Covering k-uniform hypergraphs by monochromatic loose paths

A conjecture of Gyárfás and Sárközy says that in every 2-coloring of the edges of the complete k-uniform hypergraph Kk n, there are two disjoint monochromatic loose paths of distinct colors such that they cover all but at most k − 2 vertices. Recently, the authors affirmed the conjecture. In the note we show that for every 2-coloring of Kk n, one can find two monochromatic paths of distinct colors to cover all vertices of Kk n such that they share at most k − 2 vertices. Omidi and Shahsiah conjectured that R(Pk t ,Pk t ) = t(k − 1) + b t+1 2 c holds for k > 3 and they affirmed the conjecture for k = 3 or k > 8. We show that if the conjecture is true, then k−2 is best possible for our result.


Introduction
A hypergraph H = (V, E) consists of a set V of vertices and a set E of edges, where each edge is a subset of V .If all the edges of H have same size k, then the hypergraph H is said to be k-uniform.Let K k n denote the complete k-uniform hypergraph on n vertices (the family of all k-element subsets of an n-element set).For any integer m 1, let [m] = {1, 2, . . ., m}.A k-uniform loose (or linear) path of length , denoted by P k , is a k-uniform hypergraph with edges e 1 , e 2 , . . ., e such that for any i ∈ [ − 1], |e i ∩ e i+1 | = 1 and |e i ∩ e j | = 0 for all other pairs {i, j}, i = j.For a vertex v ∈ V (P k ), we call v an endpoint of P k if v ∈ e i for i ∈ {1, } and v / ∈ e j for each j = i.For k = 2 we obtain the usual definition of a path P with edges.
The subject of edge-coloring of graphs has been a classical topic of graph theory.A famous result which is called Ramsey's theorem showed that: Given an integer n 2, whenever the edges of a sufficiently large complete graph are colored with two colors, red and blue, there is a monochromatic copy of the complete graph K n .Then many authors focused on the related research and expanded several branches in the subject.For example, in generalized graph Ramsey theory, it is of interest and well motivated to find desired monochromatic structures (cycles, paths, trees and so on) in edge-colored graphs.In this paper, r-coloring always means edge-coloring with r colors (traditionally red and blue when r = 2).
A simple and basic proposition, introduced by Gerencsér and Gyárfás in [8], says that Proposition 1. ( [8]) In any 2-coloring of a finite complete graph the vertices can be partitioned into a red and a blue path.Here the empty set and a single vertex are accepted as a path of any color.
In fact, Proposition 1 subsequently gave birth to the area of partitioning edge-colored complete graphs into monochromatic subgraphs.In 1979, Lehel conjectured that Proposition 1 remains true if paths are replaced by cycles.Here the empty set, a single vertex and a single edge are accepted as a cycle of any color.Gyárfás [9] proved the following weaker statement: The vertices of a 2-coloring of K n can be covered by a red and a blue cycle such that the two cycles have at most one common vertex.It took a long time to get rid of the common vertex.Using the Regularity Lemma and some new techniques, Luczac, Rödl and Szemerédi [16] first succeeded to prove the conjecture for large n.Then Bessy and Thomassé [1] found an elementary and elegant proof of Lehel's conjecture.Another famous conjecture (usually called Ryser's conjecture), appeared in the Ph.D. thesis of Henderson [5], states that for any r-coloring of the edges of a graph G, the vertex set V (G) can be covered by at most α(G)(r − 1) monochromatic trees, where α(G) is the independence number of G.For the case α(G) = 1, i.e. for complete graphs, the cases r = 3, 4, 5 were proved by Gyárfás [6], Duchet [3] and Tuza [22], respectively.However, all remaining cases of Ryser's conjecture are still open.There have been many further results, questions and conjectures in this area, many of which generalize Proposition 1 in graphs.We refer to two surveys [10,14].
However, in contrast to the graph case, there are only a few results on covering the vertices of hypergraphs with monochromatic pieces.Note that for any 2-coloring of complete graph K n , we cannot be guaranteed to obtain a monochromatic cycle of length greater than 2n 3 .However, in the case for hypergraphs, Gyárfás et al. [13] proved that every 2-coloring of K 3 n contains a monochromatic C 3 n and conjectured that for k 2 and sufficiently large n, every (k − 1)-coloring of K k n contains a monochromatic C k n , where the Hamiltonian Berge cycle C k n is a sequence of distinct vertices v 1 , v 2 , . . ., v n , together with a set of distinct edges e 1 , e 2 , . . ., e n such that v i , v i+1 ∈ e i (v n+1 = v 1 ).Bustamante, Hàn and Stein [2] showed that for every η > 0 there exists an integer n 0 such that every 2-coloring of the 3-uniform complete hypergraph on n n 0 vertices contains two disjoint monochromatic tight cycles of distinct colors that together cover all but at most ηn vertices.The same result holds if we replace tight cycles with loose cycles.Gyárfás and Sárközy [12] conjectured that for every k-uniform hypergraph H, the vertex set V (H) can be partitioned into at most α(H) loose cycles.Ergemlidze, Györi and Methuku [4] recently affirmed a weaker version of the conjecture for k = 3.If the conjecture is true then it extends Pósa theorem to k-uniform hypergraphs.To the best of our knowledge, even for α(H) = k − 1, that is, H is a complete k-uniform hypergraph, the conjecture is still open.Somewhat surprisingly, Király [15] proved an analogue of Ryser's conjecture for hypergraphs: For k 3 and every r-coloring of K k n , the vertex set can be covered by at most r k monochromatic connected subhypergraphs.Gyárfás and Sárközy [11] also proved that for every (k − 1)-coloring of K k n , there is a partition of the vertex set into monochromatic loose cycles such that their number depends only on r and k.In [12] they proved that in any r-coloring of a k-uniform hypergraph with independence number α, there is a partition of the vertex set into monochromatic loose cycles such that their number depends only on r, k and α.There are other results about partition and covering of hypergraphs, many of the proofs use the hypergraph Regularity Lemma, we refer to [7,11,20,21].
In the note, we only consider covering hypergraphs by loose paths.A loose path is proper if the path contains at least one edge.Similar to the graph case, a set of less than k vertices in an edge-colored k-uniform hypergraph is accepted as a loose path of any color.However, it is not always possible to extend Proposition 1 to loose paths of hypergraphs.In 2013, Gyárfás and Sárközy [11] posed the following conjecture: Conjecture 2. ( [11]) In every 2-coloring of the complete k-uniform hypergraph K k n , there are two disjoint monochromatic loose paths of distinct colors such that they cover all but at most k − 2 vertices.This estimate is sharp for sufficiently large n.
Recently, the authors affirmed the conjecture in [17].Based on Conjecture 2 and the proof, we focus on the following natural question: For every 2-coloring of K k n , can we find two monochromatic paths of distinct colors to cover all vertices of V (K k n ) such that the two paths share as few vertices as possible?
Note that any result on covering the vertices of edge-colored hypergraphs by monochromatic loose paths will imply a Ramsey-type result as a corollary (even though these Ramsey-type corollaries are often not sharp).Conversely, the values of Ramsey numbers of loose paths also imply that we can cover all vertices of edge-colored graphs hypergraphs by a small number of monochromatic loose paths.
Omidi and Shahsiah posed the following conjecture on the values of Ramsey numbers of loose paths for k 3 in [19].
For k = 3, Omidi and Shahsiah affirmed the conjecture in [18].For k 8 and s = t (the diagonal case), they also gave a positive answer in [19].
Our main result is as follows: the electronic journal of combinatorics 24(4) (2017), #P4.23 Theorem 4. For each 2-coloring of K k n , all the vertices can be covered by two monochromatic loose paths of distinct colors such that the two paths share at most k − 2 vertices.Besides, if Conjecture 3 is true for diagonal case then k − 2 is best possible.
First of all we show that if Conjecture 3 is true then k − 2 is best possible.By Proposition 1, we only need to consider k 3. Let n = R(P k t , P k t ), where t 3 is an integer satisfying that t+1 2 = k + 2. Then there is a 2-coloring of K k n such that it contains no copy of a monochromatic P k t+1 .Let P R and P B be two paths of distinct colors covering all vertices of the colored K k n .To show the sharpness of the result, it suffices to prove |P R ∩ P B | k − 2. It is obvious that the two paths are proper, since the longest monochromatic path covers exactly t(k −1)+1 vertices and there are k +1 vertices remained.Note that n = t(k −1)

Some Lemmas
The greatest common divisor (gcd) of two or more positive integers is the largest positive integer that is a divisor of them.We first give two basic and well-known lemmas of number theory.We modify slightly the latter lemma such that it is more suitable for us, so we present its proof for the completeness.
Lemma 5. Let a 1 , a 2 and n be three positive integers.Then the indeterminate equation has an integer solution if and only if gcd(a 1 , a 2 )|n.
Lemma 6.Let a 1 and a 2 be two co-prime positive integers.For any integer n > a 1 a 2 , the indeterminate equation ( 1) has a positive integer solution (x, y) with y a 1 .
By Lemma 6 we have the following corollary: Proof.By Lemma 6, the indeterminate equation n The authors proved the following slightly stronger result than Conjecture 2 in [17].Lemma 8. ([17]) Suppose that the edges of the complete k-uniform hypergraph K k n are colored with two colors, where n ≡ 2 mod (k − 1).Then V (K k n ) can be partitioned into monochromatic loose paths of distinct colors.
In contrast to the graph case, there are few known results about the Ramsey numbers of hypergraphs.Concerning to exact values of diagonal Ramsey numbers of loose paths we have the following result, which is useful to our main proof of Theorem 4. Lemma 9. ( [7]) For t = 2, 3, 4 and k 3, R(P k t , P k t ) = t(k − 1) + t+1 2 .Lemma 10. ( [19]) For every t 3 and k 8, R(P k t , 3 The proof of Theorem 4 In this section we will prove Theorem 4. For simplicity, we partition the proof into the following lemmas. Lemma 11.Let n−2 = x 1 (k−1)+x 2 (k−2), where x 1 2, x 2 k−2 are two nonnegative integers.For any 2-coloring of K k n , the vertices can be covered by two monochromatic loose paths of distinct colors such that the two paths have at most k − 2 common vertices.In particular, the result holds for all n k(k − 2) + 4.
Proof.If all the edges of K k n are colored with the same color, then the result is trivial.So we assume that both colors are used at least once.
We partition , where x 1 2 and x 2 k − 2. Note that the initial case for x 2 = 0 is easy to check by Lemma 8. Now we consider the case for x 2 1.
Denote by H 1 the complete k-uniform subhypergraph induced by V 1 .Then by Lemma 8, V (H 1 ) can be partitioned into a red path P R and a blue path P B .Case 1 Both P R and P B are proper.
We partition 2 , we consider the edge consisting of V 1 2 together with any two endpoints picked from P R and P B respectively.If the edge is red, then adding it to P R will form a longer red path, otherwise adding it to P B will form a longer blue path.
For 2 i x 2 − 1, suppose that there is a red path P R and a blue path P B such that they cover exactly Then consider the edge consisting of the (i + 1)th (k − 2)-element vertex subset V i+1 2 of V 2 and two new endpoints picked from P R and P B respectively.Note that x 2 k − 2 ensures the two endpoints distinct from any endpoint picked before.For any case (the edge is red or blue), we can obtain a longer red or blue path.Then the two paths of distinct colors will cover . By induction on i we complete the proof.Case 2 One of P R and P B is not proper.
Without loss of generality suppose that P B is not proper.Then P B is induced by a singleton vertex, say v, since P R contains x(k − 1) + 1 vertices for some integer x and then we are done since a set consisting of any k − 1 vertices can be regarded as an edge of any color.Now suppose that x 2 2. Let e = {v 1 , . . ., v k } be the last edge and v 2 , . . ., v k the k − 1 endpoints of P R .Denote by H 2 the complete subhypergraph induced by V 2 ∪ {v, v 2 , . . ., v k }.Subcase 2.1 H 2 has a blue edge f .Let V 1 consist of all vertices of the red path P R − e and the blue edge f .Let be partitioned into two proper paths of distinct colors.The rest discussion is similar to Case 1. Subcase 2.2 H 2 is a red complete subhypergraph.
Denote H 3 = H 2 − {v 2 , . . ., v k−1 }.Then we can choose a red path P R (as long as possible) in H 3 starting at v k such that the number of uncovered vertices by P R in H 3 is at most k − 2. Then P R together with P R form a new red path, such that the path covers all but at most k − 2 vertices of K k n .By Corollary 7, the result holds for all n k(k − 2) + 4.This completes the proof.
Lemma 12. Let n 5k − 3 be a positive integer.For any 2-coloring of K k n , the vertices can be partitioned into two monochromatic loose paths of distinct colors.
Proof.If n 2k−2 then the result is trivial.Now consider the case for 2k−1 n 5k−4.Using Lemma 9, there is a monochromatic loose path P k t of length t ∈ {2, 3, 4} such that P k t covers all but at most k − 1 vertices of K k n .Then P k t together with the path consisting of all remaining vertices will cover all vertices of K k n .If n = 5k − 3 = 5(k − 1) + 2, then by Lemma 8, V (K k n ) can be covered by two disjoint monochromatic loose paths of distinct colors.This completes the proof.
The following result is easy to verify, but plays an important role in our proofs.Lemma 13.For each 2-coloring of K k n , either V (K k n ) can be covered by two monochromatic disjoint loose paths of distinct colors, or for each integer 1 x k − 1 there are two edges of distinct colors such that they have exactly x common vertices.
Proof.By Lemma 12 we only need to consider the case for n 5k − 2. We prove the result by induction on x.We first show that if two colors are used at least once for the 2-coloring of K k n , then there are two edges of distinct colors such that they have exactly one common vertex.Otherwise, we can order randomly the vertices of K k n such that all but at most k − 2 ordered vertices forms a monochromatic path, but this means that K k n can be partitioned into two disjoint paths of distinct colors.Now suppose that the result holds for smaller x, that is, there are two edges f and g of distinct colors such that they have exactly x common vertices. Let Then either e and f are of distinct colors or e and g are.This completes the proof.Proof.Note that the case for x 1 = 0 is proved in Lemma 11.Now consider x 1 1.By Lemma 13, we pick two edges f and g of distinct colors and |f ∩ g| = x 1 .Similar to the proof of Lemma 11, either we find two desired paths or we fail in the final step.If we fail, then all preceding x 2 − 1 new edges are of common color, and the x 2 − 1 edges together with f or g, say f , form a monochromatic path P. Since we fail in the final step, x 2 = k − x 1 + 1.Note that now all k vertices of g are in P (x 1 vertices are in f and the other k − x 1 are in x 2 − 1 new edges).Hence P and the path consisting of the last (k − 2)-element vertex subset are two desired paths.
For two positive integers n and 3 k 7, we have the following corollary.Proof.For each given k we consider all possible couples (s, t).For a couple (s, t), if (2) or (3) has a desired solution, then we say that the couple (s, t) fits (2) or (3) for short.It is easy to check that the initial couple (s, t) = (2, 2) fits (2).
This completes the proof.
Lemma 17.Let k 8.For any 2-coloring of K k n , the vertices can be covered by two monochromatic loose paths such that they share at most k − 2 vertices.Remark.We will spend the remainder of this paper discussing some problems on covering edge-colored hypergraphs.
We proved Theorem 4 by partitioning the vertex set of K k n into two subsets of desired sizes such that the resulting two paths covering K k n share at most k − 2 vertices.However, for each edge of the two paths, the edge contains at most one common vertex.That is, it is possible that the two paths intersect at many edges.It would be interesting to know if there are two paths covering K k n such that the two paths intersect at most one edge?Lemma 8 says that if n ≡ 2 mod (k − 1) then the answer is yes.

Corollary 7 .
Let k 3 and n k(k − 2) + 4 be positive integers.Then there are nonnegative integers x 1 2 and

Proof.
Let n = t(k − 1) + s, where t 0 and 0 s k − 2. If t 2 then n 2(k − 1) + s < 5k − 3.By Lemma 12, we are done.If t k, then either k = 4 or n k(k − 1) + s > k(k − 2) + 4. By Corollary 15 or Lemma 11, we are done as well.So we only need to consider the case for 3 t k − 1.By Lemma 10, if s t+1 2 , then there is a monochromatic path P k t covering n − s + 1 vertices of K k n .Hence P k t and the path consisting of the s − 1 uncovered vertices cover all vertices of K k n .If s < t+1 2 , then there is a monochromatic pathP k t−1 covering n − s − k + 2 vertices of K k n .Let U = V (K k n ) \ V (P k t−1 )and W be the endpoint set of P k t−1 .Then |U | = s + k − 2 and |W | = 2(k − 1).If s = 0 or s = 1, then P k t−1 and the path consisting of U cover all vertices of K k n .For s 2, by Lemma 16, we have n = a(k −1)+b(k −2)+2 or n = 2k −a+b(k −2) for desired (a, b).Using Lemmas 11 and 14, we complete the proof.Now combining Corollary 15 and Lemma 17 together, we complete the proof of Theorem 4.