The size-Ramsey number of 3-uniform tight paths

Given a hypergraph $H$, the size-Ramsey number $\hat{r}_2(H)$ is the smallest integer $m$ such that there exists a graph $G$ with $m$ edges with the property that in any colouring of the edges of $G$ with two colours there is a monochromatic copy of $H$. We prove that the size-Ramsey number of the $3$-uniform tight path on $n$ vertices $P^{(3)}_n$ is linear in $n$, i.e., $\hat{r}_2(P^{(3)}_n) = O(n)$. This answers a question by Dudek, Fleur, Mubayi, and R\"odl for $3$-uniform hypergraphs [On the size-Ramsey number of hypergraphs, J. Graph Theory 86 (2016), 417-434], who proved $\hat{r}_2(P^{(3)}_n) = O(n^{3/2} \log^{3/2} n)$.

The systematic investigation of size-Ramsey questions for hypergraphs was initiated by Dudek, La Fleur, Mubayi, and Rödl [8]. Besides cliques and trees, they studied generalisations of paths.
We say that an r-uniform hypergraph is an -path if there exists an ordering of its vertices such that every edge is composed of r consecutive vertices, two (vertex-wise) consecutive edges share exactly vertices, and every vertex is contained in an edge. For 1 ≤ ≤ r − 1, let P (r) n, denote the r-uniform -path on n vertices and for the tight path, where = r − 1, we write P (r) n . Dudek, La Fleur, Mubayi, and Rödl [8] deduced from Beck's result [3] thatr 2 (P (r) n, ) = O(n), when 1 ≤ ≤ r/2. Furthermore, they proved that r 2 (P (r) n ) = O r (n r−1−α log 1+α n) with α = (r − 2)/( r−1 2 + 1), which givesr 2 (P (3) n ) = O(n 3/2 log 3/2 n). This was improved and extended to more colours by Lu and Wang [15], who showed thatr s (P (r) n ) = O r (s r (n log n) r/2 ) for s ≥ 2 colours. Dudek, La Fleur, Mubayi, and Rödl [8] asked ifr 2 (P (r) n ) = O r (n) for r ≥ 3. We answer this question for 3-uniform hypergraphs by proving the following result. Trivially, we need at least n edges, so this is asymptotically optimal. As observed in [8], bounds on size-Ramsey numbers for some uniformity can be used to obtain bounds for larger uniformities. We obtain the following corollary. To see this, take the graph given by Theorem 1 and replace every vertex by a set of r/3 vertices. Then each 3-edge naturally gives an r-edge, and every 3-uniform tight path becomes an r-uniform (2r/3)-path.
Our proof combines new ideas and the method developed by Clemens, Jenssen, Kohayakawa, Morrison, Mota, Reding, and Roberts [7] for estimating the size-Ramsey number of powers of paths (see also [5,12]). It is plausible that ideas from [12,5] may provide a strategy to solve the case with s ≥ 3 colours. However, the question whether the size-Ramsey number of a tight path is linear for hypergraphs with uniformity r ≥ 4 remains open and requires additional ideas.

Preliminaries
In this short section, we give a sketch of our proof of Theorem 1 and state two simple lemmas about random graphs.

Sketch of the proof of Theorem 1
We will first sketch a proof forr 2 (P n ) = O(n). It is not hard (cf. Lemmas 3 and 4 below) to obtain a graph G with O(n) edges such that for any two sufficiently large and disjoint sets of vertices A and B there is a path of length n alternating between A and B. Given such a graph G, we show that G → (P n ) 2 . Consider an arbitrary 2-colouring of the edges of G with colours blue and red. If there is no blue P n in G we can show (cf. Lemma 6 below) that there are two sets A and B of size at least n with no blue edges in between. By the property of G mentioned above there exists a P n alternating between A and B, which unequivocally has to be red.
For the proof of Theorem 1 we follow, in principle, the same strategy. Based on a blow-up of a power of a similar graph G, we define a 3-uniform hypergraph H and claim that H → (P (3) n ) 2 . We define an auxiliary (generalised) graph F on V (G), which has 2-and 3-edges, such that a long path in F gives a blue P (3) n in H. If F does not contain a long path, then we find a family of disjoint sets such that no edge of F lies between these sets (cf. Lemma 6). Then by the properties of G there exists a path in G alternating through these sets. As there are no edges of F 'interfering' with this path, we are able to turn it into a red P (3) n in H.
In the next section we provide the lemmas needed to obtain G. Afterwards, in Section 3 we introduce the notion of (2, 3)-graphs, which, as can be seen above, plays a key role in our argument. Finally, we prove Theorem 1 in Section 4.

Sparse graphs with many long paths
The following two lemmas are proved in [7]. Basically, together they imply that for every k and n there exists a graph G with O k (n) edges such that, for any disjoint sets of vertices A 1 , . . . , A k+1 that are large enough, there exists a path of length n 'alternating' through these sets.
Note that the hypothesis on H in Lemma 4 is (P1 m ) from Lemma 3. Therefore, roughly speaking, Lemma 4 tells us that (P1 m ) implies (P2 m ).

(2, 3)-graphs
In this section we introduce a structure that helps us to transfer some ideas from the graph case to the hypergraphs setting. A (2, 3)-graph F = (V, E) consists of a set of vertices V and a set E of 2-edges of the form {u, v} and 3-edges of the form ({u, v}, w), for distinct vertices u, v, w ∈ V . For simplicity we will write uv for {u, v} and uv(w) for ({u, v}, w). A sequence of vertices P = (x 1 , . . . , Given pairwise disjoint sets V 1 , . . . ,V k+1 , we say that an edge uv ∈ E(F) (uv(w) ∈ E(F)) is a transversal with respect to V 1 , . . . ,V k+1 , if u and v (u, v, and w) are in different sets V i . When the sets V 1 , . . . ,V k+1 are clear from the context we say that the edge is a transversal.
We want to prove that if a sufficiently large (2, 3)-graph F = (V, E) contains no (2, 3)-path with n vertices, then there exist large disjoint sets V 1 , . . . ,V k ⊆ V such that E contains no transversals and that there is no edge uv(w) with u ∈ V 1 ∪ · · · ∪V k−1 and v, w ∈ V k . The last property is only required to support our inductive proof. To prove this we use a Depth First Search (DFS) algorithm. For example, Ben-Eliezer, Krivelevich, and Sudakov [4] used a DFS algorithm to find long paths in expanding graphs to obtain bounds on the size-Ramsey number of directed paths. Their algorithm traverses the vertices of the input graph and maintains a set S of vertices that are fully dealt with, a set U of currently active vertices, and a set J of vertices that were not considered so far. The set U always spans a path and in every step, if at all possible, this path is extended by adding a vertex from J. Otherwise, the last vertex of the path is removed and added to S. It is immediate that there cannot be any edges between S and J and if U stays small, then at some point during the execution both S and J are large.
We adapt this algorithm to the setting of (2, 3)-graphs (see Algorithm 1 below). As in the graph case we greedily extend a (2, 3)-path (preferring 2-edges over 3-edges) and backtrack if it gets stuck. We will now give the details of our algorithm. The input is a (2, 3)-graph F = (V, E), disjoint subsets of vertices V 1 , . . . ,V k , and an ordering of the vertices V = {v 1 , . . . , v N }. During the algorithm we maintain sets S, T , W S , W U , T i for i ∈ [k] and a (2, 3)-path U as follows: • S ⊆ V is the set of vertices that are fully dealt with.
• W S ⊆ V is the set of vertices w that were 'used' by vertices from S.
• U contains the currently active vertices in a (2, 3)-path.
• W U ⊆ V is the set of vertices w that are 'used' by the path U.
In every step of the algorithm, either the (2, 3)-path U is extended by adding a vertex from T k to it or this is not possible, and the last vertex from U is removed and put into S. While the algorithm runs, after each execution of the while loop, we have the following invariants, where m is the length of the (2, 3)-path U: This process is described in Algorithm 1.
Let v be the vertex with smallest index from T k ; Let v be the vertex with the smallest index from T ext ; Let w ∈ T 1 ∪ · · · ∪ T k be the vertex of smallest index such that u m u m+1 (w) ∈ E; // There is one because u m+1 ∈ T ext . Proof of Lemma 5. Observe that (A1) and (A2) hold when we initialise the sets and put m = 0 on line 1. Assume that we are in some step of the algorithm, where (A1) and (A2) hold and we have vertices U = (u 1 , . . . , u m ) forming a (2, 3)-path (this is true because of (A1)). Now we consider the next execution of the while loop. We know from (A1) that W U contains exactly the vertices used in the edges u i u i+1 (w) for i = 1, . . . , m − 1. Either we extend the path by an edge u m v or u m v(w) (lines 7 and 10) where w is added to W U if needed (line 13), or we remove an edge u m−1 u m or u m−1 u m (w) (lines 16 and 20) where w is removed from W U if needed (line 20). Therefore, (A1) still holds.
For (A2) it is easy to see that , as in the beginning of the execution we have T i = V i and no vertex is added to T i . Also, since every w in W U comes from T 1 ∪ · · · ∪ T k , we have W U ⊆ T 1 ∪ · · · ∪ T k−1 (see lines 7 and 20). Since every vertex of S comes from U (line 16) and every vertex of U comes from T k ⊂ V k (lines 5, 7 and 10), which implies that S, U ⊂ V k . To prove that |W S | ≤ |S|, it is enough to observe that line 20 can only be executed after an execution of line 16. Similarly, we have |W U | ≤ m − 1 with m ≥ 2, because line 13 can only be executed after an execution of line 10, and |W U | = 0 with m = 1, because on line 5 nothing is added to W U . Thus, (A2) also remains true.
It remains to show that the algorithm terminates. In every execution of the while loop, either one vertex from T k ⊆ V k is added to the path U (lines 5 and 10) or moved from U to S (line 16). Therefore, after at most 2|V k | steps we have T = / 0, and the algorithm terminates.
We are ready to prove the aforementioned result on (2, 3)-graphs F with no long (2, 3)-paths.
Lemma 6. Let k, c and n be positive integers and let F = (V, E) be a (2, 3)-graph on at least 5 k−1 cn vertices. If F contains no (2, 3)-path with n vertices, then there exist disjoint sets V 1 , . . . ,V k ⊆ V of size at least cn such that no edge from E is a transversal and there is no edge uv(w) with u ∈ V 1 ∪ · · · ∪V k−1 and v, w ∈ V k .
Proof. We prove the result by induction on k. For k = 1 the result follows by putting V 1 = V (F). Thus let k ≥ 1 and assume the statement holds for k.
To prove the result for k + 1, let F be a (2, 3)-graph on 5 k cn vertices which does not have a path of length n. In particular, F does not have a path of length 5n, and by the assumption on k, there exist disjoint sets V 1 , · · · ,V k , each of size 5cn, such that no edge is a transversal, and there is no edge uv(w) with u ∈ V 1 ∪ · · · ∪V k−1 and v, w ∈ V k . We run Algorithm 1 with input F, k and V 1 , . . . ,V k .
First, we prove that at any point in the execution of the algorithm, no edge is a transversal with respect to T 1 ∪ · · · ∪ T k , S. Suppose for a contradiction that at some point there is an edge which is a transversal. Note that T k ⊆ V k and S ⊆ V k . If uv is this edge, then by the induction hypothesis and without loss of generality we have u ∈ S and v ∈ T k . This implies that when u was moved from U to S (line 16), the set U could have been extended, which means that T ext = / 0 and line 16 would not have been executed, a contradiction. Now, assume uv(w) is the transversal. Since T i ⊆ V i for i ∈ [k] and S ⊆ V k , we have w ∈ T 1 ∪ · · · ∪ T k−1 . Again, by the induction hypothesis and without loss of generality we have u ∈ S and v ∈ T k . Similarly as when we have an edge uv, at the time u was moved from U to S, the set U could have been extended, a contradiction. Now we prove that at any point in the execution of the algorithm, there is no edge uv(w) with u ∈ T 1 , . . . , T k−1 , S and v, w ∈ T k . Suppose for a contradiction that at some point there is such edge uv(w).
By the induction hypothesis we have u ∈ S and v, w ∈ T k , which again gives a contradiction as U could have been extended.
Note that since F has at least 5 k−1 cn vertices and no (2, 3)-path with n vertices, we have |S| = cn at some point of the execution of Algorithm 1. Let U, W U , S, W S and T 1 , . . . , T k be the sets at that moment. Note that |W S | ≤ |S| = cn and, since there is no n-vertex (2, 3)-path in F, we have |U|, |W U | ≤ n. Therefore, The sets V 1 , . . . ,V k+1 satisfies the requirements of the lemma.

Proof of Theorem 1
In this section we prove our main theorem. We first define the following constants: := 17, k := 2 , ε := 1/(k + 1), t := 8k + 40k 2 + 5, and t := r 2 (K Let a L.6 = 5 k and note that a L.6 is large enough to apply Lemma 6 with k + 1, c and n and set a := 2a L.6 . Lemma 3 applied with ε and a provides a constant b. Let n be sufficiently large. Then, from Lemma 3 we know that there is a graph G on an vertices with maximum degree b such that (P1 n ) holds. Fix such a graph G. Now let G k (t ) be the graph obtained from G k -the k-th power of G -by replacing every vertex by a K t and every edge by a K t ,t . Finally, H is the 3-uniform hypergraph with vertex set V (G k (t )) and a triple of vertices xyz is an edge in H if and only if xyz forms a triangle in G k (t ). For every v ∈ V (G) we denote by H(v) the corresponding cluster consisting of a K t , which we denote by H (v). We let H ⊆ H be the 3-graph induced by the clusters We will define an auxiliary (2, 3)-graph F on the vertex set V , whose edges will indicate that we can walk between the clusters using blue triples of H . Formally, for u, v ∈ V a (2, 2)-connector between the clusters H (u) and H (v) consists of four vertices x 1 , x 2 ∈ H (u) and y 1 , y 2 ∈ H (v) such that x 1 x 2 y 1 and y 1 y 2 x 1 are triples of H. Similarly, for u, v, w ∈ V a (2, 1, 2)-connector between the clusters H (u) and H (v) through H (w) consists of five vertices x 1 , x 2 ∈ H (u), z ∈ H (w), and y 1 , y 2 ∈ H (v) such that x 1 x 2 z, x 1 zy 1 , and zy 1 y 2 are triples of H; see Figure 1. We then define a (6, 6)-connector ((6, 3, 6)-connector) between Suppose that F contains a (2, 3)-path on n vertices v 1 , . . . , v n with the w i vertices all distinct for the 3-edges. We can turn this (2, 3)-path into a blue tight path P (3) n in H as follows. First, by following the (2, 3)-path for i = 1, . . . , n − 1, we choose a (2, 2)-connector between H (v i ) and in such a way that they are all pairwise vertex-disjoint. This is possible, because we have (6, 6)-connectors and (6, 3, 6)connectors available and two vertices y 1 and y 2 from the previous connector can occupy at most two out of the three disjoint copies of connectors that are provided. Then, for each of the clusters with y 1 , y 2 and x 1 , x 2 the vertices of the connectors within that cluster, we use the edges y 1 y 2 x 2 and y 2 x 2 x 1 to connect both connectors. As the connectors only use blue edges and all edges within the clusters are blue this is a tight path only using blue edges. Thus, in this case, we are able to obtain a blue P (3) n in H and we are done.
We assume that F contains no (2, 3)-path on n vertices. From Lemma 6, there exist pairwise disjoint sets V 1 , . . . ,V k+1 ⊆ V of size at least cn such that no edge from E(F) is a transversal. We may assume that all these sets V j have size exactly cn. Let G = G[V 1 ∪ · · · ∪V k+1 ] and set m := n.
We now want to find a path P m alternating through V 1 , . . . ,V k+1 with edges in G ⊆ G using Lemma 4. Since c = εa L.4 and ε = 1/(k + 1), we have |V (G )| = (k + 1)cn = a L.4 n = a L.4 m. Also, we have |V i | = cn = εa L.4 m for 1 ≤ i ≤ k + 1. As G is an induced subgraph of G and property (P1 n ) holds in G, property (P1 m ) does hold for G . Therefore, by Lemma 4, we conclude that there is a path P m = P n with vertices alternating through V 1 , . . . ,V k+1 and with edges in G ⊆ G. This path P n gives us the kth power P k n in G k . By the choice of V 1 , . . . ,V k+1 no edge of P k n is from E(F) and also no triangle in P k n induces an edge uv(w) ∈ E(F). It remains to turn this P k n into a red P (3) n in H .

Claim 7.
If there is a P k n in G k that does not contain any edges from F, then there is a red P n in H .
Let P k n = (v 1 , . . . , v n ) and recall H (v i ) is the cluster in H corresponding to the vertex v i for i = 1, . . . , n. We want to remove all vertices of H which belong to blue (2, 2)-connectors and (2, 1, 2)connectors from clusters along edges and triangles of P k n . In P k n every vertex v i is incident to at most 2k other vertices in {v 1 , . . . , v n } (2k is the maximum degree of the v i in P k n ). Also, every v i is contained in at most 4k 2 triangles of P k n together with two other vertices in {v 1 , . . . , v n }. Let u and v be neighbours in P k n . Since there is no blue (6, 6)-connector between H (u) and H (v), there are at most two (2, 2)-connectors that do not overlap between H (u) and H (v), which can both be deleted by removing at most 4 vertices in each cluster. Let u, v, and w be vertices that form a triangle in P k n . Since there is no blue (6, 3, 6)-connector between H (u), H (v), and H (w), there are at most six (2, 1, 2)-connectors that do not overlap, two for each possibility to place the single vertex. These can be deleted by removing at most 10 vertices from each cluster.
A tuple (u, v) is an end-tuple of a tight path with at least 4 vertices if u and v are consecutive vertices in the path and u is contained in exactly two edges and v is contained in exactly one. The two tuples (u, v) and (v, w) are the end-tuples of the tight path (u, v, w) of length 3. Furthermore, every tuple (u, v) is an end-tuple of the tight path (u, v) of length 2.
To prove Claim 7, i.e., construct P n in red, it is then sufficient to construct a quadruple satisfying property Q n−1 . We will construct this quadruple inductively. The base case Q 1 asks for two paths of length 2 and, therefore, it is enough to choose any pair u 1 , u 2 from H * (v 1 ) and w 1 , w 2 from H * (v 2 ). Therefore, the following is immediate: There exists a quadruple (u 1 , u 2 , w 1 , w 2 ) for which Q 1 holds. (4.1) We will inductively find a quadruple (u 1 , u 2 , w 1 , w 2 ) satisfying property Q i for every i = 2, . . . , n − 1. Ultimately, after n steps, this gives us P (3) n in red in H . Suppose (u 1 , u 2 , w 1 , w 2 ) satisfies property Q i−1 for some 1 < i ≤ n − 1. In the inductive step, we obtain (x 1 , x 2 , y 1 , y 2 ) satisfying property Q i by extending one of the paths ending in (u 1 , u 2 ) or (w 1 , w 2 ) to get two longer paths ending in (x 1 , x 2 ) and (y 1 , y 2 ). This mainly relies on the absence of (2, 2)-connectors and (2, 1, 2)-connectors in blue and that there are clusters H * (v (i−1) +1 ), . . . , H * (v i ) to choose x 1 , x 2 , y 1 , y 2 from. As k ≥ 2 , all edges are present between these clusters and the clusters H r and H s containing u 1 , u 2 and w 1 , w 2 , respectively. Fact 9. Let 1 < i ≤ n − 1 and suppose (u 1 , u 2 , w 1 , w 2 ) satisfies property Q i−1 . Then there is a quadruple (x 1 , x 2 , y 1 , y 2 ) that satisfies property Q i .
Proof. We will find (x 1 , x 2 , y 1 , y 2 ) following the strategy sketched above. Since there are no blue (2, 2)connectors between the clusters corresponding to u 1 , u 2 and w 1 , w 2 , and all possible triples between these clusters are edges in H , then either the triple u 1 u 2 w 2 or the triple w 1 w 2 u 2 is red, say, w.l.o.g., u 1 u 2 w 2 is red. We let the red path of length at least i that ends in (u 1 , u 2 ) be called P red and note that the triple u 1 u 2 w 2 already extends this path. We will show that it is possible to further extend this path to obtain two longer red tight paths with ends (x 1 , x 2 ) and (y 1 , y 2 ), respectively, such that (x 1 , x 2 , y 1 , y 2 ) satisfies property Q i .
Notice that, as ≥ 17, by the pigeonhole principle there are nine sets X 1 , . . . , X 9 each contained in a different cluster from H * (v (i−1) +1 ), . . . , H * (v i ) and of size |X j | ≥ 3 for j ∈ [9] (here we use that |H * (v i )| ≥ 5) such that either for all j ∈ [9] and every x ∈ X j the triple u 2 w 2 x is red (4.2) or for all j ∈ [9] and every x ∈ X j the triple u 2 w 2 x is blue. We first consider the case where (4.2) holds. It is enough to assume that we have X 1 , . . . , X 5 and |X j | ≥ 2 for j ∈ [5]. If there are two sets X,Y from X 1 , . . . , X 5 and x 1 , x 2 ∈ X, y 1 , y 2 ∈ Y such that the triples w 2 x 1 x 2 and w 2 y 1 y 2 are red, then the quadruple (x 1 , x 2 , y 1 , y 2 ) satisfies property Q i as we can obtain two longer red paths by extending P red following u 1 u 2 w 2 x 1 x 2 and u 1 u 2 w 2 y 1 y 2 , respectively. Otherwise, there are two sets X,Y and x 1 , x 2 ∈ X, y 1 , y 2 ∈ Y such that the triples w 2 x 1 x 2 and w 2 y 1 y 2 are blue. As there is no blue (2, 1, 2)-connector, the triple w 2 x 1 y 1 is red. There is no blue (2, 2)-connector between the corresponding clusters, so either the triple x 1 y 1 y 2 or y 1 x 1 x 2 is red, say w.l.o.g. x 1 y 1 y 2 is red. This extends P red by following u 1 u 2 w 2 x 1 y 1 y 2 and gives the end-tuple (y 1 , y 2 ). Repeating the same argument, which is possible, because there were five sets available (sets X 1 , . . . , X 5 ), we get an end-tuple (z 1 , z 2 ) that extends P red to a longer red path and, thus, a quadruple (y 1 , y 2 , z 1 , z 2 ) satisfying Q i .
In the case where (4.3) holds we proceed as follows. As for all j ∈ [9] there is no blue (2, 1, 2)connector between the clusters of u 1 , u 2 and w 1 , w 2 and X j , we have for every x ∈ X j that either the triple w 1 w 2 x or the triple u 1 u 2 x is red. Then we can assume by the pigeonhole principle w.l.o.g. (we will not use the triple u 1 u 2 w 1 ) that there are sets X j ⊆ X j with |X j | ≥ 2 for j ∈ [5] (here we use that |X j | ≥ 3) such that for all j ∈ [5] and every x ∈ X j the triple w 1 w 2 x is red. Now we can continue exactly as in the case where (4.2) holds, with u 1 u 2 replaced by w 1 w 2 throughout and extending the path with end-tuple (w 1 , w 2 ). Observing that the red tight paths that we have constructed have length at least i + 1, we see that Fact 9 is proved. Fact 9 together with (4.1) finishes the proof of Claim 7 and hence the proof of Theorem 1 is complete.