Anti-Ramsey Threshold of Cycles for Sparse Graphs
1,2G. F. Barros B. P. Cavalar
Institute of Mathematics and Statistics, University of S˜ao Paulo, S˜ao Paulo, Brazil
G. O. Mota
Center for Mathematics, Computing and Cognition, Federal University of ABC, Santo Andr´e, S˜ao Paulo, Brazil
O. Parczyk
Institut f¨ur Mathematik, Technische Universit¨at Ilmenau, Ilmenau, Germany
Abstract
For graphs GandH, letG−→rbp H denote the property that for everyproper edge colouring ofGthere is a rainbow copy ofH inG. Extending a result of Nenadov, Person, ˇSkori´c and Steger (2017), we prove thatn−1/m2(C) is the threshold forG(n, p)−→rbp Cwhen≥5. Thus our result together with a result of the third author which says that the threshold forG−→rbp C4isn−3/4settles the problem of determining the threshold forG−→rbp Cfor all values of.
Keywords: anti-Ramsey, proper colouring, random graphs, threshold.
1 Introduction
For a positive integer r and graphs G and H, let G → (H)r denote the following Ramsey property: for every edge colouring of G with at most r colours, there is a
1 G. F. Barros was supported by CAPES. B. P. Cavalar was supported by FAPESP (Proc. 2018/05557-7).
G. O. Mota was supported by FAPESP (Proc. 2018/04876-1) and CNPq (Proc. 304733/2017-2). O. Parczyk was partially supported by the Carl Zeiss Foundation. The collaboration of the authors was supported by CAPES/DAAD PROBRAL (Proc. 430/15). This study was financed in part by the Coordena¸c˜ao de Aperfei¸coamento de Pessoal de N´ıvel Superior, Brasil (CAPES), Finance Code 001.
2 Emails:gbarros@ime.usp.br,brunopc@ime.usp.br,g.mota@ufabc.edu.br, olaf.parczyk@tu-ilmenau.de
Electronic Notes in Theoretical Computer Science 346 (2019) 89–98
1571-0661/© 2019 The Author(s). Published by Elsevier B.V.
www.elsevier.com/locate/entcs
https://doi.org/10.1016/j.entcs.2019.08.009
This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/).
monochromatic copy of H in G. In random graph theory, we are interested in a function ˆp: N → [0,1] such that with p pˆthe probability that G(n, p) satisfies some graph property tends to 1 as n tends to infinity. We then say that G(n, p) satisfies this property asymptotically almost surely (a.a.s.) and if additionally we have that forppˆwe a.a.s. do not have this property, then we call ˆpthethreshold for this property. Note that often there are more precise statements known, but for simplicity we will only discuss this kind of threshold.
A classical result in extremal combinatorics by R¨odl and Ruci´nski[9]determines the threshold forG(n, p)→(H)r, whereG(n, p) is the binomial random graph. In particular, when H contains a cycle, the threshold is expressed in terms of the maximum 2-density m2(H) = max{(e(J)−1)/(v(J)−2) :J⊆H, v(J)≥3}. Theorem 1.1 ([9]) Let H be a graph which contains a cycle. Then the thresh- old function pH = pH(n) for the property G(n, p) → (H)r is given by pH(n) = n−1/m2(H).
In this paper we investigate ananti-Ramsey property of sparse random graphs.
Given graphsGandH, we denote byG−→rbp H the following anti-Ramsey property:
everyproper edge colouring ofGcontains arainbow copy ofH inG, i.e. a subgraph of G isomorphic to H in which all edges have distinct colours. Since G−→rbp H is an increasing property, there exists a threshold prbH =prbH(n) for any fixed graph H (see[1]).
R¨odl and Tuza [10] already studied anti-Ramsey properties of random graphs in the early 90’s, obtaining an upper bound for the threshold prbC for G(n, p)−→rbp C, whereC is a fixed cycle. In[4], an upper bound for the thresholdprbH for any fixed graphH was obtained.
Theorem 1.2 ([4]) Let H be a fixed graph. Then there exists a constant C > 0 such that a.a.s. G(n, p)−→rbp H whenever p = p(n) ≥ Cn−1/m2(H). In particu- lar,prbH ≤n−1/m2(H).
The threshold for a graph H is not always given by n−1/m2(H), as in [5] it was proved that there are infinitely many graphs H for which the threshold prbH is asymptotically smaller than n−1/m2(H). However, Nenadov, Person, ˇSkori´c and Steger [8] proved that at least for sufficiently large cycles and complete graphs H the lower bound forprbH matches the upper boundn−1/m2(H) of Theorem1.2.
Theorem 1.3 ([8]) If H is a cycle on at least 7 vertices or a complete graph on at least19 vertices, thenprbH =n−1/m2(H).
In [6], extending Theorem 1.3, it was proved that for complete graphs Kk
withk ≥5 the threshold prbKk is in factn−1/m2(Kk) and that forK4 we have prbK4 = n−7/15 n−1/m2(K4). Our result determines the threshold prbC for every cycle C
on≥5 vertices.
Theorem 1.4 IfH is a cycle on at least5 vertices, thenprbH =n−1/m2(H).
In Section 2 we prove Theorem 1.4. Similarly to what happens for complete graphs, the situation forC4is different. In a short abstract of the third author [7], it is proved thatprbC4 =n−3/4 n−1/m2(C4). For completeness, we exhibit in Section3 the proof of thresholdprbC4obtained in[7]. We use standard notation and terminology (see e.g.[2]and[3]). In particular,C denotes a cycle onvertices,v(G) ande(G) denote the number of vertices and edges of G, respectively, and we write G−H for G−V(H).
2 Cycles on at least five vertices
The maximum density of a graph G is denoted by m(G) = max{e(J)/v(J) :J ⊆G, v(J)≥1}. The results of Nenadov, Person, ˇSkori´c, and Steger[8]provide a general framework that allows for a transference of random Ramsey problems into questions for deterministic graphs with bounded density.
The proof of Theorem 1.3for cycles relies on the following lemma.
Lemma 2.1 ([8]) Let ≥ 7 be an integer and G be a graph such that m(G) <
m2(C). ThenG−→rbp C.
Note that in fact they prove a slightly stronger statement for which they need a non-strict inequality for the density [8, Corollary 13]. The condition≥7 above is simply a consequence of the proof of the lemma, as observed by the authors [8]. We extend Lemma 2.1,proving the following result, which we state in words for clarity.
It is easy to see that m2(C) = (−1)/(−2).
Lemma 2.2 Let ≥ 5 be an integer and G be a graph such that m(G) < (− 1)/(−2). Then there exists a proper edge colouring of G such that every -cycle has two non-adjacent edges with the same colour.
Theorem 1.4 thus follows by replacing Lemma 2.1 with our Lemma 2.2 in the proof of Nenadov et al.[8]. We remark that the proof of Lemma2.2considers all the cycle lengths in the range≥5; it is not a proof only for the cases= 5 and= 6.
Throughout this section let≥5 be an integer andG be a graph with m(G)<
(−1)/(−2). For the proof of our lemma, we shall define a partial proper edge colouring of G such that every -cycle has two non-adjacent edges with the same colour. Clearly, having defined such a partial edge colouring, we can extend it to a proper edge colouring (for instance, the uncoloured edges may be assigned distinct colours).
Let C(G) be the set of all -cycles of G. The edge intersection graph of C(G) is the graph whose vertex set is C(G) and whose edges correspond to pairs {C, C}, C = C, such that E(C)∩E(C) = ∅. A subgraph H ⊆G is aC- component of G if it is the union of all the -cycles corresponding to the vertices of some component of the edge intersection graph of C(G). ClearlyH can be con- structed from an-cycleH1inGas follows. Suppose we have definedH1⊆ · · · ⊆Hi, i ≥ 1. If there is an -cycle C in G such that C ⊆ Hi and E(C)∩E(C) = ∅ for some -cycle C ⊆ Hi then we put Hi+1 = Hi∪C; otherwise we terminate the
construction and set H = Hi. Note that H can be reconstructed by this proce- dure starting from any -cycle H1 ⊆ H. Also note that two-cycles belonging to distinctC-components may share vertices (obviously they do not share edges).
We start the colouring procedure in someC-componentH ofG. Once we have the partial colouring of H, we proceed to assign colours different from those used inH to edges of aC-component ofG−E(H), according to the same steps as those taken for the colouring ofH. We continue in this manner (taking an uncolouredC- component, colouring it and removing its edges), until we have considered all the -cycles ofG. So our aim is to describe the colouring procedure of an arbitraryC- componentH ofG.
Let t be such that H = Ht, i.e. t is the number of steps taken for the con- struction of H starting from some -cycle H1 in G. We may assume t ≥ 2, since the case t= 1 is trivial. The following proposition describes the possibilities for a step of construction ofH. Some configurations are impossible because they imply thate(H)/v(H)≥(−1)/(−2), which contradicts m(G)<(−1)/(−2).
Proposition 2.3 Let 1 ≤i ≤ t−1. Let C be an -cycle which is added to Hi to form Hi+1. There exists a labelling C =u1u2· · ·uu1 such that exactly one of the following configurations of Hi+1 (with respect to C) occurs, where 2 ≤ k ≤ and 3≤j≤:
(Ak) u1u2· · ·uk is a k-path in Hi and uk+1, . . . , u∈/ V(Hi);
(Bj) u1u2 ∈E(Hi), u2u3∈/ E(Hi), {u3, . . . , u} \ {uj} ⊆ V(Hi+1)\V(Hi), uj ∈ V(Hi).
With this proposition at hand the proof of Lemma2.2proceeds as follows: For configurations(Ak)with 2≤k ≤−2 we assign a new colour to two non-adjacent new edges. All other configurations appear at most twice and in these cases we will colour all previous configurations more carefully so that we are able to proceed.
To prove Proposition 2.3 and that bad configurations are rare, we heavily use m(G) < (−1)/(−2). For each 1 ≤ j ≤ i, let rj be the number of edges inE(Hj+1)\E(Hj),sj be the number of vertices inV(Hj+1)\V(Hj), andcj be the number of components of Hj+1−Hj. We have thatrj ≥sj+ 1. In fact, if sj = 0 then rj ≥ 1, and if sj ≥ 1 then the components of Hj+1−Hj are paths and the number of edges ofHj+1with at least one end inHj+1−Hj isrj =sj+cj ≥sj+ 1.
We have that sj ≤ −2, because an -cycle is added to Hj with at least one edge in E(Hj). By an adequate number of applications of the inequality a/b >
(a+ 1)/(b+ 1) (which is equivalent toa > b), we obtain −1
−2 > e(Hi+1)
v(Hi+1) = e(Hi) +ri
v(Hi) +si ≥ +ri+i−1
j=1(sj+ 1) +si+i−1
j=1sj ≥ +ri+ (i−1)(−1) +si+ (i−1)(−2),
(1) which is equivalent tori <((−1)si+)/(−2).
Proof of Proposition2.3 Using this last inequality, one can easily check thatsi=
−3 implies ri ≤ −1. We get that for si = −3 exactly one of the following three alternatives holds:Hi+1−Hi has one component and no edge with both ends in V(Hi) is added to Hi, or Hi+1− Hi has one component and one edge with both ends in V(Hi) is added to Hi, or Hi+1 −Hi has two components and no edge with both ends in V(Hi) is added to Hi. These alternatives correspond to Configurations (A3) and (Bj) with 3 ≤ j ≤ , respectively. The inequality also gives us that 0≤ si ≤−4 implies ri ≤si+ 1; we have thus Configurations(Ak) with 4≤k≤. Configuration (A2) is clearly the only possibility forsi =−2. 2 For arguments involving similar calculations to (1) we will refer to this as the density argument. For example, whenHi+1has Configuration(A),thensi= 0 and ri= 1, which with (1) immediately implies that there cannot be another occurrence of(A). Similarly, we can show that Configuration(A−1),wheresi= 1 andri= 2, appears at most twice and any (Bj), where si = −3 and ri = −1, at most once. Furthermore, when one of the configurations appears, the occurrence of the other (Ak)with 3 ≤k≤ −2 is restricted, while only(A2) can appear arbitrarily often.
Proof of Lemma 2.2 Choose an arbitrary -cycle H1 and assign a colour c1 to some pair of non-adjacent edges of H1. LetH = Ht, t≥ 2, be the C-component ofG constructed from H1.
Next we consider the cases according to which configurations given by Proposi- tion2.3occur during the construction ofH. For each 1≤i≤t−1 and each-cycle which is added to Hi to get Hi+1, we shall choose two non-adjacent edges of the cycle and a colour c, and assign cto the chosen edges.
The -cycles added to Hi are the ones of Hi+1 which pass through edges of components ofHi+1−Hi(recall that these components are paths) or through edges in E(Hi+1)\E(Hi) with both ends in V(Hi). Thus, if all the edges in E(Hi+1)\ E(Hi) belong to components ofHi+1−Hiand each of these components has at least two vertices, then the task is easy: for each component we choose two non-adjacent edges of it and assign a new colour to them, and this guarantees that Hi+1contains no rainbow copy of C. If nothing else is explicitly stated we will always do this whenHi+1has Configuration(Ak)with 2≤k≤−2. Hence the effort in the proof consists in dealing with the other configurations. These will receive colours first.
By the density argument Configuration (A)appears at most once,(A−1)at most twice, and any (Bj) at most once.
Case 1 For all 1≤i≤t−1,Hi+1 has Configuration(Ak) with2≤k ≤−2.
For 1≤i≤t−1, assign a new colourci+1to two non-adjacent edges inE(Hi+1)\ E(Hi).
Case 2 There is 1≤i1 ≤t−1such thatHi1+1 has Configuration(A)with respect to some C.
In this case, for all i = i1, Hi+1 has Configuration (Ak) with 2 ≤ k ≤ −2, by the density argument. Moreover, for at most one 1 ≤ i2 ≤ t−1, Hi2+1 has Configuration (A3) with respect to someC.
Let C = u1u2· · ·uu1, where P = u1u2· · ·u is an -path in Hi1 and uu1 ∈/ E(Hi1). The number of -cycles in Hi1+1 which are not in Hi1 equals that of - paths inHi1 with endsu1 andu. First, we consider the case in which the number of such paths is greater than one, and, for this case, we define the colouring ofH in such a way that each of these paths contains two non-adjacent edges with the same colour.
Suppose thatHi1 contains someP=u1x2· · ·x−1u,P =P. We have thatP∪ P contains an even cycle with length at most 2−2, and inHi1 the only such cycles are-cycles (whenis even), or (2−4)-cycles, or (2−2)-cycles. A (2−2)-cycle appears during the construction only if, for somei,Hihas two-cyclesCandCsuch thatC∩C is a 2-path. Similarly a (2−4)-cycle appears during the construction only if, for somei, Hi has two-cyclesC andC such thatC∩C is a 3-path.
First consider that P ∪P is a (2−2)-cycle in Hi1 (P and P are internally disjoint). So we may assume without loss of generality thati1= 2,H2 has Configu- ration(A2) andP∪P ⊆H2. Assign a colourc1 to the edges ofP∪P alternately.
Note that, if is even then there may be a third-path in H2 betweenu1 andu, and one can easily check that such a path will have two edges with the same colour.
Consider that P ∪P contains a (2−4)-cycle. We may assume that there is no-pathPinHi1 betweenu1andusuch thatP∪P orP∪Pis a (2−2)-cycle.
Without loss of generalityx2=u2, H2 has Configuration(A3)and (P∪P)−u1⊆ H2. Alternately colour the edges of (P ∪P)−u1 with two coloursc1 andc2. If is even then there may be a third (−1)-path in H2 between u2 and u5, and one can easily check that such a path will have two edges with the same colour.
Now consider thatP∪Pcontains an-cycle,even. We may assume that there is no -path P in Hi1 between u1 and u such that P∪P or P∪P contains a cycle with length at least 2−4. Without loss of generality H1 is an -cycle contained inP ∪P. and we alternately colour the edges ofH1 with two coloursc1 andc2.
Finally, assume that Hi1 contains no -path linking u1 and u other than P. Suppose that P has two consecutive edges in some -cycle C. Obviously, C can- not contain both u1u2 and u−1u. Assign a colour c1 to two non-adjacent edges in E(C) in such a way that, if {u1u2, u2u3} ⊆ E(C), then c1 is not assigned to u2u3, and, if {u−2u−1, u−1u} ⊆ E(C), then c1 is not assigned to u−2u−1. Without loss of generality H1 = C. For 1 ≤ i ≤ t− 1, i = i1, assign a new colourci+1to two non-adjacent edges in E(Hi+1)\E(Hi). We have that some edge inE(P)\ {u2u3, u−2u−1}is uncoloured. Assign a new colourci1+1touu1 and to an uncoloured edge inE(P)\ {u2u3, u−2u−1}.
Suppose now that no two consecutive edges ofP lie in the same-cycle inHi1. We have thatP cannot have two non-consecutive edges in same-cycle, otherwiseP would have length greater than −1. So any -cycle in Hi1 has at most one edge ofP. For 1≤i≤t−1,i=i1, i2, assign a new colourci+1to two non-adjacent edges inE(Hi+1)\(E(Hi)∪E(P)). Colour two non-adjacent edges inE(Hi2+1)\E(Hi2) (when= 5 there are exactly two such edges) with a new colourci2+1. Again some edge in E(P)\ {u2u3, u−2u−1} is uncoloured. Assign a new colour ci1+1 to uu1
and to an uncoloured edge in E(P)\ {u2u3, u−2u−1}.
Case 3 There are 1≤ i1 < i2≤ t−1 such that Hi1+1 and Hi2+1 have Configura- tion (A−1) with respect to some C and to some C, respectively.
By the density argument, this case occurs only if = 5 and we have thatHi+1
has Configuration (A2) for all i = i1, i2. Let C = u1u2u3u4u5u1, and let C = v1v2v3v4v5v1, where P = u1u2u3u4 and P = v1v2v3v4 are 4-paths in Hi1, u5 ∈/ V(Hi1) and v5 ∈/ V(Hi2). We see that P is the only 4-path between u1 and u4
in Hi1 and thus C is the only 5-cycle added toHi1 to formHi1+1. As forP, there may be a 4-path P in Hi2 between v1 and v4 other than P. If such is the case then we have that P∪Pcontains a 4-cycle or a 6-cycle.
One of the following three alternatives holds forP in Hi1: the three edges ofP lie in the same 5-cycle, or two consecutive edges of P lie in the same 5-cycle but the third one does not, or any 5-cycle in Hi1 contains at most one edge ofP.
Consider that the first alternative holds for P. Without loss of generality all the edges of P lie in H1 and i1 = 1. Hence H1 is of the form H1 =u1u2u3u4x5u1 for some x5. Note thatC =u1x5u4u5u1 is a 4-cycle inH2. Suppose that all the edges of P lie in H2. Therefore, without loss of generality i2 = 2. If the ends ofP are two adjacent vertices inV(C) then we colouru1u2andu3u4 withc1 and we colour two non-adjacent edges of C with a new colour c2. If the ends of P are u1 andu4 then we colour u1u2 andu3u4 withc1. If the ends ofP are x5 and a vertex in{u2, u3}then we assignc1tou1u2 andu3u4and we assign a new colourc2 to v5x5 andu2u3. The case in which the ends of P are u5 and a vertex in{u2, u3} is symmetric. Suppose that the ends ofPareu1andu3. Note that there are two 4- paths betweenu1 andu3 (and thus two possibilities forP):u1u5u4u3 oru1x5u4u3. We assign c1 to u4x5, u1u2 and u3v5, and assign c2 to u2u3, u4u5 and v5u1. The case in which the ends of P areu4 andu2 is symmetric.
Now suppose that P has at most two edges in H2. If P has two edges in H2
then these must be consecutive. Hence we may assume without loss of generality that v3v4 ∈/ E(H2). Since there is no 6-cycle in Hi2 and the unique 4-cycle inHi2
has its edges in H2, a 4-pathP=P betweenv1 andv4 must pass through v3v4. Colour u1u2 andu3u4 withc1, and v3v4 andv5v1 with a new colourci2.
Let us consider the second alternative for P. Without loss of generality H1 contains the edges u1u2 and u2u3 but does not contain u3u4. Thus H1 is of the form H1 = u1u2u3x4x5u1 for some x4 and x5. Note that C = u1x5x4u3u4u5u1 is a 6-cycle in Hi1+1. We see that C is the unique 6-cycle in Hi2, and that Hi2
contains no 4-cycle. Hence, the number of 4-paths linkingv1andv4is at most two. If there are two such paths, these correspond to two internally disjoint paths alongC. Suppose thatE(P)⊆E(C). Alternately colour the edges ofCwith two coloursc1 andc2 and, for 1≤i≤t−1,i=i1, i2, assign a new colourci+2to two non-adjacent edges in E(Hi+1)\(E(Hi)∪ {u3u4}). Assume thatE(P)⊆E(C). ThusP is the unique 4-path betweenv1 andv4. IfE(P)⊆E(H1) thenE(P)∩ {u1u2, u2u3} =∅ (by assumption P cannot be u1x5x4u3), and we colour u4u5 and the two non- adjacent edges inE(P) withc1. Assign a new colourci+1to two non-adjacent edges
inE(Hi+1)\E(Hi), for 1≤i≤t−1,i=i1, i2. Thus assume thatE(P)⊆E(H1) (possiblyP=P). ThereforePhas an edgevjvj+1which does not belong toE(H1).
Colouru2u3, x4x5 and an edge in {u5u1, u5u4} \ {vjvj+1} withc1, and give a new colour ci2+1 to vjvj+1 and to some edge in {v5v1, v5v4} not incident with vj nor withvj+1.
Finally, consider the third alternative forP. In Hi2 there are neither 4-cycles nor 6-cycles, and therefore P is the unique 4-path between v1 and v4. We may assume without loss of generality thatH1containsu2u3and thatu2u3is uncoloured.
Suppose that P = P. Colour u2u3, u5u1 and v5u4 with a new colour ci1+1, and assign a new colourci+1to two non-adjacent edges inE(Hi+1)\E(Hi), for 1≤i≤ t−1, i = i1, i2. Now suppose that P = P. Since P is the unique 4-path inHi2
linking v1 and v4, P and P cannot have both ends in common. Without loss of generality v1 =u1. Colour u2u3 andu5u1 with a new colourci1+1. Ifv2v3 = u2u3
then colourv5v1withci1+1; otherwise colourv2v3 andv5v1 with a new colourci2+1. Case 4 There is exactly one1≤i1≤t−1such thatHi1+1has Configuration(A−1) with respect to some C.
By the density argument, Hi+1 has Configuration (Ak) with 2 ≤ k ≤ 4 for all i = i1. Let C = u1u2· · ·uu1, where P = u1· · ·u−1 is an (−1)-path inHi1
andu ∈/ V(Hi1). The number of-cycles inHi1+1which are not inHi1 equals that of (−1)-paths with ends u1 and u−1. Next we consider the case in which the number of such paths is greater than one, and, for this case, we define the colouring ofH in such a way that each of these paths contains two non-adjacent edges with the same colour.
Suppose thatHi1contains someP=u1x2· · ·x−2u−1,P=P. We see thatP∪ P contains an even cycle with length at most 2−4, and inHi1 the only such cycles are-cycles (whenis even), or (2−6)-cycles, or (2−4)-cycles. A (2−4)-cycle appears during the construction only if, for somei,Hihas two-cyclesCandCsuch thatC∩C is a 3-path. Similarly a (2−6)-cycle appears during the construction only if, for somei, Hi has two-cyclesC andC such thatC∩C is a 4-path.
First, consider that P ∪P is a (2−4)-cycle in Hi1 (P andP are internally disjoint). So we may assume without loss of generality that i1 = 2,H2 has Con- figuration(A3) andP ∪P ⊆H2. Alternately colour the edges ofP ∪P with two colours c1 and c2. Note that if is even then there may be a third (−1)-path inH2 betweenu1 andu−1, and one can easily check that such a path will have two edges with the same colour.
Consider that P ∪P contains a (2−6)-cycle. We may assume that there is no (−1)-path P in Hi1 between u1 and u−1 such that P∪P or P∪P is a (2−4)-cycle. Without loss of generality x2 = u2, H2 has Configuration (A4) and (P ∪P)−u1 ⊆ H2. Alternately colour the edges of (P ∪P)−u1 with two colours c1 and c2, and colour the two non-adjacent edges of C ∩H1 with a new colour c3. If is even then there may be a third (−2)-path in H2 between u2 andu−1. Such a path passes through the edges ofC∩H1, and therefore will have two edges with the same colour.
Now consider that P ∪P contains an -cycle, even. We may assume that there is no (−1)-pathPin Hi1 betweenu1 andu such thatP∪P or P∪P contains a cycle with length at least 2−6. Without loss of generality H1 is an- cycle contained in P ∪P. Alternately colour the edges of H1 with two coloursc1 andc2, and assign a new colourci+2to two non-adjacent edges inE(Hi+1)\E(Hi) for 1≤i≤t−1,i=i1.
Finally, assume thatP is the only (−1)-path inHi1betweenu1andu−1. LetC be an-cycle containing the edgeu2u3. Assign a colourc1to two non-adjacent edges in E(C)\ {u2u3}. Without loss of generality H1=C. For 1≤i≤i1−1, assign a new colourci+1to two non-adjacent edges inE(Hi+1)\E(Hi). We have thatu2u3
is uncoloured. Assign a new colourci1+1to uu1 andu2u3.
Case 5 There is 1 ≤ i1 ≤ t−1 such that Hi1+1 has Configuration (Bj) with 3≤j≤with respect to some C.
By the density argument, Hi+1 has Configuration (A2) for all i = i1. Let C = u1u2· · ·uu1, whereuj ∈V(Hi) for some 3≤j≤,{u3, . . . , u}\{uj} ⊆V(Hi+1)\ V(Hi). If there is a pathP inHi1 betweenu1anduj such thatV(P)∪{uj+1, . . . , u} induces an -cycle in Hi1+1 or there is a path P in Hi1 between u2 and uj such that V(P)∪ {u3, . . . , uj−1} induces an -cycle in Hi1+1, then Hi1+1 can be con- structed in such a way that each of the last two steps has Configuration (A−j+3) and(Aj),respectively, and therefore Case1, 2, 3,or4happens. So we may suppose that Hi1 contains none of these paths and colouru2u3 anduu1 with colourci1+1.
2
3 Cycle on four vertices
In this section we give a sketch of the proof that prbC4 =n−3/4. By a classical result of Bollob´as (see[3]), we know that ifpn−3/4, then a.a.s.G(n, p) contains a copy ofK2,4. It is not hard to see that in any proper colouring of the edges ofK2,4 there is a rainbow copy ofC4, which implies thatprbC4 ≤n−3/4.
For the lower bound, define a sequence F =C41, . . . , C4 of copies of C4 in G as aC4-chain if for any 2≤i≤we haveE(C4i)∩ i−j=11E(C4i)
=∅. Letpn−3/4 andG=G(n, p). We want to show that a.a.s. there exists a proper colouring of G that contains no rainbow copy of C4. It is enough to consider C4-chains that are maximal with respect to the number of C4’s. The first step is to properly colour some edges in all maximal C4-chains so that all C4’s in G will be non-rainbow.
Then, since all C4’s are coloured we can just give a new colour for each one of the remaining uncoloured edges. For the first step, we use Markov’s inequality and a union bound to obtain that a.a.s.Gdoes not contain any graphHwithm(H)≥4/3 and|V(H)| ≤12.
LetF =C41, . . . , C4be an arbitraryC4-chain inGwithm(F)≥4/3. Let 2≤i≤ be the smallest index such that F =C41, . . . , C4i has densitym(F)≥4/3. Then, since F = C41, . . . , C4i−1 has density m(F) < 4/3, we can explore the structure of G to conclude that |V(F)| ≤ 10, which implies |V(F)| ≤ 12, a contradiction.
Therefore, a.a.s. Gcontains no copies ofC4-chainsF withm(F)≥4/3. SinceF is not too dense, a careful analysis of such chains shows that it is possible to obtain the desired colouring, which proves that prbC4 ≥n−3/4.
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