1 Introduction
For a uniform hypergraph , the Ramsey number is the smallest such that any 2coloring of , the complete uniform hypergraph on vertices, contains a monochromatic copy of . Let denote the analogous Ramsey number for colorings, so that .
It is a major open problem to determine the growth of , the Ramsey number of the complete 3uniform hypergraph on vertices. It is known [6, 7] that there are constants such that
(1) 
Erdős conjectured that , i.e. the upper bound is closer to the truth. Erdős and Hajnal gave some evidence that this conjecture is true by showing that , i.e. the four color Ramsey number of is doubleexponential in (see, for example [9]).
Definition 1.1.
The hedgehog is a 3uniform hypergraph on vertices such that, for each , there exists a unique vertex such that is an edge, and there are no additional edges.
We sometimes refer to the first vertices as the body of the hedgehog. For any , one can also define a uniform hedgehog on , with a body of size and a unique hyperedge for every sized subset of the body. In this notation, we have .
Hedgehogs are interesting because their 2color Ramsey number is polynomial in , while their 4color Ramsey number is exponentially large in [10, 5]. This suggests that the bound by Erdős and Hajnal may not be such strong evidence that .
Hedgehogs are also interesting because they are a natural family of hypergraphs with degeneracy 1. Degeneracy is a notion of sparseness for graphs and hypergraphs. For graphs, the degeneracy is defined as the minimum such that every subgraph induced by a set of vertices has a vertex of degree at most . The BurrErdős conjecture [2] states that there exists a constant depending only on such that the Ramsey number of any degenerate graph on vertices satisfies . Building on the work of Kostochka and Sudakov [11] and Fox and Sudakov [8], Lee [12] recently proved this conjecture. We can similarly define the degeneracy of a hypergraph as the minimum such that every subhypergraph induced by a subset of vertices has a vertex of degree at most . Under this definition, Conlon, Fox, and Rödl [5] observe that the 4uniform analogue of the BurrErdős conjecture is false: the 4uniform hedgehog , which is 1degenerate, satisfies . They also observe that the 3uniform analogue of the BurrErdős conjecture is false for 3 or more colors: the 3uniform hedgehog, which is 1degenerate, satisfies .
However, the analogue of the BurrErdős conjecture for 3uniform hypergraphs and 2 colors remains open. In particular, it was not known whether the Ramsey number of the hedgehog is linear, or even nearlinear, in the number of vertices, . Conlon, Fox and Rodl [5] show , and, with the above in mind, ask if . We answer this question affirmatively.
Theorem 1.2.
If and , then every twocoloring of the complete uniform hypergraph on vertices contains a monochromatic copy of the hedgehog . That is,
We make no attempt to optimize the absolute constants here.
2 Ramsey number of hedgehogs
Throughout this section, we assume , and that we have a fixed twocoloring of the edges of a complete 3uniform hypergraph on vertex set with vertices. Let
(2) 
Let denote the set of pairs of elements of . For integer , let . For vertices and of , we write as an abbreviation for the unordered pair .
For , let
(3) 
For a set of pairs , let
(4) 
Here, and throughout, we use and to refer to the colors blue and red, respectively. For a vertex and set , let
(5) 
If is omitted, take . We define to be sets of such that is small, rather than those such that is small, because we wish to think of ’s as sets helpful for finding a blue hedgehog. Similarly, we think of ’s as sets helpful for finding a red hedgehog.
Lemma 2.1.
For any , and ,
(6) 
Proof.
Fix and . For convenience, let and . Assume for contradiction that . For every , we have , so and are disjoint. Consider the set of edges of containing , one element of , and one element of . On one hand, . On the other hand, for every , the pair is in at most such red triples, so the number of red triples of is at most . Additionally, for every , the pair is in at most such blue triples, so the number of blue triples of is at most . Hence, , a contradiction of . ∎
The following “matching condition” for hedgehogs is useful.
Lemma 2.2.
Let be a set of vertices. If, for all nonempty sets , we have , then there exists a blue hedgehog with body . Similarly, if, for all nonempty sets , we have , then there exists a red hedgehog with body .
Proof.
By symmetry, it suffices to prove the first part. Consider the bipartite graph between pairs in and vertices of , where is connected with if and only if triple is blue. If, for all nonempty , we have , then any such has at least neighbors in . By Hall’s marriage lemma on , there exists a matching in using every element of . Taking triples where and is the vertex matched with pair gives a blue hedgehog with body . ∎
2.1 Special Cases
We start by finding monochromatic hedgehogs in two specific classes of colorings on . We base our proof of Theorem 1.2 on the argument for the first class of colorings, which we call simple colorings. We use the result for the second class of colorings, which we call balanced colorings, as a specific case in the general argument.
2.1.1 Simple colorings
Consider hypergraphs that are colored the following way:

Start with a graph on .

Color a complete hypergraph on by coloring the triple blue if at least one of is in , and red otherwise.
Lemma 2.3.
If , any hypergraph colored as above has a monochromatic .
Proof.
Set . For , pick a vertex whose degree in is at least and let be an arbitrary set of neighbors of . Remove from . We call this the peeling step of . Figure 1 shows the first three peeling steps of this process for . If this process succeeds, we have found a set of vertices and disjoint sets of vertices also disjoint from , from which we can greedily embed a bluehedgehog in with body : for each with , pick an arbitrary unused element of for the third vertex of the hedgehog’s edge containing .
Now suppose this process finds vertices but fails to find for some . After picking , we remove and of it’s neighbors from , for a total of vertices. Then we have removed exactly vertices from . Hence, , and every vertex has degree at most in the subgraph of induced . Thus, there exists an independent set in of size at least . Furthermore, any vertex has at most neighbors in , so any two vertices share at least red triples in the subhypergraph of induced by , so we can greedily find a red hedgehog with body . ∎
2.1.2 Balanced colorings
In this section, we consider the case where our coloring is “balanced”. Lemma 2.1 tells us that, for every vertex and every nonnegative integer less than , one of and is at most . In “balanced” colorings, we assume, for all and all , both of and are . We show, in this case, there is a monochromatic hedgehog. The proof is by choosing a random subset of approximately
vertices, and showing that, with positive probability, we can remove vertices so that the remaining set of
vertices is the body of some red hedgehog.Lemma 2.4.
Let . Consider a twocolored hypergraph on vertices. Suppose that for all and all , we have
(7) 
Then has a red hedgehog .
Proof.
It suffices to prove for , so assume without loss of generality that . Pick a random set by including each vertex of in independently with probability . By the Chernoff bound, .
Fix such that and is a multiple of . Let be the pairs such that for all , and let
the indicator random variables for these pairs being in
. Let . By (7), we have . Each for is a Bernoulli random variable. Consider a graph on where and are adjacent (written ) if and share a vertex. This is a valid dependency graph for as is independent of all such that is vertex disjoint from . Furthermore, by the condition (7), each endpoint of any pair is in at most pairs, so each has degree at most in the dependency graph, and the total number of pairs such that is at most . We have(8)  
(9) 
Hence,
(10) 
The first inequality is by (8) and the second is by Chebyshev’s inequality. By the union bound over the multiples of in , of which there are less than , the probability there exists some a multiple of with
(11) 
is less than . Again by the union bound, with probability more than over the randomness of , we have (i) , and (ii) for all a multiple of in , (11) holds. Hence, there exists an such that (i) and (ii) hold, so consider such an . Remove vertices from , at least one from each of the pairs with smallest , to obtain a set of vertices such that, for all a multiple of in , we have
(12) 
Then, for all with , set to be the smallest multiple of larger than , so that
(13) 
Now, we show our matching condition holds. Setting in (13), we have for all . Hence, for any nonempty subset of size at most , any satisfies . If has size greater than , then, by setting in (13), we know that there are at most pairs such that , so again there exists such that . We conclude that, for all nonempty subsets of pairs , there exists such that . By Lemma 2.2, there exists a red hedgehog with body . ∎
2.2 Proof of Theorem 1.2
2.2.1 Proof outline
To prove Theorem 1.2, we follow the proof of Lemma 2.3. First, “peel off” vertices into a set to try to find a blue or red hedgehog.^{1}^{1}1 For technical reasons, we peel vertices to find both blue and red hedgehogs, as opposed to Lemma 2.3 where we only peeled vertices to find a blue hedgehog. If we succeed, we are done. If we fail, we end up with an induced twocolored hypergraph that is “balanced” in the sense of Lemma 2.4. In this case, we simply apply Lemma 2.4.
In the proof of Lemma 2.3, we started with and iteratively removed from a vertex and a set of size such that, for all , vertices and share many blue triples. This deletes vertices per round, which is small enough for the argument to succeed. For general hypergraphs, we peel off vertices with many “blueheavy neighbors”, meaning there exists some such that .^{2}^{2}2 For technical reasons, we peel vertices in increasing order of the corresponding . However, can be , so if we simply deleted along with of its blueheavy neighbors , we could delete vertices for every , which is too many. Instead, when we peel off , we delete from , add a penalty of to each , accumulated as , and delete from every vertex with . With these penalties, we guarantee that, on average, we delete vertices from per peeled vertex .
However, we need more care. In Lemma 2.3, we can find a hedgehog with body because, for any peeled vertices , the edges are blue for every . However, in our procedure, for a chosen with corresponding of size , there are some vertices such that is blue for few (at most ) vertices . We denote this set of “bad” vertices by . As much as possible, we wish to avoid choosing both and, at some later step, for the body of our blue hedgehog. Ideally, we simply delete all vertices in the step we peel off . However, can have vertices, which again could be too many if . Instead, for each we add a penalty of , accumulated as , and delete from every vertex with . We guarantee that, on average, we delete vertices from per peeled vertex (Lemma 2.9).
To finish the proof, we show, if our peeling produces a set (where is chosen before ), then, because we track the penalties and carefully, the matching condition of Lemma 2.2 holds. On the other hand, if the peeling procedure fails, the subhypergraph induced by is large and balanced, in which case we apply Lemma 2.4.
2.2.2 The peeling procedure
We now describe the procedure formally. Start with , and . For all , initialize . If, at any point, or has vertices, stop.
Recall that . For , do the following, which we refer to as .

While there exists a vertex and a color such that :

Let be the set truncated to vertices arbitrarily.

Let

Add to .

For all , add to .

For all , add to .

Delete from all vertices with or .

Delete from .

Note that and are only defined for . We refer to steps 1(a)1(g) as the peeling step for , denoted . We let denote the value such that the peeling step for occurred during , and call the peeling parameter of . Throughout the analysis, let denote the set immediately before . For any , let denote the set immediately after , so that is the set at the end of the peeling procedure.
The above process terminates in one of two ways. Either we “get stuck”, i.e. we complete Stage and and , or we “finish”, i.e. we terminate earlier with or . We show there is a monochromatic hedgehog in each case. In Subsection 2.2.5, we handle the case where we “get stuck”. In Subsection 2.2.6, we handle the case where we “finish”.
2.2.3 Basic facts about peeling
We first establish the following facts about the procedure.
Lemma 2.5.
For any such that , for any time in the procedure after , the following holds: for all colors , for all with , and for all vertices , we have .
Proof.
Fix with . We have for all : if not, then there exists a vertex with , in which case we would have peeled vertex during , and we would have deleted from during , which is a contradiction. Throughout the procedure, is nonincreasing. Thus, at any point in the procedure after , we have , so for all , we have and . ∎
Lemma 2.6.
For all colors and all vertices , we have .
Proof.
We prove this for , and the case follows from symmetry. We doublecount the number of red triples such that and . On one hand, every is in at most red triples because we chose as a subset of , so the total number of red triples is at most . On the other hand, by definition of , each is in at least such red triples. Thus, the number of such triples is at least . Hence, so as desired. ∎
Lemma 2.7.
For all colors and all vertices , we have .
Proof.
Assume for sake of contradiction that . Without loss of generality, was added to before . We have , so during , vertex is included in . Hence, is added to during 1(e) of , so during 1(f) of , vertex is deleted from if it hasn’t been deleted already. Thus, we could not have added to after , which is a contradiction, so , as desired. ∎
2.2.4 Bounding the number of deleted vertices
Lemma 2.8.
For all colors and all vertices , during , the total increase in over all is exactly .
Proof.
Fix . We have by definition, and, for , each increases by exactly , for a total increase of . ∎
Lemma 2.9.
For all colors and all vertices , during , the total increase in over all is at most .
Proof.
By symmetry, it suffices to prove the lemma for . Let . For , let
(14)  
(15) 
is after . Hence, by Lemma 2.5, for , we have . We know
(16) 
where the second inequality holds because was chosen to be peeled in . Hence, by Lemma 2.1, . As is nondecreasing in , we conclude for .
For , for any with , the peeling of increases by exactly . Thus, for many , the penalty increases by . Furthermore, increases only for , which has at most vertices by Lemma 2.6. For vertices , increases by less than , giving a total increase in of less than from those vertices. The total increases in is thus less than
(17) 
The coefficients of in (17) are nonincreasing, so (17) is plus a positive linear combination of . Subject to for , all of are simultaneously maximized when and for , so (17) is maximized there as well. Hence,
Total increase in  
(18) 
where, for the last inequality, we used and . This is what we wanted to show. ∎
Lemma 2.10.
The total number of vertices deleted from in the peeling procedure is at most .
Proof.
A vertex is deleted either for being added to or , having or at least 1/2, or having or at least 1/4. At the end of the procedure, we have the following inequalities. For all and all , we have and are initially 0 and increase only during the peeling step of some vertex . Hence, by Lemma 2.8, for ,
(19) 
Furthermore, by Lemma 2.9, for ,
(20) 
We conclude that, at the end of the procedure,
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