On Ramsey numbers of hedgehogs

TL;DR

This paper proves that the two-color Ramsey number of the hedgehog hypergraph grows nearly linearly with the number of vertices, specifically as O(t^2 log t), answering a question about its growth rate.

Contribution

The authors establish an upper bound of O(t^2 log t) for the two-color Ramsey number of hedgehogs, improving understanding of its growth and confirming near-linear behavior.

Findings

Two-color Ramsey number of hedgehogs is O(t^2 log t)

Answers affirmatively to the near-linear growth question

Provides bounds contrasting with exponential four-color Ramsey numbers

Abstract

The hedgehog $H_t$ is a 3-uniform hypergraph on vertices $1,\dots,t+\binom{t}{2}$ such that, for any pair $(i,j)$ with $1\le i<j\le t$, there exists a unique vertex $k>t$ such that $\{i,j,k\}$ is an edge. Conlon, Fox, and R\"odl proved that the two-color Ramsey number of the hedgehog grows polynomially in the number of its vertices, while the four-color Ramsey number grows exponentially in the number of its vertices. They asked whether the two-color Ramsey number of the hedgehog $H_t$ is nearly linear in the number of its vertices. We answer this question affirmatively, proving that $r(H_t) = O(t^2\ln t)$.

Equations

$$2^{c{t}^2}\le r(K_t^{(3)})\le 2^{2^{c^{\prime }t}}.$$

$$r(H_t)<200t^2\ln t+400t^2+1.$$

$$m_{max}:=2t+\left(\genfrac{}{}{0ex}{}{t}{2}\right).$$

$$\displaystyle d^{(r)}_{uv}\$$

$$\displaystyle d^{(b)}_{uv}\$$

$$\displaystyle N^{(b)}(F)\$$

$$\displaystyle N^{(r)}(F)\$$

$$\displaystyle U^{(b)}_{\leq m}(v,X)\$$

$$\displaystyle U^{(r)}_{\leq m}(v,X)\$$

$$\min \left(|U_{\le m}^{(b)}(v)|,|U_{\le m}^{(r)}(v)|\right)\le 2m.$$

$$\left|U_{\le m}^{(b)}(v)\right|\le cm.$$

$$\displaystyle\mathop{\bf E\/}[X]\$$

$$\displaystyle\mathop{\bf Var\/}[X]\$$

$$=\sum _{\ell \sim {\ell }^{\prime }}\mathbf{E}\text{/}[X_{\ell }{X}_{{\ell }^{\prime }}]-\mathbf{E}\text{/}[X_{\ell }]\mathbf{E}\text{/}[X_{{\ell }^{\prime }}]$$

$$\le 2cmp\cdot \left({\left(\frac{4t}{n}\right)}^3-{\left(\frac{4t}{n}\right)}^4\right)$$

$$<\frac{128t^{3}cmp}{n^3}\le \frac{64t^3{c}^2{m}^2}{n^2}=\frac{m^2}{25t}.$$

$$\displaystyle\mathop{\bf Pr\/}\left[\#\left\{uv\in\tbinom{S}{2}:d^{(r)}_{uv}\leq m\right\}>m-t\right]\$$

$$=\mathbf{Pr}\text{/}[X-\mathbf{E}\text{/}[X]\ge m-t-\mathbf{E}\text{/}[X]]$$

$$\le \mathbf{Pr}\text{/}[X-\mathbf{E}\text{/}[X]\ge m/4]$$

$$\le \frac{\mathbf{Var}\text{/}[X]}{(m/4{)}^2}<\frac{16}{25t}.$$

$$\#\left\{uv\in \left(\genfrac{}{}{0ex}{}{S}{2}\right):d_{uv}^{(r)}\le m\right\}\le m-t$$

$$\#\left\{uv\in \left(\genfrac{}{}{0ex}{}{T}{2}\right):d_{uv}^{(r)}\le m\right\}\le \max \left(0,\#\left\{uv\in \left(\genfrac{}{}{0ex}{}{S}{2}\right):d_{uv}^{(r)}\le m\right\}-2t\right)\le \max (0,m-3t).$$

$$\#\left\{uv\in \left(\genfrac{}{}{0ex}{}{T}{2}\right):d_{uv}^{(r)}\le m\right\}\le \#\left\{uv\in \left(\genfrac{}{}{0ex}{}{T}{2}\right):d_{uv}^{(r)}\le m^{\prime }\right\}\le \max (0,m^{\prime }-3t)\le m-2t.$$

$$\displaystyle a_{m}\$$

$$\displaystyle a_{\leq m}\$$

$$|U_{\le 4m_v}^{(b)}(v,X_v)|\ge |U_{\le m_v}^{(b)}(v,X_v)|\ge 10m_v>8m_v,$$

$$\frac{1}{4}\left(a_0+a_1+\cdots +a_{4t}\right)+\frac{a_{4t+1}t}{4t+1}+\cdots +\frac{a_{4m_v}t}{4m_v}+t.$$

$${\beta }^{(b)}(w)$$

$$\le t+\frac{1}{4}\cdot 40t+\frac{10t}{4t+1}+\frac{10t}{4t+2}+\cdots +\frac{10t}{4m_v}$$

$$\le 11t+10t\ln (4m_v/4t)<20t\ln t,$$

$$\sum _{u\in V}{\alpha }^{(\chi )}(u)=10t\cdot |S^{(\chi )}|\le 10t^2.$$

$$\sum _{u\in V}{\beta }^{(\chi )}(u)\le 20t\ln t\cdot |S^{(\chi )}|\le 20t^2\ln t.$$

$$\displaystyle\#\{\text{deleted $u$}\}\$$

$$+\#\{u:\text{β(b)(u)≥1/4}\}+\#\{u:\text{β(r)(u)≥1/4}\}$$

$$<2t+\sum _{u\in V}\left(2{\alpha }^{(b)}(u)+2{\alpha }^{(r)}(u)+4{\beta }^{(b)}(u)+4{\beta }^{(r)}(u)\right)$$

$$\le 2t+2\cdot 10t^2+2\cdot 10t^2+4\cdot 20t^2\ln t+4\cdot 20t^2\ln t$$

$$<200t^2\ln t.\text{qed}$$

$$|U_{\le m}^{(b)}(v,X)|<10m,|U_{\le m}^{(r)}(v,X)|<10m.$$

$$\sum _{v_i{v}_j\in E_{bad}}\frac{1}{d_{v_i{v}_j}^{(b)}}<\frac{1}{4}.$$

Full text

1 Introduction

For a $k$-uniform hypergraph $H$, the Ramsey number $r(H)$ is the smallest $n$ such that any 2-coloring of $K_{n}^{(k)}$, the complete $k$-uniform hypergraph on $n$ vertices, contains a monochromatic copy of $H$. Let $r(H;q)$ denote the analogous Ramsey number for $q$-colorings, so that $r(H)=r(H;2)$.

It is a major open problem to determine the growth of $r(K_{t}^{(3)})$, the Ramsey number of the complete 3-uniform hypergraph on $t$ vertices. It is known [6, 7] that there are constants $c,c^{\prime}>0$ such that

[TABLE]

Erdős conjectured that $r(K_{t}^{(3)})=2^{2^{\Theta(t)}}$, i.e. the upper bound is closer to the truth. Erdős and Hajnal gave some evidence that this conjecture is true by showing that $r_{3}(K_{t}^{(3)};4)\geq 2^{2^{ct}}$, i.e. the four color Ramsey number of $K_{t}^{(3)}$ is double-exponential in $t$ (see, for example [9]).

Definition 1.1**.**

The hedgehog $H_{t}$ is a 3-uniform hypergraph on $t+\binom{t}{2}$ vertices $1,\dots,t+\binom{t}{2}$ such that, for each $1\leq i<j\leq t$, there exists a unique vertex $k>t$ such that $\{i,j,k\}$ is an edge, and there are no additional edges.

We sometimes refer to the first $t$ vertices as the body of the hedgehog. For any $k\geq 4$, one can also define a $k$-uniform hedgehog $H_{t}^{(k)}$ on $t+\binom{t}{k-1}$, with a body of size $t$ and a unique hyperedge for every $k-1$-sized subset of the body. In this notation, we have $H_{t}=H_{t}^{(3)}$.

Hedgehogs are interesting because their 2-color Ramsey number $r(H_{t};2)$ is polynomial in $t$, while their 4-color Ramsey number $r(H_{t};4)$ is exponentially large in $t$ [10, 5]. This suggests that the bound $r(K_{t}^{(3)};4)\geq 2^{2^{ct}}$ by Erdős and Hajnal may not be such strong evidence that $r(K_{t}^{(3)})=2^{2^{\Theta(t)}}$.

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 $d$ such that every subgraph induced by a set of vertices has a vertex of degree at most $d$. The Burr-Erdős conjecture [2] states that there exists a constant $c(d)$ depending only on $d$ such that the Ramsey number of any $d$-degenerate graph $G$ on $n$ vertices satisfies $r(G)\leq c(d)\cdot n$. 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 $d$ such that every subhypergraph induced by a subset of vertices has a vertex of degree at most $d$. Under this definition, Conlon, Fox, and Rödl [5] observe that the 4-uniform analogue of the Burr-Erdős conjecture is false: the 4-uniform hedgehog $H^{(4)}_{t}$, which is 1-degenerate, satisfies $r(H^{(4)}_{t})\geq 2^{ct}$. They also observe that the 3-uniform analogue of the Burr-Erdős conjecture is false for 3 or more colors: the 3-uniform hedgehog, which is 1-degenerate, satisfies $r(H_{t};3)\geq\Omega(t^{3}/\log^{6}t)$.

However, the analogue of the Burr-Erdős conjecture for 3-uniform hypergraphs and 2 colors remains open. In particular, it was not known whether the Ramsey number of the hedgehog $H_{t}$ is linear, or even near-linear, in the number of vertices, $t+\binom{t}{2}$. Conlon, Fox and Rodl [5] show $r(H_{t};2)\leq 4t^{3}$, and, with the above in mind, ask if $r(H_{t};2)=t^{2+o(1)}$. We answer this question affirmatively.

Theorem 1.2**.**

If $t\geq 10$ and $n\geq 200t^{2}\ln t+400t^{2}$, then every two-coloring of the complete $3$-uniform hypergraph on vertices contains a monochromatic copy of the hedgehog $H_{t}$. That is,

[TABLE]

We make no attempt to optimize the absolute constants here.

2 Ramsey number of hedgehogs

Throughout this section, we assume $t\geq 10$, and that we have a fixed two-coloring of the edges of a complete 3-uniform hypergraph $\mathcal{H}$ on vertex set $V$ with $n\geq 200t^{2}\ln t+400t^{2}$ vertices. Let

[TABLE]

Let $\binom{S}{2}$ denote the set of pairs of elements of $S$. For integer $a$, let $[a]=\{1,2,\dots,a\}$. For vertices $u$ and $v$ of $\mathcal{H}$, we write $uv$ as an abbreviation for the unordered pair $\{u,v\}$.

For $u,v\in V$, let

[TABLE]

For a set of pairs $F\subset\binom{V}{2}$, let

[TABLE]

Here, and throughout, we use $b$ and $r$ to refer to the colors blue and red, respectively. For a vertex $v$ and set $X$, let

[TABLE]

If $X$ is omitted, take $X=V$. We define $U^{(b)}_{\leq m}(v,X)$ to be sets of $u$ such that $d^{(r)}_{uv}$ is small, rather than those such that $d^{(b)}_{uv}$ is small, because we wish to think of $U^{(b)}$’s as sets helpful for finding a blue hedgehog. Similarly, we think of $U^{(r)}$’s as sets helpful for finding a red hedgehog.

Lemma 2.1**.**

For any $0\leq m<\frac{|V|}{2}-1$, and $v\in V$,

[TABLE]

Proof.

Fix $m$ and $v$. For convenience, let $A=U^{(b)}_{\leq m}(v)$ and $B=U^{(r)}_{\leq m}(v)$. Assume for contradiction that $|A|,|B|\geq 2m+1$. For every $u$, we have $d^{(r)}_{uv}+d^{(b)}_{uv}=|V|-2>2m$, so $A$ and $B$ are disjoint. Consider the set $E^{\prime}$ of edges of $\mathcal{H}$ containing $v$, one element of $A$, and one element of $B$. On one hand, $|E^{\prime}|=|A|\cdot|B|$. On the other hand, for every $u\in A$, the pair $uv$ is in at most $m$ such red triples, so the number of red triples of $E^{\prime}$ is at most $|A|\cdot m$. Additionally, for every $u\in B$, the pair $uv$ is in at most $m$ such blue triples, so the number of blue triples of $E^{\prime}$ is at most $|B|\cdot m$. Hence, $(|A|+|B|)\cdot m\leq|E^{\prime}|=|A|\cdot|B|$, a contradiction of $|A|,|B|\geq 2m+1$. ∎

The following “matching condition” for hedgehogs is useful.

Lemma 2.2**.**

Let $S\subset V$ be a set of $t$ vertices. If, for all nonempty sets $F\subset\binom{S}{2}$, we have $|N^{(b)}(F)|\geq|F|+t$, then there exists a blue hedgehog with body $S$. Similarly, if, for all nonempty sets $F\subset\binom{S}{2}$, we have $|N^{(r)}(F)|\geq|F|+t$, then there exists a red hedgehog with body $S$.

Proof.

By symmetry, it suffices to prove the first part. Consider the bipartite graph $G$ between pairs in $\binom{S}{2}$ and vertices of $V\setminus S$, where $uv\in\binom{S}{2}$ is connected with $w\in V\setminus S$ if and only if triple $\{u,v,w\}$ is blue. If, for all nonempty $F\subset\binom{S}{2}$, we have $|N^{(b)}(F)|\geq|F|+t$, then any such $F$ has at least $|F|+t-|S|=|F|$ neighbors in $G$. By Hall’s marriage lemma on $G$, there exists a matching in $G$ using every element of $\binom{S}{2}$. Taking triples $\{u,v,w\}$ where $uv\in\binom{S}{2}$ and $w\in V\setminus S$ is the vertex matched with pair $uv$ gives a blue hedgehog with body $S$. ∎

2.1 Special Cases

We start by finding monochromatic hedgehogs in two specific classes of colorings on $\mathcal{H}$. 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 $G$ on $[n]$. 2. 2.

Color a complete hypergraph $\mathcal{H}$ on $[n]$ by coloring the triple $\{u,v,w\}$ blue if at least one of $uv,uw,vw$ is in $G$, and red otherwise.

Lemma 2.3**.**

If $n\geq t^{2}+t$, any hypergraph colored as above has a monochromatic $H_{t}$.

Proof.

Set $X=V(G)$. For $i=t-1,t-2,\dots,0$, pick a vertex $v_{i}\in X$ whose degree in $G$ is at least $i$ and let $\hat{U}(v_{i})\subset X$ be an arbitrary set of $i$ neighbors of $v_{i}$. Remove $v_{i}\cup\hat{U}(v_{i})$ from $X$. We call this the peeling step of $v_{i}$. Figure 1 shows the first three peeling steps of this process for $t=5$. If this process succeeds, we have found a set $S=\{v_{t-1},\dots,v_{0}\}$ of $t$ vertices and disjoint sets of vertices $\hat{U}(v_{0}),\dots,\hat{U}(v_{t-1})$ also disjoint from $S$, from which we can greedily embed a blue-hedgehog in $\mathcal{H}$ with body $\{v_{0},\dots,v_{t-1}\}$: for each $v_{i}v_{j}$ with $i<j$, pick an arbitrary unused element of $\hat{U}(v_{j})$ for the third vertex of the hedgehog’s edge containing $v_{i}v_{j}$.

Now suppose this process finds vertices $v_{t-1},v_{t-2},\dots,v_{i+1}$ but fails to find $v_{i}$ for some $i\leq t-1$. After picking $v_{j}$, we remove $v_{j}$ and $j$ of it’s neighbors from $X$, for a total of $j+1$ vertices. Then we have removed exactly $t+(t-1)+\cdots+(i+2)=\binom{t+1}{2}-\binom{i+2}{2}$ vertices from $X$. Hence, $|X|\geq(t^{2}+t)-\binom{t+1}{2}+\binom{i+2}{2}=\binom{t+1}{2}+\binom{i+2}{2}>\frac{t^{2}+i^{2}}{2}\geq ti$, and every vertex has degree at most $i-1$ in the subgraph of $G$ induced $X$. Thus, there exists an independent set $S\subset X$ in $G$ of size at least $|X|/i\geq t$. Furthermore, any vertex has at most $i-1$ neighbors in $X$, so any two vertices $u,v\in S$ share at least $|X|-2i\geq t+\binom{t}{2}+\binom{i+2}{2}-2i>t+\binom{t}{2}$ red triples in the subhypergraph of $\mathcal{H}$ induced by $X$, so we can greedily find a red hedgehog with body $S$. ∎

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 $v$ and every nonnegative integer $m$ less than $\frac{|V|}{2}-1$, one of $|U_{\leq m}^{(b)}(v)|=\#\{u:d^{(r)}_{uv}\leq m\}$ and $|U_{\leq m}^{(r)}(v)|=\#\{u:d^{(b)}_{uv}\leq m\}$ is at most $2m$. In “balanced” colorings, we assume, for all $v\in V$ and all $2t\leq m\leq m_{max}\vcentcolon=2t+\binom{t}{2}$, both of $|U_{\leq m}^{(b)}(v)|$ and $|U_{\leq m}^{(r)}(v)|$ are $O(m)$. We show, in this case, there is a monochromatic hedgehog. The proof is by choosing a random subset of approximately $4t$ vertices, and showing that, with positive probability, we can remove vertices so that the remaining set of $t$ vertices is the body of some red hedgehog.

Lemma 2.4**.**

Let $c\geq 1$. Consider a two-colored hypergraph $\mathcal{H}=(V,E)$ on $n\geq 40ct^{2}$ vertices. Suppose that for all $2t\leq m\leq m_{max}$ and all $v\in V$, we have

[TABLE]

Then $\mathcal{H}$ has a red hedgehog $H_{t}$.

Proof.

It suffices to prove for $n=40ct^{2}$, so assume without loss of generality that $n=40ct^{2}$. Pick a random set $S$ by including each vertex of $V$ in $S$ independently with probability $4t/n$. By the Chernoff bound, $\mathop{\bf Pr\/}[|S|\leq 3t]\leq e^{-t/8}$.

Fix $m$ such that $2t\leq m\leq m_{max}$ and $m$ is a multiple of $t$. Let $e_{1},\dots,e_{p}$ be the pairs such that $d^{(b)}_{e_{\ell}}\leq m$ for all $\ell\in[p]$, and let $X_{1},\dots,X_{p}$ the indicator random variables for these pairs being in $\binom{S}{2}$. Let $X=X_{1}+\cdots+X_{p}$. By (7), we have $p\leq cmn/2$. Each $X_{\ell}$ for $\ell\in[p]$ is a Bernoulli$(16t^{2}/n^{2})$ random variable. Consider a graph on $[p]$ where $\ell$ and $\ell^{\prime}$ are adjacent (written $\ell\sim\ell^{\prime}$) if $e_{\ell}$ and $e_{\ell^{\prime}}$ share a vertex. This is a valid dependency graph for $\{X_{\ell}\}$ as $X_{\ell}$ is independent of all $X_{\ell^{\prime}}$ such that $e_{\ell^{\prime}}$ is vertex disjoint from $e_{\ell}$. Furthermore, by the condition (7), each endpoint of any pair $e_{\ell}$ is in at most $cm$ pairs, so each $\ell\in[p]$ has degree at most $2cm$ in the dependency graph, and the total number of pairs $(\ell,\ell^{\prime})$ such that $\ell\sim\ell^{\prime}$ is at most $2cmp$. We have

[TABLE]

Hence,

[TABLE]

The first inequality is by (8) and the second is by Chebyshev’s inequality. By the union bound over the multiples of $t$ in $[2t,m_{max}]$, of which there are less than $t$, the probability there exists some $m\in[2t,m_{max}]$ a multiple of $t$ with

[TABLE]

is less than $t\cdot\frac{16}{25t}=\frac{16}{25}$. Again by the union bound, with probability more than $1-(\frac{16}{25}+e^{-t/8})>0$ over the randomness of $S$, we have (i) $|S|\geq 3t$, and (ii) for all $m$ a multiple of $t$ in $[2t,m_{max}]$, (11) holds. Hence, there exists an $S$ such that (i) and (ii) hold, so consider such an $S$. Remove $|S|-t\geq 2t$ vertices from $S$, at least one from each of the $2t$ pairs with smallest $d^{(r)}_{uv}$, to obtain a set of $t$ vertices $T$ such that, for all $m$ a multiple of $t$ in $[2t,m_{max}]$, we have

[TABLE]

Then, for all $m$ with $2t\leq m\leq m_{max}-t$, set $m^{\prime}$ to be the smallest multiple of $t$ larger than $m$, so that

[TABLE]

Now, we show our matching condition holds. Setting $m=2t$ in (13), we have $x_{uv}>2t$ for all $uv\in\binom{T}{2}$. Hence, for any nonempty subset $F\subset\binom{T}{2}$ of size at most $t$, any $uv\in F$ satisfies $x_{uv}>t+|F|$. If $F\subset\binom{T}{2}$ has size greater than $t$, then, by setting $m=t+|F|$ in (13), we know that there are at most $m-2t=|F|-t$ pairs $uv\in F$ such that $d^{(r)}_{uv}\leq t+|F|$, so again there exists $uv\in F$ such that $d^{(r)}_{uv}>t+|F|$. We conclude that, for all nonempty subsets of pairs $F\subset\binom{T}{2}$, there exists $uv\in F$ such that $|N^{(r)}(F)|\geq d^{(r)}_{uv}\geq t+|F|$. By Lemma 2.2, there exists a red hedgehog with body $T$. ∎

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 $v$ into a set $S$ to try to find a blue or red hedgehog.111 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 two-colored 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 $X=V$ and iteratively removed from $X$ a vertex $v$ and a set $\hat{U}(v)$ of size $t$ such that, for all $u\in\hat{U}(v)$, vertices $u$ and $v$ share many blue triples. This deletes $O(t)$ vertices per round, which is small enough for the argument to succeed. For general hypergraphs, we peel off vertices $v$ with many “blue-heavy neighbors”, meaning there exists some $m$ such that $|U^{(b)}_{\leq m}(v,X)|\geq 10m$.222 For technical reasons, we peel vertices $v$ in increasing order of the corresponding $m$. However, $m$ can be $\Theta(t^{2})$, so if we simply deleted $v$ along with $10m$ of its blue-heavy neighbors $\hat{U}^{(b)}(v)\subset U^{(b)}_{\leq m}(v,X)$, we could delete $\Theta(t^{2})$ vertices for every $v$, which is too many. Instead, when we peel off $v$, we delete $v$ from $X$, add a penalty of $t/m$ to each $u\in\hat{U}^{(b)}(v)$, accumulated as $\alpha^{(b)}(u)$, and delete from $X$ every vertex $u$ with $\alpha^{(b)}(u)\geq 1/2$. With these penalties, we guarantee that, on average, we delete $O(t)$ vertices from $X$ per peeled vertex $v$.

However, we need more care. In Lemma 2.3, we can find a hedgehog with body $S$ because, for any peeled vertices $v,v^{\prime}\in S$, the edges $\{u,v,v^{\prime}\}$ are blue for every $u\in\hat{U}(v)$. However, in our procedure, for a $v$ chosen with corresponding $\hat{U}^{(b)}(v)$ of size $10m$, there are some vertices $w$ such that $\{u,v,w\}$ is blue for few (at most $4m$) vertices $u\in\hat{U}^{(b)}(v)$. We denote this set of “bad” vertices by $B^{(b)}(v)$. As much as possible, we wish to avoid choosing both $v$ and, at some later step, $w\in B^{(b)}(v)$ for the body $S^{(b)}$ of our blue hedgehog. Ideally, we simply delete all vertices $u\in B(v)$ in the step we peel off $v$. However, $B^{(b)}(v)$ can have $\Omega(m)$ vertices, which again could be too many if $m=\Theta(t^{2})$. Instead, for each $w\in B^{(b)}(v)$ we add a penalty of $t/d^{(b)}_{wv}$, accumulated as $\beta^{(b)}(w)$, and delete from $X$ every vertex $w$ with $\beta^{(b)}(w)\geq 1/4$. We guarantee that, on average, we delete $O(t\ln t)$ vertices from $X$ per peeled vertex $v$ (Lemma 2.9).

To finish the proof, we show, if our peeling produces a set $S^{(b)}=\{v_{1},\dots,v_{t}\}$ (where $v_{i}$ is chosen before $v_{i+1}$), then, because we track the penalties $\alpha^{(b)}(u)$ and $\beta^{(b)}(w)$ carefully, the matching condition of Lemma 2.2 holds. On the other hand, if the peeling procedure fails, the subhypergraph induced by $X$ 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 $S^{(b)}=S^{(r)}=\emptyset$, and $X=V$. For all $u\in V$, initialize $\alpha^{(r)}(u)=\alpha^{(b)}(u)=\beta^{(r)}(u)=\beta^{(b)}(u)=0$. If, at any point, $S^{(b)}$ or $S^{(r)}$ has $t$ vertices, stop.

Recall that $m_{max}=2t+\binom{t}{2}$. For $m=2t,2t+1,\cdots,m_{max}$, do the following, which we refer to as $\textnormal{Stage}(m)$.

While there exists a vertex $v\in X$ and a color $\chi\in\{b,r\}$ such that $|U^{(\chi)}_{\leq m}(v,X)|\geq 10m$:

1. (a)

Let $\hat{U}^{(\chi)}(v)$ be the set $U^{(\chi)}_{\leq m}(v,X)$ truncated to $10m$ vertices arbitrarily. 2. (b)

Let B^{(\chi)}(v)=\left\{w:\left|u\in\hat{U}^{(\chi)}(v):\{u,v,w\}\text{ is color \chi}\right|\leq 4m\right\}. 3. (c)

Add $v$ to $S^{(\chi)}$. 4. (d)

For all $u\in\hat{U}^{(\chi)}(v)$, add $t/m$ to $\alpha^{(\chi)}(u)$. 5. (e)

For all $w\in B^{(\chi)}(v)$, add $\min(1/4,t/d^{(\chi)}_{vw})$ to $\beta^{(\chi)}(w)$. 6. (f)

Delete from $X$ all vertices $u$ with $\alpha^{(\chi)}(u)\geq 1/2$ or $\beta^{(\chi)}(u)\geq 1/4$. 7. (g)

Delete $v$ from $X$.

Note that $B^{(\chi)}(v)$ and $\hat{U}^{(\chi)}(v)$ are only defined for $v\in S^{(\chi)}$. We refer to steps 1(a)-1(g) as the peeling step for $v$, denoted $\textnormal{Peel}(v)$. We let $m_{v}$ denote the value such that the peeling step for $v$ occurred during $\textnormal{Stage}(m_{v})$, and call $m_{v}$ the peeling parameter of $v$. Throughout the analysis, let $X_{v}$ denote the set $X$ immediately before $\textnormal{Peel}(v)$. For any $m\in[2t,m_{max}]$, let $X_{m}$ denote the set $X$ immediately after $\textnormal{Stage}(m)$, so that $X_{m_{max}}$ is the set $X$ 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$(m_{max})$ and $|S^{(b)}|<t$ and $|S^{(r)}|<t$, or we “finish”, i.e. we terminate earlier with $|S^{(b)}|=t$ or $|S^{(r)}|=t$. 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 $m$ such that $2t\leq m\leq m_{max}$, for any time in the procedure after $\textnormal{Stage}(m)$, the following holds: for all colors $\chi\in\{b,r\}$, for all $m^{\prime}$ with $2t\leq m^{\prime}\leq m$, and for all vertices $v\in X$, we have $|U^{(\chi)}_{\leq m^{\prime}}(v,X)|<10m^{\prime}$.

Proof.

Fix $m$ with $2t\leq m\leq m_{max}$. We have $|U^{(\chi)}_{\leq m}(v,X_{m})|<10m$ for all $v\in X_{m}$: if not, then there exists a vertex $v\in X_{m}$ with $|U^{(\chi)}_{\leq m}(v,X_{m})|\geq 10m$, in which case we would have peeled vertex $v$ during $\textnormal{Stage}(m)$, and we would have deleted $v$ from $X_{m}$ during $\textnormal{Peel}(v)$, which is a contradiction. Throughout the procedure, $X$ is nonincreasing. Thus, at any point in the procedure after $\textnormal{Stage}(m)$, we have $X\subset X_{m}$, so for all $v\in X$, we have $v\in X_{m}$ and $|U^{(\chi)}_{\leq m}(v,X)|\leq|U^{(\chi)}_{\leq m}(v,X_{m})|<10m$. ∎

Lemma 2.6**.**

For all colors $\chi\in\{b,r\}$ and all vertices $v\in S^{(\chi)}$, we have $|B^{(\chi)}(v)|\leq 2m_{v}$.

Proof.

We prove this for $\chi=b$, and the case $\chi=r$ follows from symmetry. We double-count the number $Z$ of red triples $\{u,v,w\}$ such that $u\in\hat{U}^{(b)}(v)$ and $w\in B^{(b)}(v)$. On one hand, every $u\in\hat{U}^{(b)}(v)$ is in at most $m_{v}$ red triples because we chose $\hat{U}^{(b)}(v)$ as a subset of $U^{(b)}_{\leq m_{v}}(v,X_{v})$, so the total number of red triples is at most $m_{v}\cdot|\hat{U}^{(b)}(v)|=10m_{v}^{2}$. On the other hand, by definition of $B^{(b)}(v)$, each $w\in B^{(b)}(v)$ is in at least $|\hat{U}^{(b)}(v)|-4m_{v}=6m_{v}$ such red triples. Thus, the number of such triples is at least $|B^{(b)}(v)|\cdot 6m_{v}$. Hence, $10m_{v}^{2}\geq Z\geq 6m_{v}|B^{(b)}(v)|$ so $|B^{(b)}(v)|\leq 2m_{v}$ as desired. ∎

Lemma 2.7**.**

For all colors $\chi\in\{b,r\}$ and all vertices $v,v^{\prime}\in S^{(\chi)}$, we have $d^{(\chi)}_{vv^{\prime}}\geq 4t$.

Proof.

Assume for sake of contradiction that $d^{(\chi)}_{vv^{\prime}}<4t$. Without loss of generality, $v$ was added to $S^{(\chi)}$ before $v^{\prime}$. We have $d^{(\chi)}_{vv^{\prime}}<4t<4m_{v}$, so during $\textnormal{Peel}(v)$, vertex $v^{\prime}$ is included in $B^{(\chi)}(v)$. Hence, $\min(1/4,t/d^{(\chi)}_{vv^{\prime}})=1/4$ is added to $\beta^{(\chi)}_{\leq 4t}(v^{\prime})$ during 1(e) of $\textnormal{Peel}(v)$, so during 1(f) of $\textnormal{Peel}(v)$, vertex $v^{\prime}$ is deleted from $X$ if it hasn’t been deleted already. Thus, we could not have added $v^{\prime}$ to $S^{(\chi)}$ after $\textnormal{Peel}(v)$, which is a contradiction, so $d^{(\chi)}_{vv^{\prime}}\geq 4t$, as desired. ∎

2.2.4 Bounding the number of deleted vertices

Lemma 2.8**.**

For all colors $\chi\in\{b,r\}$ and all vertices $v\in S^{(\chi)}$, during $\textnormal{Peel}(v)$, the total increase in $\alpha^{(\chi)}(u)$ over all $u\in V$ is exactly $10t$.

Proof.

Fix $v\in S^{(\chi)}$. We have $|\hat{U}^{(\chi)}(v)|=10m_{v}$ by definition, and, for $u\in\hat{U}^{(\chi)}(v)$, each $\alpha^{(\chi)}(u)$ increases by exactly $t/m_{v}$, for a total increase of $10m_{v}\cdot(t/m_{v})=10t$. ∎

Lemma 2.9**.**

For all colors $\chi\in\{b,r\}$ and all vertices $v\in S^{(\chi)}$, during $\textnormal{Peel}(v)$, the total increase in $\beta^{(\chi)}(w)$ over all $w\in V$ is at most $20t\ln t$.

Proof.

By symmetry, it suffices to prove the lemma for $\chi=b$. Let $v\in S^{(b)}$. For $m=0,\dots,4m_{v}$, let

[TABLE]

$\textnormal{Peel}(v)$ is after $\textnormal{Stage}(m_{v}-1)$. Hence, by Lemma 2.5, for $2t\leq m\leq m_{v}-1$, we have $a_{\leq m}\leq 10m$. We know

[TABLE]

where the second inequality holds because $v$ was chosen to be peeled in $\textnormal{Stage}(m_{v})$. Hence, by Lemma 2.1, $a_{\leq 4m_{v}}=|U^{(r)}_{\leq 4m_{v}}(v,X_{v})|\leq|U^{(r)}_{\leq 4m_{v}}(v)|\leq 8m_{v}$. As $a_{\leq m}$ is non-decreasing in $m$, we conclude $a_{\leq m}\leq 10m$ for $2t\leq m\leq 4m_{v}$.

For $m=0,\dots,4m_{v}$, for any $w$ with $d^{(b)}_{vw}=m$, the peeling of $v$ increases $\beta^{(b)}(w)$ by exactly $\min(1/4,t/m)$. Thus, for $a_{m}$ many $w$, the penalty $\beta^{(b)}(w)$ increases by $\min(1/4,t/m)$. Furthermore, $\beta^{(b)}(w)$ increases only for $w\in B^{(b)}(v)$, which has at most $2m_{v}$ vertices by Lemma 2.6. For $2m_{v}-a_{\leq 4m_{v}}$ vertices $w$, $\beta^{(b)}(w)$ increases by less than $t/4m_{v}$, giving a total increase in $\beta^{(b)}(w)$ of less than $t$ from those vertices. The total increases in $\beta^{(b)}(w)$ is thus less than

[TABLE]

The coefficients of $a_{0},\dots,a_{4m_{v}}$ in (17) are nonincreasing, so (17) is $t$ plus a positive linear combination of $a_{\leq 4t},a_{\leq 4t+1},\cdots,a_{\leq 4m_{v}}$. Subject to $a_{\leq m}\leq 10m$ for $2t\leq m\leq 4m_{v}$, all of $a_{\leq 4t},a_{\leq 4t+1},\dots,a_{\leq 4m_{v}}$ are simultaneously maximized when $a_{0}=0$ and $a_{m}=10$ for $m=1,\dots,4m_{v}$, so (17) is maximized there as well. Hence,

[TABLE]

where, for the last inequality, we used $m_{v}\leq t^{2}$ and $t\geq 10$. This is what we wanted to show. ∎

Lemma 2.10**.**

The total number of vertices deleted from $X$ in the peeling procedure is at most $200t^{2}\ln t$.

Proof.

A vertex is deleted either for being added to $S^{(b)}$ or $S^{(r)}$, having $\alpha^{(b)}(\cdot)$ or $\alpha^{(r)}(\cdot)$ at least 1/2, or having $\beta^{(b)}(\cdot)$ or $\beta^{(r)}(\cdot)$ at least 1/4. At the end of the procedure, we have the following inequalities. For all $\chi\in\{b,r\}$ and all $u\in V$, we have $\alpha^{(\chi)}(u)$ and $b^{(\chi)}(u)$ are initially 0 and increase only during the peeling step of some vertex $v\in S^{(\chi)}$. Hence, by Lemma 2.8, for $\chi\in\{b,r\}$,

[TABLE]

Furthermore, by Lemma 2.9, for $\chi\in\{b,r\}$,

[TABLE]

We conclude that, at the end of the procedure,

[TABLE]

2.2.5 Case 1: Peeling procedure gets stuck

By Lemma 2.10, the number of vertices deleted in the peeling process is at most $200t^{2}\ln t$, so, at the end of the peeling procedure, $|X|\geq(200t^{2}\ln t+400t^{2})-200t^{2}\ln t=400t^{2}$.

Consider the complete 2-colored subhypergraph $\mathcal{H}^{\prime}$ of $\mathcal{H}$ induced by the vertex set $X$. By Lemma 2.5, at the end of the procedure, for all $m=2t,2t+1,\dots,m_{max}$ and all $v\in X$,

[TABLE]

Applying Lemma 2.4 to $\mathcal{H}^{\prime}$ with $c=10$, we conclude $\mathcal{H}^{\prime}$ (and hence $\mathcal{H}$) has a red hedgehog $H_{t}$.333By the same reasoning $\mathcal{H}^{\prime}$ also has a blue hedgehog.

2.2.6 Case 2: Peeling procedure finishes

Suppose we finish with $|S^{(b)}|=t$. The analysis for $|S^{(r)}|=t$ is symmetrical. We try to find a blue hedgehog. For brevity, in the rest of this section, let $S=S^{(b)}$. Let $S=\{v_{1},\cdots,v_{t}\}$, where the $v_{i}$ were chosen in the order $v_{1},\dots,v_{t}$. For $i=1,\dots,t$, let $m_{i}=m_{v_{i}}$ be the peeling parameter for $v_{i}$, so that $m_{1}\leq m_{2}\leq\cdots\leq m_{t}$.

Definition 2.11**.**

Call a pair $v_{i}v_{j}\in\binom{S}{2}$ with $i<j$ bad if $v_{j}\in B^{(b)}(v_{i})$. Otherwise, call $v_{i}v_{j}\in\binom{S}{2}$ good. Let $E_{bad}\subset\binom{S}{2}$ be the set of all bad pairs and let $E_{good}\subset\binom{S}{2}$ be the set of all good pairs, so that $\binom{S}{2}=E_{bad}\cup E_{good}$ is a partition.

Lemma 2.12**.**

[TABLE]

Proof.

Fix $2\leq j\leq t$. Consider all bad pairs $v_{i}v_{j}$ with $i<j$. At the peeling of $v_{j}$, $\beta(v_{j})<1/4$, otherwise $v_{j}$ would have been deleted from $X$ and we could not have peeled $v_{j}$. Hence, at the peeling of $v_{j}$,

[TABLE]

The first equality is by definition of $\beta^{(b)}(v_{j})$, the second is by definition of $E_{bad}$, and the last is because $d^{(b)}_{v_{i}v_{j}}\geq 4t$ for all $i<j$ by Lemma 2.7. Thus,

[TABLE]

We prove that there is a blue hedgehog with body $S$, by showing the matching condition of Lemma 2.2 holds. Consider an arbitrary $F\subset\binom{S}{2}$. Partition $F=F_{bad}\cup F_{good}$, where $F_{bad}=F\cap E_{bad}$ and $F_{good}=F\cap E_{good}$. We wish to show that $N^{(b)}(F)\geq|F|+t$.

Subcase 1: $|F_{bad}|\geq|F_{good}|$.

By Lemma 2.12,

[TABLE]

Thus, there exists some $v_{i}v_{j}\in F_{bad}$ such that $d^{(b)}_{v_{i}v_{j}}>4|F_{bad}|$. Furthermore, this $v_{i}v_{j}$ satisfies $d^{(b)}_{v_{i}v_{j}}\geq 4t$ by Lemma 2.7, so $d^{(b)}_{v_{i}v_{j}}\geq 2|F_{bad}|+2t$. Hence,

[TABLE]

as desired. The first inequality is because the blue edges containing $v_{i}v_{j}$ are all elements of $N^{(b)}(F)$. The second inequality is because $d^{(b)}_{v_{i}v_{j}}$ is at least $4|F_{bad}|$ and at least $4t$ by above. The third inequality is by the assumption $|F_{bad}|\geq|F_{good}|$. The fourth inequality is because $|F|=|F_{bad}|+|F_{good}|$ and $2t>t$.

Subcase 2: $|F_{bad}|<|F_{good}|$.

In particular, $|F_{good}|>0$, so $|F|$ has some good pair $v_{i}v_{j}$ with $i<j$. This pair is in at least $4m_{i}\geq 8t$ blue triples, so $|N^{(b)}(F)|\geq 8t$.

Let $I$ be the set of all indices $i$ such that there exists $j$ with $i<j\leq t$ with $v_{i}v_{j}\in F_{good}$. For each $i$, there are less than $t$ indices $j$ such that $i<j\leq t$, so

[TABLE]

For each $i\in I$, arbitrarily fix $j_{i}>i$ such that $v_{i}v_{j_{i}}$ is good. For $i\in I$, define

[TABLE]

so that $U^{*}_{I}\subset N^{(b)}(F)$. For all $i\in I$, the pair $v_{i}v_{j_{i}}$ is good, so $v_{j_{i}}\notin B^{(b)}(v_{i})$. Hence, by the definition of $B^{(b)}(v_{i})$, there are more than $4m_{i}$ vertices $u\in\hat{U}^{(b)}(v_{i})$ such that $\{u,v_{i},v_{j_{i}}\}$ is blue. Thus, for all $i\in I$, the set $U^{*}_{i}$ has at least $4m_{i}$ vertices. In the peeling of $v_{i}$, the penalty $\alpha^{(b)}(u)$ increases by $t/m_{i}$ for each $u\in U^{*}_{i}$. Hence, in peeling $v_{i}$, the sum of penalties $\sum_{u\in U^{*}_{I}}\alpha^{(b)}(u)$, increases by at least $4m_{i}\cdot t/m_{i}=4t$. Thus,

[TABLE]

On the other hand, the vertex $u$ is deleted from $X$ whenever $\alpha^{(b)}(u)\geq 1/2$, the penalty $\alpha^{(b)}(u)$ increases by at most $t/2t=1/2$ in any peeling step, and the penalty $\alpha^{(b)}(u)$ never changes after $u$ is deleted from $X$. Thus, for all vertices $u\in V$, we have

[TABLE]

We conclude

[TABLE]

The first inequality is by the assumption $|F_{bad}|<|F_{good}|$, the second is by (26), the third is by (28), the fourth is by (29), the fifth is by $U_{I}^{*}\subset N^{(b)}(F_{good})$, and the sixth is by $F_{good}\subset F$. Combining with $\left|N^{(b)}(F)\right|\geq 8t$, we conclude $\left|N^{(b)}(F)\right|\geq|F|+t$, as desired.

This covers all subcases, so we’ve proven that, for any nonempty subset $F\subset\binom{S}{2}$, we have $N^{(b)}(F)\geq|F|+t$. Hence, the matching condition of Lemma 2.2 holds, so there is a blue hedgehog with body $S$, as desired. This completes the proof of Theorem 1.2. ∎