances of such problems, it turns out that we can find subgraphs with very interesting structure only assuming very weak bounds on the number of edges.
For example, Janzer and Sudakov [8] showed that any n-vertex graph with average degree at least Ωplog log nq contains a k-regular subgraph, which is optimal up to a constant factor and answers an old question of Erdős and Sauer.Liu and Montgomery [10] recently solved several open problems using methods related to sublinear expansion.In particular, they showed that any graph with a large enough constant average degree contains a cycle whose length is a power of 2. Another result of similar flavour by Bucić, Gishboliner, and Sudakov [2] shows that for k ě 3, every k-regular Hamiltonian graph has cycles of n 1´op1q many lengths, asymptotically solving a problem of Jacobson and Lehel.Furthermore, Fernández and Liu [7] proved a conjecture of Thomassen [14], showing that large enough constant average degree forces the existence of a pillar (two vertex-disjoint cycles of the same length, along with vertex-disjoint paths of the same length which connect matching vertices in order around the cycles).
Many of the problems of this sort also deal with conditions which force the existence of cycles with chords.Answering a question of Erdős [5], Bollobás [1] proved that a large enough constant average degree is enough to force the existence of a cycle whose chords also contain a cycle.Extending this result, Chen, Erdős and Staton [4] proved that for every k ě 2 there is a constant c k such that any graph with average degree at least c k contains k cycles C 1 , . . ., C k , such that the edges of C i`1 are chords of the cycle C i .This answered a question of Bollobás [1].More recently, Fernández, Kim, Kim and Liu [6] strengthened the result of Bollobás, showing that large enough constant average degree is enough to force the existence of a cycle whose chords contain a cycle whose vertices follow the orientation of the first cycle.Another similar result was shown by Thomassen [13], who proved that for every k ě 1, there exists g k such that any graph with minimum degree 3 and girth at least g k contains a cycle with at least k chords.
In 1996, Chen, Erdős and Staton [4] also considered the following natural question: how many edges force the existence of a cycle with as many chords as it has vertices?They showed that if an n-vertex graph has minimum degree at least 2 ?n then it contains a cycle which has n chords, thus showing that 2n 3{2 edges are enough.In this paper, we significantly improve this old result of Chen, Erdős and Staton, by showing that Ωpn log 8 nq edges are enough to force a cycle with at least as many chords as it has vertices.
Theorem 1.1.If n is sufficiently large, then every n-vertex graph with at least n log 8 n edges contains a cycle C with at least |C| chords.
Overview of the proof.Initially, we undertake a process of cleaning the graph, i.e. finding a subgraph that is nearly regular (with a constant factor difference between the minimum and maximum degrees), good expansion properties, and a sufficiently high average degree.Subsequently, we investigate a random walk of an appropriate length within this subgraph.We consider two critical events: firstly, we analyze the probability that the random walk is self-avoiding, meaning that it does not revisit any of the previously visited vertices.Secondly, we assess what is the likelihood of the set of vertices visited by the random walk to span at least chords.While the occurrence of the first event is characterized by an exponentially small probability q, we carefully establish that the second event still holds with probability more than 1 ´q.Crucially, for bounding the probability for the first event we use the fact that the obtained graph has good expansion properties.For the second event, directly applying standard concentration inequalities does not yield a strong enough bound on the required probability.To remedy the situation, we prove an edge-decomposition result in almost-regular graphs, which combined with concentration inequalities produces the required bound.
Notation.We use standard graph theoretic notation throughout the paper.In particular, for a graph G, we denote by dpGq its average degree, and by δpGq, ∆pGq its minimum degree and maximum degree, respectively.By epGq, we denote the number of edges of G, and for S Ď V pGq, by e G pSq we denote the number of edges induced by S. For two disjoint sets A, B Ď V pGq, e G pA, Bq is the number of edges of G which are incident to both A and B. We omit the subscripts if it is clear from the context which graph we refer to.Given an event E in a probability space, we denote by 1 E the indicator random variable of E, which is equal to 1 when E holds, and 0 otherwise.

Preliminaries
In this section, we collect several useful definitions and results used in our proofs.Definition 2.1.Let K ą 0 and let G be a graph.We say that G is K-almost-regular if ∆pGq ď KδpGq.
We use the following lemma from [3], which states that every graph contains an almost regular subgraph whose average degree is at most by a logarithmic factor smaller than that of the original graph.
Lemma 2.3.Every graph G on n vertices contains a 6-almost regular subgraph G 1 Ď G with average degree at least dpGq 100 log n .

Finding an almost regular expanding subgraph
The goal of this subsection is to prove the following standard statement which allows us to find a (weakly) expanding subgraph in any graph with large enough degree.Its proof is a standard application of the density increment method.
Lemma 2.4.Let G be an n-vertex graph with average degree d ě log 2 n and let n be large enough.
Then there exists a bipartite subgraph G 1 Ď G with the following properties.
Proof.First, let G 0 be a bipartite subgraph G with average degree at least d 3 .Then, we apply Lemma 2.3 to G 0 in order to find a (bipartite) 6-almost regular subgraph G 1 Ď G 0 with dpG 1 q ě d 300 log n .Now let λ :" 1  2 log n and let d 1 :" dpG 1 q.We now perform a procedure which finds the desired subgraph G 1 in G 1 .At every step, we consider a subgraph H and show that either G 1 :" H satisfies the desired properties and we finish the procedure or we will find a certain subgraph H 1 Ď H and continue the procedure with H 1 .We will then show that at some point this procedure must finish.
Let us now describe a step in this procedure.Consider a subgraph H with average degree dpHq.If H has a vertex v with degree less than dpHq{2 we remove it, and define H 1 :" Hzv and proceed to the next step with H 1 .Note that H 1 has average degree at least dpHq.such that epU, Ū q ă λ 3 dpHq|U | then either dpHrU sq ě dpHq or dpHrU sq ě p1 ´λqdpHq.If dpHrU sq ě dpHq, we define H 1 :" HrU s and proceed to the next step with H 1 .On the other hand, if dpHrU sq ě p1 ´λqdpHq, we define H 1 :" HrU s and proceed to the next step with H 1 .Now, note that at any step where the procedure does not terminate, the following always holds: if |V pH 1 q| ě |V pHq|{2 then dpH 1 q ě dpHq; if |V pH 1 q| ă |V pHq|{2 then dpH 1 q ě p1 ´λqdpHq.Also, we have |V pH 1 q| ă |V pHq|.Furthermore, since |V pH 1 q| ă |V pHq|{2 can only occur for at most log n steps, at any step the subgraph H we consider satisfies dpHq ě p1´λ log nqdpG 1 q ě dpG 1 q{2 ą 0, and therefore, the procedure must eventually stop with a non-empty subgraph G 1 and dpG 1 q ě dpG 1 q{2.
We now show that this final subgraph G 1 satifies the desired properties.Firstly, the discussion above implies that dpG 1 q ě dpG 1 q 2 ě d 600 log n .Secondly, note that since the procedure removes every vertex of low degree, we have δpG 1 q ě dpG 1 q{2.Since G 1 was 6-almost regular and dpG 1 q ě dpG 1 q{2 we have that ∆pG 1 q ď 100δpG 1 q as desired.Finally, the procedure also implies that every set U Ď V pG 1 q of size at most |V pG 1 q|{2 satisfies epU, Ū q ě λ 3 dpG 1 q|U | ě

Random walks in expanders
In this subsection we compute the mixing time of a random walk in an almost regular expander.
The notation and results that we cite in this subsection can be found in [9] and [11].
Let G be a connected graph on the vertex set rns.Consider a random walk on V pGq, where we start at some vertex v 0 and at the i-th step we move from v i to one of its neighbours, say v i`1 , where each neighbour of v i is chosen as v i`1 with probability 1 dpv i q .Let M be an n ˆn matrix defined as follows.Let M v,u be the probability of stepping from v to u; so M v,u " 1 dpvq if vu P EpGq, and M v,u " 0 otherwise.Denote by D the n ˆn diagonal matrix with D v,v " 1 dpvq for v P rns, and let A be the adjacency matrix of G. Then M " DA.So the probability that a random walk starting at vertex v ends in u after t steps is pM t q v,u .Definition 2.6.Let the graph G and matrices M, D, A be as above and define N pGq " D 1{2 AD 1{2 .Note that the matrix N pGq is symmetric, so it has n real eigenvalues.Let λ 1 pN q ě λ 2 pN q ě ¨¨¨ě λ n pN q denote the eigenvalues of N :" N pGq.
Lemma 2.7 (Lemma 5.2 in [9]).Let G be a connected n-vertex bipartite graph, with the bipartition tX, Y u with m edges.Let M " DpGqApGq and N " N pGq.Then for every v, u P V pGq and integer k ě 1, we have Note that Lemma 2.7 says that when k is even and both v, u are in the same part or when k is odd and v, u are in different parts then ˇˇpM k q v,u ´dpuq m ˇˇď b dpuq dpvq ¨`λ 2 pN q ˘k.Also observe that when k is even and v and u are in different parts or when k is odd and v and u are in the same part then pM k q v,u " 0. Definition 2.8 (Conductance).For a graph G with m edges, let πpvq " dpvq 2m , and for any S Ď V pGq, let πpSq :" ř sPS πpsq; observe that πpSq ď 1 for every S Ď V pGq.Define the conductance of a set S, denoted by ΦpSq, as ΦpSq :" epS, Sq 2m ¨πpSqπpSq , and let the conductance of a graph G, denoted by Φ G , be defined as ΦpSq.
Proof.Suppose all the vertices in G have their degrees between d and Kd.Let S Ď V pGq.The above inequality thus implies Φ G ě λ K .
Combining Lemma 2.10 and Theorem 2.9, we obtain that if G is a K-almost-regular λ-expander and λ 2 " λ 2 pN pGqq, then λ 2 ď 1 ´1 8 p λ K q 2 .Therefore, Lemma 2.7 implies the following.Corollary 2.11.Let λ ą 0, K ě 1, and let G be a bipartite graph on n vertices which is a Kalmost-regular λ-expander.Let tX, Y u be the bipartition of G with m edges and no isolated vertices.Let M " DpGqApGq and N " N pGq.Then for every v, u P V pGq and integer k ě 1, the probability pM k q v,u that a random walk starting at vertex v ends in u after k steps satisfies We shall utilize the following definition of mixing time in bipartite graphs.
Definition 2.12 (Mixing time).Let G be a bipartite graph on n vertices.Let tX, Y u be the bipartition of G with m edges and no isolated vertices.We say that G has mixing time k if for every u, v P G, the probability pM k q v,u that a random walk starting at vertex v ends in u after k steps satisfies ˇˇˇp The following is a corollary of the previous statements and succinctly summarizes a few pertinent properties of mixing time that are essential for our proofs.piq For any given set S Ď V pGq, the probability that a random walk starting at a given vertex ends in a vertex of S after at least k steps is at most 4K n |S|.
piiq If k 1 is even (if k 1 is odd), then the probability that a random walk starting at a vertex of X ends in any given vertex of X (of Y respectively) after k 1 ě k steps is at least 1 Kn .
piiiq If G is a λ-expander for some λ ą 0, then it has mixing time k ď 30K 2 λ 2 log n.Proof.Suppose G has m edges.Since G is K-almost-regular and k is its mixing time, the probability that a random walk starting at a given vertex ends in a vertex of S after k steps is at most Similarly, the required probability in piiq is at least Finally, by Corollary 2.11, if G is a λ-expander, then it has mixing time at most 30K 2 λ 2 log n.

Proof
As mentioned earlier, our strategy for proving Theorem 1.1 is to first pass to an almost-regular expander (using Lemma 2.4).In Section 3.1, we show that one can find a collection of star-forests in almost-regular graphs.In Section 3.2, we prove a concentration inequality that allows us to show that a random walk must contain many vertices from any large enough set with high probability.Using this result and the star-forests that we found, we show that the random walk must contain many chords and that it can be closed into a cycle with high enough probability in Section 3.3.In Section 3.4, we compute the probability that a random walk is self-avoiding, and we put everything together and prove Theorem 1.1 in Section 3.5.

Finding star forests in an almost-regular graph
Given disjoint sets A and B, an AB-star-forest F is a set of vertex-disjoint stars such that the root of each of the stars is in A and the leaves are in B. Two star forests F and F 1 are called root-disjoint if the set of root vertices of the stars in F is disjoint from the set of root vertices of the stars in F 1 .Lemma 3.1.Let G " pA, Bq be a 1000-almost-regular bipartite graph on n vertices, and let d :" δpGq ě 10 6 .Then, there exists an AB-star-forest F Ď G consisting of n 100d stars of size d 10 6 .Proof.Note that since G is 1000-almost-regular, by double counting the edges we have |B| ď 1000|A|.Then n " |A| `|B| ď 1001|A|, and so |A| ě n{1001.Let F be a maximal AB-star-forest F Ď G consisting of stars of size d{10 6 and for the sake of contradiction assume that F contains less than n 100d stars.Consider A 1 :" AzV pF q, A 2 :" A X V pF q and B 1 :" BzV pF q, B 2 :" B X V pF q.By assumption, note that |A 2 | ă n{100d ď n{10 5 ď |A|{4 and |B 2 | ď n{10 8 .Every vertex in A 1 must have less than d{10 6 neighbours in B 1 , as otherwise F would not be maximal.Hence epA 1 , B 2 q ě |A 1 |p1 ´10 ´6qd ě |A|d{4.Therefore there is a vertex in B 2 with degree at least a contradiction with the fact that G is 1000-almost-regular.Repeated application of the lemma above produces a collection of root-disjoint AB-star forests.This is shown by the following corollary.Proof.Suppose we have already found the desired AB-star-forests F 1 , F 2 , . . ., F i for i ă d 10 , we find the next one F i`1 as follows.We remove the root vertices of the stars in F 1 , F 2 , . . ., F i from A Ď V pGq.This removes at most i ¨n 10 5 d ¨100d ă nd 10 4 edges from G as ∆pGq ď 100d (since G is 100-almost-regular), let the resulting graph be G 1 .Hence, G 1 still has at least p 1 2 ´10 ´4qnd edges, so by repeatedly removing vertices of degree less than d 4 , we obtain a subgraph G 2 of G 1 with minimum degree d 4 ě 10 6 , while its maximum degree is still at most 100d, so G 2 is 1000-almostregular.Moreover, G 2 contains at least p 1 2 ´10 ´4qnd ´nd 4 ě p 1 4 ´10 ´4qnd edges, so G 2 has at least ě n 250 vertices (as ∆pG 2 q ď 100d).So by Lemma 3.1, G 2 has an AB-star-forest F i`1 consisting of n 10 5 d stars of size d 10 7 , as desired.

Intersection of random walks with arbitrary sets
For a random walk W " tX i u on a graph G with mixing time k, the set of vertices tX ik : i P rtsu for some t ě 1 behaves almost like a random set of t vertices chosen uniformly at random with repetition from G. We exploit this fact in this subsection.More precisely, let G be a 100-almost-regular bipartite graph on n vertices with parts A, B and consider a random walk R of length t starting at some vertex v 0 P A and take a k 1 P tk, k `1u which is odd, where k is the mixing time of G. Let S be a random set obtained by the following procedure which consists of tt{k 1 u steps: • In each step 1 ď i ď tt{k 1 u, with probability 10 ´5 we either choose a uniformly random vertex v i from A (if i is even) or from B (if i is odd), or we do nothing (with probability 1 ´10 ´5).
• The set S " tv i : 1 ď i ď tt{k 1 uu consists of all of the chosen vertices.Now, consider the set of vertices Rpkq :" tu 1 , u 2 , . ..u were for each 1 ď i ď tt{k 1 u, u i is the ik 1 -th vertex of the random walk R. We then have the following property given by the definition of mixing time and Corollary 2.13.
Observation 3.3.Conditioning on the choice of u 0 , u 1 , . . ., u i´1 , we have that for every a P A and b P B it holds that: By definition of S this implies that for any fixed set X Ď V pGq, the random variable |Rpkq X X| stochastically dominates |S X X|.Therefore, we have the following.Proof.Let R " tX i : 0 ď i ď tu be a random walk, and without loss of generality, suppose X 0 P A.
As noticed before, |S X X| is stochastically dominated by |Rpkq X X|, so it is enough to show the statement with |S X X| instead of |Rpkq X X|.Note that either X X A or X X B has size at least |X|{2.Suppose without loss of generality that |X X A| ě |X|{2, the other case is very similar.Let C :" 10 ´9 |X|t kn , and consider the procedure that was used to define S, where in each step 1 ď j ď tt{k 1 u, a vertex v j is (randomly) added to S. Suppose that a new vertex from X is added to S in only at most i ď C steps; we are interested in the probability that this event occurs.Fix such a choice of i steps.In any such step, the probability that a new vertex from X is added to S is at most maxt10 ´5 |X| |A| , 10 ´5 |X| |B| u ď |X| 100n (since G is 100-almost-regular).Moreover, note that since at most C ď |X| 4 vertices from X have been added to S, the probability that a new vertex from X X A is added to S in any step j (with j even) is at least 10 ´5 |X|{2´|C| |A| ě 10 ´6 |X| n .Therefore, as there are at least 1  2 tt{k 1 u ´C ě 1 4 tt{k 1 u steps j (with j even) where no new vertex is added to S, the required probability is at most Since the common ratio satisfies

Chords in random walks
In this subsection, we show that with very high probability, the graph induced by the vertices of two random walks contains many edges.Recall that for a random walk R of length t, we denote Rpkq " tu 1 , u 2 , . ..u were for each 1 ď i ď tt{k 1 u, u i is the ik 1 -th vertex of the random walk R and k 1 P tk, k `1u is odd.Lemma 3.5.Let G be a 100-almost-regular bipartite graph on n vertices with δpGq " d ě 10 8 and mixing time k.Let R 1 and R 2 be random walks in G of length t for n ě t ě maxt10 17 kn d , 10 25 k log nu, starting at arbitrary vertices v 1 and v 2 , respectively.Then, Proof.Consider some F i and let A, B be the bipartition of G, and denote by m i " n 10 5 d the number of stars in F i .Fix a collection of m i {2 of stars in F i and note that the probability that every star in this collection has less than dp{10 7 leaves in R 1 pkq is at most e ´mi dp 2¨10 7 by applying Lemma 3.4 to the set of leaves of all the m i {2 stars in the collection.
Hence, by the union bound over all such collections of m i {2 stars of F i , we have that the event from the statement of the claim does not hold with probability at most where we used that m i ď dpm i 10 8 , since p ě 10 8 {d.
By a simple union bound and since pn ě 10 16 log n, we then have that with probability at least 1 ´ne ´pn 10 13 ě 1 ´e´p n 10 14 the following holds: for every star-forest F i , more than half of its stars each have at least dp{10 7 leaves in R 1 pkq.Suppose this event occurs.Then for each F i , let A i denote the set of vertices in F i which are the roots of stars with more than dp{10 7 leaves in R 1 pkq.Then, we have

Self-avoiding walks in expanders
In this subsection we show that a random walk with small mixing time in an almost-regular graph is self-avoiding with a certain positive probability.The exact details are given in Theorem 3.7, whose proof uses the ideas from [12], with the necessary changes to fit our setting.
Let G be a graph with mixing time k.Denote by tX v t u the nearest neighbour random walk in G which starts at a vertex v.For a vertex set A Ď V pGq, let Q v t pAq denote the probability that X v t P A, and let E v A be the event that X v t R A for all t P rks, i.e. the random walk starting at v avoids the set A in the first k steps.Theorem 3.7.Let β :" 10 ´28 , and let G be a 100-almost-regular graph with mixing time k with δpGq ě 10 3 k 2 β .Then the probability that a random walk starting at any given vertex of G and of length β 2 n 10 6 k is self-avoiding is at least e ´β3 n 10 5 k 2 .
Proof.The following claim allows us to show that most vertices v are such that if we start a random walk at the vertex v, it is likely to avoid a given set.
Claim 3.8.For every set A Ď V pGq, it holds that the set B of vertices v such that PpE v A q ě β is of size at most 100k|A| β .

Proof of claim. Notice first that
For every pair of vertices v, u P G we have that Q v t puq ď 100 ¨Qu t pvq because of our assumption that G is 100-almost-regular.Indeed, for every walk P " v 0 , v 1 , . . ., v t we know that Since Q v t puq " , where the sum is over all walks P " v 0 , v 1 , . . ., v t of length t with v 0 " v and v t " u, we conclude that t pvq " dpuq dpvq ď ∆pGq δpGq ď 100.Using this, we obtain the following: This immediately implies the claim as The following claim gives the probability that a random walk of length k is self-avoiding and additionally avoids a fixed set of k vertices.
Claim 3.9.Let S Ď V pGq with |S| ď k, and let v P V pGq.The probability that X v i ‰ X v j for every 1 ď i ă j ď k and that X v i R S for all i P rks, is at least 1 ´200k 2 ∆pGq .
Proof of claim.For each i ď k, out of the dpX v i q neighbours of X v i , only at most |S| `i vertices are contained in S Y tX v 0 , X v 1 , . . ., X v i´1 u.Hence the required probability is at least For every t ď β 2 n 10 6 k 2 , let A t :" tX v j u jďtk be the set of vertices visited by the random walk in the first tk steps, and let Z t be the set of vertices u for which α u pA t´1 q ě 1 ´β ´200k 2  ∆pGq .Let us now show by induction on i that with probability at least p1 ´2βq i our random walk is self-avoiding after ik steps and moreover, it ends in a vertex of Z i .By setting i " β 2 n 10 6 k 2 we can then complete our proof of Theorem 3.7.As p1 ´2βq ě e ´10β for small β ą 0, we get p1 ´2βq To that end, suppose that with probability at least p1 ´2βq i´1 our random walk is self-avoiding after pi ´1qk steps and moreover, it ends in a vertex u P Z i´1 i.e., X v pi´1qk " tuu.Now, we claim that the probability that our random walk is self-avoiding in the next k steps, avoids tX v j u jďpi´1qk and satisfies that X v ik P Z i is at least Indeed, by Lemma 3.10, the probability that X v ik P Z i is at least 1´β 10 and the probability that the random walk tX v pi´1qk`1 , X v pi´1qk`2 , . . ., X v ik u does not avoid A i´2 YtX v pi´2qk`1 , X v pi´2qk`2 , . . ., X v pi´1qk u " A i´1 is at most 1 ´αu pA i´2 q (which is at most β `200k 2  ∆pGq since u P Z i´1 ), and by Claim 3.9, the probability that the random walk tX v pi´1qk`1 , X v pi´1qk`2 . . ., X v ik u is not self-avoiding is at most 200k 2 ∆pGq .Putting all of this together and using that ∆pGq ě 10 3 k 2 β , the above inequalities hold, as desired.This completes the proof of Theorem 3.7.

Putting everything together
Proof of Theorem 1.1.Let G be a graph on n vertices for n large enough, and with average degree dpGq :" d ě log 8 n.First we use Lemma 2.4 to find a 100-almost-regular bipartite subgraph G 1 on n 1 vertices with average degree at least d 600 log n which is a 1 10 log n -expander.Now, by Corollary 2.13, we have that G 1 has mixing time at most k :" 10 10 log 2 n log n 1 .Let β " 10 ´28 .
Consider now a random walk R " tX v 0 j u jďt starting at an arbitrary vertex v 0 P V pG 1 q and of length t :" β 2 n 1 10 6 k .Let E 1 be the event that R is self-avoiding, and let E 2 be the event that there is an edge between the first t{4 and last t{4 vertices of R.
As we have several parameters, we now collect several simple inequalities which hold between them, and which we use to complete our proof.Note first that since the average degree in G 1 is at least d 600 log n and G 1 is 100-almost-regular, we have δpG 1 q ě d 10 5 log n ě log 7 n 10 5 .Note further that trivially n 1 ě δpG 1 q ě log 7 n 10 5 , and also that δpG 1 q " k 2 " Oplog 6 nq.Now, by Theorem 3.7 the event E 1 occurs with probability at least e ´β3 n 1 10 5 k 2 " e ´10βt k .Now we want to show that t 4 ě maxt 10 17 kn 1 δpG 1 q , 10 25 k log n 1 u, so that we can apply Lemma 3.5 to obtain that with probability at least 1 ´e´t {4 10 24 k the event E 2 occurs.Indeed, the first inequality follows from the fact that t " Θp n 1 k q and δpG 1 q " k 2 .The second inequality follows from k 2 log n 1 " opn 1 q.To see why k 2 log n 1 " opn 1 q holds, we have two simple cases.If n 1 ě log 8 n then this trivially holds as k 2 " Oplog 6 nq, and otherwise log n 1 " Oplog log nq, so again we are done because n 1 " Ωplog 7 nq.
Finally, for each i P rks, let W i be the random walk starting at the p t 4 `iq-th step of the random walk R and finishing at step 3t 4 of R. For each W i , we will consider the set W i pkq, and show that it spans many edges with high probability.Again, we can easily check that t{2´i 2 ě t 5 ě maxt 10 17 kn 1 δpG 1 q , 10 25 k log n 1 u, so by Lemma 3.5 for every i P rks we have that epW i pkqq ě pt{5q 2 δpG 1 q 10 32 k 2 n 1 ą 2t k with probability at least 1 ´e´t {5 10 24 k , as we can split the random walk W i into two random walks of length at least t{5.Let E 3 be the event that for all i P rks, epW i pkqq ě 2t k .Since k " Oplog 3 nq and t k " Ωplog nq, by a union bound, E 3 occurs with probability at least 1 ´ke ´t 10 25 k ě 1 ´e´t 10 26 k .Since β " 10 ´28 , we have PpE 1 XE 2 XE 3 q ě PpE 1 q´PpE 2 q´PpE 3 q ě e ´10βt k ´e´t 10 26 k ´e´t 10 25 k ą 0.Moreover, the event E 1 X E 2 X E 3 implies the existence of a cycle of length t with at least t chords, since if the random walk R is self-avoiding, then the edges spanned by the sets W i pkq are mutually disjoint for i P rks.This completes our proof of Theorem 1.1.

Concluding remarks
In this paper we have shown that every n-vertex graph with Ωpn log 8 nq edges contains a cycle C with at least |C| chords.Although this is a significant improvement upon the previous bound [4] of Θpn 3{2 q edges, we believe that the truth is closer to Θpnq.It would be interesting to show an upper bound of this order (which would be optimal), or to show any lower bound which is super-linear.
Another avenue towards understanding this problem is to consider the following closely related question.What is the largest t " tpe, nq so that any n-vertex graph G with e " epnq edges is guaranteed to contain some cycle C with at least t|C| chords?Let us note that our proof gives tpe, nq " Ω ´e n log 7 n ¯for e " Ωpn log 8 nq, and that the question from the previous paragraph is whether tpe, nq ě 1 when e ě cn for a large enough absolute constant c.
Let us note that we did not make an attempt to improve the power of the logarithmic factor or the used absolute constants in our result, in order to keep the presentation clean.We expect that one can save a few logarithmic factors by being more careful, but new ideas are certainly required to push the bound very close to Θpnq, even if we assume the original graph is almost-regular and expanding.Roughly speaking, the reason is that we can only guarantee that the random walk is self-avoiding up to length O `n k ˘, where k is the mixing time of the graph (which is at least of order log 2 n in our proof).Now, if we assume the set of vertices in the random walk behaves like a random set of vertices of size Θ ´n log 2 n ¯, then the expected number of edges spanned by the set is Θ ´e n 2 p n log 2 n q 2 ¯" Θ ´e log 4 n ¯, which is at least Θ ´n log 2 n ¯only when we have e " Ωpn log 2 nq edges in our graph.Additional logarithmic factors are used in our proof for cleaning the graph to find an almost-regular expander in it and because the random walk is not exactly a random set.

Corollary 2 . 13 .
Let K ě 1, let G be a connected K-almost-regular bipartite graph on n vertices with mixing time k and suppose n is large enough.Let tX, Y u be the bipartition of G. Then the following holds:

Lemma 3 . 4 .9 kn ˙ď e ´|X|t 10 9
Let G be an n-vertex 100-almost regular bipartite graph with parts A, B with mixing time k and let R be a random walk in G of length t ď 10n starting at a given vertex.Then, for any set X Ď V pGq we have that P ˆ|Rpkq X X| ď |X|t 10 kn .
ˆepR 1 pkq, R 2 pkqq ď t 2 d 10 32 k 2 n ˙ď e ´t 10 24 k .Proof.Let p :" t 10 9 kn , and note that by the bound on t we have p ě maxt 10 8 First, we apply Corollary 3.2 to find d 10 root-disjoint AB-star-forests F i Ď G, each consisting of n 10 5 d stars of size d 10 7 .Claim 3.6.For each i, with probability at least 1 ´e´np{10 13 , there is a set of at least n 2¨10 5 d stars in F i such that R 1 pkq contains at least dp 10 7 leaves of each of those stars.
d , 10 16 log n n u.
1 2 ¨d 10 ¨n 10 5 d ě n 10 7 and so, by Lemma 3.4, with probability at least 1 ´e´p n 10 7 we have that |R 2 pkq X Ť i A i | ě pn 10 7 .Hence, by the choice of vertices in Ť i A i we have epR 1 pkq, R 2 pkqq ě |R 2 pkqX Ť i A i |¨d p 10 7 ě dp 2 n 10 14 with probability at least p1´e Lemma 3.10.Let A Ď V pGq with |A| ď β 2 n 10 6 k , and let X be the set of vertices v P V pGq such that α v pAq ě 1 ´β ´200k2∆pGq .Then |X| ě n ´100k|A| β .In particular, for every v P V pGq we have Note that by Claim 3.9, we have α v pAq ě P Now, for every v P V pGq and every set A Ď V pGq define α v pAq :" min