On the Matching Problem in Random Hypergraphs

Peter Frankl¹¹1Rényi Institute, Budapest, Hungary. Email: [email protected]. Jiaxi Nie²²2School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332, USA. Email: [email protected]. Jian Wang³³3Department of Mathematics, Taiyuan University of Technology, Taiyuan, 030024, China. Research supported by National Natural Science Foundation of China No. 12471316. Email: [email protected].

Abstract

We study a variant of the Erdős Matching Problem in random hypergraphs. Let $\mathcal{K}_{p}(n,k)$ denote the Erdős-Rényi random $k$ -uniform hypergraph on $n$ vertices where each possible edge is included with probability $p$ . We show that when $n\gg k^{2}s$ and $p$ is not too small, with high probability, the maximum number of edges in a sub-hypergraph of $\mathcal{K}_{p}(n,k)$ with matching number $s$ is obtained by the trivial sub-hypergraphs, i.e. the sub-hypergraph consisting of all edges containing at least one vertex in a fixed set of $s$ vertices.

1 Introduction

In this paper, we consider the so-called Turán problem concerning matchings in random hypergraphs. Let $\mathcal{K}(n,k)$ denote the complete $k$ -graph on $n$ vertices–often denoted as $\binom{[n]}{k}$ . For a fixed $p$ , $0<p<1$ let $\mathcal{K}_{p}=\mathcal{K}_{p}(n,k)$ denote the random $k$ -graph that we obtain by choosing each edge of $\binom{[n]}{k}$ independently and with probability $p$ . By the law of large numbers with (very) high probability $|\mathcal{K}_{p}|/\binom{n}{k}$ is very close to $p$ .

Similarly, for every $x\in[n]$ , the full star $\mathcal{K}_{p}[x]=\{K\in\mathcal{K}_{p}\colon x\in K\}$ has size very close to $p\binom{n-1}{k-1}$ . E.g., by Chernoff’s inequality

\displaystyle Pr\left(\left|\frac{|\mathcal{K}_{p}[x]|}{p\binom{n-1}{k-1}}-1\right|>\varepsilon\right)<2e^{-\frac{\varepsilon^{2}}{3}p\binom{n-1}{k-1}}.

(1)

The central problem that we consider is the following. Let $s$ be a fixed positive integer. Try and determine the size and structure of largest subhypergraphs of $\mathcal{K}_{p}$ which do not contain $s+1$ pairwise disjoint edges.

To put this problem into context let us first recall the corresponding problem for the complete $k$ -graph, $\binom{[n]}{k}$ , that is, the case $p=1$ .

Erdős Matching Conjecture [6]. Suppose that $n,k,s$ are positive integers, $n\geq k(s+1)$ and $\mathcal{E}\subset\binom{[n]}{k}$ contains no $s+1$ pairwise disjoint edges. Then

\displaystyle|\mathcal{E}|\leq\max\left\{\binom{k(s+1)-1}{k},\binom{n}{k}-\binom{n-s}{k}\right\}.

(2)

Note that the simple constructions giving rise to the formulae on the right hand side are

\binom{[k(s+1)-1]}{k}\mbox{ and }\mathcal{E}_{T}=\left\{E\in\binom{[n]}{k}\colon E\cap T\neq\emptyset\right\},

where $T$ is some fixed $s$ -subset of $[n]$ . There has been a lot of research going on concerning this conjecture. Let us recall some of the results.

The current best bounds establish (2) for $n>(2s+1)k-s$ [8] and for $s>s_{0}$ , $n>\frac{5}{3}sk$ [10].

The $s=1$ case has received the most attention. A family $\mathcal{F}\subset\binom{[n]}{k}$ is called intersecting if $\mathcal{F}$ contains no two pairwise disjoint members (edges). Say that $\mathcal{F}$ is a star if there is a vertex that lies in all edges.

Theorem 1.1 (Erdős-Ko-Rado Theorem, [7]).

Suppose that $\mathcal{F}\subset\binom{[n]}{k}$ is intersecting and $n\geq 2k$ . Then

|\mathcal{F}|\leq\binom{n-1}{k-1}.

Further, when $n\geq 2k+1$ , the equality holds only for a star.

Concerning the probabilistic versions of EKR, Balogh, Bohman and Mubayi[2] have proved several strong results. Here we mention some of them. Let $f(n)$ , $g(n)$ be functions of $n$ . The notation $f(n)\ll g(n)$ means $f(n)/g(n)\rightarrow 0$ as $n\rightarrow\infty$ .

Theorem 1.2 ([2]).

When $k\ll n^{1/4}$ and $p\in[0,1]$ , w.h.p. the maximum sized intersecting subhypergraph in $\mathcal{K}_{p}(n,k)$ is a star. Furthermore, the same conclusion also holds when $n^{1/4}\ll k\ll n^{1/3}$ , and $p\gg n^{-1/4}/\binom{n-1}{k-1}$ or $p\ll k^{-1}/\binom{n-1}{k-1}$ .

For more on the probabilistic EKR, see also [12, 13, 11, 3].

Let us recall the notation $\nu(\mathcal{F})$ for the matching number, that is, $\nu(\mathcal{F})$ is the maximum number of pairwise disjoint edges in $\mathcal{F}$ . In particular, $\mathcal{F}$ is intersecting if and only if $\nu(\mathcal{F})\leq 1$ .

One natural question is to investigate for which range of the values $n,k,s,p$ does

\displaystyle|\mathcal{F}|\leq\max\{|\mathcal{K}_{p}(\hat{S})|\colon|S|=s\}

(3)

hold with high probability, where $\nu(\mathcal{F})=s$ and $\mathcal{H}(\hat{S}):=\{H\in\mathcal{H}\colon H\cap S\neq\emptyset\}$ . The fact that $|\mathcal{H}(\hat{S})|$ is asymptotically $p\left(\binom{n}{k}-\binom{n-s}{k}\right)$ when $p$ is not too small follows from the law of large numbers.

Let $\tau(\mathcal{H})$ be the covering number of $\mathcal{H}$ , which is the minimum size of a set $W$ of vertices such that $E\cap W\not=\emptyset$ for each $E\in\mathcal{H}$ . Clearly, $\nu(\mathcal{H})\leq\tau(\mathcal{H})$ always. We say that $\mathcal{H}$ is trivial if $\nu(\mathcal{H})=\tau(\mathcal{H})$ . Otherwise $\mathcal{H}$ is called non-trivial. We prove the following result.

Theorem 1.3.

For any integers $s=s(n)$ , $k=k(n)$ and real number $p=p(n)$ with

(i)

$\frac{64ks\log n}{\binom{n-1}{k-1}}\leq p\leq 1$ , and
(ii)

$n\geq 200k^{3}s$ ,

with high probability (w.h.p.), every non-trivial $\mathcal{F}\subseteq\mathcal{K}_{p}(n,k)$ with $\nu(\mathcal{F})\leq s$ satisfies

|\mathcal{F}|<|\mathcal{K}_{p}(\hat{S})|,

for arbitrary $S\in\binom{[n]}{s}$ .

We also prove that the lower bound for $n$ can be improved to $O(k^{2}s)$ at the expense of a worse lower bound for $p$ .

Theorem 1.4.

Let $t$ be an integer with $2\leq t\ll k$ . For any integers $s=s(n)$ , $k=k(n)$ and real number $p=p(n)$ with

(i)

$\frac{8(t+1)k^{t+1}s^{t}\log n}{\binom{n-1}{k-1}}\leq p\leq 1$ , and
(ii)

$56k^{2+1/t}s\leq n\leq 56k^{2+1/(t-1)}s$ ,

with high probability, every non-trivial $\mathcal{F}\subseteq\mathcal{K}_{p}(n,k)$ with $\nu(\mathcal{F})\leq s$ satisfies

|\mathcal{F}|<|\mathcal{K}_{p}(\hat{S})|,

for arbitrary $S\in\binom{[n]}{s}$ .

Let $0<\epsilon<1$ and $t=k^{\epsilon}$ in Theorem 1.4. Let $f(x):=\frac{\log x}{x^{\epsilon}}$ . Then

f^{\prime}(x)=\frac{1}{x^{1+\epsilon}}-\epsilon\frac{\log x}{x^{1+\epsilon}}=\frac{1-\epsilon\log x}{x^{1+\epsilon}}.

Clearly $x^{1+\epsilon}>0$ for $x\geq 2$ and $1-\epsilon\log x\geq 0$ for $x\leq e^{1/\epsilon}$ , $1-\epsilon\log x\leq 0$ for $x\geq e^{1/\epsilon}$ . It follows that

f(x)\leq f\left(e^{1/\epsilon}\right)=\frac{1}{e\epsilon}.

Thus $k^{1/t}=k^{1/k^{\epsilon}}=e^{f(k)}<e^{{1}/{(e\epsilon)}}$ and we have the following corollary.

Corollary 1.5.

For $0<\epsilon<1$ , $n\geq 56e^{1/(e\epsilon)}k^{2}s$ and $p\geq\frac{10(ks)^{k^{\epsilon}}k^{1+\epsilon}}{\binom{n-1}{k-1}}$ , every non-trivial $\mathcal{F}\subseteq\mathcal{K}_{p}(n,k)$ with $\nu(\mathcal{F})=s$ satisfies $|\mathcal{F}|<|\mathcal{K}_{p}(\hat{S})|$ for arbitrary $S\in\binom{[n]}{s}$ .

Our next result shows that Theorem 1.4 does not hold for the range $p>\frac{8(t+1)k^{t+1}s^{t}\log n}{\binom{n-1}{k-1}}$ when $k^{2}s\geq 2(t+3)n\log n$ .

Proposition 1.6.

\frac{s\log n}{\binom{n-s}{k}}\ll p\ll\frac{e^{\frac{k^{2}s}{2n}}}{\binom{n}{k}},

then with high probability, $\mathcal{K}_{p}(n,k)$ has matching number at most $s$ and is non-trivial.

Note that for $k^{2}s\geq 2(t+3)n\log n$ and $n>ks$ we have

\frac{k}{n}e^{\frac{k^{2}s}{2n}}\geq\frac{k}{n}e^{(t+3)\log n}=kn^{t+2}>k(ks)^{t+2}=(k^{t+1}s^{t})k^{2}s^{2}\gg(k^{t+1}s^{t})8(t+1)\log n.

It follows that

\frac{e^{\frac{k^{2}s}{2n}}}{\binom{n}{k}}=\frac{e^{\frac{k^{2}s}{2n}}}{\binom{n-1}{k-1}}\frac{k}{n}\gg\frac{8(t+1)k^{t+1}s^{t}\log n}{\binom{n-1}{k-1}}.

Thus Proposition 1.6 implies that Theorem 1.4 does not hold for the range $n\log n\ll k^{2}s$ and $\frac{8(t+1)k^{t+1}s^{t}\log n}{\binom{n-1}{k-1}}\ll p\ll\frac{e^{\frac{k^{2}s}{2n}}}{\binom{n}{k}}$ with some constant integer $t<k$ .

For $k=2$ , we obtain the following result. We write $a=(1\pm\varepsilon)b$ for $(1-\varepsilon)b\leq a\leq(1+\varepsilon)b$ .

Theorem 1.7.

Let $0<\varepsilon<1$ and $n\geq 2s+2$ . Let $X$ be the maximum number of edges in a subgraph $F$ of $G(n,p)$ with $\nu(F)\leq s$ . If $p\geq\frac{250\log n}{\varepsilon^{2}n}$ , then with probability at least $1-2n^{-s}$ ,

X=(1\pm\varepsilon)p\max\left\{\binom{2s+1}{2},\binom{s}{2}+s(n-s)\right\}.

Let $\mathcal{F}\subset\binom{[n]}{k}$ . For $R,Q\subset[n]$ with $R\cap Q=\emptyset$ , define

\mathcal{F}(Q)=\{F\setminus Q\colon Q\subset F\in\mathcal{F}\},\ \mathcal{F}(\bar{R},\hat{Q})=\{F\in\mathcal{F}\colon F\cap R=\emptyset,F\cap Q\neq\emptyset\}.

For $R=\emptyset$ we write $\mathcal{F}(\bar{R},\hat{Q})$ as $\mathcal{F}(\hat{Q})$ . For $Q=\emptyset$ we write $\mathcal{F}(\bar{R},\hat{Q})$ as $\mathcal{F}(\bar{R})$ .

2 A weaker result and the outline of the main proof

Let $\binom{[n]}{\leq k}$ denote the collection of all subsets of $[n]$ of size at most $k$ . For $\mathcal{G}\subset\binom{[n]}{\leq k}$ , let

\langle\mathcal{G}\rangle=\left\{F\in\binom{[n]}{k}\colon\mbox{ there exists }G\in\mathcal{G}\mbox{ such that }G\subset F\right\}.

We need the following version of the Chernoff’s bound.

Theorem 2.1 ([14]).

Let $X\in Bi(n,p)$ , $\mu=np$ . If $0<\varepsilon\leq 3/2$ then

\displaystyle Pr(|X-\mu|\geq\varepsilon\mu)\leq 2e^{-\frac{\varepsilon^{2}}{3}\mu}.

(4)

If $x\geq 7\mu$ then

\displaystyle Pr(X\geq x)\leq e^{-x}.

(5)

We say that a $k$ -graph $\mathcal{H}$ is resilient if for any vertex $u\in V(\mathcal{H})$ , $\nu(\mathcal{H}-u)=\nu(\mathcal{H})$ ; see [9] for the maximum size of a resilient hypergraph with given matching number.

Consider $\mathcal{F}\subset\mathcal{K}_{p}$ with $\nu(\mathcal{F})\leq s$ and of maximal size. The ultimate goal is to prove

\displaystyle|\mathcal{F}|\leq\max\left\{|\mathcal{K}_{p}(\hat{S})|\colon|S|=s\right\}.

(6)

For the complete graph, i.e., for $p=1$ ,

|\mathcal{K}_{1}(\hat{S})|=\binom{n}{k}-\binom{n-s}{k},\mbox{ for all $s$-sets }S.

For $p<1$ the expected size of $\mathcal{K}_{1}(\hat{S})$ is $p\left(\binom{n}{k}-\binom{n-s}{k}\right)$ but it is impossible to know its exact value.

How to prove (6) without this piece of information? Since (6) is evident for $\mathcal{F}=\mathcal{K}_{p}(\hat{S_{0}})$ with $S_{0}$ an $s$ -element set, to start with we may assume that $\mathcal{F}$ is not of this form. That is, $\tau(\mathcal{F})>\nu(\mathcal{F})$ , meaning that $\mathcal{F}$ is non-trivial.

With this assumption we want to prove a stability result, i.e., a bound considerably smaller than $p\left(\binom{n}{k}-\binom{n-s}{k}\right)$ . For $p=1$ this was done by Bollobás, Daykin and Erdős [4] who generalized the Hilton-Milner Theorem ( $\nu=1$ case) to this situation. Unfortunately, their argument does not seem to work for the $p<1$ case.

Here is what we can do. Choose a set $R$ of maximal size satisfying

\displaystyle\nu(\mathcal{F}(\bar{R}))=s-|R|.

(7)

Set $q=s-|R|$ . By our assumption $1\leq q\leq s$ . Moreover, for all elements $x\in[n]\setminus R$ , by the maximal choice of $R$ , $\nu(\mathcal{F}(\overline{R\cup\{x\}}))=q$ . That is, $\mathcal{F}(\bar{R})$ is resilient. Now it suffices to show that $|\mathcal{F}(\bar{R})|$ is smaller than $\frac{1}{2}pq\binom{n-1}{k-1}$ , which is a lower bound for $\max\left\{|\mathcal{K}_{p}(\hat{S})|\colon|S|=q\right\}$ .

Lemma 2.2.

Suppose that $\mathcal{F}$ is a resilient $k$ -graph, $\nu(\mathcal{F})=q$ . Then there is a collection $\mathcal{B}$ of 2-element sets satisfying $\mathcal{F}\subset\langle\mathcal{B}\rangle$ and $|\mathcal{B}|\leq(kq)^{2}$ .

Proof.

Let $G_{1},\ldots,G_{q}\in\mathcal{F}$ form a maximal matching and set $Y=G_{1}\cup\ldots\cup G_{q}$ . Then $F\cap Y\neq\emptyset$ for all $F\in\mathcal{F}$ . For each $y\in Y$ , $\nu(\mathcal{F}(\bar{y}))=q$ by resilience. Define $Z_{y}=H_{1}^{(y)}\cup\ldots\cup H_{q}^{(y)}$ where $H_{1}^{(y)},\ldots,H_{q}^{(y)}$ form a maximal matching in $\mathcal{F}(\bar{y})$ . Now $\mathcal{B}=\{(y,z_{y})\colon y\in Y,z_{y}\in Z_{y}\}$ satisfies the requirements. ∎

For the case $n\gg k^{3}q$ and $p\gg\frac{k^{2}q\log n}{\binom{n-1}{k-1}}$ Lemma 2.2 implies

\displaystyle|\mathcal{F}(\bar{R})|\leq\frac{1}{2}pq\binom{n-1}{k-1}\mbox{ with high probability.}

(8)

Indeed, since $|\mathcal{B}|\leq(kq)^{2}$ we only need to show

\displaystyle|\mathcal{F}(\{x_{1},x_{2}\})|\leq|\mathcal{K}_{p}(\{x_{1},x_{2}\})|\leq\frac{p}{2k^{2}q}\binom{n-1}{k-1}

(9)

for any 2-subset $\{x_{1},x_{2}\}\subset[n]\setminus R$ . Note that, since $n\gg k^{3}q$ , the expected size of $|\mathcal{K}_{p}(\{x_{1},x_{2}\})|$ is

p\binom{n-2}{k-2}\leq\frac{pk}{n}\binom{n-1}{k-1}\ll\frac{p}{k^{2}q}\binom{n-1}{k-1}.

Thus (9) is true by combining the Chernoff bound, the union bound and $p\gg\frac{k^{2}q\log n}{\binom{n-1}{k-1}}$ . (there are “only” $\binom{n-r}{2}$ choices for $\{x_{1},x_{2}\}$ ).

How to improve on the bound for $p$ ? Note that we do not need (9) for each individual pair but only an upper bound for $|\mathcal{F}\cap\langle\mathcal{B}\rangle|$ . Let us use the special structure of $\mathcal{B}$ . By our proof of Lemma 2.2, $\mathcal{B}$ is the (not necessarily disjoint) union of $qk$ “fans” $\{\{y,z_{y}\}\colon z_{y}\in Z\}=:\mathcal{D}_{y,Z}$ . In fact, by a more careful analysis we can reduce the number of fans to $q$ (at the expense of adding some cliques, which is even easier to control, see Section 3). Thus, instead of (9) it is sufficient to require

\displaystyle|\mathcal{F}\cap\langle\mathcal{D}_{y,Z}\rangle|\leq\frac{1}{2}p\binom{n-1}{k-1}\mbox{ for each fan }\mathcal{D}_{y,Z}.

(10)

Here we’ll have $n\times\binom{n-1}{qk}$ requirements–much more than $\binom{n}{2}$ but the probability that (10) is violated for any fixed fan is going down exponentially and we should get bounds better than requiring (9). We will elaborate on this idea in Section 3, which eventually gives Theorem 1.3.

3 Proof of Theorem 1.3

Lemma 3.1.

Let $\mathcal{K}_{p}=\mathcal{K}_{p}(n,k)$ . For $k,s\geq 2$ , $n\geq 200k^{3}s$ , $p\geq\frac{64ks\log n}{\binom{n-1}{k-1}}$ , properties (i), (ii) and (iii) hold for every $1\leq q\leq s$ with high probability.

(i)

For $R,Q\subset[n]$ with $R\cap Q=\emptyset$ , $|R|=s-q$ and $|Q|=q$ ,

$\displaystyle|\mathcal{K}_{p}(\bar{R},\hat{Q})|>\frac{1}{2}pq\binom{n-1}{k-1}.$ (11)
(ii)

For every $Q\subset[n]$ with $|Q|<3kq\leq 3ks$ ,

$\displaystyle\left|\mathcal{K}_{p}\cap\left\langle\binom{Q}{2}\right\rangle\right|<\frac{1}{4}pq\binom{n-1}{k-1}.$ (12)
(iii)

For every $x\in[n]$ , $Q\subset[n]\setminus\{x\}$ with $|Q|=kq\leq ks$ ,

$\displaystyle|\mathcal{K}_{p}\cap\langle\mathcal{D}_{x,Q}\rangle|<\frac{1}{4}p\binom{n-1}{k-1}.$ (13)

Proof.

(i)

Let $\mathcal{K}=\binom{[n]}{k}$ . It is easy to see that

|\mathcal{K}(\bar{R},\hat{Q})|=\binom{n-|R|}{k}-\binom{n-|R|-|Q|}{k}\geq q\binom{n-s}{k-1}.

Note that

\frac{\binom{n-s}{k-1}}{\binom{n-1}{k-1}}\geq\left(\frac{n-k-s}{n-k}\right)^{k-1}\geq 1-\frac{s(k-1)}{n-k}\geq\frac{3}{4}.

Thus we have

|\mathcal{K}(\bar{R},\hat{Q})|\geq\frac{3}{4}q\binom{n-1}{k-1}.

Applying (4) with $\epsilon=1/4$ , we infer that

Pr\left(|\mathcal{K}_{p}(\bar{R},\hat{Q})|\leq\frac{1}{2}pq\binom{n-1}{k-1}\right)\leq 2e^{-\frac{1}{64}pq\binom{n-1}{k-1}}\leq 2e^{-\frac{1}{64}p\binom{n-1}{k-1}}.

Since for all $1\leq q\leq s$ the number of such pairs $(R,Q)$ is at most $n^{s}2^{s}$ , by the union bound and $p\geq\frac{64ks\log n}{\binom{n-1}{k-1}}$ , the probability that (i) does not hold for some $1\leq q\leq s$ is at most

n^{s}2^{s}2e^{-\frac{1}{64}p\binom{n-1}{k-1}}\leq n^{s}2^{s+1}n^{-ks}\leq 2^{s+1}n^{-s}\rightarrow 0~{}\text{as}~{}n\rightarrow\infty.

(ii)

It suffices to show that with high probability (12) holds for every $1\leq q\leq s$ and $Q\subset[n]$ with $|Q|=3kq$ . By $n\geq 200k^{3}s\geq 200k^{3}q$ ,

\left|\left\langle\binom{Q}{2}\right\rangle\right|\leq\binom{|Q|}{2}\binom{n-2}{k-2}\leq\frac{9k^{2}q^{2}}{2}\frac{k}{n}\binom{n-1}{k-1}\leq\frac{q}{28}\binom{n-1}{k-1}.

By (5),

Pr\left(\left|\mathcal{K}_{p}\cap\left\langle\binom{Q}{2}\right\rangle\right|\geq\frac{1}{4}pq\binom{n-1}{k-1}\right)\leq e^{-\frac{1}{4}pq\binom{n-1}{k-1}}.

Since for fix $q$ the number of such $Q$ is at most $\binom{n}{3kq}$ and $p\geq\frac{64ks\log n}{\binom{n-1}{k-1}}$ , by the union bound, the probability that (12) doesn’t hold for some $1\leq q\leq s$ is at most

\sum_{q=1}^{s}\binom{n}{3kq}e^{-\frac{1}{4}pq\binom{n-1}{k-1}}\leq\sum_{q=1}^{s}n^{3kq-16ksq}\leq\sum_{q=1}^{s}n^{-16kq}\leq sn^{-16k}\rightarrow 0~{}\text{as}~{}n\rightarrow\infty.

(iii)

Finally for every $x\in[n]$ , $Q\subset[n]\setminus\{x\}$ with $|Q|=kq\leq ks$ , by $n\geq 200k^{3}s$ ,

|\langle\mathcal{D}_{x,Q}\rangle|\leq kq\binom{n-2}{k-2}\leq kq\frac{k}{n}\binom{n-1}{k-1}\leq\frac{1}{28}\binom{n-1}{k-1}.

By (5),

Pr\left(|\mathcal{K}_{p}\cap\langle\mathcal{D}_{x,Q}\rangle|\geq\frac{1}{4}p\binom{n-1}{k-1}\right)\leq e^{-\frac{1}{4}p\binom{n-1}{k-1}}.

Since for fix $1\leq q\leq s$ the number of such pair $(x,Q)$ is at most $n\binom{n}{kq}$ and $p\geq\frac{64ks\log n}{\binom{n-1}{k-1}}$ , by the union bound, the probability that (13) doesn’t hold for some $1\leq q\leq s$ is at most

\sum_{q=1}^{s}n\binom{n}{kq}e^{-\frac{1}{4}p\binom{n-1}{k-1}}\leq\sum_{q=1}^{s}n^{1+kq-16ks}\leq sn^{-10ks}\rightarrow 0~{}\text{as}~{}n\rightarrow\infty.

∎

Proof of Theorem 1.3.

By Lemma 3.1, with high probability, $\mathcal{K}_{p}$ satisfies (i), (ii) and (iii) of Lemma 3.1 for every $1\leq q\leq s$ . Let us fix a hypergraph $\mathcal{F}$ that satisfies (i), (ii) and (iii) for every $1\leq q\leq s$ and let $\mathcal{H}\subset\mathcal{F}$ be a sub-hypergraph with $\nu(\mathcal{H})=s$ and $|\mathcal{H}|$ maximal. We have to show that $\mathcal{H}$ is trivial. Arguing indirectly assume that $\mathcal{H}$ is not trivial. Hence there exists $R$ , $|R|=r$ , $0\leq r<s$ such that $\mathcal{H}(\bar{R})$ is resilient with $\nu(\mathcal{H}(\bar{R}))=s-r$ .

Let $q=s-r$ . Note that for an arbitrary $Q\in\binom{[n]\setminus R}{q}$ the family $\mathcal{F}(\bar{R},\hat{Q})$ can be used to replace $\mathcal{H}(\bar{R})$ and the resulting family has matching number $s$ . To get a contradiction we need to show

\displaystyle|\mathcal{F}(\bar{R},\hat{Q})|>|\mathcal{H}(\bar{R})|.

(14)

By (11),

|\mathcal{F}(\bar{R},\hat{Q})|>\frac{q}{2}p\binom{n-1}{k-1}.

Let $H_{1},H_{2},\ldots,H_{q}$ be a maximal matching in $\mathcal{H}(\bar{R})$ and let $X=H_{1}\cup H_{2}\cup\ldots\cup H_{q}$ . Define

\mathcal{H}_{0}=\{H\in\mathcal{H}(\bar{R})\colon|H\cap X|\geq 2\}

and

\mathcal{H}_{i}=\{H\in\mathcal{H}(\bar{R})\colon|H\cap X|=1,H\cap X=H\cap H_{i}\},\ i=1,2,\ldots,q.

By (12),

\displaystyle|\mathcal{H}_{0}|<\frac{1}{4}pq\binom{n-1}{k-1}.

(15)

We claim that $\mathcal{H}_{i}$ is intersecting for $i=1,2,\ldots,q$ . Indeed, otherwise if there exist $H_{i}^{\prime},H_{i}^{\prime\prime}\in\mathcal{H}_{i}$ such that $H_{i}^{\prime}\cap H_{i}^{\prime\prime}=\emptyset$ then $H_{1},\ldots,H_{i-1},H_{i}^{\prime},H_{i}^{\prime\prime},H_{i+1},\ldots,H_{q}$ form a matching of size $q+1$ in $\mathcal{H}(\bar{R})$ , a contradiction. Thus $\mathcal{H}_{i}$ is intersecting. Without loss of generality, assume that $\mathcal{H}_{1},\ldots,\mathcal{H}_{\ell}$ are stars and $\mathcal{H}_{\ell+1},\ldots,\mathcal{H}_{q}$ are non-trivial intersecting. Let $x_{i}$ be the center of the star $\mathcal{H}_{i}$ , $i=1,2,\ldots,\ell$ . For each $i\in[\ell]$ , since $\mathcal{H}(\bar{R})$ is resilient, there exists a matching $L_{1}^{i},L_{2}^{i},\ldots,L_{q}^{i}$ in $\mathcal{H}(\overline{R\cup\{x_{i}\}})$ . Let

Q_{i}=L_{1}^{i}\cup L_{2}^{i}\cup\ldots\cup L_{q}^{i}.

Clearly $x_{i}\in H$ and $H\cap Q_{i}\neq\emptyset$ for any $H\in\mathcal{H}_{i}$ with $i\in[\ell]$ . Thus by (13) we have

\displaystyle\sum_{1\leq i\leq\ell}|\mathcal{H}_{i}|\leq\sum_{1\leq i\leq\ell}|\mathcal{F}\cap\langle\mathcal{D}_{x_{i},Q_{i}}\rangle|<\frac{1}{4}p\ell\binom{n-1}{k-1}.

(16)

Since $\mathcal{H}_{i}$ is non-trivial intersecting for $\ell+1\leq i\leq q$ , there exist $H_{i}^{\prime},H_{i}^{\prime\prime}\in\mathcal{H}_{i}$ such that $H_{i}^{\prime}\cap H_{i}=\{x_{i}\}$ and $x_{i}\notin H_{i}^{\prime\prime}$ . It follows that $|H\cap(H_{i}\cup H_{i}^{\prime}\cup H_{i}^{\prime\prime})|\geq 2$ for any $H\in\mathcal{H}_{i}$ . Let $Q=\cup_{\ell+1\leq i\leq r}(H_{i}\cup H_{i}^{\prime}\cup H_{i}^{\prime\prime})$ . Clearly $|Q|<3k(q-\ell)$ . Then by (12) we infer that

\displaystyle\sum_{\ell+1\leq i\leq q}|\mathcal{H}_{i}|<\left|\mathcal{F}\cap\left\langle\binom{Q}{2}\right\rangle\right|\leq\frac{1}{4}p(q-\ell)\binom{n-1}{k-1}.

(17)

Adding (15), (16) and (17), we obtain that

|\mathcal{H}(\bar{R})|=\sum_{0\leq i\leq q}|\mathcal{H}_{i}|<\frac{1}{2}pq\binom{n-1}{k-1}<|\mathcal{F}(\bar{R},\hat{Q})|,

the desired contradiction. ∎

4 Proof of Theorem 1.4 and Proposition 1.6

In this section, we prove Theorem 1.4 and Proposition 1.6. Note that Theorem 1.4 improves the lower bound for $n$ in Theorem 1.3 at the expense of a worse lower bound for $p$ .

We say that a $k$ -graph $\mathcal{H}$ is $t$ -resilient if $\nu(\mathcal{H}-T)=\nu(\mathcal{H})$ for any $T\subset V(\mathcal{H})$ with $|T|\leq t$ . Clearly $t\leq k-1$ . Indeed, deleting $k$ vertices in an edge decreases the matching number by at least one.

Let us prove the following simple fact.

Fact 4.1.

For every $\mathcal{H}\subset\binom{[n]}{k}$ and $t\leq k-1$ , there exist $\ell\leq\nu(\mathcal{H})$ and $T_{1},T_{2},\ldots,T_{\ell}$ such that

(i)

$|T_{i}|\leq t$ for all $i=1,2,\ldots,\ell$ .
(ii)

For $i=1,2,\ldots,\ell$ , $\mathcal{H}-U_{i-1}$ is a $(|T_{i}|-1)$ -resilient $k$ -graph with matching number $\nu(\mathcal{H})-i+1$ , where $U_{i-1}=T_{1}\cup T_{2}\cup\ldots\cup T_{i-1}$ .
(iii)

$\mathcal{H}-U$ is a $t$ -resilient $k$ -graph with matching number $\nu(\mathcal{H})-\ell$ , where $U=T_{1}\cup T_{2}\cup\ldots\cup T_{\ell}$ .

Proof.

Let $\mathcal{H}_{0}=\mathcal{H}$ . We obtain $T_{1},T_{2},\ldots,T_{\ell}$ by a greedy algorithm. For $i=0,1,\ldots$ , repeat the following procedure. If $\mathcal{H}_{i}$ is $t$ -resilient then let $\ell=i$ and stop. Otherwise find $T_{i+1}$ of the minimum size so that $\nu(\mathcal{H}_{i}-T_{i+1})=\nu(\mathcal{H}_{i})-1$ and let $\mathcal{H}_{i+1}=\mathcal{H}_{i}-T_{i+1}$ . Note that in each step the matching number of $\mathcal{H}_{i}$ decreases by 1. The procedure will terminate in at most $\nu(\mathcal{H})$ steps.

Since $\mathcal{H}_{i-1}$ is not $t$ -resilient for $1\leq i\leq\ell$ , we have $|T_{i}|\leq t$ and (i) holds. Since $\mathcal{H}_{\ell}=\mathcal{H}-U$ is empty or $t$ -resilient, (iii) follows.

We are left with (ii). As $T_{i}$ is of the minimum size satisfying $\nu(\mathcal{H}_{i-1}-T_{i})=\nu(\mathcal{H}_{i-1})-1$ , we infer that $\nu(\mathcal{H}_{i-1}-R)=\nu(\mathcal{H}_{i-1})$ for all $R\subset V(\mathcal{H}_{i-1})$ with $|R|<|T_{i}|$ . That is, $\mathcal{H}_{i-1}=\mathcal{H}-U_{i-1}$ is $(|T_{i}|-1)$ -resilient with matching number $\nu(\mathcal{H})-i+1$ . Thus (ii) holds. ∎

Lemma 4.2.

Let $\mathcal{H}$ be a $t$ -resilient $k$ -graph with matching number $s$ . Then there exists $\mathcal{B}\subset\binom{V(\mathcal{H})}{t+1}$ with $|\mathcal{B}|\leq(ks)^{t+1}$ such that $\mathcal{H}\subset\langle\mathcal{B}\rangle$ . Moreover, for any $X\subset V(\mathcal{H})$ with $|X|\leq x<ks$ , there exists $\mathcal{B}^{\prime}\subset\binom{V(\mathcal{H})}{t+1}$ with $|\mathcal{B}^{\prime}|\leq x(ks)^{t}$ such that $\mathcal{H}(\hat{X})\subset\langle\mathcal{B}^{\prime}\rangle$ .

Proof.

Let us prove the lemma by a branching process of $t+1$ stages. A sequence $S=(x_{1},x_{2},\ldots,x_{\ell})$ is an ordered sequence of distinct elements of $V(\mathcal{H})$ and we use $\widehat{S}$ to denote the underlying unordered set $\{x_{1},x_{2},\ldots,x_{\ell}\}$ .

At the first stage, let $H_{1}^{1},\ldots,H_{s}^{1}$ be a maximal matching in $\mathcal{H}$ . Make $ks$ sequences $(x_{1})$ with $x_{1}\in H_{1}^{1}\cup\ldots\cup H_{s}^{1}$ . At the $p$ th stage for $p=1,2,3,\ldots,t$ , for each sequence $(x_{1},\ldots,x_{p})$ of length $p$ , since $\mathcal{H}$ is $t$ -resilient, then there exists a maximal matching $H_{1}^{p},\ldots,H_{s}^{p}$ in $\mathcal{H}-\{x_{1},\ldots,x_{p}\}$ . We replace $(x_{1},\ldots,x_{p})$ by $ks$ sequences $(x_{1},\ldots,x_{p},x_{p+1})$ with $x_{p+1}\in H_{1}^{p}\cup\ldots\cup H_{s}^{p}$ . Eventually by the branching process we shall obtain at most $(ks)^{t+1}$ sequences of length $t+1$ .

Claim 4.3.

For each $H\in\mathcal{H}$ , there is a sequence $S$ of length $t+1$ with $\widehat{S}\subset H$ .

Proof.

Let $S=(x_{1},\ldots,x_{\ell})$ be a sequence of maximal length that occurred at some stage of the branching process satisfying $\widehat{S}\subset H$ . Suppose indirectly that $\ell<t+1$ . Since $H\cap(H_{1}^{1}\cup\ldots\cup H_{s}^{1})\neq\emptyset$ at the first stage, there is a sequence $(x_{1})$ with $x_{1}\in H_{1}^{1}\cup\ldots\cup H_{s}^{1}$ such that $\{x_{1}\}\subset H$ . Thus $\ell\geq 1$ . Since $\ell<t+1$ , at the $\ell$ th stage there is a maximal matching $H_{1}^{\ell},\ldots,H_{s}^{\ell}$ in $\mathcal{H}-\widehat{S}$ and $S$ was replaced by $ks$ sequences $(x_{1},\ldots,x_{\ell},y)$ with $y\in H_{1}^{\ell}\cup\ldots\cup H_{s}^{\ell}$ . Since $H\cap(H_{1}^{\ell}\cup\ldots\cup H_{s}^{\ell})\neq\emptyset$ , there is some $y_{0}\in H\cap(H_{1}^{\ell}\cup\ldots\cup H_{s}^{\ell})$ . Then $(x_{1},\ldots,x_{\ell},y_{0})$ is a longer sequence satisfying $\{x_{1},\ldots,x_{\ell},y_{0}\}\subset H$ , contradicting the maximality of $\ell$ . ∎

Let $\mathcal{B}$ be the collection of $\widehat{S}$ over all sequences $S$ of length $t+1$ . By Claim 4.3, we infer that $\mathcal{H}\subset\langle\mathcal{B}\rangle$ .

Similarly, if we make a branching process starting at $x$ sequences $(x_{1})$ with $x_{1}\in X$ , then we shall obtain a $\mathcal{B}^{\prime}\subset\binom{V(\mathcal{H})}{t+1}$ with $|\mathcal{B}^{\prime}|\leq x(ks)^{t}$ such that $\mathcal{H}(\hat{X})\subset\langle\mathcal{B}^{\prime}\rangle$ . ∎

Lemma 4.4.

Let $\mathcal{K}_{p}=\mathcal{K}_{p}(n,k)$ and let $t$ be an integer with $1\leq t\ll k$ . If $n\geq 56k^{2+1/t}s$ and $p\geq\frac{8(t+1)k^{t+1}s^{t}\log n}{\binom{n-1}{k-1}}$ , then with high probability (i), (ii) and (iii) hold.

(i)

For $R,Q\subset[n]$ with $R\cap Q=\emptyset$ , $|R|=s-q$ and $|Q|=q$ ,

$\displaystyle|\mathcal{K}_{p}(\bar{R},\hat{Q})|>\frac{1}{2}pq\binom{n-1}{k-1}.$ (18)
(ii)

For every $R\subset[n]$ with $2\leq|R|\leq t$ ,

$|\mathcal{K}_{p}(R)|\leq\frac{1}{4|R|(ks)^{|R|-1}}p\binom{n-1}{k-1}.$
(iii)

For every $T\subset[n]$ with $|T|=t+1$ ,

$|\mathcal{K}_{p}(T)|\leq\frac{1}{4k^{t+1}s^{t}}p\binom{n-1}{k-1}.$

Proof.

Clearly, (i) follows from Lemma 3.1 (i). Thus we only need to show (ii) and (iii).

(ii)

Let $|R|=r\leq t$ . Note that $n\geq 56k^{2}s$ implies

|\mathcal{K}(R)|=\binom{n-r}{k-r}\leq\frac{k^{r-1}}{n^{r-1}}\binom{n-1}{k-1}\leq\frac{1}{28r(ks)^{r-1}}\binom{n-1}{k-1}.

By (5), we infer that

Pr\left(|\mathcal{K}_{p}(R))|>\frac{1}{4r(ks)^{r-1}}p\binom{n-1}{k-1}\right)<e^{-\frac{1}{4r(ks)^{r-1}}p\binom{n-1}{k-1}}<e^{-\frac{1}{4t(ks)^{t-1}}p\binom{n-1}{k-1}}.

Since the number of choices for $R$ is at most $\sum_{2\leq i\leq t}\binom{n}{i}\leq n^{t}$ , by $p\geq\frac{8t^{2}(ks)^{t-1}\log n}{\binom{n-1}{k-1}}$ and the union bound, the probability that condition (ii) is violated is at most

n^{t}e^{-\frac{1}{4t(ks)^{t-1}}p\binom{n-1}{k-1}}\leq e^{t\log n-\frac{1}{4t(ks)^{t-1}}p\binom{n-1}{k-1}}\leq n^{-t}\rightarrow 0\text{~{}as~{}}n\rightarrow\infty.

Thus (ii) holds.

(iii)

Similarly, $n\geq 56k^{2+1/t}s$ implies

|\mathcal{K}(T)|=\binom{n-t-1}{k-t-1}\leq\frac{k^{t}}{n^{t}}\binom{n-1}{k-1}\leq\frac{1}{28k^{t+1}s^{t}}\binom{n-1}{k-1}.

By (5), we infer that

Pr\left(|\mathcal{K}_{p}(T))|>\frac{1}{4k^{t+1}s^{t}}p\binom{n-1}{k-1}\right)<e^{-\frac{1}{4k^{t+1}s^{t}}p\binom{n-1}{k-1}}.

Since the number of choices for $T$ is at most $\binom{n}{t+1}\leq n^{t+1}$ , by $p\geq\frac{8(t+1)k^{t+1}s^{t}\log n}{\binom{n-1}{k-1}}$ and the union bound, the probability that condition (iii) is violated is at most

n^{t+1}e^{-\frac{1}{4k^{t+1}s^{t}}p\binom{n-1}{k-1}}\leq e^{(t+1)\log n-\frac{1}{4k^{t+1}s^{t}}p\binom{n-1}{k-1}}\leq n^{-t-1}\rightarrow 0\text{~{}as~{}}n\rightarrow\infty.

Thus (iii) holds.

∎

Proof of Theorem 1.4.

By Lemma 4.4, with high probability, $\mathcal{K}_{p}$ satisfies (i), (ii) and (iii) of Lemma 4.4. Let us fix a hypergraph $\mathcal{F}$ that satisfies (i), (ii), (iii) and let $\mathcal{H}\subset\mathcal{F}$ be a sub-hypergraph with $\nu(\mathcal{H})=s$ and $|\mathcal{H}|$ maximal. We have to show that $\mathcal{H}$ is trivial.

Suppose indirectly that $\mathcal{H}$ is non-trivial. By Fact 4.1, there exist $\ell\leq\nu(\mathcal{H})$ and $T_{1},T_{2},\ldots,T_{\ell}$ such that (i), (ii), (iii) of Fact 4.1 hold. Let $a_{i}=|T_{i}|$ , $i=1,2,\ldots,\ell$ . Recall that $\mathcal{H}_{0}=\mathcal{H}$ and $\mathcal{H}_{i}=\mathcal{H}-(T_{1}\cup\ldots\cup T_{i})$ , $i=1,2,\ldots,\ell$ . Let $\mathcal{H}_{1}^{\prime}=\mathcal{H}_{0}(\hat{T_{1}})$ , $\mathcal{H}_{2}^{\prime}=\mathcal{H}_{1}(\hat{T_{2}})$ , $\ldots$ , $\mathcal{H}_{\ell}^{\prime}=\mathcal{H}_{\ell-1}(\hat{T_{\ell}})$ and $\mathcal{H}_{0}^{\prime}=\mathcal{H}_{\ell}$ . Since $\mathcal{H}_{0}^{\prime}$ is $t$ -resilient, by Lemma 4.2 there exists $\mathcal{B}_{0}$ of size at most $(k(s-\ell))^{t+1}$ such that $\mathcal{H}_{0}^{\prime}\subset\langle\mathcal{B}_{0}\rangle$ . Similarly, for $i\geq 1$ , since $\mathcal{H}_{i}^{\prime}=\mathcal{H}_{i-1}(\hat{T_{i}})$ and $\mathcal{H}_{i-1}$ is $(a_{i}-1)$ -resilient, by Lemma 4.2 there exists $\mathcal{B}_{i}$ of size at most $a_{i}(k(s-i+1))^{a_{i}-1}$ such that $\mathcal{H}_{i}^{\prime}\subset\langle\mathcal{B}_{i}\rangle$ . Thus $\mathcal{H}=\mathcal{H}_{0}^{\prime}\cup\mathcal{H}_{1}^{\prime}\cup\ldots\cup\mathcal{H}_{\ell}^{\prime}$ is contained in $\cup_{0\leq i\leq\ell}\langle\mathcal{B}_{i}\rangle$ .

By relabeling the indices if necessary, we may assume that $a_{\ell}=a_{\ell-1}=\ldots=a_{\ell-r+1}=1$ and $a_{1}\geq a_{2}\geq\ldots\geq a_{\ell-r}\geq 2$ . Let $\mathcal{B}_{i}=\{\{x_{i}\}\}$ for $i=\ell-r+1,\ldots,\ell$ and let $R=\{x_{i}\colon i=\ell-r+1,\ldots,\ell\}$ . Let $q=s-r$ . Note that for an arbitrary $Q\in\binom{[n]\setminus R}{q}$ the family $\mathcal{F}(\bar{R},\hat{Q})$ can be used to replace $\mathcal{H}(\bar{R})$ and the resulting family has matching number $s$ . To get a contradiction we need to show

\displaystyle|\mathcal{F}(\bar{R},\hat{Q})|>|\mathcal{H}(\bar{R})|.

(19)

By Lemma 4.4 (i) we have $|\mathcal{F}(\bar{R},\hat{Q})|>\frac{1}{2}pq\binom{n-1}{k-1}$ . By Lemma 4.4 (ii) and $|\mathcal{B}_{i}|\leq a_{i}(k(s-i+1))^{a_{i}-1}$ , we have

\sum_{1\leq i\leq\ell-r}|\mathcal{H}_{i}^{\prime}|\leq\sum_{1\leq i\leq\ell-r}|\mathcal{F}\cap\langle\mathcal{B}_{i}\rangle|\leq\sum_{1\leq i\leq\ell-r}|\mathcal{B}_{i}|\frac{1}{4a_{i}(ks)^{a_{i}-1}}p\binom{n-1}{k-1}\leq\frac{\ell-r}{4}p\binom{n-1}{k-1}.

By Lemma 4.4 (ii) and $|\mathcal{B}_{0}|\leq(k(s-\ell))^{t+1}$ ,

|\mathcal{H}_{0}^{\prime}|\leq|\mathcal{F}\cap\langle\mathcal{B}_{0}\rangle|\leq|\mathcal{B}_{0}|\frac{1}{4k^{t+1}s^{t}}p\binom{n-1}{k-1}\leq\frac{s-\ell}{4}p\binom{n-1}{k-1}.

Thus,

|\mathcal{H}(\bar{R})|\leq\sum_{0\leq i\leq\ell-r}|\mathcal{H}_{i}^{\prime}|<\frac{s-\ell+\ell-r}{4}p\binom{n-1}{k-1}=\frac{q}{4}p\binom{n-1}{k-1}<|\mathcal{F}(\bar{R},\hat{Q})|,

as desired. Thus the theorem holds. ∎

Proof of Proposition 1.6.

Let $N$ denote the number of $(s+1)$ -matchings in $\binom{[n]}{k}$ . Then

N=\binom{n}{(s+1)k}\frac{((s+1)k)!}{k!^{s+1}}.

Since the probability that a fixed $(s+1)$ -matching is in $\mathcal{K}_{p}$ is $p^{s+1}$ , by the union bound we have

Pr(\mathcal{K}_{p}\mbox{ contains an $(s+1)$-matching})\leq Np^{s+1}\leq\binom{n}{(s+1)k}\frac{((s+1)k)!}{k!^{s+1}}p^{s+1}.

We use $(n)_{k}$ to denote $n(n-1)\ldots(n-k+1)$ . Note that

\binom{n}{(s+1)k}\frac{((s+1)k)!}{(k!)^{s+1}}=\frac{n(n-1)\ldots(n-(s+1)k+1)}{(k!)^{s+1}}\leq\frac{(n)_{k}(n-k)_{k}\ldots(n-sk)_{k}}{(k!)^{s+1}}

and

(n-ik)_{k}\leq\left(1-\frac{ik}{n}\right)^{k}(n)_{k}\leq e^{-ik^{2}/n}(n)_{k}.

Using $p\ll e^{\frac{k^{2}s}{2n}}/\binom{n}{k}$ , we obtain that

\frac{(n)_{k}(n-k)_{k}\ldots(n-sk)_{k}}{(k!)^{s+1}}p^{s+1}\leq e^{-(1+2+\ldots+s)k^{2}/n}\binom{n}{k}^{s+1}p^{s+1}\leq e^{-s(s+1)k^{2}/(2n)}\binom{n}{k}^{s+1}p^{s+1}\ll 1.

On the other hand, by the union bound and $p\gg\frac{s\log n}{\binom{n-s}{k}}$ we have

Pr(\mathcal{K}_{p}\mbox{ is trivial})\leq\binom{n}{s}(1-p)^{\binom{n-s}{k}}<\binom{n}{s}\exp\left(-p\binom{n-s}{k}\right)\ll 1.

Thus with high probability, $\mathcal{K}_{p}$ has matching number at most $s$ and is non-trivial. ∎

5 Proof of Theorem 1.7

In this section, we prove Theorem 1.7, which is a result specifically for graphs.

Let $(B,A_{1},A_{2},\ldots,A_{m})$ be a partition of $[n]$ . We say that $(B,A_{1},A_{2},\ldots,A_{m})$ is an $s$ -partition if the following hold:

(i)

Let $|A_{i}|=a_{i}$ , $i=1,2,\ldots,m$ . Each $a_{i}$ is odd and $a_{1}\geq a_{2}\geq\ldots\geq a_{m}\geq 1$ ;
(ii)

$b+\sum\limits_{1\leq i\leq m}\frac{a_{i}-1}{2}=s$ ;
(iii)

$b+\sum\limits_{1\leq i\leq m}a_{i}=n$ .

For an $s$ -partition $(B,A_{1},A_{2},\ldots,A_{m})$ of $[n]$ , define a graph $G(B,A_{1},A_{2},\ldots,A_{m})$ on the vertex set $[n]$ so that $G[B]$ and $G[A_{i}]$ , $i=1,2,\ldots,m$ , are all cliques, and $G[B,A_{1}\cup\ldots\cup A_{m}]$ is complete bipartite.

Theorem 5.1 (Tutte-Berge theorem or the Edmonds-Gallai Theorem, see [15]).

Let $G$ be a graph on the vertex set $[n]$ with $\nu(G)\leq s$ and $n\geq 2s+2$ . Then there is an $s$ -partition $(B,A_{1},A_{2},\ldots,A_{m})$ of $[n]$ such that $G$ is a subgraph of $G(B,A_{1},A_{2},\ldots,A_{m})$ .

By applying Theorem 5.1, Erdős and Gallai determined the maximum number of edges in a graph with matching number at most $s$ , which is the $k=2$ case of the Erdős Matching Conjecture.

Theorem 5.2 ([5]).

Let $G$ be a graph on $n$ vertices. If $\nu(G)\leq s$ and $n\geq 2s+2$ , then

e(G)\leq\max\left\{\binom{2s+1}{2},\binom{s}{2}+s(n-s)\right\}.

Let us mention that very recently by combining the Tutte-Berge theorem and other techniques, Alon and the first author [1] determined the maximum number of edges in a $K_{r+1}$ -free graph with matching number at most $s$ .

Note that

\binom{2r_{1}+1}{2}+\binom{2r_{2}+1}{2}\leq\binom{2(r_{1}+r_{2})+1}{2}.

Let $a_{i}=2r_{i}+1$ , $i=1,2,\ldots,m$ and let $r_{\ell}\geq 1$ and $r_{\ell+1}=\ldots=r_{m}=0$ . By (ii) we have $r_{1}+r_{2}+\ldots+r_{\ell}+b=s$ . Then

$\displaystyle e(G(B,A_{1},A_{2},\ldots,A_{m}))$	$\displaystyle=\binom{b}{2}+\sum_{1\leq i\leq\ell}\binom{a_{i}}{2}+b\sum_{1\leq i\leq m}a_{i}$
	$\displaystyle\leq\binom{b}{2}+\binom{2(r_{1}+r_{2}+\ldots+r_{\ell})+1}{2}+b(n-b)$
	$\displaystyle\leq\binom{b}{2}+\binom{2(s-b)+1}{2}+b(n-b)$
	$\displaystyle\leq\max\left\{\binom{2s+1}{2},\binom{s}{2}+s(n-s)\right\}.$	(20)

Proof of Theorem1.7.

Let

f(n,s):=\max\left\{\binom{2s+1}{2},\binom{s}{2}+s(n-s)\right\}.

Let $\mathcal{G}$ be the family of all graphs with $\nu(G)\leq s$ on the vertex set $[n]$ . Then

		$\displaystyle Pr\left(X\geq(1+\varepsilon)pf(n,s)\right)$
	$\displaystyle=$	$\displaystyle Pr\left(\exists G\in\mathcal{G},e(G\cap G(n,p))\geq(1+\varepsilon)pf(n,s)\right)$
	$\displaystyle=$	$\displaystyle Pr\left(\exists\mbox{ an $s$-parition }(B,A_{1},A_{2},\ldots,A_{m}),e(G(B,A_{1},A_{2},\ldots,A_{m})\cap G(n,p))\geq(1+\varepsilon)pf(n,s)\right).$

Assume that $|A_{\ell+1}|=\cdots=|A_{m}|=1$ for some $\ell\in[m]$ , the $s$ -partition $(B,A_{1},A_{2},\ldots,A_{m})$ is determined by $(B,A_{1},A_{2},\ldots,A_{\ell})$ . By (ii) we have

b+\sum\limits_{1\leq i\leq\ell}\frac{a_{i}-1}{2}=s.

It follows that $\ell\leq s-1$ and $b+a_{1}+a_{2}+\ldots+a_{\ell}<3s$ . Thus the total number of $s$ -partitions is at most $s^{3s}n^{3s}$ .

Let $(B,A_{1},A_{2},\ldots,A_{m})$ be an $s$ -partition of $[n]$ and let $G=G(B,A_{1},A_{2},\ldots,A_{m})$ . If $e(G)\geq\frac{1}{7}f(n,s)$ , then by (4),

Pr\left(e(G\cap G(n,p))\geq(1+\varepsilon)pf(n,s)>(1+\varepsilon)pe(G)\right)\leq e^{-\frac{\varepsilon^{2}pe(G)}{3}}\leq e^{-\frac{\varepsilon^{2}pf(n,s)}{21}}.

If $e(G)<\frac{1}{7}f(n,s)$ , then $(1+\varepsilon)pf(n,s)\geq 7pe(G)$ . By (5) we have

Pr\left(e(G\cap G(n,p))\geq(1+\varepsilon)pf(n,s)\right)\leq e^{-(1+\varepsilon)pf(n,s)}\leq e^{-\frac{\varepsilon^{2}pf(n,s)}{21}}.

Thus, by the union bound we conclude that

Pr\left(\exists G\in\mathcal{G},e(G\cap G(n,p))\geq(1+\varepsilon)pf(n,s)\right)\leq s^{3s}n^{3s}e^{-\frac{\varepsilon^{2}pf(n,s)}{21}}=e^{-\frac{11f(n,s)\log n}{n}+3s\log(sn)}.

If $n\geq\frac{5s+3}{2}$ , then $s<\frac{2n}{5}$ and $f(n,s)=\binom{s}{2}+s(n-s)>\frac{4sn}{5}$ . If $2s+2\leq n<\frac{5s+3}{2}$ , then $f(n,s)=\binom{2s+1}{2}>\frac{4sn}{5}$ as well. It follows that

\frac{11f(n,s)\log n}{n}-3s\log(sn)>7s\log n-6s\log n=s\log n.

Thus with probability $1-n^{-s}$ ,

X\leq(1+\varepsilon)p\max\left\{\binom{2s+1}{2},\binom{s}{2}+s(n-s)\right\}.

For the lower bound, let us consider the graph $G_{1}=\binom{[2s+1]}{2}$ and $G_{2}=\binom{[s]}{2}\cup([s]\times[s+1,n])=\binom{[n]}{2}\setminus\binom{[s+1,n]}{2}$ . Then by (4) we have

Pr\left(e(G_{1}\cap G(n,p))\leq(1-\varepsilon)pe(G_{1})\right)\leq e^{-\frac{\varepsilon^{2}pe(G_{1})}{3}}.

and

Pr\left(e(G_{2}\cap G(n,p))\leq(1-\varepsilon)pe(G_{2})\right)\leq e^{-\frac{\varepsilon^{2}pe(G_{2})}{3}}.

Since $\max\{e(G_{1}),e(G_{2})\}\geq\frac{4sn}{5}$ , we infer that for $i=1$ or $2$ ,

\frac{\varepsilon^{2}pe(G_{i})}{3}\geq\frac{250\log n}{3n}\times\frac{4sn}{5}\geq 64s\log n.

It follows that with probability $1-n^{-64s}$ , we have

X\geq(1-\varepsilon)p\max\left\{\binom{2s+1}{2},\binom{s}{2}+s(n-s)\right\}.

Thus the theorem is proven. ∎

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