The game chromatic number of a random hypergraph

TL;DR

This paper investigates the game chromatic number of random hypergraphs, establishing probabilistic bounds on the minimum number of colors needed for a player to guarantee a proper coloring in a two-player game.

Contribution

It provides the first probabilistic bounds on the game chromatic number for random hypergraphs, extending understanding of coloring games in probabilistic combinatorics.

Findings

Established upper bounds on the game chromatic number w.h.p.

Derived lower bounds for the game chromatic number w.h.p.

Analyzed the behavior of the game in the context of random hypergraphs.

Abstract

We consider the following game, played on a $k$-uniform hypergraph $H$. There are $q$ colors available and two players take it in turns to color vertices. A partial coloring is proper if no edge is mono-chromatic. One player, A, wishes to color all the vertices and the other player, B, wishes to prevent this. The {\em game chromatic number} $\chi_g(H)$ is the minimum number of colors for which A has a winning strategy. We consider this in the context of a random $k$-uniform hypergraph and prove upper and lower bounds that hold w.h.p.

Equations

$$\chi (H_{n,p;k})\approx {\left(\frac{d}{k(k-2)!\log d}\right)}^{1/(k-1)}$$

$${\chi }_g(H_{n,p;k})\ge (1+\epsilon ){\left(\frac{d}{k(k-2)!\log d}\right)}^{1/(k-1)},\text{ if d is sufficiently large}.$$

$${\chi }_g(H_{n,p;3})\le d^{2/3+\delta },\text{ if d is sufficiently large}.$$

$$\mathrm{Pr}(X\le (1-\epsilon )np)$$

$$\mathrm{Pr}(X\ge (1+\epsilon )np)$$

$$\mathrm{Pr}(X\ge \mu np)$$

$$D={\left(\frac{d}{k!\log d}\right)}^{1/(k-1)}$$

$$2(2\beta +\gamma {)}^k-(2\beta {)}^k>\frac{2(\beta +\gamma )}{k-1}$$

$$\beta =\frac{1-2\epsilon }{2(k-1{)}^{1/(k-1)}},\gamma =\frac{\epsilon }{(k-1{)}^{1/(k-1)}}$$

$$\frac{\ell (2\beta +\gamma )}{D}+\frac{2(q-\ell )\beta }{D}\ge 2\theta \text{ or }\frac{\ell \gamma +2q\beta }{D}\ge 2\theta .$$

$$\frac{q}{D}\ge \frac{2\theta }{2\beta +\gamma }=\alpha .$$

$$\mathrm{Pr}(\exists i,C_i,T)$$

$${\le }_{\mathcal{O}}q\left(\genfrac{}{}{0ex}{}{n}{\frac{\beta n}{D}}\right)\left(\genfrac{}{}{0ex}{}{n}{\frac{\gamma n}{D}}\right)\sum _{|S|=\beta n/D}P(S_i=S)(1-p{)}^{\frac{(2\beta +\gamma {)}^k{n}^k}{k!D^k}}$$

$$\le q\left(\genfrac{}{}{0ex}{}{n}{\frac{\beta n}{D}}\right)\left(\genfrac{}{}{0ex}{}{n}{\frac{\gamma n}{D}}\right)\sum _{|S|=\beta n/D}\left(\frac{\beta n}{D}\right)!(1-p{)}^{\frac{(2\beta +\gamma {)}^k{n}^k}{k!D^k}}\prod _{j=1}^{\beta n/D}\frac{7}{(1-2\theta )n(1-p{)}^{\left(\frac{\beta n}{D}-j+1\right)\frac{2(2j-1{)}^{k-2}}{(k-2)!}}}$$

$${\le }_{\mathcal{O}}q{\left(\genfrac{}{}{0ex}{}{n}{\frac{\beta n}{D}}\right)}^2\left(\genfrac{}{}{0ex}{}{n}{\frac{\gamma n}{D}}\right)\left(\frac{\beta n}{D}\right)!\frac{7^{\frac{\beta n}{D}}(1-p{)}^{\frac{(2\beta +\gamma {)}^k{n}^k}{k!D^k}}}{((1-2\theta )n{)}^{\beta n/D}(1-p{)}^{{\sum }_{j=1}^{\beta n/D}\left(\frac{\beta n}{D}-j+1\right)\frac{2(2j-1{)}^{k-2}}{(k-2)!}}}$$

$${\le }_{\mathcal{O}}q{\left(\genfrac{}{}{0ex}{}{n}{\frac{\beta n}{D}}\right)}^2\left(\genfrac{}{}{0ex}{}{n}{\frac{\gamma n}{D}}\right)\left(\frac{\beta n}{D}\right)!\frac{7^{\frac{\beta n}{D}}(1-p{)}^{\frac{(2\beta +\gamma {)}^k{n}^k}{k!D^k}}}{((1-2\theta )n{)}^{\beta n/D}(1-p{)}^{\frac{(2\beta n{)}^k}{2k!D^k}}}$$

$${\le }_{\mathcal{O}}q{\left(\genfrac{}{}{0ex}{}{n}{\frac{\beta n}{D}}\right)}^2\left(\genfrac{}{}{0ex}{}{n}{\frac{\gamma n}{D}}\right)\left(\frac{\beta n}{D}\right)!(1-p{)}^{\left(\frac{2n^k((2\beta +\gamma {)}^k)-(2\beta {)}^k}{2k!D^k}\right)}\frac{7^{\frac{\beta n}{D}}}{((1-2\theta )n{)}^{\frac{\beta n}{D}}}$$

$${\le }_{\mathcal{O}}q{n}^{1/2}{\left(\frac{(eD{)}^{2\beta +\gamma }}{{\beta }^{2\beta }{\gamma }^{\gamma }}{\left(\frac{7\beta }{De(1-2\theta )}\right)}^{\beta }\exp \left\{-\frac{d}{D^{k-1}}\left(\frac{2(2\beta +\gamma {)}^k-(2\beta {)}^k}{2k!}\right)\right\}\right)}^{\frac{n}{D}}$$

$$\le {\left(c{D}^{\beta +\gamma }\exp \left\{-\left(\frac{2(2\beta +\gamma {)}^k-(2\beta {)}^k}{2}\log d\right)\right\}\right)}^{\frac{n}{D}},$$

$$c=c(\theta ,\beta ,\gamma )=O(1)$$

$$=o(1/n),$$

$$\sum _{j=1}^{\beta n/D}\left(\frac{\beta n}{D}-j+1\right)\frac{2(2j-1{)}^{k-2}}{(k-2)!}$$

$$\approx {\int }_{x=1}^{\beta n/D}\left(\frac{\beta n}{D}-x+1\right)\frac{2(2x-1{)}^{k-2}}{(k-2)!}dx$$

$$=\frac{1}{(k-2)!}{\int }_{x=1}^{\beta n/D}\left(\left(\frac{2\beta n}{D}+1\right)(2x-1{)}^{k-2}-(2x-1{)}^{k-1}\right)dx$$

$$=\left(\frac{2\beta n}{D}+1\right)\left(\frac{(2\beta n/D-1{)}^{k-1}-1}{2(k-1)!}\right)-\frac{(2\beta n/D-1{)}^k-1}{2k(k-2)!}$$

$$\le \frac{(2\beta n/D{)}^k}{2k!}.$$

$$\mathrm{Pr}(S_i=S)$$

$$\le \mathbf{E}\left(\frac{7}{X_1\cdots X_{t-1}(X_{t-1}+1)p_t}\right)$$

$$\le \mathbf{E}\left(\frac{7^2}{X_1\cdots X_{t-2}(X_{t-2}+1)(X_{t-2}+2)p_{t-1}^2{p}_t}\right)$$

$$⋮$$

$$\le \prod _{j=1}^t\frac{7}{(N+j-1)(1-p{)}^{(t-j+1)\frac{2(2j-1{)}^{k-2}}{(k-2)!}}}$$

$$\le \prod _{j=1}^t\frac{7}{N(1-p{)}^{(t-j+1)\frac{2(2j-1{)}^{k-2}}{(k-2)!}}}.$$

$$e_2(S)=\left\{\left\{x,y\right\}\subseteq S:\exists y\notin S\text{ such that }\left\{x,y,z\right\}\in E\right\}.$$

$${\left(\frac{\sigma ed}{2\theta }\right)}^{\theta }\le \frac{\sigma }{2e}$$

$$\mathrm{Pr}(\exists S,|S|\le \sigma n,e_2(S)\ge \theta |S|)$$

$$\le \sum _{s=2\theta }^{\sigma n}{\left(\frac{ne}{s}\right)}^s{\left(\frac{es}{2\theta }\right)}^{\theta s}{\left(1-\left(1-\frac{d}{n}\right)\right)}^{\theta s}$$

$$\le \sum _{s=2\theta }^{\sigma n}{\left(\left(\frac{ne}{s}\right){\left(\frac{es}{2\theta }\frac{d}{n}\right)}^{\theta }\right)}^s$$

$$=\sum _{s=2\theta }^{\sigma n}{\left(e{\left(\frac{s}{n}\right)}^{\theta -1}{\left(\frac{ed}{2\theta }\right)}^{\theta }\right)}^s$$

$$=O\left(\frac{d^{\theta }}{n^{\theta -1}}\right)=o(1)$$

$${\left(\frac{{\sigma }^{2}ed}{6\theta }\right)}^{\theta }\le \frac{\sigma }{2e}$$

$$\mathrm{Pr}(\exists S,|S|\le \sigma n,e_3(S)\ge \theta |S|)$$

$$\le \sum _{s=\sqrt{\theta }}^{\sigma n}{\left(\frac{ne}{s}\cdot {\left(\frac{e{s}^{2}d}{6\theta n^2}\right)}^{\theta }\right)}^s$$

$$=\sum _{s=\sqrt{\theta }}^{\sigma n}{\left(e{\left(\frac{s}{n}\right)}^{2\theta -1}\cdot {\left(\frac{ed}{6\theta }\right)}^{\theta }\right)}^s$$

Full text

1 Introduction

Let $G=(V,E)$ be a graph and let $q$ be a positive integer. Consider the following game in which two players A(lice) and B(ob) take turns in coloring the vertices of $G$ with $q$ colors. Each move consists of choosing an uncolored vertex of the graph and assigning to it a color from $\{1,\ldots,k\}$ so that the resulting coloring is proper, i.e., adjacent vertices get different colors. A wins if all the vertices of $G$ are eventually colored. B wins if at some point in the game the current partial coloring cannot be extended to a complete coloring of $G$, i.e., there is an uncolored vertex such that each of the $q$ colors appears at least once in its neighborhood. We assume that A goes first (our results will not be sensitive to this choice). The game chromatic number $\chi_{g}(G)$ is the least integer $q$ for which A has a winning strategy.

This parameter is well defined, since it is easy to see that A always wins if the number of colors is larger than the maximum degree of $G$. Clearly, $\chi_{g}(G)$ is at least as large as the ordinary chromatic number $\chi(G)$, but it can be considerably more. The game was first considered by Brams about 25 years ago in the context of coloring planar graphs and was described in Martin Gardner’s column [13] in Scientific American in 1981. The game remained unnoticed by the graph-theoretic community until Bodlaender [5] re-invented it. For a survey see Bartnicki, Grytczuk, Kierstead and Zhu [4].

The papers by Bohman, Frieze and Sudakov [6], Frieze, Haber and Lavrov [11] and by Keusch and Steger [16] discuss the game chromatic number of random graphs. In this paper we discuss the game chromatic number of random hypergraphs. Given a hypergraph $H=(V,E)$ we can consider basically the same game. Here A, B color the vertices of $H$ consecutively and a coloring is proper if there is no $e\in E$ such that all vertices in $e$ have the same color. This problem has hardly been studied, even in a deterministic setting, but we feel it is of interest to extend the results of [6], [11] and [16] to this setting.

In this paper we will restrict our attention to the random $k$-uniform hypergraph $H_{n,p;k}$ where each of the $\binom{n}{k}$ edges appear independently with probability $p=\frac{d}{n^{k-1}}$ where $d$ is a large constant. Now Krivelevich and Sudakov [18] have shown that

[TABLE]

Here we use the notation $A_{n}\approx B_{n}$ for sequences $A_{n},B_{n},n\geq 1$ to mean that $A_{n}=(1+o(1))B_{n}$ as $n\to\infty$.

Our first theorem shows that w.h.p. the game chromatic number $\chi_{g}$ is significantly larger than the chromatic number.

Theorem 1.1**.**

There exists a constant $\varepsilon>0$ such that w.h.p.,

[TABLE]

We also prove an upper bound in the case $k=3$ that is somewhat far from that implied by (1).

Theorem 1.2**.**

Let $\delta>0$ be arbitrary. Then w.h.p.,

[TABLE]

It is natural to state the following:

Conjecture: W.h.p. $\chi_{g}(H_{n,p;k})=O\left(\frac{d}{k!\log d}\right)^{1/(k-1)}$.

We often refer to the following Chernoff-type bounds for the tails of binomial distributions (see, e.g., [3] or [14]). Let $X=\sum_{i=1}^{n}X_{i}$ be a sum of independent indicator random variables such that $\operatorname{\bf Pr}(X_{i}=1)=p_{i}$ and let $p=(p_{1}+\cdots+p_{n})/n$. Then

[TABLE]

2 Lower Bound

Let

[TABLE]

and suppose that there are $q=\alpha D$ colors available.

Bob’s strategy is to choose the same color as Alice, $i$ say, but assign it randomly to one of the set of available vertices for color $i$.

Notation: Let $C_{i}=C_{i}(t)$ be the set of vertices that have been colored $i$ after $t$ rounds. Let $S_{i}=S_{i}(t)$ be the set of vertices that were colored by B. Let $C=C(t)=\cup_{i=0}^{q}C_{i}$ denote the partial coloring of the vertex set.

Lemma 2.1**.**

Suppose we run this process for $t=\theta n,\theta<1/2$ many rounds and that $|C_{i}(t)|=2\beta n/D$. We show that if $d$ is sufficiently large and

[TABLE]

then with probability $1-o(1/n)$ there exists no set $T$ such that (i) $C_{i}\cap T=\emptyset$, (ii) $C_{i}\cup T$ is independent and $|T|=\gamma n/D$.

The reader can easily check that (5) is satisfied for $k\geq 3$ and

[TABLE]

when $\varepsilon>0$ is sufficnetly small. The proof of the lemma is deferred to Section 2.1.

If the event $\left\{\exists i:C_{i},T\right\}$ does not occur then because no color class has size greater than $(2\beta+\gamma)n/D$ the number $\ell$ of colors $i$ for which $|S_{i}|\geq\beta n/D$ by this time satisfies

[TABLE]

We choose $\alpha=(1+\varepsilon)(k-1)^{1/(k-1)}$ and $\theta=\alpha(2\beta+\gamma)/2<1/2$. Since $q\geq\ell$, this implies that

[TABLE]

This completes the proof of Theorem 1.1, after replacing $(1+\varepsilon)$ by $(1+\varepsilon)^{1/(k-1)}$ for aesthetic purposes.

2.1 Proof of Lemma 2.1

For expressions $X,Y$ we sometimes use the notation $X\leq_{\mathcal{O}}Y$ in place of $X=O(Y)$ when the bracketing is “ugly”.

Now, (explanations for (6), (7), (8) and (9) below), if $d$ is sufficiently large then

[TABLE]

assuming (5).

Justifying (8):

[TABLE]

Justifying (9): We used the asymptotic formula for summation of $k$-th power of first $n$ natural numbers, i.e. $\sum_{i=1}^{n}i^{k}\approx\frac{n^{k+1}}{k+1}$

Justifying (6): There are $q$ choices for color $i$. Then we take the union bound over all $\binom{n}{\beta n/D}\binom{n}{\gamma n/D}$ possible choices of $C_{i}\setminus S_{i}$ and $T$. In some sense we are allowing Alice to simultaneously choose all possible sets of size $\beta n/D$ for $C_{i}\setminus S_{i}$. The union bound shows that w.h.p. all choices fail. We do not sum over orderings of $C_{i}\setminus S_{i}$. We instead compute an upper bound on $\operatorname{\bf Pr}(S_{i}=S)$ that holds regardless of the order in which Alice plays. We consider the situation after $\theta n$ rounds. That is, we think of the following random process: pick a $k$-uniform hypergraph $H\sim H(n,p;k)$, let Alice play the coloring game on $H$ with $q$ colors against a player who randomly chooses an available vertex to be colored by the same color as Alice. Stop after $\theta n$ moves. At this point Alice played with color $i$ and there are $\beta n/D$ vertices that were colored $i$ by Alice and the same number that were colored $i$ by Bob. We bound the probability that at this point there are $\gamma n/D$ vertices that form an independent set with the color class $C_{i}$. We take a union bound over all the possible sets for Alice’s vertices and for the vertices in $T$. The probability of Bob choosing a certain set is computed next.

Justifying (7): Consider a sequence of random variables $X_{1}=N=(1-2\theta)n,X_{j}=Bin(X_{j-1},p_{j})$ where $p_{j}=(1-p)^{\frac{2(2j-1)^{k-2}}{(k-2)!}},2\leq j\leq t$. $X_{j}$ is a lower bound for the number of vertices that Bob can color $i$ and $p_{j}$ is a lower bound on the probability that a vertex $v$ that was $i$-available at step $j-1$ is also $i$-available after step $j$. The probability that a vertex $v$ was $i$-available at time $j-1$ and is still $i$-available now is at least $(1-p)^{\binom{2j-2}{k-3}}(1-p)^{2\binom{2j-2}{k-2}}\geq p_{j}$. Our estimate for $p_{j}$ arises as follows: there are at most $2(j-1)$ vertices $x_{1},\dots,x_{2(j-1)}$ of color $i$ and each of the vertices $z=y,y^{\prime}$ colored $i$ in round $j$ yield possible edges $\left\{v,z,x_{i},\ldots\right\}$ that could remove $v$ from $C_{i}$. There are also the edges $\left\{v,y,y^{\prime},x_{i},\ldots\right\}$ to account for.

We need to estimate ${\bf E}(Y_{t})$ where $Y_{t}=1/(X_{1}X_{2}...X_{t})$. $1/X_{j}$ is an upper bound for the probability that Bob chooses a particular vertex at step $j$ and then $Y_{\beta n/D}$ is an upper bound on the probability that Bob’s sequence of choices is $x_{1},x_{2},....x_{\beta n/D}$ where $S=\{x_{1},x_{2},...x_{\beta n/D}\}$. The following lemma is proven in [11]:

Lemma 2.2**.**

If $B=Bin(v,\rho)$, then ${\bf E}(\prod_{i=1}^{q}\frac{1}{B+i-1})\leq\frac{7}{\rho^{q}}\prod_{i=1}^{q}\frac{1}{v+i}$.

Using Lemma 2.2 we see that

[TABLE]

This completes the justification of (7).

3 Upper Bound

3.1 Simple density properties

For $S\subseteq[n]$ and $k=2,3$ we let $e_{3}(S)=|\left\{f\in E:\;f\subseteq S\right\}$ and

[TABLE]

Lemma 3.1**.**

If $\theta>1$ and

[TABLE]

then w.h.p there does not exist $S\subset$ $[n],|S|\leq\sigma n$ such that $e_{2}(S)\geq\theta|S|$.

Proof.

[TABLE]

∎

Lemma 3.2**.**

If $\theta>1/2$ and

[TABLE]

then w.h.p. there does not exists $S\subseteq[n],|S|\leq\sigma n$ such that $e_{3}(S)\geq\theta|S|$.

Proof.

[TABLE]

∎

For $S\subseteq[n]$ and $k=1,2$ and vertex $v$, we let $d_{S,k}(v)$ denote the number of edges $\left\{v,x,y\right\}$ such that $|\left\{x,y\right\}\cap S|=k$.

Lemma 3.3**.**

Let $\sigma$ and $\theta$ be as in Lemma 3.1. If $(\Delta-2\theta)\tau>1$ and

[TABLE]

then w.h.p there do not exist $S\supseteq T$ such that $|S|=s\leq\sigma n,|T|\geq\tau s$ and $d_{S,1}(v)\geq\Delta,\forall v\in T$.

Proof.

In the light of Lemma 3.1, the assumptions imply that w.h.p. $|e_{1}(T:S\setminus T)|\geq(\Delta-2\theta)\tau s$. In which case,

[TABLE]

∎

Lemma 3.4**.**

Let $\sigma$ and $\theta$ be as in Lemma 3.2 and $(\Delta-3\theta)\tau>1$ and

[TABLE]

then w.h.p there do not exist $S\supseteq T$ such that $|S|=s\leq\sigma n,|T|\geq\tau s$ and $d_{S,2}(v)\geq\Delta,\forall v\in T$

Proof.

In the light of Lemma 3.2, the assumptions imply that w.h.p. $|e_{2}(T:S\setminus T)|\geq(\Delta-3\theta)\tau s$. In which case,

[TABLE]

∎

Now let

[TABLE]

for some small absolute constant $\delta>0$.

We will now argue that w.h.p. A can win the game if $q$ colors are available.

A’s initial strategy will be the same as that described in [6]. Let ${\mathcal{C}}=(C_{1},C_{2},\ldots,C_{q})$ be a collection of pairwise disjoint subsets of $[n]$, i.e. a (partial) coloring. Let $\bigcup{\mathcal{C}}$ denote $\bigcup_{i=1}^{q}C_{i}$. For a vertex $v$ let

[TABLE]

and set

[TABLE]

Note that $A(v,{\mathcal{C}})$ is the set of colors that are available at vertex $v$ when the partial coloring is given by the sets in ${\mathcal{C}}$ and $v\not\in\bigcup{\mathcal{C}}$. A’s initial strategy can now be easily defined. Given the current color classes ${\mathcal{C}}$, A chooses an uncolored vertex $v$ with the smallest value of $a(v,{\mathcal{C}})$ and colors it by any available color.

As the game evolves, we let $u$ denote the number of uncolored vertices in the graph. So, we think of $u$ as running “backward” from $n$ to [math].

We show next that w.h.p. every $q$-coloring (proper or improper) of the full vertex set has the property that there are at most $\gamma n$ vertices with less than $\beta/2$ available colors. For this we need the following lemma.

Lemma 3.5**.**

$p=d/n^{2}$* and let $x_{0}=(2/(-\log(1-p)))^{1/2}\approx(2/d)^{1/2}n$ and*

[TABLE]

Then we have that for $n$ sufficiently large,

[TABLE]

Proof.

We have that the function $\phi(x)=(1-p)^{x^{2}/2}$ is convex in the interval $[x_{0},\infty]$. It follows from convexity that if $I=\left\{i:c_{i}\leq x_{0}\right\}$ then

[TABLE]

Suppose now that $x_{0}\leq n/q$. Then $\frac{n-|I|x_{0}}{q-|I|}\geq\frac{n}{q}$. And then convexity and (12) implies

[TABLE]

If $x_{0}>n/q$ then $\frac{n-|I|x_{0}}{q-|I|}<\frac{n}{q}$ and then (12) implies that

[TABLE]

∎

Let

[TABLE]

Lemma 3.6**.**

W.h.p., for all collections ${\mathcal{C}}$,

[TABLE]

Proof.

We first note that if $|S|=\gamma n$ then w.h.p. $e_{2}(S)\leq 4\gamma^{2}dn$. This follows from Lemma 3.1 with $\sigma=\gamma$ and $\theta=4\gamma d$. It follows that for any $\varepsilon>0$ that there is a set $S_{1}^{\prime}\subseteq S$ of size at least $(1-\varepsilon)\gamma n$ such that if $v\in S_{1}^{\prime}$ then $d_{S,1}(v)\leq 8\varepsilon^{-1}\gamma d$. Furthermore, Lemma 3.2 with $\sigma=\gamma$ and $\theta=3$ implies that w.h.p. $e_{3}(S)\leq 3\gamma n$. Therefore there is a set $S_{1}^{\prime\prime}\subseteq S$ of size at least $(1-\varepsilon)\gamma n$ such that if $v\in S_{1}^{\prime\prime}$ then $d_{S,2}(v)\leq\varepsilon^{-1}$. Let $S_{1}=S_{1}^{\prime}\cup S_{1}^{\prime\prime}$.

Fix ${\mathcal{C}}$ and suppose that $|B({\mathcal{C}})|\geq\gamma n$. Choose $S\subseteq B({\mathcal{C}})$ and let $S_{1}$ be as defined above. For $v\in S_{1}$ let

[TABLE]

Thus $a(v,{\mathcal{C}})\geq b(v,{\mathcal{C}})-8\varepsilon^{-1}\gamma d-\varepsilon^{-1}$. $b(v,{\mathcal{C}})$ is the sum of independent indicator variables $X_{i}$, where $X_{i}=1$ if $v$ is not in a hyperedge $(v,x,y)$ such that $x,y\in C_{i}\setminus S$ in $G_{n,p}$. Then $\operatorname{\bf Pr}(X_{i}=1)\geq(1-p)^{\binom{|C_{i}|}{2}}$ and since $(1-p)^{t}$ is a convex function of $t$ and using the Lemma 3.5 we get

[TABLE]

It follows from the Chernoff bound (2) that

[TABLE]

Now, when ${\mathcal{C}}$ is fixed, the events $\left\{b(v,{\mathcal{C}})\leq 0.51\beta\right\},v\in S_{1}$ are independent. Thus, because $a(v,{\mathcal{C}})\leq\beta/2$ implies that $b(v,{\mathcal{C}})\leq 0.51\beta$ we have

[TABLE]

for large $d$ and small enough $\varepsilon$. ∎

Let $u_{0}$ to be the last time for which A colors a vertex with at least $\beta/2$ available colors, i.e.,

[TABLE]

where ${\mathcal{C}}_{u}$ denotes the collection of color classes when $u$ vertices remain uncolored.

If $u_{0}$ does not exist then A will win. It follows from Lemma 3.6 that w.h.p. $u_{0}\leq 2\gamma n$ and that at time $u_{0}$, every vertex still has at least $\beta/2$ available colors. Indeed, consider the final coloring ${\mathcal{C}}^{*}$ in the game that would be achieved if A follows her current strategy, even if she has to improperly color an edge. Let $U=\left\{v\notin{\mathcal{C}}_{u_{0}}:a(v,{\mathcal{C}}^{*})<\beta/2\right\}$. Now we can assume that $|U|\leq\gamma n$. Because the number of colors available to a vertex decreases as vertices get colored, from $u_{0}$ onward, every vertex colored by $A$ is in $U$. Therefore $u_{0}\leq 2\gamma n$. Next let $G_{U}=(U,F)$ be the graph with vertex set $U$ and edges $F$ where $\left\{x,y\right\}\in F$ if there exists $z$ such that $\left\{x,y,z\right\}\in E$.

Now let $u_{1}$ be the first time that there are at most $2\gamma n$ uncolored vertices and $a(v,{\mathcal{C}}_{u})\geq\beta/2,\mbox{ for all }v\not\in\bigcup{\mathcal{C}}_{u}$. By the above, w.h.p. $u_{1}\leq u_{0}$, so in particular w.h.p. $u_{1}$ exists. A can determine $u_{1}$ but not $u_{0}$, as $u_{0}$ depends on the future.

A will follow a more sophisticated strategy from $u_{1}$ onward. A will however play the remainder of the game on the graph $G_{U}$. By this we mean that she will ensure that if $\left\{x,y\right\}$ is a $G_{U}$-edge and $x$ has color $i$ at some stage, then she will not color $y$ with color $i$ even though this is strictly admissible.

This weakens A and explains why our upper bound does not match our lower bound. On the other hand, if she can properly color $G_{U}$, then she will have succeeded in properly coloring $H=H_{n,p;3}$. B of course, does not play by these rules. We will show next that we can find a sequence $U=U_{0}\supseteq U_{1}\supseteq\cdots\supseteq U_{\ell}$ with the following properties: The $G_{U}$-edges of $U_{i}:(U_{i-1}\setminus U_{i})$ between $U_{i}$ and $U_{i-1}\setminus U_{i}$ will be divided into two classes, heavy and light. Vertex $w$ is a heavy (resp. light) $G_{U}$-neighbor of vertex $v$ if the edge $(v,w)$ is $G_{U}$-heavy (resp. $G_{U}$-light).

1. (P1)

Each vertex of $U_{i}\setminus U_{i+1}$ has at most one light $G_{U}$-neighbor in $U_{i+1}$, for $0\leq i<\ell$. 2. (P2)

All $U_{i}:(U_{i-1}\setminus U_{i})$ $G_{U}$-edges are light for $i\geq 3$. 3. (P3)

Each vertex of $U_{i}$ has at most $3\beta/50$ $G_{U}$-heavy neighbors in $U_{i-1}\setminus U_{i}$ for $i=1,2$. 4. (P4)

Each vertex of $U_{i}\setminus U_{i+1}$ has at most $\beta/3$ $G_{U}$-neighbors in $U_{i}$, for $0\leq i<\ell$. 5. (P5)

$U_{\ell}$ contains at most one $G_{U}$-cycle.

From this, we can deduce that the $G_{U}$-edges of $U_{0}$ can be divided up into the $G_{U}$-heavy edges $E_{H}$, $G_{U}$-light edges $F_{L}$, the $G_{U}$-edges inside $U_{\ell}$ and the rest of the $G_{U}$-edges. Assume first that $U_{\ell}$ does not contain a $G_{U}$-cycle. $\Phi=(U,F_{L})$ is a forest and the strategy in [10] can be applied. When attempting to color a vertex $v$ of $\Phi$, there are never more than three $\Phi$-neighbors of $v$ that have been colored. Since there are at most $\beta/3+2\cdot 3\beta/50$ non-$\Phi$ neighbors, A will succeed since she has an initial list of size $\beta/2$.

If $U_{\ell}$ contains a $G_{U}$-cycle $C$ then A can begin by coloring a vertex of $C$. This puts A one move behind in the tree coloring strategy, in which case we can bound the number of $\Phi$-neighbors by four.

It only remains to prove that the construction P1–P5 exists w.h.p. Remember that $d$ is sufficiently large here.

We can assume without loss of generality that $|U_{0}|=2\gamma n$. This will not decrease the sizes of the sets $a(v,U_{0})$.

3.2 The verification of P1–P4: Constructing $U_{1}$

The general strategy will be as follows : We will consider two separate types of edge listed below. To tackle each type, we will formulate corresponding lemmas that will be presented subsequently.

1. Type 1:

The edges $\{x,y\}$ in $G_{U}$ such that $\{x,y,z\}\not\in E$ for all $z\in U$. 2. Type 2:

The remaining edges where for $\{x,y\}$ in $G_{U}$, there is $z\in U$ such that $\{x,y,z\}\in E$.

Note that $d_{U}(v)\leq d_{U,1}(v)+2d_{U,2}(v)$. Recall their definition just before Lemma 3.3.

Let $L=100$. Applying Lemmas 3.3 and 3.4 separately with

[TABLE]

we see that w.h.p.

[TABLE]

[TABLE]

[TABLE]

We then let $U_{1,a}\supseteq U_{1,a}^{\prime}$ be the subset of $U_{0}$ consisting of the vertices with the $4\tau\gamma n$ largest values of $d_{U_{0}}$.

Let $A_{0}=U_{0}\setminus U_{1,a}$ and $B_{0}=\left\{v\in U_{1,a}:|d_{A_{0}}(v)|\geq\frac{3\beta}{L}\right\}$. Iteratively we define

[TABLE]

[TABLE]

Lemma 3.7**.**

W.h.p., $\not\exists$ disjoint sets $S,T\subseteq V(G),G=G_{U}$ such that

[TABLE]

Proof.

We observe that if $S,T$ exist then one of the following two cases must occur:

1. C1:

$f(v)=|\left\{u\in T:\exists w\in V(G)\setminus(S\cup T),\{u,w,v\}\in E(G)\right\}|\geq\frac{\beta}{2L}$ for at least $\frac{s_{0}}{2}$ vertices $v\in S$. 2. C2:

There are at least $t=\frac{\beta s_{0}}{8L}$ hyperedges $\{u,v,w\}$ such that $u,v\in S$ and $w\in T$.

[TABLE]

[TABLE]

∎

Thus if $B^{\prime}=\left\{v\in U_{1,a}:d_{A_{0},1}(v)\geq\frac{\beta}{L}\right\}$ then w.h.p.

[TABLE]

Lemma 3.8**.**

W.h.p $\not\exists$ disjoint $S,T$ s.t

[TABLE]

Proof.

[TABLE]

∎

Thus if $B^{\prime\prime}=\left\{v\in U_{1,a}:d_{A_{0},2}(v)\geq\frac{\beta}{L}\right\}$ then w.h.p.

[TABLE]

Clearly, $B_{0}\subseteq B^{\prime}\cup B^{\prime\prime}$. Hence,

[TABLE]

Let $D_{2}(S)=\left\{v:d_{S,1}(v)\geq 2\right\}$ for $S\subseteq V(G)$.

Lemma 3.9**.**

W.h.p., $|D_{2}(S)|<3K|S|,\;\;\forall\;|S|\leq s_{0}=\frac{3000L\log d}{d^{\frac{4}{3}+2\delta}}n$ where $K=d^{\frac{2}{3}-2\delta}\log^{2}d$.

Proof.

Suppose that there exist $S$ and $T$ with $|S|\leq s_{0}$ and $|T|=3\cdot K|S|$ such that $d_{s,1}(v)\geq 2$ for all $v\in T$. Then for $v\in T$, one of the following can occur.

1. D1:

There are $x,y\in S$ and $a,b\in T$ such that $\{v,x,a\},\{v,y,b\}\in E(G)$. 2. D2:

There are $x,y\in S$ and $a\in T$, $b\in V(G)\setminus(S\cup T)$ such that $\{v,x,a\},\{v,y,b\}\in E(G)$. 3. D3:

There are $x,y\in S$ and $a,b\in V(G)\setminus(S\cup T)$ such that $\{v,x,a\},\{v,y,b\}\in E(G)$.

Now we construct $T^{\prime}\subseteq T$ with $|T^{\prime}|\geq K|S|$ such that if $v\in T$ then there exist $x,y\in S$ and $a,b\in V(G)\setminus(S\cup T^{\prime})$ such that D1 holds. First, for every vertex of type D1, put $v$ in $T^{\prime}$ and remove $a,b$ from further consideration. Second, for every vertex of type D2, put $v$ in $T^{\prime}$ and remove $a$ from further consideration. Finally, for every vertex of type D3, put $v$ in $T^{\prime}$. We observe that for every $v\in T^{\prime}$ we have thrown away at most 2 vertices of $T$ and hence $|T^{\prime}|\geq K|S|$. We will now estimate the probability of the existence of $S,T^{\prime}$.

[TABLE]

∎

Thus if $A^{\prime}=\left\{v\in A_{0}:v\in D_{2}(B_{0})\right\}$ then w.h.p.

[TABLE]

Let $D^{\prime}_{2}(S)=\left\{v:d_{S,2}(v)\geq 1\right\}$ for $S\subseteq V(G)$.

Lemma 3.10**.**

W.h.p. $|D^{\prime}_{2}(S)|\leq K|S|,\;\forall\;|S|\leq\frac{3000L\log d}{d^{\frac{4}{3}+2\delta}}n$ where $K=d^{\frac{2}{3}-2\delta}\log^{2}d$.

Proof.

[TABLE]

∎

So w.h.p. $|A_{1}|=|A^{\prime}\cup A^{\prime\prime}|\leq\frac{12000\log^{3}d}{d^{\frac{2}{3}+4\delta}}n$. From (15), (16), Lemma 3.7 and Lemma 3.8 we see that

[TABLE]

Using Lemmas 3.9 and Lemma 3.10,

[TABLE]

where $\frac{KL}{\beta}=150d^{-3\delta}\log^{2}d$.

Let $\zeta=\left\lceil\frac{2}{\delta}\right\rceil$ and let $Y=N(B_{\zeta})\cap A_{\zeta}$. Then,

[TABLE]

Let $U_{1}=U_{1,a}\cup Y$. Then

[TABLE]

We now define the light and heavy edges in the following fashion,

1. Q1:

The edges between $B_{i}$ and $A_{i}\setminus A_{i+1}$ are light 2. Q2:

The edges between $B_{i}\setminus B_{i+1}$ and $A_{i+1}$ are heavy 3. Q3:

The edges between $U_{1}\setminus U_{1,a}$ and $U_{0}\setminus U_{1}$ are heavy

We now check that P1-P4 hold. First consider the light edges. For every vertex $v\in U_{0}\setminus U_{1}$ there is at most one light neighbour in $U_{1}$. Because if $v\in A_{i}$ and $v\not\in A_{i+1}$ and there are 2 light neighbors $x,w$ of $v$ in $U_{1}$, by Q1, $x,w\in B_{i}$ and that would contradict the fact that $v\not\in A_{i+1}$. This implies that P1 holds.

We will argue next that for all $v\in U_{1}$, $d_{U_{0}\setminus U_{1}}(v)\leq 3\Delta\leq\frac{3\beta}{50}$. For $v\not\in U_{1,a}$ this is true from the definition of $U_{1,a}$. Similarly, for $v\notin B_{0}$. Now consider $v\in B_{i}\setminus B_{i+1},i\geq 0$. It only has light neighbors in $A_{i}\setminus A_{i+1}$ and if $v$ has more than $\frac{3\beta}{L}$ heavy neighbors in $A_{i+1}$ then $v$ should be in $B_{i+1}$, which is a contradiction. Because it is also in $B_{j},j\leq i-1$ it only has light neighbors in $(A_{i-1}\setminus A_{i})\cup(A_{i-2}\setminus A_{i-3})\cup\cdots=A_{0}\setminus A_{i}$. Clearly P3, P4 hold.

3.3 The verification of P1–P4: Constructing $U_{2}$

Applying Lemma 3.3 and 3.4 separately with

[TABLE]

we see that w.h.p.

[TABLE]

[TABLE]

[TABLE]

We then let $U_{2,a}\supseteq U_{2,a}^{\prime}$ be the subset of $U_{1}$ consisting of the vertices with the $2\tau\sigma n$ largest values of $d_{U_{1}}$. As in Section 3.2, define $A_{0}=U_{1}\setminus U_{2,a}$ and let $B_{0}=\left\{v\in U_{2,a}:|d_{A_{0}}(v)|\geq\frac{3\beta}{L}\right\}$. Iteratively we define

[TABLE]

[TABLE]

Let $B^{\prime}=\left\{v\in U_{2,a}:d_{A_{0},1}(v)\geq\frac{\beta}{L}\right\}$. Using Lemma 3.7, we see that w.h.p., $|B^{\prime}|\leq\frac{8L|U_{1}|}{\beta}\leq\frac{360eL\;\log^{3}d}{d^{\frac{5}{3}+4\delta}}n$.

Let $B^{\prime\prime}=\left\{v\in U_{2,a}:d_{A_{1},2}(v)\geq\frac{\beta}{L}\right\}$. Using Lemma 3.8, we see that w.h.p. $|B^{\prime\prime}|\leq\frac{90eL\;\log^{3}d}{d^{\frac{5}{3}+4\delta}}n$.

Clearly, $B_{0}\subseteq B^{\prime}\cup B^{\prime\prime}$. Therefore, w.h.p.,

[TABLE]

Arguing as in Section 3.2 we see that w.h.p.

[TABLE]

Remember that $\zeta=\left\lceil\frac{2}{\delta}\right\rceil$ and let $Y=N(B_{\zeta})\cap A_{\zeta}$. Then,

[TABLE]

Let $U_{1}=U_{1,a}\cup Y$. Then w.h.p.

[TABLE]

We now define the light and heavy edges in the following fashion,

1. Q1:

The edges between $B_{i}$ and $A_{i}\setminus A_{i+1}$ are light 2. Q2:

The edges between $B_{i}\setminus B_{i+1}$ and $A_{i+1}$ are heavy 3. Q3:

The edges between $U_{1}\setminus U_{1,a}$ and $U_{0}\setminus U_{1}$ are heavy

We now check that P1-P4 hold. First consider the light edges. For every vertex $v\in U_{0}\setminus U_{1}$ there is at most one light neighbour in $U_{1}$. Because if $v\in A_{i}$ and $v\not\in A_{i+1}$ and there are 2 light neighbors $x,w$ of $v$ in $U_{1}$, by Q1, $x,w\in B_{i}$ and that would contradict the fact that $v\not\in A_{i+1}$. This implies that P1 holds.

We will argue next that for all $v\in U_{1}$, $d_{U_{0}\setminus U_{1}}(v)\leq 3\Delta\leq\frac{3\beta}{50}$. For $v\not\in U_{1,a}$ this is true from the definition of $U_{1,a}$. Similarly, for $v\notin B_{0}$. Now consider $v\in B_{i}\setminus B_{i+1},i\geq 0$. It only has light neighbors in $A_{i}\setminus A_{i+1}$ and if $v$ has more than $\frac{3\beta}{L}$ heavy neighbors in $A_{i+1}$ then $v$ should be in $B_{i+1}$, which is a contradiction. Because it is also in $B_{j},j\leq i-1$ it only has light neighbors in $(A_{i-1}\setminus A_{i})\cup(A_{i-2}\setminus A_{i-3})\cup\cdots=A_{0}\setminus A_{i}$. Clearly P3, P4 hold.

3.4 The verification of P1–P4: Constructing $U_{2}$

Applying Lemma 3.3 and 3.4 separately with

[TABLE]

we see that w.h.p.

[TABLE]

[TABLE]

[TABLE]

We then let $U_{2,a}\supseteq U_{2,a}^{\prime}$ be the subset of $U_{1}$ consisting of the vertices with the $2\tau\sigma n$ largest values of $d_{U_{1}}$. As in Section 3.2, define $A_{0}=U_{1}\setminus U_{2,a}$ and let $B_{0}=\left\{v\in U_{2,a}:|d_{A_{0}}(v)|\geq\frac{3\beta}{L}\right\}$. Iteratively we define

[TABLE]

[TABLE]

Let $B^{\prime}=\left\{v\in U_{2,a}:d_{A_{0},1}(v)\geq\frac{\beta}{L}\right\}$. Using Lemma 3.7, we see that w.h.p., $|B^{\prime}|\leq\frac{8L|U_{1}|}{\beta}\leq\frac{360eL\;\log^{3}d}{d^{\frac{5}{3}+4\delta}}n$.

Let $B^{\prime\prime}=\left\{v\in U_{2,a}:d_{A_{1},2}(v)\geq\frac{\beta}{L}\right\}$. Using Lemma 3.8, we see that w.h.p. $|B^{\prime\prime}|\leq\frac{90eL\;\log^{3}d}{d^{\frac{5}{3}+4\delta}}n$.

Clearly, $B_{0}\subseteq B^{\prime}\cup B^{\prime\prime}$. Therefore, w.h.p.,

[TABLE]

Arguing as in Section 3.2 we see that w.h.p.

[TABLE]

With $\zeta=\left\lceil\frac{2}{\delta}\right\rceil$ and $K=d^{\frac{2}{3}-2\delta}\log^{2}d$ as before and $Y=N(B_{\zeta})\cap A_{\zeta}$ we get

[TABLE]

Letting $\gamma_{2}=\frac{500Le\log^{3}d}{\delta d^{5/3+4\delta}}$ and $U_{2}=U_{2,a}\cup Y$ we see that w.h.p. $|U_{2}|\leq\gamma_{2}n$.

We can define heavy and light edges as in $U_{1}$ and P1-P4 follows.

3.5 The verification of P1–P4: Constructing $U_{3}$

Applying Lemma 3.3 and 3.4 separately with

[TABLE]

we see that w.h.p.

[TABLE]

[TABLE]

[TABLE]

We now construct $U_{3}\supseteq U_{3}^{\prime}$ by repeatedly adding vertices $y_{1},y_{2},..y_{s}$ of $U_{2}\setminus U_{3}^{\prime}$ such that $y_{j}$ is the lowest numbered vertex not in $Y_{j}=U_{3}^{\prime}\cup\left\{y_{1},y_{2}....y_{j-1}\right\}$ that has at least two neighbors in $Y_{j}$ in $G$. W.h.p., this process terminates with $j=t\leq 39|U_{3}^{\prime}|$. We can apply Lemma 3.2 to see that w.h.p. this does not happen. Indeed, let $S_{0}=U_{3}^{\prime}$. We add vertices to $S_{j-1}$ to create the set $S_{j}$ iteratively. In this procedure we encounter two cases.

• •

$\exists x,w\in Y_{j}$ such that $(y_{j},x,w)\in E(G)$. Then $S_{j}=S_{j-1}\cup\{y_{j}\}$.

• •

$\exists x,w\in Y_{j}\;\mbox{and}\;a,b\not\in Y_{j}$ such that $(x,a,y_{j}),(w,b,y_{j})\in E(G)$. Then $S_{j}=S_{j-1}\cup\left\{y_{j},a,b\right\}$

Note that we are adding at least 2 hyper-edges for every 3 vertices added to $S_{t}$. If $s\geq 39|U_{3}^{\prime}|$, then $e_{3}(S_{s})\geq\frac{13}{20}|S_{s}|$. Apply Lemma 3.2 with $\theta=\frac{13}{20}$ and $\sigma=120\cdot\frac{2500Le\log^{3}d}{\delta d^{7/3+5\delta}}$ to conclude that $t\leq 39|U_{3}^{\prime}|$.

Putting $U_{3}=U_{3}^{\prime}\cup S_{t}$ we see that each vertex in $U_{2}\setminus U_{3}$ has at most one $G$-neighbor in $U_{3}$. We can therefore make the $U_{3}:(U_{2}\setminus U_{3})$ edges light and satisfy P1, P2 and P4.

3.6 The verification of P1-P5 : Construction of $U_{i},i\geq 4$

We repeat the argument of Section 3.5 to construct the rest of the sequence $U_{0}\supseteq U_{1}\supseteq U_{2}\supseteq...\supseteq U_{l}$. One can check that $|U_{i}|\leq\frac{200L}{\beta}|U_{i-1}|$. We choose $l$ so that $|U_{l}|\leq\log n$. We can then easily prove that w.h.p. $S$ contains at most $|S|$ edges of $G$ whenever $|S|\leq\log n$, implying P5.

4 Final remarks

We have shown lower bounds for the game chromatic number of random $k$-uniform hypergraphs and upper bounds for random 3-uniform hypergraphs. The lower bound is satisfactory in that it is within a constant factor of the chromatic number. The upper bound is most likely not tight, but it is still non-trivial in that it is much smaller than $d$.

We conjecture that the upper bound can be reduced to within a constant factor of the lower bound. It would also be of interest to consider upper bounds for $k$-uniform hypergraphs, $k\geq 4$.