On arithmetic progressions of cycle lengths in graphs

On Arithmetic Progressions of Cycle Lengths in

Graphs

Jacques Verstra¨e te

Department of Pure Mathematics and Mathematical Statistics

Centre for Mathematical Sciences

Wilberforce Road,Cambridge CB3OWB

England August1999.

jbav@185fc662ddccda38376baf7a

Abstract

A recently posed question of H¨a ggkvist and Scott’s asked whether

or not there exists a constant c such that if G is a graph of minimum

degree ck then G contains cycles of k consecutive even lengths.In

this paper we answer the question by proving that for k≥2,a

bipartite graph of average degree at least4k and girth g contains

cycles of(g/2?1)k consecutive even lengths.We also obtain a

short proof of the theorem of Bondy and Simonovits,that a graph

of order n and size at least8(k?1)n1+1/k has a cycle of length2k.

Erd?o s and Burr[4]conjectured that for every odd number k,there is a constant c k such that for every natural number m,every graph of average degree at least c k contains a cycle of length m modulo k.Erd?o s and Burr[4] settled their conjecture in the case m=2and Robertson(see[4])settled the case m=0.The full conjecture was resolved by Bollob′a s[1],who proved the conjecture with c k=2[(k+1)k?1]/k.In this paper,we show that c k=8k will do.Thomassen[11]later showed cycles of all even lengths modulo k are obtained under the hypothesis that the average degree is at least4k(k+1),without requiring k to be odd.Thomassen[10]also proved that if G is a graph of minimum degree at least three and girth at least 2(k2+1)(3·2k2+1+(k2+1)2?1),then G contains cycles of all even lengths modulo k.

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Bondy and Vince[3]proved that in a graph in which all but at most two vertices have degree at least three,there exist two cycles whose lengths di?er by at most two.This answered a conjecture of Erd?o s and was also studied by H¨a ggkvist and Scott[8].Recently,H¨a ggkvist and Scott[7], considered extending this to considering arithmetic progressions of cycle lengths in graphs.H¨a ggkvist and Scott[7]proved that if G is a graph of minimum degree at least300k2then G contains k+1consecutive even cycle lengths.The same authors asked if a linear bound on the minimum degree is possible.In this paper,we answer the question of H¨a ggkvist and Scott in the following theorem.

Theorem1Let k≥2be a natural number and G a bipartite graph of average degree at least4k and girth g.Then there exist cycles of(g/2?1)k consecutive even lengths in G.Moreover,the shortest of these cycles has length at most twice the radius of G.

This result generalises the above-mentioned result of Bollob′a s and partly generalises that of Thomassen,insofar as Thomassen’s result is valid for graphs of minimum degree at least three,whereas the result above requires average degree at least eight.The graph K k,n?k shows that we require the average degree to be at least about2k to ensure the conclusion of Theorem 1.

The following lemma lies at the heart of the proof of Theorem1.It was originally inspired by methods used by Gy′a rf′a s,Koml′o s and Szemer′e di[6]. Whilst this paper was being written,the lemma was discovered to be implicit in a lemma of Bondy and Simonovits[2].Nevertheless,the proof is short and is retained here for completeness.

Lemma2Let H be a graph comprising a cycle with a chord.Let(A,B) be a non-trivial partition of V(H).Then H contains A–B paths of every length less than|H|,unless H is bipartite with bipartition(A,B).

185fc662ddccda38376baf7abel the vertices of the cycle0,1,...,n?1where n=|H|. Suppose H does not contain A–B paths of every length less than n,and let

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m be the smallest integer for which there is no A–B path of length m not using the chord;m>1since(A,B)is a non-trivial partition of V(H).We remark also that m≤n/2,or H would contain A–B paths of all lengths less than n.

Nowχ(j)=χ(j+m)for every j∈V(H),whereχis the characteristic function of A(label arithmetic is modulo n).Let d=hcf(n,m).Then there are integers p and q such that pm+qn=d;henceχ(j)=χ(j+d)for every j.But then there is no A–B path of length d round the cycle;thus d=m and m|n.In particular,A–B paths of every length less than m exist by the de?nition of m,so A–B paths of every length other than multiples of m exist by periodicity ofχ.

We?nd paths of the remaining lengths km,1≤k≤n/m?1,using the chord.Suppose?rst that the chord joins two vertices within distance m on the cycle,say0and r where1

So we may suppose the chord is0r,where m

Thus it follows,asχ(j)=χ(m+j),that for?m

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We conclude,therefore,that |H |is even and the vertices of the cycle are alternately in A and in B .It is immediately seen that,under these circumstances,if the chord joins two vertices in the same class then H contains A –B paths of all lengths less than |H |.Consequently,the chord joins A to B ,so H is bipartite,with bipartition (A,B ).

Lemma 3Let k ≥2be a natural number and let G be a graph of average degree at least 2k and girth g .Then G contains a cycle of length at least (g ?2)k +2,with at least one chord.

Proof.It is easily seen that a graph G of average degree at least 2k contains a subgraph H of minimum degree at least k +1.If P is a longest path in H ,then an endvertex v of P has all its neighbours on P .Some neighbour u of v is at distance at least (g ?2)k +1from v on P .Hence P +uv is a cycle of length at least (g ?2)k +2.As k +1≥3,this cycle has at least one chord.

Proof of Theorem 1.We may assume that G is connected and let the radius of G be rad(G ).Choose a central vertex v 0∈V (G ),and let V i denote the set of vertices a distance i from v 0in G .Then there exists l such that V l ∪V l +1spans a graph G with at least k |V l ∪V l +1|edges.By Lemma 3,?nd H ?G comprising a cycle,of length at least (g ?2)k +2,with a chord.Let T be a minimal subtree of T ,restricted to i ≤l V i ,such that T contains V (H )∩V l .The minimality of T ensures that it branches at its root.Now let A be the set of vertices of H in one of these branches and let B =V (H )\A .By Lemma 2,and as (A,B )is not the bipartition of H ,there are A –B paths of all lengths up to (g ?2)k +1,all disjoint from T ?end(T ).Each A –B path of even length s ,together with a subpath of T between the ends of such a path,gives rise to a cycle of length s +2r ,where r is the distance from V l to the root of T .Note that,as G is bipartite,all paths of even length with one end in A have their other end in V l .This gives cycles C 2r +2,C 2r +4,..,C 2r +(g ?2)k ,of (g/2?1)k consecutive even lengths,and since v

0is a central vertex,2r +2≤rad(G ),as required.

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We de?ne the even girth of a graph G to be the length of a shortest even cycle in G.Theorem1easily extends to general graphs,as is shown by the following corollary.

Corollary4Let k≥2be a natural number,and let G be a graph of average degree at least8k and even girth g.Then there are cycles of(g/2?1)k consecutive even lengths in G.

Proof.This follows from the observation that a graph of average degree at least8k has a spanning bipartite subgraph of average degree at least4k, and then applying Theorem1to this bipartite subgraph.

In the case of graphs of average degree at least6k,we may also argue as follows.Given a vertex v0,let V l denote the vertices a distance l from v0.Then either there exists l such that V l∪V l+1spans a bipartite graph of average degree at least2k,or there exists l such that V l spans a graph of average degree at least2k.In the former case,the method of Theorem1gives cycles of all even lengths in an integer interval of form[2r+1,2r+(g?2)k+1] and the latter case gives(also following the proof of Theorem1)cycles of all odd lengths in an integer interval of form[2r+1,2r+(g?2)k+1].So we have the following theorem:

Theorem5Let k≥2be a natural number,and let G be a graph of average degree at least6k and girth g.Then,for some odd number r≥3,there exist cycles of all even lengths or all odd lengths in the interval[r,r+(g?2)k].

The above result recalls the result of Bondy and Vince[3],that if G is a graph with at most two vertices of degree at most two,then G contains cycles of two consecutive lengths or two consecutive even lengths.This was proved using a technique of Thomassen and Toft[12].In comparison,note that Theorem5requires average degree at least6k where k≥2.Therefore, to ensure two cycles of consecutive lengths or consecutive even lengths,we require average degree at least twelve,which is higher than what is required in the context of Bondy and Vince’s results.

The following result is proved in the same was as Theorem1:

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Corollary6Let k≥2be a natural number,and suppose that G has chro-matic number at least2k+2and girth at least g.Then G contains cycles of k(g?2)consecutive lengths.

The idea is that some level of a breadth-?rst search tree induces a graph of chromatic number at least k+1.Such a graph contains an odd cycle of length at least k(g?2)+1with a chord.We then apply the colouring lemma to deduce that G contains cycles of k(g?2)consecutive lengths. In a sense,this generalizes a result of Gy′a rf′a s who showed that if G has chromatic number at least2k+2,then G contains cycles of k distinct odd lengths.These ideas may also be used to give a relatively short proof that r(C2k+1,K n) n1+1/(k+1).However,better results are easily obtain—for example,it is possible to show that r(C2k+1,K n) n1+1/(k+1)(log2n)?1/k.

The cycles we have obtained are all very close together in the sense that they share many vertices.H¨a ggkvist and Scott[7]asked if it was possible, under an appropriate bound on the size of the graph,to?nd disjoint cycles of k consecutive even lengths.This question also remains open,noting that a bound of order at least k2on the average degree would be required for disjoint cycles of k consecutive even lengths.This is shown,for example,by K l,n?l where l<2k+k(k?1)/2and n is su?ciently large.

We remark that from Theorem1we may obtain a result on extremal numbers for even cycles,that slightly improves the result obtained by Bondy and Simonovits[2](see Corollary9).From their paper,it follows that a graph of order n and size at least90kn1+1/k contains a cycle of length2k. Two simple lemmas are required before proving our result.The?rst lemma is a special case of a lemma of Kostochka and Pyber[9].

Lemma7Let G be a graph of order n and size at least cn1+1/k,where c≥1.Then G contains a subgraph of average degree at least c and radius at most k.

Proof.We may assume that G has minimum degree at least cn1/k.Let v0 be an arbitrary vertex in G and de?ne H i to be the subgraph of G induced by

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vertices at distance at most i from v0.De?ne r=min{i:e(H i)≥1

c|H i|}.

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cn1/k|H i?1|for all i and so,by de?nition of r,|H r?1|> Clearly e(H i)≥1

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n1/k|H r?2|which gives|H r?1|>n(r?1)/k.Since|H r?1|≤n,r?1

r is the desired subgraph.

Lemma8Let G be a graph of order n with e(G)≥2n1+1/k where k≥2. Then G has girth at most2k+1.

Proof.By Lemma6,G has a subgraph H of radius at most k and average degree at least two.So H contains a cycle and some cycle in H has length at most2k+1.

In particular,if G is bipartite and k=2in Lemma7,then G has a cycle of length four.For comparison,a standard result states that a graph which has at least n3/2/2+n/4edges contains a cycle of length four.We are able, using Theorem1,to show the existence of longer even cycles:

Theorem9Let G be a bipartite graph of order n and girth g,and of size at least4 2(k?1)/(g?2) n1+1/k,where k≥2is an integer.Then G has a cycle of length2k.

Proof.By Lemma7,either G contains a cycle of length2k or g<2k and k≥3.In the latter case, 2(k?1)/(g?2) ≥2.Lemma6shows that G contains a subgraph of average degree at least4 2(k?1)/(g?2) and of radius at most k.By Theorem1,there are cycles of at least k?1 consecutive even lengths in H,the shortest length being at most2k.So one of these cycles must have length exactly2k.

As a corollary to Theorem8,we slightly improve the result of Bondy and Simonovits[2]:

Corollary10Let G be a graph of order n and size at least8(k?1)n1+1/k, where k≥2.Then G contains a cycle of length2k.

Acknowledgements

I would like to thank Andrew Thomason for his many helpful suggestions.

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