2.1. Preliminaries

Let $p=\mathrm{const}\in (0,1)$ ,

\begin{equation*} k_+=k_+(n,p)=\lceil 2\log _{1/p}n\rceil, \quad k_-=k_-(n,p)=\lceil 2\log _{1/p}n-9\log _{1/p}\log _{1/p} n\rceil . \end{equation*}

We will use the following well-known bounds on the maximum size of a clique in $G_n$ (see, e.g. [Reference Janson, Łuczak and Ruciński23, Theorem 7.1, Lemma 7.13]).

Also, for technical reasons, we need to estimate the probability that a fixed vertex sends $s=(\gamma +o(1))\log _{1/p} n$ edges to a fixed set of size about $(2+o(1))\log _{1/p}n$ . Note that, as follows from the claim below, $\gamma$ is defined in (1) in such a way that $\mathbb{P}[\mathrm{Bin}(k,p)=s]=n^{-1+o(1)}$ .

Claim 2.2. Let $x\neq \frac{1}{2(1-\log _{1/p}2/\gamma )}$ be an arbitrary (not necessarily positive) real number,

\begin{equation*} k=2\log _{1/p}n-x\log _{1/p}\log n +O(1),\quad s=\gamma \log _{1/p}n-x\log _{1/p}\log n +O(1) \end{equation*}

be integers. Then

\begin{equation*} \mathbb {P}[\mathrm {Bin}(k,p)=s]=\frac {1}{n}\exp \biggl [\ln \ln n\left (x\left (1-\log _{1/p}\frac {2}{\gamma }\right )-\frac {1}{2}\right )(1+o(1))\biggr ]. \end{equation*}

Proof. Set $\varepsilon =x\frac{\ln \ln n}{\ln n}$ . Then

\begin{align*}\tiny\left(\begin{array}{c}k\\s\end{array}\right) & =\Theta \left (\frac{1}{\sqrt{\ln n}}\left (\frac{k^k}{s^s(k-s)^{k-s}}\right )\right ) =\Theta \left (\frac{1}{\sqrt{\ln n}}\left (\frac{k}{s}\right )^s\left (\frac{k}{k-s}\right )^{k-s}\right )\\ &= \Theta \left (\frac{1}{\sqrt{\ln n}}\left (\frac{2-\varepsilon }{\gamma -\varepsilon }\right )^{(\gamma -\varepsilon )\log _{1/p} n}\left (\frac{2-\varepsilon }{2-\gamma }\right )^{(2-\gamma )\log _{1/p} n}\right )\\ &=\exp \left [\log _{1/p} n\left ((\gamma -\varepsilon )\ln \frac{2-\varepsilon }{\gamma -\varepsilon }+(2-\gamma )\ln \frac{2-\varepsilon }{2-\gamma }\right )-\frac{1}{2}\ln \ln n(1+o(1)))\right ]. \end{align*}

Since

\begin{align*} (\gamma -\varepsilon )\ln \frac{2-\varepsilon }{\gamma -\varepsilon } & +(2-\gamma )\ln \frac{2-\varepsilon }{2-\gamma } = (\gamma -\varepsilon )\ln \frac{2}{\gamma }+(\gamma -\varepsilon )\ln \left (1-\frac{\varepsilon }{2}\right )\\& -(\gamma -\varepsilon )\ln \left (1-\frac{\varepsilon }{\gamma }\right )+(2-\gamma )\ln \frac{2}{2-\gamma }+(2-\gamma )\ln \left (1-\frac{\varepsilon }{2}\right )\\[4pt]& =\gamma \ln \frac{2}{\gamma }+(2-\gamma )\ln \frac{2}{2-\gamma }-\varepsilon \ln \frac{2}{\gamma }+O(\varepsilon ^2), \end{align*}

we get that

\begin{align*}\tiny\left(\begin{array}{c}k\\s\end{array}\right) &=\exp \left [\log _{1/p} n\left (\gamma \ln \frac{2}{\gamma }+(2-\gamma )\ln \frac{2}{2-\gamma }-\varepsilon \ln \frac{2}{\gamma }\right )-\frac{1}{2}\ln \ln n(1+o(1))\right ]\\[4pt] &\stackrel{(1)}=\exp \left [\log _{1/p} n\left ((2-\gamma )\ln \frac{1}{1-p}-(1-\gamma )\ln \frac{1}{p}-\varepsilon \ln \frac{2}{\gamma }\right )-\frac{1}{2}\ln \ln n(1+o(1))\right ].\\ \end{align*}

Thus,

\begin{align*} \mathbb{P}[\mathrm{Bin}(k,p)=s]&=\tiny\left(\begin{array}{c}k\\s\end{array}\right)p^s(1-p)^{k-s}\\[4pt] &=\exp \biggl [\ln n\left ((2-\gamma )\log _{1/p}\frac{1}{1-p}+\gamma -\varepsilon \log _{1/p}\frac{2}{\gamma }-1\right )\\[4pt] &\quad \quad \quad \quad \quad -s\ln \frac{1}{p}-(k-s)\ln \frac{1}{1-p}-\left (\frac{1}{2}+o(1)\right )\ln \ln n\biggr ]\\[4pt] &=\frac{1}{n}\exp \biggl [\varepsilon \left (1-\log _{1/p}\frac{2}{\gamma }\right )\ln n-\frac{1}{2}\ln \ln n(1+o(1))\biggr ], \end{align*}

as needed.

Remark 2.1. In order to justify the choice of the constant factor in the second-order term in the definition of $k_-$ , let us observe that (1) implies

\begin{equation*} \log _{1/p}\frac {2}{\gamma }=1-\frac {1}{\gamma }-\frac {2-\gamma }{\gamma }\log _{1/p}\frac {2(1-p)}{2-\gamma }\lt \frac {1}{2} \end{equation*}

since $\gamma \lt 2$ and $\frac{2(1-p)}{2-\gamma }\gt 1$ . Thus, we get that there exists $\varepsilon =\varepsilon (p)\gt 0$ such that

\begin{align*} \mathbb{P}[\mathrm{Bin}(k,p)=s]\geq \frac{1}{n}\exp [\ln \ln n(x+\varepsilon -1-o(1))/2],\quad \textrm{if }x\gt 0,\\ \mathbb{P}[\mathrm{Bin}(k,p)=s]\leq \frac{1}{n}\exp [\ln \ln n(x-\varepsilon -1-o(1))/2],\quad \textrm{if }x\lt 0. \end{align*}

2.2. Lower bound

We let $n^{\prime}= \lfloor n/\ln n\rfloor$ and $k=k_-(n^{\prime},p)$ . Note that $k=(2-o(1))\log _{1/p}n$ . We then divide $[n]$ into two parts $V\sqcup U$ , where $V$ has size $n^{\prime}$ . Expose first the edges of $G_n$ spanned by $V$ . Due to Claim 2.1, whp there exists a set $V_0\subset V$ of size at most $n/\ln ^2 n$ such that $G(n,p)[V\setminus V_0]$ can be partitioned into cliques of size $k$ . Let $K_1,\ldots,K_m$ be such cliques, where

\begin{equation*} m=(1+o(1))\frac {n}{2\ln n\log _{1/p}n}. \end{equation*}

Let us now expose edges of $G_n$ between $V$ and $U$ . Let $s$ be the maximum integer such that

\begin{equation*} \mathbb {P}[\mathrm {Bin}(k,p)\geq s]\geq \mathbb {P}[\mathrm {Bin}(k,p)=s]\geq \frac {\ln ^4 n}{n}. \end{equation*}

Then a vertex $u\in U$ has at least $s$ neighbours in some $K_j$ , $j\in [m]$ , with probability at least

\begin{equation*} 1-\left (1-\frac {\ln ^4 n}{n}\right )^m=1- e^{-\Omega (\ln ^2n)}=1-o(1/n). \end{equation*}

By the union bound, whp, for every vertex $u\in U$ , there is $j\in [m]$ such that $u$ has at least $s$ neighbours in $K_j$ .

Consider a subgraph $H$ of $G_n$ obtained from the disjoint union of cliques $K_1,\ldots,K_m$ with vertices from $U$ , each sending $s$ edges to exactly one of the cliques. This graph has

\begin{equation*} m\tiny\left(\begin{array}{c}k\\2\end{array}\right)+s|U|\geq \gamma n\log _{1/p}n-9n\log _{1/p}\log n+O(n) \end{equation*}

edges due to Claim 2.2. Indeed, letting $x=9$ , we get from Claim 2.2, Remark 2.1, and the definition of $s$ that $s\gt \gamma \log _{1/p}n-9\log _{1/p}\log n +O(1)$ .

It remains to prove that $H$ is chordal. Consider any ordering of vertices of $H$ , where each vertex of $V$ precedes any vertex of $U$ . Obviously, this is a perfect elimination ordering implying that $H$ is chordal.

2.3. Upper bound

Let $k=k_+(n,p)$ . Letting

\begin{equation*} s=\gamma \log _{1/p}n+3\log _{1/p}\log n +O(1) \end{equation*}

to be an integer, we get that

(2) \begin{equation} \mathbb{P}[\mathrm{Bin}(k,p)=s]\leq \mathbb{P}[\mathrm{Bin}(\lfloor k+3\log _{1/p}\log n\rfloor,p)=s]\leq \frac{1}{n}\exp [-2\ln \ln n]. \end{equation}

The first inequality holds true since $\gamma \gt 2p$ and thus $\mathbb{P}[\mathrm{Bin}(y,p)=s]$ increases in $y$ on $[s,k+\delta (k)]$ for any choice of $\delta (k)=o(k)$ : indeed

\begin{equation*} \frac {\mathbb {P}[\mathrm {Bin}(y+1,p)=s]}{\mathbb {P}[\mathrm {Bin}(y,p)=s]}=\frac {(y+1)(1-p)}{y+1-s}\gt \frac {k(1+o(1))(1-p)}{k(1+o(1))-s}=\frac {2(1-p)+o(1)}{2-\gamma }\gt 1. \end{equation*}

The second inequality in (2) holds true due to Claim 2.2 and Remark 2.1. Let

\begin{equation*} \ell =sn=\gamma n\log _{1/p}n+3n\log _{1/p}\log n +O(n). \end{equation*}

We will prove that whp the maximum number of edges in a chordal subgraph of $G_n$ is less than $\ell$ , and this immediately implies the desired upper bound.

Let $H$ be a chordal graph, and assume we are given a perfect elimination ordering $v_1\prec \ldots \prec v_n$ of the vertices of $H$ . For every vertex $v_i$ , let $d(v_i)$ be the outdegree of $v_i$ , that is, the number of outgoing neighbours of $v_i$ , and let $K_{v_i}$ be the clique induced by $v_i$ and its outgoing neighbours. For every $i$ such that $d(v_i)\gt 0$ , let $\nu (v_i)$ be the last outgoing neighbour of $v_i$ in the given perfect elimination ordering. Consider a graph $T_{\prec }(H)$ on the vertex set $V(H)$ consisting of all edges $\{v_i,\nu (v_i)\}$ . Note that $T_{\prec }(H)$ is 1-degenerate; thus it is a forest.

We further assume that $H$ is connected. In this case, for any perfect elimination ordering $\prec$ , we have that $T=T_{\prec }(H)$ is a tree. Indeed, it is sufficient to show that, for every pair of vertices $v_j\prec v_i$ forming an edge of $H$ , there is a path in $T$ from $v_i$ to $v_j$ . If $\nu (v_i)=v_j$ , then $\{v_j,v_i\}$ is an edge of $T$ itself. Otherwise, $v_j\prec \nu (v_i)\prec v_i$ , $\{v_i,\nu (v_i)\}$ is an edge of both $T$ and $H$ , and thus $\{v_j,\nu (v_i)\}$ is an edge of $H$ . By induction, we eventually will get a path from $v_i$ to $v_j$ in $T$ . Let us call $T$ a perfect elimination tree of $H$ , and let $v_1$ be the root of $T$ . Note that any layer-preserving ordering of the vertices of $T$ (i.e. vertices that are further from the root in $T$ occur later in this ordering) is a perfect elimination ordering of $H$ .

For any rooted tree $T$ on $[n]$ , there is at least one connected chordal graph $H$ such that $T$ is its perfect elimination tree (actually we may take $H=T$ ). In order to recover an $H$ (uniquely) from $T$ , we also equip $T$ with additional data: assume that $v_1\prec \ldots \prec v_n$ is a layer-preserving ordering of the vertices of $T$ and assign to each vertex $v_i$ , $i\geq 2$ , a vector $\textbf{e}_i\in \{0,1\}^{k_i}$ that encodes the outgoing neighbourhood of the vertex $i$ in $H$ , where $k_i=|K_{\nu (v_i)}|-1$ (see Fig. 1). Thus, in $H$ , the vertex $i$ is adjacent to a vertex $j\lt \nu (v_i)$ if and only if $j\in K_{\nu (v_i)}$ and the respective coordinate of $\textbf{e}_i$ equals 1. Note that $k_2=0$ , and for every $i\geq 3$ , vectors $\textbf{e}_j$ , $j\leq i-1$ , define $k_i$ uniquely.

Figure 1.

Reconstruction of $H$ from $T$ .

We can now estimate the expected number of rooted trees $T$ on $[n]$ such that in $G_n$ there exists a connected chordal subgraph $H$ with the following properties:

• $H$ has at least $\ell$ edges,

• $H$ does not have cliques of size $k$ ,

• $T$ is a perfect elimination tree of $H$ .

For our goal, it is sufficient to prove that this expectation approaches 0 due to Claim 2.1.1. In particular, since whp $G_n$ is connected, a chordal subgraph with the maximum number of edges is also connected (any disconnected chordal subgraph on $[n]$ can be supplemented by an edge between any pair of connected components); thus the connectivity assumption does not cause a loss of generality. There are $n^{n-1}$ ways to construct a rooted tree $T$ . Take a rooted tree $T$ and consider any layer-preserving ordering $\prec$ . Without loss of generality, we assume that this ordering is defined by the identity permutation on $[n]$ . Thus, the desired expectation is bounded from above by $\rho n^{n-1}$ , where $\rho$ is the maximum (over $T$ ) probability that $G_n$ has a chordal subgraph $H$ with $T=T_{\prec }(H)$ , at least $\ell$ edges, and without cliques of size $k$ . Let us expose the edges of $G_n$ in the following order that $\prec$ induces on the pairs $u\lt v$ of vertices in $[n]$ : $(u,v)\lt (u^{\prime},v+1)$ for any $u^{\prime}$ , and $(u,v)\lt (u+1,v)$ . For any $v\geq 2$ , as soon as $G_n[\leq v-1]$ is exposed, the outdegree of $v$ to the clique $K_{\nu (v)}$ (note that $\nu (v)$ is defined by $T$ ) has binomial distribution with $k_{\nu }(v)+1$ trials, and $k_{\nu }(v)+1$ is uniquely defined by $G_n[\leq v-1]$ . Let us also recall that each $k_{\nu (v)}+1$ should be at most $k$ . We then get that $d(2)+\ldots +d(n)$ is stochastically dominated by the sum of $n-1$ independent $\mathrm{Bin}(k,p)$ random variables implying

\begin{align*} \rho \leq \mathbb{P}[d(2)+\ldots +d(n)\geq \ell ]\leq \mathbb{P}[\mathrm{Bin}(kn,p)\geq \ell ]&\stackrel{(*)}\leq kn\mathbb{P}[\mathrm{Bin}(kn,p)=\ell ]\\ &= kn\tiny\left(\begin{array}{c}kn\\\ell\end{array}\right)p^{\ell }(1-p)^{kn-\ell }. \end{align*}

The inequality (*) holds since $\ell \gt (1+\varepsilon )knp$ for a sufficiently small constant $\varepsilon \gt 0$ (that follows immediately from $p\lt \gamma/2$ ), and so $\tiny\left(\begin{array}{c}kn\\x\end{array}\right)\left (\frac{p}{1-p}\right )^x$ decreases on $[\ell,kn]$ (see, e.g. [Reference Bollobás4, Chapter 1.2]). Let us note that, for all $n$ large enough,

\begin{align*}\tiny\left(\begin{array}{c}kn\\\ell\end{array}\right)=\tiny\left(\begin{array}{c}kn\\sn\end{array}\right) &=\frac{(1+o(1))\sqrt{k}}{\sqrt{2\pi s(k-s)n}}\cdot \frac{(kn)^{kn}}{(sn)^{sn}(kn-sn)^{n(k-s)}}\\ & \leq \left (\frac{k^k}{s^s(k-s)^{k-s}}\right )^n \\ &=\left ((1+o(1))\frac{\sqrt{2\pi s(k-s)}}{\sqrt{k}}\tiny\left(\begin{array}{c}k\\s\end{array}\right)\right )^n \leq \left (\tiny\left(\begin{array}{c}k\\s\end{array}\right)\ln n\right )^n \end{align*}

implying that

\begin{equation*} \rho \leq kn(\ln n)^n\biggl (\mathbb {P}[\mathrm {Bin}(k,p)=s]\biggr )^n. \end{equation*}

Thus, the desired expectation is upper bounded by

\begin{equation*} k (n\ln n)^n \biggl (\mathbb {P}[\mathrm {Bin}(k,p)=s]\biggr )^n\leq k\exp [-n\ln \ln n]=o(1). \end{equation*}

The last inequality follows from (2), completing the proof.

4.1. $\alpha \gt 1/2$

If $\frac{1}{n}\ll p\ll n^{-2/3}$ , then whp $G_n$ has a connected component of size $n(1-o(1))$ (see, e.g. [Reference Bollobás4, Reference Janson, Łuczak and Ruciński23]) implying that whp $X_n\geq n-o(n)$ (a tree that covers almost all vertices of $G_n$ is chordal). On the other hand, the expected number of triangles is $o(n)$ , and the expected number of 4-cliques is $o(1)$ implying that whp there are $o(n)$ triangles and no 4-cliques by Markov’s inequality. Then, whp for any chordal graph $H\subset G_n$ and its perfect elimination ordering, no vertices have outdegree 3, and $o(n)$ vertices have outdegree 2. Thus, whp $X_n\leq n+o(n)$ . We eventually get that $X_n/n\stackrel{\mathbb{P}}\to 1$ .

Other $\alpha \gt \frac{1}{2}$ can be handled similarly: let $n^{-\frac{1+k}{1+2k}}\ll p \ll n^{-\frac{2+k}{3+2k}}$ for some integer $k\geq 1$ . Let $H$ be the square of a path of length $k+1$ . Consider a graph $F=F(M)$ obtained by gluing sequentially $M$ copies of $H$ : the first vertex of the $i$ -th copy is glued with the last vertex of the $(i-1)$ -th copy. Note that $F$ is chordal and its maximum 1-density equals $\rho ^*(F)=\rho ^*(H)=\frac{1+2k}{1+k}$ . By Theorem 4.1, whp there exists an almost $F$ -tiling in $G_n$ . In other words, whp there is a disjoint union of copies of $F$ (which is chordal) that cover $n(1-o(1))$ vertices of $G_n$ . Clearly, for every $\varepsilon \gt 0$ , there exists $M$ such that an almost $F$ -tiling has at least $(\frac{1+2k}{1+k}-\varepsilon )n$ edges for $n$ large enough. Thus, whp $X_n/n\geq \frac{1+2k}{1+k}-o(1)$ .

It remains to show that whp $X_n/n\leq \frac{1+2k}{1+k}+o(1)$ . For every positive integer $i$ , let $\mathcal{H}_i$ be the family of all chordal graphs on $0\sqcup [i]$ that can be obtained in the following way:

• $0$ and $1$ are adjacent,

• for every $j\in [i-1]$ , the vertex $j+1$ is adjacent to at least 2 vertices in $0\sqcup [j]$ ,

• for every $j\in [i-1]$ , the neighbours of $j+1$ in $0\sqcup [j]$ , form a clique.

Note that each graph from $\mathcal{H}_i$ has $i+1$ vertices and at least $2i-1$ edges. Set $\mathcal{H}=\sqcup _{i\geq 1}\mathcal{H}_i$ .

Note that, for $i\gt i^{\prime}$ , if $H\in \mathcal{H}_i$ , then $H[\leq i^{\prime}]\in \mathcal{H}_{i^{\prime}}$ . Recall that every graph from $\mathcal{H}_i$ has at least $2i-1$ edges. Thus, by Markov’s inequality, there exists $M=M(k)$ such that whp $G_n$ does not have a copy of any graph from $\mathcal{H}_M$ , and, therefore, whp $G_n$ does not have a copy of any graph from $\mathcal{H}_{\geq M}$ . Let $\mathcal{H}^{\prime}\subset \sqcup _{i=1}^{k+1}\mathcal{H}_i$ be the set of all graphs $H$ from $\sqcup _{i=1}^{k+1}\mathcal{H}_i$ that have at least $2|V(H)|$ edges (i.e. $H\in \mathcal{H}_i$ belongs to $\mathcal{H}^{\prime}$ if and only if there exists $j\leq i$ sending at least 3 edges to $0\sqcup [j-1]$ in $H$ ). Let $\mathcal{H}^{\prime\prime}=\sqcup _{i=k+2}^{M-1}\mathcal{H}_i$ . We then observe that, for every graph $H\in \mathcal{H}^{\prime}\sqcup \mathcal{H}^{\prime\prime}$ , the expected number of subgraphs isomorphic to $H$ in $G_n$ is $o(n)$ : a graph $H\in \mathcal{H}^{\prime}$ on $i+1$ vertices has at least $2i$ edges, and thus the expected number of copies of $H$ in $G_n$ is $O(n^{i+1}n^{-2i\alpha })=O(n^{1+i(1-2\alpha )})=o(n)$ . In the same way, a graph $H\in \mathcal{H}^{\prime\prime}$ on $i+1$ vertices has at least $2i-1$ edges; thus the expected number of copies of $H$ in $G_n$ is $O(n^{i+1}n^{-(2i-1)\alpha })=O(n^{1+(k+2)-(2k+3)\alpha })=o(n)$ . By Markov’s inequality, whp there are $o(n)$ subgraphs isomorphic to a graph from $H\in \mathcal{H}^{\prime}\sqcup \mathcal{H}^{\prime\prime}$ in $G_n$ .

Now, let us consider an arbitrary chordal subgraph $F\subset G_n$ , and let $\Sigma$ be the set of all its non-empty blocks (i.e. inclusion-maximal 2-connected subgraphs and edges that do not belong to any cycle, see [Reference Diestel13, Chapter 3.1]). Without loss of generality, we may assume that $F$ is connected (in particular, there are no empty blocks) since whp $G_n$ is connected, and so the same is true for any its inclusion-maximal chordal subgraph. Note that every edge of $F$ belongs to some graph in $\Sigma$ . By Claim 4.2, we get that whp all graphs from $\Sigma$ have copies in $\sqcup _{i=1}^{M-1}\mathcal{H}_i$ . Moreover, whp the total number of edges in the graphs from $\Sigma$ that have copies in $\mathcal{H}^{\prime}\sqcup \mathcal{H}^{\prime\prime}$ is $o(n)$ . For every $H\in \Sigma$ that has a copy in $\sqcup _{i=1}^{k+1}\mathcal{H}_i\setminus \mathcal{H}^{\prime}$ , we get that $H$ has at most $k+2$ vertices, and $|E(H_i)|=2|V(H_i)|-3$ . Since the block-cutpoint graph is a tree, we get that there is an ordering of graphs from $\Sigma =\{H_1,\ldots,H_K\}$ such that, for every $i$ , $H_i$ has at most 1 common vertex with $H_1\cup \ldots \cup H_{i-1}$ . So, $|E(H_i)\setminus E(H_1\cup \ldots \cup H_{i-1})|=|E(H_i)|$ while $|V(H_i)\setminus V(H_1\cup \ldots \cup H_{i-1})|\geq |V(H_i)|-1=i$ . Letting $x_H$ be the number of graphs from $\Sigma$ isomorphic to $H$ , we get that whp

\begin{equation*} \rho (F)\leq \frac {o(n)+\sum _{i=1}^{k+1}\sum _{H\in \mathcal {H}_i\setminus \mathcal {H}^{\prime}}(2i-1)x_H}{\sum _{i=1}^{k+1}\sum _{H\in \mathcal {H}_i\setminus \mathcal {H}^{\prime}}ix_H}\leq \frac {1+2k}{1+k}+o(1). \end{equation*}

The latter inequality hods true since $F$ is connected implying that the denominator in the left-hand side fraction is linear in $n$ . Thus, the number of edges in $F$ is at most $\frac{1+2k}{1+k}n+o(n)$ , completing the proof.

4.2. $\alpha \leq 1/2$

We first prove the upper bound: let us fix $\varepsilon \gt 0$ and prove that whp $X_n\leq (1/\alpha +\varepsilon )n$ . Since whp $G_n$ does not have $(\lceil 2/\alpha \rceil +1)$ -cliques, whp in a chordal subgraph of $G_n$ every vertex has outdegree at most $\lceil 2/\alpha \rceil$ . Moreover, arguing similarly to the proof of the upper bound of Theorem 1 from Section 2.3, we see that the expected number of chordal subgraphs of $G_n$ with at least $(1/\alpha +\varepsilon )n$ edges and outdegrees at most $\lceil 2/\alpha \rceil$ (for some perfect elimination ordering) can be upper bounded by

\begin{equation*} n^{n-1}A^n p^{(1/\alpha +\varepsilon )n}\leq (An^{1-1-\alpha \varepsilon +o(1)})^n=o(1) \end{equation*}

for some constant $A\gt 0$ . Here, $n^{n-1}$ is the number of rooted trees that can play a role of a perfect elimination tree and $A^n$ is the bound for the number of ways to choose vectors $\textbf{e}_i$ (see Section 2.3). Thus, by Markov’s inequality, we get that whp $X_n/n\leq 1/\alpha +o(1)$ as needed.

We then prove the lower bound. Let us first assume that $1/\alpha =\ell \geq 2$ is an integer. For every $j\in \mathbb{N}$ , consider the following chordal graph $F_j$ on $\{0\}\sqcup [j]$ : the vertex $i\in [j]$ is adjacent to its $\min \{j-1,\ell \}$ predecessors, that is, $F_j$ is the $\ell$ -th power of a path of length $\ell$ . Note that $F_j$ is a strictly 1-balanced graph with 1-density strictly less than $\ell$ and approaching $\ell$ with growing $j$ . By Theorem 4.1, for every $j$ , whp $G_n$ has an almost $F_j$ -tiling, implying that $X_n/n\geq (1/\alpha -o(1))$ whp.

Now, let $1/\alpha \notin \mathbb{Z}$ . Let $\ell$ be the minimum positive integer greater than $1/\alpha$ . Note that $\ell \geq 3$ . Let us consider the (unique) sequence $x_i$ , $i\in \mathbb{N}$ , satisfying the following conditions:

• $(x_1,\ldots,x_{\ell })=(1,2,\ldots,\ell )$ ;

• for every $i\geq \ell +1$ , we define recursively $x_i$ to be the maximum integer in $[2,\ell ]$ such that

  \begin{equation*} \rho _i:=\frac {x_1+\ldots +x_i}{i}\lt \frac {1}{\alpha }. \end{equation*}

Let us consider the sequence $(s_j)_{j\in \mathbb{N}}$ of all $s$ such that

• $\rho _s\gt \ell -1$ ,

• $\rho _s\gt \rho _i$ for all $i\lt s$ .

Since $\rho _i\uparrow 1/\alpha \gt \ell -1$ , there are infinitely many such $s$ . For every $j$ , set $\textbf{x}_j=(x_1,\ldots,x_{s_j})$ . Consider the following graph $F_j$ on $\{0\}\sqcup [s_j]$ : the vertex $i\in [s_j]$ is adjacent to its $x_i$ immediate predecessors. Note that $\rho _{s_j}$ is the 1-density of $F_j$ . Let us prove that $F_j$ is chordal. It is sufficient to show that the natural order of integers is perfect elimination. We first observe that, for every $i\gt \ell$ , we have $x_i\in \{\ell -1,\ell \}$ . Indeed, the inequality $x_1+\ldots +x_{i-1}\lt \frac{1}{\alpha }(i-1)$ that holds for all $i\gt \ell$ , implies $x_1+\ldots +x_{i-1}+\ell -1\lt \frac{1}{\alpha }i$ since $\ell -1\leq \frac{1}{\alpha }$ . Also, the first $\ell +1$ vertices of $F_j$ compose a clique. Proceeding by induction, we get that any set of $\ell$ consecutive vertices in $F_j$ induces a clique. Thus, every $i\geq \ell +1$ is adjacent to at most its $\ell$ predecessors, and these predecessors induce a clique. So the considered order is indeed perfect elimination.

Since the 1-density of $F_j$ approaches $1/\alpha$ as $j$ grows, then, for every $\varepsilon \gt 0$ and large enough $j$ , an almost $F_j$ -tiling has at least $(1/\alpha -\varepsilon )n$ edges. Therefore, we conclude that $X_n/n\geq (1/\alpha -o(1))$ by applying Theorem 4.1 which is possible since all $F_j$ are strictly 1-balanced due to the following claim. This completes the proof of Theorem 2.

Proof. Fix $j$ and assume that $F_j$ is not strictly 1-balanced. Let $\tilde F$ be a proper subgraph of $F_j$ that has 1-density $\rho (\tilde F)\geq \rho _{s_j}$ . First of all, let us note that without loss of generality $\tilde F$ is connected since the 1-density of $\tilde F$ does not exceed 1-densities of all its connected components. Also, we may assume that $\tilde F$ is an induced subgraph.

Let us assume that in $\tilde F$ some intermediate vertices are missing, that is, there exist $i_-,i,i_+\in V(F_j)$ such that $i_-\lt i\lt i_+$ and $i_-,i_+\in V(\tilde F)$ , while $i\notin V(\tilde F)$ . Note that the number of consecutive missing vertices could not be bigger than $\ell -1$ since otherwise $\tilde F$ is not connected. Let $i\geq 1$ be the minimum number such that, for some $\mu \in [\ell -1]$ ,

\begin{align*} i,i+\mu +1\in V(\tilde F),\,\,\textrm{while }\,i+1,\ldots,i+\mu \notin V(\tilde F). \end{align*}

If $i\geq \ell -1$ , then the missing vertices from $[i+1,i+\mu ]$ add at least $\mu \ell \gt \mu/\alpha$ edges to $\tilde F$ . Indeed, every missing vertex sends at least $\ell -1$ edges to its immediate predecessors in $F_j$ and is adjacent to $i+\mu +1$ . Thus, the inequality $\rho (F_j)\leq \rho (\tilde F)$ implies $\rho (F_j)\lt \rho (\tilde F^{\prime})$ , where the graph $\tilde F^{\prime}$ is obtained from $\tilde F$ by adding back the vertices $i+1,\ldots,i+\mu$ . If $i\lt \ell -1$ , then $i+1$ vertices $i^{\prime}\leq i$ contribute at most $(\ell -1)(i+1)$ edges to $E(\tilde F)$ . Indeed, every vertex $i^{\prime}\leq i$ is adjacent to at most $\ell$ and at least $\ell -1$ its immediate successors in $F_j$ . Since the missing vertex $i+1$ is adjacent to all $i^{\prime}\leq i$ in $F_j$ , we get that every $i^{\prime}\leq i$ is adjacent to at most $\ell -1$ its successors in $\tilde F$ . Thus, the deletion of vertices $i^{\prime}\leq i$ from $\tilde F$ leads to the graph $\tilde F^{\prime}$ with $\rho (\tilde F^{\prime})\gt \rho (F_j)$ . We conclude that there is a proper subgraph in $F_j$ with the 1-density at least $\rho (F_j)$ and without missing intermediate vertices. Thus, without loss of generality, we may assume that $\tilde F$ is induced by $[i_-,i_+]$ for certain $0\lt i_-\lt i_+\leq s_j$ . Note that the first inequality is strict since otherwise we get a contradiction with the definition of $F_j$ .

Let $\nu \gt \ell$ be the minimum number such that $x_{\nu }=\ell -1$ . We have that, among the first $\nu -1$ elements of the sequence $\textbf{x}_j$ , there are exactly $\nu -\ell$ numbers equal to $\ell$ . Assume first that $\nu -\ell +1\leq i_-\leq \nu -1$ . Then the missing vertices $0,1,\ldots,i_-1$ add at least $(\ell -1)(i_-(\nu -\ell ))+\ell (\nu -\ell )$ edges to $\tilde F$ . Note that $\frac{(\ell -1)(i_-(\nu -\ell ))+\ell (\nu -\ell )}{i_-}$ is minimised at $i_-=\nu -1$ :

\begin{equation*} \frac {(\ell -1)(i_-(\nu -\ell ))+\ell (\nu -\ell )}{i_-}\geq \frac {(\ell -1)^2+\ell (\nu -\ell )}{\nu -1}. \end{equation*}

We also recall that

\begin{equation*} \frac {|E(F_j[\leq \nu -1])|+\ell }{(|V(F_j[\leq \nu -1])|-1)+1}=\frac {\ell (\ell -1)/2+(\nu -\ell +1)\ell }{\nu }\geq \frac {1}{\alpha }, \end{equation*}

implying that

\begin{align*} (\ell -1)^2+\ell (\nu -\ell ) & =\nu \ell -2\ell +1\geq \frac{\nu }{\alpha }+\frac{\ell ^2-\ell }{2}-2\ell +1\\ &\gt \frac{1}{\alpha }(\nu -1)+(\ell -1)+\frac{\ell ^2-\ell }{2}-2\ell +1\\ &=\frac{\nu -1}{\alpha }+\frac{\ell ^2-3\ell }{2}\geq \frac{\nu -1}{\alpha }, \end{align*}

since $\frac{1}{\alpha }\gt \ell -1$ and $\ell \geq 3$ . So, the “additional density” recovered by the vertices $0,1,\ldots,i_-1$ equals

\begin{equation*} \frac {(\ell -1)(i_-(\nu -\ell ))+\ell (\nu -\ell )}{i_-}\gt \frac {1}{\alpha }. \end{equation*}

If $i_-\leq \nu -\ell$ , then the missing vertices $0,1,\ldots,i_-1$ add at least $i_-\ell$ edges to $\tilde F$ . Thus, in both cases, the addition of vertices $0,1,\ldots,i_-1$ leads to the graph $\tilde F^{\prime}$ satisfying the inequality $\rho (\tilde F^{\prime})\gt \rho (F_j)$ – contradiction with the property of $F_j$ that all $F_j[\leq s]$ , $s\lt s_j$ , have smaller 1-densities.

It remains to consider the case $i_-\geq \nu$ . Clearly, we may assume that $\tilde F$ has at least $\ell +1$ vertices since otherwise $\rho (\tilde F)\leq \frac{\ell }{2}\lt \ell -1\lt \rho _{s_j}$ due to the fact that $\ell \geq 3$ and the choice of $s_j$ . Due to the definition of $\textbf{x}_j$ , we have that, for every $i\geq \nu$ ,

\begin{equation*} \frac {i}{\alpha }-1\leq x_1+\ldots +x_i\lt \frac {i}{\alpha }. \end{equation*}

Let $\delta _i:=\frac{i}{\alpha }-(x_1+\ldots +x_i)$ . We then get

(3) \begin{align} \frac{i_+-i_-}{\alpha }-(x_{i_-+1}+\ldots +x_{i_+}) &= \frac{i_+}{\alpha }-(x_1+\ldots +x_{i_+})- \left (\frac{i_-}{\alpha }-(x_1+\ldots +x_{i_-})\right )\notag \\ &\geq \delta _{i_+}-1. \end{align}

On the other hand, the vertex $x_{i_-+1}$ sends $1\leq (x_{i_-+1}-(\ell -2))$ edge to $x_{i_-}$ in $\tilde F$ , the vertex $x_{i_-+2}$ sends $2\leq (x_{i_-+2}-(\ell -3))$ edges to $\{x_{i_-},x_{i_+}\}$ in $\tilde F$ , etc., implying

\begin{equation*} \rho (\tilde F) \leq \frac {x_{i_-+1}+\ldots +x_{i_+}-(1+\ldots +\ell -2)}{i_+-i_-} \leq \frac {x_{i_-+1}+\ldots +x_{i_+}-1}{i_+-i_-}. \end{equation*}

Therefore, due to (3), we get

\begin{align*} \rho (\tilde F) & \leq \frac{x_{i_-+1}+\ldots +x_{i_+}-1}{i_+-i_-}\\ &\leq \frac{(i_+-i_-)/\alpha -\delta _{i_+}+1-1}{i_+-i_-}\\ &=\frac{1}{\alpha }-\frac{\delta _{i_+}}{i_+-i_-}\lt \frac{1}{\alpha }-\frac{\delta _{i_+}}{i_+}\\ &=\rho (F_j[\{0,1,\ldots,i_+\}])\leq \rho (F_j) \end{align*}

by the definition of $F_j$ – a contradiction.