Note on the multicolour size-Ramsey number for paths

The size-Ramsey number $\hat{R}(F,r)$ of a graph $F$ is the smallest integer $m$ such that there exists a graph $G$ on $m$ edges with the property that any colouring of the edges of $G$ with $r$ colours yields a monochromatic copy of $F$. In this short note, we give an alternative proof of the recent result of Krivelevich that $\hat{R}(P_n,r) = O((\log r)r^2 n)$. This upper bound is nearly optimal, since it is also known that $\hat{R}(P_n,r) = \Omega(r^2 n)$.


Introduction
Following standard notation, we write G → (F ) r if any r-edge colouring of G (that is, any colouring of the edges of G with r colours) yields a monochromatic copy of F . We define the size-Ramsey number of F asR(F, r) = min{|E(G)| : G → (F ) r }; that is,R(F, r) is the smallest integer m such that there exists a graph G on m edges such that G → (F ) r . For two colours (that is, for r = 2) the size-Ramsey number was first studied by Erdős, Faudree, Rousseau and Schelp [9].
In this note, we are concerned with the size-Ramsey number of the path P n on n vertices. It is obvious thatR(P n , 2) = Ω(n) and it is easy to see thatR(P n , 2) = O(n 2 ); for example, K 2n → (P n ) 2 . The exact behaviour ofR(P n , 2) was not known for a long time. In fact, Erdős [8] offered $100 for a proof or disproof thatR(P n , 2)/n → ∞ andR(P n , 2)/n 2 → 0. This problem was solved by Beck [1] in 1983 who, quite surprisingly, showed thatR(P n , 2) < 900n. (Each time we refer to inequality such as this one, we mean that the inequality holds for sufficiently large n.) A variant of his proof, provided by Bollobás [5], givesR(P n , 2) < 720n. Recently, the authors of this paper [6] used a different and more elementary argument that shows thatR(P n , 2) < 137n. The argument was subsequently tuned by Letzter [12] who showed thatR(P n , 2) < 91n, and then further refined by the authors of this paper [7] who showed thatR(P n , 2) ≤ 74n. On the other hand, the first nontrivial lower bound was provided by Beck [2] and his result was subsequently improved by Bollobás [4] who showed thatR(P n , 2) Let us now move to the multicolour version of this graph parameter. It was proved in [7] that (r+3)r 4 n − O(r 2 ) ≤R(P n , r) ≤ 33r4 r n. It follows thatR(P n , r) is linear for any fixed value of r but the two bounds are quite apart from each other in terms of their dependence on r. Subsequently, Krivelevich [11] showed that in fact the dependence on r is (nearly) quadratic; that is,R(P n , r) = r 2+or(1) n. Here is the precise statement of his result: 11]). For any C > 5, r ≥ 2, and all sufficiently large n we havê R(P n , r) < 400 5 Cr 2+ 1 C−4 n.
The first author was supported in part by Simons Foundation Grant #522400. The second author was supported in part by NSERC.
It is straightforward to see that C = C(r) that minimizes the upper bound in this theorem is of order log r. As a result we get thatR(P n , r) = O((log r)r 2 n). In this note, we give an alternative proof of this fact.
Theorem 1.2. For any integer r ≥ 2 and all sufficiently large n we havê R(P n , r) < 600(log r)r 2 n.
It will follow from the proof that the constant 600 is not optimal. Since we believe that the factor log r is not necessary, we do not attempt to optimize it.

Proof
Before we move to the proof of Theorem 1.2, we need one, straightforward, auxiliary result.
Proposition 2.1. For any integer r ≥ 2 there exists an integer N = N(r) such that the following holds. For any integer n ≥ N, there exists a graph G = (V, E) such that (i) |V | = 7rn, (ii) 500(log r)r 2 n < |E| < 600(log r)r 2 n, and (iii) for every two disjoint sets S, T ⊆ V , |S| = |T | = n, the number of edges induced by S ∪ T with at least one endpoint in S is at most 70(log r)n.
Proof. The proof is an easy application of random graphs. Recall that the binomial random graph G(n, p) is a distribution over the class of graphs with vertex set [n] in which every pair {i, j} ∈ [n] 2 appears independently as an edge in G with probability p, which may (and usually does) tend to zero as n tends to infinity. Furthermore, we say that events A n in a probability space hold asymptotically almost surely (or a.a.s.), if the probability that A n holds tends to 1 as n goes to infinity.
Fix any integer r ≥ 2. It suffices to show that the random graph G ∈ G(7rn, p) with p = 22(log r)/n a.a.s. satisfies properties (ii) and (iii). (Property (i) trivially holds.) Indeed, if this is the case, then there exists an integer N = N(r) such that the desired properties hold with probability at least 1/2 for G ∈ G(7rn, p) for all n ≥ N. This implies that for each n ≥ N, there exists at least one graph with these properties.
Thus, the probability that there exist S and T such that X S,T ≥ 70(log r)n is, by the union bound, at most

Property (ii):
This property is straightforward to prove. Note that |E| is distributed as Bin 7rn 2 , p with E(|E|) = (539 + o(1))(log r)r 2 n. It follows immediately from Chernoff's bound that property (ii) holds a.a.s. The proof of the proposition is finished.
Proof of Theorem 1.2. The proof is based on the depth first search algorithm (DFS), applied several times, and it is a variant of the previous approach taken in [7] where it was proved thatR(P n , r) ≤ 33r4 r n. Using the DFS algorithms in a Ramsey-type problem was first successfully applied by Ben-Eliezer, Krivelevich and Sudakov [3].
Fix r ≥ 2 and suppose that n is sufficiently large so that Proposition 2.1 can be applied. Let G = (V, E) be a graph satisfying properties (i)-(iii) from Proposition 2.1. We will show that G → (P n ) r that implies the desired upper bound as |E| < 600(log r)r 2 n by property (ii). Consider any r-colouring of the edges of G. By averaging argument, there is a colour (say blue) such that the number of blue edges is at least |E(G)|/r. For a contradiction, suppose that there is no monochromatic copy of P n ; in particular, there is no blue copy of P n .
From now on, we restrict ourselves to the graph G 1 = (V 1 = V, E 1 ⊆ E), the subgraph of G induced by blue edges. We perform the following algorithm on G 1 to construct a path P . Let v 1 be an arbitrary vertex of G 1 , let P = (v 1 ), U = V \ {v 1 }, and W = ∅. If there exists an edge from v 1 to some vertex in U (say from v 1 to v 2 ), we extend the path as P = (v 1 , v 2 ) and remove v 2 from U. We continue extending the path P this way for as long as possible. Since there is no P n in the blue graph, we must reach a point of the process in which P cannot be extended, that is, there is a path from v 1 to v k (k < n) and there is no edge from v k to U (including the case when U is empty). This time, v k is moved to W and we try to continue extending the path from v k−1 , reaching another critical point in which another vertex will be moved to W , etc. If P is reduced to a single vertex v 1 and no edge to U is found, we move v 1 to W and simply re-start the process from another vertex from U, again arbitrarily chosen.
Observe that during this algorithm there is never an edge between U and W . Moreover, in each step of the process, the size of U decreases by 1 or the size of W increases by 1. The algorithm ends when U becomes empty and all vertices from P are moved to W . However, we will finish it prematurely, distinguishing 7r phases; phase i starts with graph G i = (V i , E i ) and ends when for the first time |W | = n. Before we move to the next phase, we set S i = W , T i = V (P ), and F i to be all edges incident to W . Then, we set V i+1 = V i \ W and G i+1 = G i [V i+1 ], the graph induced by V i+1 (in other words, G i+1 is formed from G i by removing vertices from W together with F i , all edges incident to them). Phase i ends now and we move to phase i + 1 where we run the algorithm on G i+1 .
There are a few important observations. Note that, by property (i), |V | = |V 1 | = 7rn so the last phase, phase 7r, finishes with U = ∅ and T 7r = ∅. As a result, family (F i : 1 ≤ i ≤ 7r) is a partition of E 1 . By construction, |S i | = n for all i and, since there is no path on n vertices in G 1 (and so also in any G i ), |T i | < n for all i. Hence, |F i | < 70(log r)n by property (iii). Putting these things together and using property (ii) in the very last inequality, we get the desired contradiction: |E|/r ≤ |E 1 | = |F 1 | + |F 2 | + · · · + |F 7r | ≤ 7r · 70(log r)n < 500(log r)rn < |E|/r.
The proof is finished.