Proof of Theorem 8. Note that, since all of the $G^{(i)}$ are non-trivial, we have $t \leq d \leq Ct$ and $|G| \geq 2^t$ , and therefore,

\begin{equation*} t^{t^{\frac {1}{4}}} = o\left ( d^{d^{\frac {3}{10}}}\right ) = o(|G|). \end{equation*}

We argue via a two-round exposure. Set $p_2=\frac{1}{d\log d}$ and $p_1=\frac{p-p_2}{1-p_2}$ , so that $(1-p_1)(1-p_2)=1-p$ . Note that $G_p$ has the same distribution as $G_{p_1}\cup G_{p_2}$ and that $p_1=\frac{1+\epsilon -o(1)}{d}$ .

Set $I\;:\!=\; \left [\left (1-\frac{1}{\log d}\right )d, \left (1+\frac{1}{\log d}\right )d\right ]$ , and denote the set of vertices whose degree in $G$ lies in the interval $I$ by $V_1\subseteq V(G)$ . Note that the degree of a uniformly chosen vertex in $G$ is distributed as the sum of $t$ random variables, each of which is bounded by $C$ . Since the expected degree of a uniformly chosen vertex is the average degree $d$ , we have by Lemma 11 that $|V\setminus V_1|\le 4\exp \left (-\frac{d}{\log ^2d}\right )|G|.$

Let $W$ be the set of vertices belonging to components of order at least $d^{d^{\frac{3}{10}}}$ in $G_{p_1}$ . Since for every $v\in V_1$ we have that $d_G(v)\cdot p_1=1+\epsilon -o(1)$ , we have by Lemma 16 that

(8) \begin{align} \text{every } v\in V_1 \text{ belongs to } W \text{ with probability at least } y(\epsilon )-o(1). \end{align}

We assume henceforth that the above stated whp statements of $G_{p_1}$ hold. Note furthermore that by adding or deleting an edge from $G_{p_1}$ we can change the number of vertices in $W$ by at most $2d^{d^{\frac{3}{10}}} = o\left ( \frac{|G|}{d}\right )$ . Hence, by Lemma 11 applied to the edge-exposure martingale on $G_{p_1}$ of length $\frac{|G| d}{2}$ , we see that whp $|W| = \left (y(\epsilon )+o(1)\right )|G|$ . Note that, by Lemma 17, whp there are at most $\exp \left (-d^{\frac{3}{2}}\right )|G|$ vertices at distance (in $G$ ) greater than $2$ from $W$ .

Let $A\cup B$ be a partition of $W$ into two parts, $A$ and $B$ , with $|A|\le |B|$ and $|A|\ge \frac{|W|}{t}$ . Let $A'$ be the set of vertices in $A$ together with all vertices in $V(G)\setminus B$ at distance at most $2$ from $A$ , and let $B'$ be the set of vertices in $B$ together with all vertices in $V(G)\setminus A'$ at distance at most $2$ from $A$ . By Lemma 8, we have

\begin{equation*}i(G)\ge \frac {1}{2}\min _{j\in [t]}i\left (G^{(j)}\right )\ge \frac {1}{2}t^{-t^{\frac {1}{4}}}.\end{equation*}

Let $E'\;:\!=\; E(A', (A')^C).$ We thus have that

(9) \begin{align} |E'| =e(A', (A')^C) \ge \frac{|W|}{t} \cdot \frac{1}{2}t^{-t^{\frac{1}{4}}} = \frac{|W|}{2}t^{-t^{\frac{1}{4}}-1} =\Omega \left (\exp \left (-t^{\frac{1}{4}}\log t\right )|G|\right ). \end{align}

Recall that whp $|V(G) \setminus V_1|\le 4\exp \left (-\frac{d}{\log ^2d}\right )|G|$ and $|V(G) \setminus N_G^2(W)| \leq \exp \left (-d^{\frac{3}{2}}\right )|G|$ , where $N_G^2(W)$ is the set of vertices at distance at most two from $W$ . Thus, whp

\begin{equation*} |V(G) \setminus \left (V_1\cap \left (A'\cup B'\right )\right )| \leq 5\exp \left (-\frac {d}{\log ^2d}\right )|G|. \end{equation*}

Therefore, whp the number of edges in $E'$ which do not meet $B'$ is at most

\begin{align*} 2\Delta (G)\cdot 5\exp \left (-\frac{d}{\log ^2d}\right )|G|\le 10Ct\exp \left (-\frac{d}{\log ^2d}\right )|G|=o(|E'|). \end{align*}

Hence, whp at least $\frac{|E'|}{2}$ of the edges in $E'$ have their end-vertices in $B'$ .

Since each vertex in $A'$ is at distance at most $2$ from $A$ , and similarly every vertex in $B'$ is at distance at most $2$ from $B$ , we can extend these edges to a family of $\frac{|E'|}{2}$ paths of length at most $5$ between $A$ and $B$ . Since every edge participates in at most $5\Delta (G)^4$ paths of length $5$ , we can greedily thin this family to a set of

\begin{equation*} \frac {|E'|}{50\Delta (G)^4} \geq {|W|}{t^{-t^{\frac {1}{4}} - 6}} \end{equation*}

edge-disjoint paths of length at most $5$ , where the inequality follows from (9) and the fact that $\Delta (G) \leq Ct$ .

We can thus apply Lemma 18, with $H=G_{p_1}, c=\frac{1}{t}, X=W, p=p_2, k=d^{d^{\frac{3}{10}}},r=t^{t^{\frac{1}{4}}+6}$ , and $m=5$ . Indeed, by our assumptions on $G_{p_1}$ , we have that whp both conditions in Lemma 18 hold, and

\begin{align*} rp^{-m} &=t^{t^{\frac{1}{4}}+6} p_2^{-5} \leq t^{t^{\frac{1}{4}}+6}\left (d\log d\right )^5 = o\left (d^{d^{\frac{3}{10}}}\right ) = o(k). \end{align*}

Hence, whp $G_p$ contains a component $L_1$ with at least $\left (1-\frac{1}{t}\right )|W|= (y(\epsilon )-o(1))|G|$ vertices from $W$ .

As for the remaining components, observe that by (2) we can couple the first $\sqrt{d}$ steps of a BFS exploration process in $G_p$ starting from $v\in V_1$ from above by a Galton-Watson branching process with offspring distribution $\text{Bin}\left (\left (1+\frac{1}{\log d}\right )d + C\sqrt{d},p\right )$ . Then, by standard results on Galton-Watson trees (see, for example, [Reference Durrett10, Theorem 4.3.12]), we have that

(10) \begin{align} \forall v\in V_1\colon \mathbb{P}\left [|C_v(G_p)|\ge \sqrt{d}\right ]\le y(\epsilon )+o(1). \end{align}

Let $W'$ be the set of vertices of $V_1$ which lie in components of order larger than $\sqrt{d}$ in $G_p$ , noting that $W \subseteq W'$ .

Then, by (10), $\mathbb{E}|W'| \leq \left (y(\epsilon )+o(1)\right )|V_1|$ . Furthermore, by (8), $\mathbb{E}(|W'|) \geq \mathbb{E}(|W|) \geq \left (y(\epsilon )-o(1)\right )|V_1|$ , and so $\mathbb{E}|W'|=\left (y(\epsilon )+o(1)\right )|V_1|$ . Hence, by a similar argument as before, by Lemma 11 we have that whp $|W'|= (y(\epsilon )+o(1))|V_1|$ .

In particular, every component of $G_p$ apart from $L_1$ either meets $V_1 \setminus W'$ , and so has order at most $\sqrt{d} = o(|G|)$ , or is contained in $(V(G) \setminus V_1) \cup (W' \setminus L_1)$ , and so has order at most

\begin{equation*} |V(G) \setminus V_1| + |W' \setminus L_1| = o(|G|), \end{equation*}

as required.

Proof of Lemma 19. Throughout the proof, we will use ideas similar to those of Lemma 16, while refining them to our particular setting. Let $M_{z-1}$ be the set of vertices with $z-1$ centre coordinates in $G$ . Let $\Gamma \;:\!=\; \Gamma _{z}$ be the bipartite biregular induced subgraph of $G$ between the layers $M_{z}$ and $M_{z-1}$ . Throughout the proof, we will in fact consider the percolated graph $\Gamma _p\subseteq G_p$ .

We prove by induction on $k$ , over all relevant values of $z$ , $\epsilon$ and $\delta$ .

For $k=1$ , let us first describe the following variant of the BFS algorithm. Here, we maintain the following sets of vertices:

• • $Q$ , the vertices are currently exploring, kept in a queue;

• • $S_1 \cup S_2$ , the vertices already processed in the exploration process;

• • $T_1 \cup T_2$ , the unvisited vertices.

We begin with $Q=\{v\}$ , $S_1, S_2=\emptyset$ and $T_1=M_z\setminus \{v\}$ , $T_2=M_{z-1}$ . At each iteration, we proceed as follows. We consider the first vertex $u\in Q$ , and explore its neighbourhood in $T_2$ in $G_p$ , that is, $N_{G_p}(u, T_2)$ . For each vertex $x\in N_{G_p}(u, T_2)$ in turn, we do the following:

1. - We move $x$ to $S_2$ ;

2. - We expose the neighbours of $x$ in $T_1$ in $G_p$ , that is, $N_{G_p}(x,T_1)$ ;

3. - We move the vertices $y\in N_{G_p}(x,T_1)$ from $T_1$ to $Q$ one by one.

Finally, after exploring all relevant vertices $x\in N_{G_p}(u, T_2)$ , we move $u$ from $Q$ to $S_1$ . See Figure 2 for an illustration. Note that, by taking an edge from the vertex $u$ to each vertex $x \in N_{G_p}(u, T_2)$ and an edge from each vertex $x \in N_{G_p}(u, T_2)$ to each vertex $y\in N_{G_p}(x,T_1)$ discovered in this way, we can think of this process as building a tree $B_1 \subseteq G_p$ which spans $S_1 \cup S_2 \cup Q$ at each stage of the process.

Figure 2. The vertex $u\in Q$ and its exposed neighbourhood in $T_2$ , and a vertex $x\in T_2$ and its exposed neighbourhood in $T_1$ . Note that in the course of the exploration process, all the vertices in $N_{G_p}(x,T_2)$ move from $T_1$ to $Q$ , the vertex $x$ (and, subsequently, all the vertices from $N_{G_p}(u,T_2)$ ) move to $S_2$ , and, finally, the vertex $u$ moves to $S_1$ .

We run the algorithm until either $Q$ is empty, or $|S_1\cup S_2\cup Q|=\sqrt{t}$ . Let us first note that at all times, $Q\cup S_1\cup T_1\subseteq M_z$ , and $S_2\cup T_2\subseteq M_{z-1}$ .

Observe that, since at all times $|S_1\cup S_2\cup Q|\le \sqrt{t}$ , we have that for every $u\in Q$ , $d(u, T_2)\ge zs-\sqrt{t}\ge \frac{zs}{2}$ $\left(\textrm{since}\; z\ge \frac{1+\delta }{10\sqrt{s}p}\ge \frac{(1+\delta )t}{10\sqrt{s}}\right),$ and for every $w\in T_2$ , $d(w, T_1)\ge t-z+1-\sqrt{t}\ge \frac{2t}{3}$ (since $z\le \frac{t}{\sqrt{s}}$ and $s$ is large enough). We can thus couple the tree $B_1$ constructed in our truncated algorithm with a Galton-Watson tree $B_2$ rooted at $v$ , such that for all $i$ , the offspring distribution at the $2i$ th generation (where we start at generation $0$ , with $v$ ) is $\text{Bin}\left (\frac{zs}{2},p\right )$ , and at the $(2i+1)$ th generation is $\text{Bin}\left (\frac{2t}{3},p\right )$ , such that $B_2$ is isomorphic to a subgraph of $B_1$ as long as $|B_1|\le \sqrt{t}$ .

Let us denote by $Z_i$ be the number of vertices at depth $i$ in $B_2$ (where $Z_0=1$ , and the vertex in the $0$ th depth is $v$ ). We then have by the above and by Lemma 10 that, for large enough $s$ ,

\begin{align*} \mathbb{P}\left [Z_{2(i+1)}\le \frac{zs^{\frac{2}{3}}x}{10t}\mid Z_{2i}=x\right ]&\le \mathbb{P}\left [Z_{2i+1}\le \frac{zs^{\frac{2}{3}}x}{t}\mid Z_{2i}=x\right ]+\mathbb{P}\left [Z_{2(i+1)}\le \frac{zs^{\frac{2}{3}}x}{10t}\mid Z_{2i+1}\ge \frac{zs^{\frac{2}{3}}x}{t}\right ]\\[10pt] &\le \mathbb{P}\left [\text{Bin}\left (\frac{zsx}{2},p\right )\le \frac{zs^{\frac{2}{3}}x}{t}\right ]+\mathbb{P}\left [\text{Bin}\left (\frac{2zs^{\frac{2}{3}}x}{3},p\right )\le \frac{zs^{\frac{2}{3}}x}{10t}\right ]\\[5pt] &\le \exp \left (\!-\!2x\right )+ \exp \left (-\frac{3x}{16}\right )\le \exp \left (-\frac{x}{100}\right ), \end{align*}

where the penultimate inequality follows since $p\ge \frac{1-\epsilon }{3t}$ and $zsp\ge \frac{(1+\delta )\sqrt{s}}{10}$ . Hence, we obtain:

\begin{align*} \mathbb{P}\left [\exists i \text{ such that } Z_{2i}\le \left (\frac{zs^{\frac{2}{3}}}{10t}\right )^i\right ]&\le \sum _{i=1}^{\infty }\mathbb{P}\left [Z_{2i}\le \left (\frac{zs^{\frac{2}{3}}}{10t}\right )^i \mid Z_{2(2i-1)}\ge \left (\frac{zs^{\frac{2}{3}}}{10t}\right )^{i-1}\right ]\\[5pt] &\le \sum _{i=1}^{\infty }\exp \left (-\frac{1}{100}\cdot \left (\frac{zs^{\frac{2}{3}}}{10t}\right )^i\right )\le \exp \left (-\frac{1}{200}\right ), \end{align*}

where the last inequality follows since $p\le \frac{1}{t}$ , and $z\ge \frac{(1+\delta )}{10\sqrt{s}p}\ge \frac{(1+\delta )t}{10\sqrt{s}}$ , and we take $s$ to be large enough. In particular, $\frac{zs^{\frac{2}{3}}}{10t}\gt 1$ , and hence the Galton-Watson tree grows to infinity with probability at least $\exp \left (-\frac{1}{200}\right )$ , and therefore with probability at least $\exp \left (-\frac{1}{200}\right )$ we have that $|B_2|=\sqrt{t}$ and hence $|S_1\cup S_2\cup Q|=\sqrt{t}$ at some point. We thus have that with the same probability, we have a connected set $W$ such that $v\in W$ , $W$ is connected in $G_p$ , and $|W|=\sqrt{t}$ . Let us argue further that, in fact, $|W\cap \left (S_1\cup Q\right )|\ge \frac{\sqrt{t}}{22}$ . Consider any vertex $w\in S_2$ . Crucially, observe that by the construction of the exploration process, unless $w$ is the last vertex we have moved such that now $|B_1|=\sqrt{t}$ , after we move $w$ from $T_2$ to $S_2$ we expose the neighbourhood of $w$ in $T_1$ (see Figure 2). If $|N_{G_p}(w, T_1)|\ge 1$ , then we added at least one child of $w$ to $B_1$ . The probability that $|N_{G_p}(w,T_1)|=0$ is at most

\begin{align*} (1-p)^{\frac{2t}{3}-\sqrt{t}}\le \left (1-\frac{1-\epsilon }{3t}\right )^{\frac{t}{2}}\le \exp \left (-\frac{1-\epsilon }{6}\right )\le \frac{9}{10}. \end{align*}

Hence the number of $w\in S_2$ which does not have at least one child, stochastically dominates $1+\text{Bin}\left (\sqrt{t},\frac{9}{10}\right )$ (where we use the fact that $|B_1|\le \sqrt{t}$ and our stochastic domination on the number of neighbours). Hence, by Lemma 11, the number of vertices $w\in T_2$ which do not have at least one child in $T_1$ in $B_1$ (that afterwards moves to $Q_1$ and perhaps subsequently to $S_1$ ) is whp at most $\frac{10\sqrt{t}}{11}$ . Therefore, there are at most $\frac{10\sqrt{t}}{11}$ vertices in $S_2$ which do not have a child in $B_1$ . Since $B_1$ is a tree, these children are distinct for distinct $w$ , and they all lie in $S_1$ or $Q$ . It follows that whp $|S_1\cup Q|\ge |S_2|-\frac{10\sqrt{t}}{11}$ , and hence with probability at least $c\gt 0$ , we have that $|S_1\cup Q|\ge \frac{\sqrt{t}}{22}$ , and hence $|W\cap M_{z}|\ge \frac{\sqrt{t}}{22}$ , completing the case of $k=1$ .

We proceed in a manner somewhat similar to the proof of Lemma 16. Let $k\ge 2$ and assume the statement holds with $c_1(z', k-1,\epsilon ',\delta ')$ and $c_2(z', k-1,\epsilon ',\delta ')$ for all relevant values of $z'$ , $\epsilon '$ and $\delta '$ . We argue via a two-stage exposure, with $p_2=\frac{1}{t\log t}$ and $p_1=\frac{p-p_2}{1-p_2}$ so that $(1-p_1)(1-p_2)=1-p$ . Note that $G_p$ has the same distribution as $G_{p_1}\cup G_{p_2}$ , and that $p_1\ge (1-o_t(1))p\ge \frac{1-\epsilon _1}{3t}$ for some $\epsilon _1\gt 0$ constant. Similarly, $z\ge \frac{1+\delta }{10\sqrt{s}p}\ge \frac{1+\delta _1}{10\sqrt{s}p_1}$ , and we can thus begin in the same manner as $k=1$ , and find with probability at least $c\gt 0$ , a set $W_0$ such that $v\in W_0$ , $W_0$ is connected in $G_p$ , $|W_0|=\sqrt{t}$ and $|W_0\cap M_z|\ge \frac{\sqrt{t}}{22}$ . We further note that $W_0\subseteq V(\Gamma )$ . We proceed under the assumption that indeed $|W_0|=\sqrt{t}$ . Let us enumerate the vertices in $W_0\cap M_z$ as $\left \{v_1,\ldots,v_{\ell }\right \}$ , where $\ell \ge \frac{\sqrt{t}}{22}$ and $\ell \le \sqrt{t}$ (since $|W_0|=\sqrt{t}$ ).

Using Lemma 13, we can find pairwise disjoint projections $H_1,\ldots, H_{\ell }$ of $G$ , each having dimension at least $t-\sqrt{t}$ , such that each $v_i\in W_0$ is in exactly one of $H_i$ ’s.

We now intend to find pairwise disjoint projection of $G$ , each with a vertex in $M_{z}$ , to which we can apply our induction hypothesis. Note that, in an arbitrary projection $H'$ of $G$ with dimension $t'$ , the set $M_z \cap V(H')$ will still be a layer of $H'$ . However, for the vertices in this layer, the number of ‘active’ coordinates in $H'$ which are centres of stars will be some number between $z$ and $z-(t-t')$ . Our approach here is then almost identical to that of the proof of Lemma 16, with the slight difference that we want our vertices to be in $M_{z}$ , and have $z'$ centre coordinates inside the projected subgraph, for some $z'$ such that $z'\ge \frac{1+\delta '}{10\sqrt{s}p_1}$ for some $\delta '\gt 0$ .

In order to obtain that, we begin by exposing some carefully chosen sets of edges with probability $p_2$ . Observe that for each $v_i\in W_0\cap M_{z}$ , we have that

\begin{equation*} |N_{H_i\cap \Gamma }(v_i)|\ge |N_\Gamma (v_i)|-s\cdot \sqrt {t}\ge \frac {zs}{3}, \end{equation*}

where the first inequality follows since the dimension of $H_i$ is at least $t-\sqrt{t}$ and the maximum degree of the base graphs is $s$ . We further note that $N_{H_i\cap \Gamma }(v_i)\subseteq M_{z-1}$ . Let

\begin{equation*} W_1=\bigcup _{i\in \left [\sqrt {\ell }\right ]}N_{H_i\cap \Gamma }(v_i)\subseteq M_{z-1}. \end{equation*}

Then, recalling that the $H_i$ ’s are pairwise disjoint, we have that

\begin{align*} |W_1|\ge \frac{\sqrt{t}}{25}\cdot \frac{zs}{3}\ge \frac{\sqrt{t}(1+\delta )\sqrt{s}}{750p}\ge t^{\frac{3}{2}}, \end{align*}

where we used our assumption that $z\ge \frac{1+\delta }{10\sqrt{s}p}$ and that $s$ is large enough.

We now expose the edges between $W_0$ and $W_1$ in $G_{p_2}$ . Let us denote the set of vertices in $W_1$ that are connected with $W_0$ in $G_{p_2}$ by $W_1'$ . Then, $|W_1'|$ stochastically dominates $\text{Bin}\left (t^{\frac{3}{2}},p_2\right )$ . Thus, by Lemma 10, we have that whp $|W_1'|\ge \frac{\sqrt{t}}{10\log t}$ . We assume in what follows that this property holds.

We now expose the edges in $G_{p_2}$ between $W_1'$ and $M_{z}\setminus W_0$ , recalling that by our assumption $|W_0|=\sqrt{t}$ . Note that for every $w_i\in W_1'$ , we have that

\begin{equation*}|N_{H_i\cap \Gamma }(w_i) \cap (M_{z}\setminus W_0)|\ge |N_\Gamma (w_1)|-2s\cdot \sqrt {t}\ge \frac {t}{3},\end{equation*}

and $N_{H_i\cap \Gamma }(w_i) \cap (M_{z}\setminus W_0) \subseteq M_{z}$ . Let

\begin{equation*} W_2=\bigcup _{i\in \left [\ell \right ]} N_{H_i\cap \Gamma }(w_i)\cap (M_{z}\setminus W_0)\subseteq M_{z}, \end{equation*}

where we note that by the projection Lemma (lemma 13) and our construction, these neighbourhoods are disjoint. Then whp $|W_2|\ge \frac{t}{3}\cdot \frac{\sqrt{t}}{10\log t}.$

Let $W_2'\subseteq M_{z}$ be the set of vertices in $W_2$ that are connected with $W_1'$ in $G_{p_2}$ . Then, $|W_2'|$ stochastically dominates $\text{Bin}\left (\frac{t^{\frac{3}{2}}}{10\log t},p_2\right ).$ Thus, by Lemma 10, we have that whp $|W_2'|\ge \frac{\sqrt{t}}{20\log ^2t}$ and $|W_2'|\le \sqrt{t}$ . We assume henceforth that these two properties hold.

The subsequent part of the proof is almost identical to that of the proof of Lemma 16, where here we only need to run the inductive step constantly many times (as opposed to $\omega (1)$ times), allowing us to analyse it in a more straightforward manner.

Indeed, let $W_{2,i}'=W_2'\cap V(H_i)\subseteq M_{z}$ . We thus have now our set of vertices in $M_{z}$ . Now, for each $i$ , we apply once again Lemma 13 to find a family of $|W_{2,i}'|\;:\!=\; \ell _i\le \sqrt{t}$ pairwise disjoint projections of $H_i$ , which we denote by $H_{i,1},\ldots, H_{i,\ell _i}$ , such that every vertex of $W_{2,i}'\subseteq M_{z}$ is in exactly one of the $H_{i,j}$ , and each of the $H_{i,j}$ is of dimension at least $t-2\sqrt{t}$ . We have that for all $i,j$ , the number of centre coordinates of $v_{i,j}$ in $H_{i,j}$ , denote it by $z_{i,j}$ , is at least $z-2s\sqrt{t}$ . Recall that $p_1\ge \frac{1-\epsilon _1}{3t}$ , and note that

\begin{align*} z_{i,j} \ge z-2s\sqrt{t}\ge \frac{1+\delta }{10\sqrt{s}p}-2s\sqrt{t} \ge \frac{1+\delta _2}{10\sqrt{s}p} \ge \frac{1+\delta _3}{10\sqrt{s}p_1} \end{align*}

for some $\delta _2,\delta _3\gt 0$ , where we used $p\le \frac{1}{t}$ and $p_1=(1-o(1))p$ . Altogether, the $v_{i,j}$ (that is, their corresponding $z_{i,j}$ ) together with $p_1$ satisfy the conditions of the induction hypothesis for some $\epsilon _{i,j},\delta _{i,j}\gt 0$ .

As in the proof of Lemma 16, note that when we ran the algorithm on $G_{p_1}$ , we did not query any of the edges inside any of the $H_{i,j}$ . By the above inequality we can apply the induction hypothesis to $v_{i,j}$ in $G_{p_1}\cap H_{i,j}$ and conclude that with probability at least $c_1(z_{i,j},k-1,\epsilon _{i,j},\delta _{i,j})$ , there exists a connected set $W_{i,j}$ such that $v_{i,j}$ is $W_{i,j}$ , $|W_{i,j}\cap M_{z}|\ge c_2(z_{i,j},k-1,\epsilon _{i,j},\delta _{i,j})t^{\frac{k-1}{6}}$ . Let $c'_1=\min _{i,j}c_1(z_{i,j},k-1,\epsilon _{i,j},\delta _{i,j})$ and $c'_{\!\!2}=\min _{i,j}c_2(z_{i,j},k-1,\epsilon _{i,j},\delta _{i,j})$ . Noting that the above described events are independent for every $H_{i,j}$ , we obtain by Lemma 10 that whp, at least $\frac{c'_1\sqrt{t}}{40\log ^2t}$ of the $H_{i,j}$ have that $|W_{i,j}\cap M_{z}|\ge c'_{\!\!2}t^{\frac{k-1}{6}}$ . Thus, with probability at least $c'_1+o(1)\;:\!=\; c_1$ , we have found a connected set $W$ such that $v$ is in $W$ , and there is some $c_2\gt 0$ such that

\begin{align*} |W\cap M_{z}|\ge \frac{c'_{\!\!1}\sqrt{t}}{40\log ^2t}\cdot c'_{\!\!2}t^{\frac{k-1}{6}}\gt c_2t^{\frac{k}{6}}, \end{align*}

completing the induction step.

Proof. Observe that in order to choose a vertex in $M_{z}$ we have $\binom{t}{z}$ choices for the $z$ -set of coordinates which are the centre of a star, and then we have $s^{t-z}$ choices for the leaf coordinates. It follows that

\begin{align*} |M_{z}|= \binom{t}{z}s^{t-z}\ge \left (\sqrt{s}\right )^{\frac{t}{\sqrt{s}}}s^{\left (1-\frac{1}{\sqrt{s}}\right )t} \ge (s+1)^{\left (1-s^{-\frac{1}{3}}\right )t} = |G|^{1-s^{-\frac{1}{3}}}, \end{align*}

where the penultimate inequality holds for any $s$ large enough.

By Lemma 19, there is some constant $c' \gt 0$ such that every $v\in M_{z}$ belongs to a connected set of order $t^k$ with probability at least $c'\gt 0$ . We thus have that

\begin{equation*} \mathbb {E}[|W|]\ge c'|M_{z}|\ge c'|G|^{1-s^{-\frac {1}{3}}}. \end{equation*}

Consider the edge-exposure martingale on $G_p$ . It has length $|E(G)| = \frac{st}{s+1} |G| \leq 2t |G|$ , and the addition or deletion of an edge can change the value of $|W|$ by at most $2t^k$ . Therefore, by Lemma 11, we have for $\rho =\frac{c'}{2}\gt 0$ that

\begin{align*} \mathbb{P}\left [|W|\le \rho |M_{z}|\right ]&\le \mathbb{P}\left [|W|\le \frac{\mathbb{E}|W|}{2}\right ] \le 2\exp \left (-\frac{|G|^{2-2s^{-\frac{1}{3}}}}{16|G|t^{2k+1}} \right ) \le 2\exp \left (-|G|^{1-s^{-\frac{1}{4}}}\right )=o(1), \end{align*}

as long as $s$ is large enough. It follows that whp

\begin{equation*} |W|\ge \rho |M_{z}|\ge \frac {c'}{2}|G|^{1-s^{-\frac {1}{4}}} \geq |G|^{1-s^{-\frac {1}{5}}}. \end{equation*}

Proof of Theorem 9. Assume that all the conditions in Theorem 9 hold.

Once again, we argue via a two-round exposure. Let $\epsilon$ be a small enough constant, such that $p_2=\frac{\epsilon }{3t}$ , $p_1=\frac{1}{3t}$ and $(1-p_1)(1-p_2)=1-p$ , noting that $G_{p_1} \cup G_{p_2}$ has the same distribution as $G_p$ .

Let $z\;:\!=\; \frac{t}{\sqrt{s}}$ . Let $M_{z}$ be the set of vertices with $z$ centre coordinates in $G$ , and let $W\subseteq M_{z}$ be the set of vertices in $M_{z}$ belonging to a component with at least $t^{15}$ vertices in $M_{z}$ in $G_{p_1}$ . Then, by Lemma 20 with $k=15$ , we have that there exists a constant $\rho \gt 0$ such that whp,

(11) \begin{align} |W|\ge \rho |M_{z}|\ge |G|^{1-s^{-\frac{1}{5}}}. \end{align}

We note further that, since the average degree $t = \Theta (\log |G|)$ , by a similar argument as in Lemma 17 applied to the vertices with $z$ centre coordinates in $G$ , we claim that whp

(12) \begin{align} \text{every vertex in $M_{z}$ is at distance at most 2 from $W$.} \end{align}

To see why, observe that given a vertex $v\in M_{z}$ , we can form $t^{\frac{4}{3}}$ pairwise disjoint projections of $G$ , each isomorphic to the product of $(1-o(1))t$ copies of $S(1,s)$ and each containing a vertex with $z$ centre coordinates at distance $2$ from $v$ . Indeed, since the number $z$ of coordinates of $v$ which are the centre of star satisfies $t^{\frac{2}{3}} \leq z \leq t-t^{\frac{2}{3}}$ , we can assume without loss of generality that the first $t^{\frac{2}{3}}$ coordinates of $v$ are the centres of stars, and the second $t^{\frac{2}{3}}$ coordinates are leaves.

Given a pair of coordinates $1 \leq i \leq t^{\frac{2}{3}} \lt j \leq 2t^{\frac{2}{3}}$ , let $H_{i,j}$ be the projection of $G$ to the first $2t^{\frac{2}{3}}$ coordinates where the first $2t^{\frac{2}{3}}$ coordinates agree with $v$ , except that the $i$ th coordinate is some arbitrary leaf and the $j$ th coordinate is the centre of the star.

Then, the graphs $H_{i,j}$ form a collection of $t^{\frac{4}{3}}$ pairwise disjoint projections of $G$ , each isomorphic to the product of $(1-o(1))t$ copies of $S(1,s)$ and each containing a vertex with $z$ centre coordinates in $G$ at distance $2$ from $v$ .

Since $p$ will still be ‘supercritical’ at these vertices in each $H_{i,j}$ , an argument as in Lemma 17 will imply that the expected number of vertices in $M_{z}$ which are not at distance at most 2 from a vertex in $W$ is at most $\exp \left ( - \Omega \left (t^{\frac{4}{3}} \right )\right )|G| = o(1)$ , and so the claim follows by Markov’s inequality. We assume in the following that claim (12) holds.

Let $A\cup B$ be a $G_{p_1}$ -respecting partition of $W$ into two parts, $A$ and $B$ , such that $|A|\le |B|$ and $|A|\ge \frac{|W|}{3}$ . Let $A'$ be the set of vertices in $A$ together with all vertices in $M_{z}\setminus B$ that are within distance $2$ to $A$ , and let $B'$ be the set of vertices in $B$ together with all vertices in $M_{z}\setminus A'$ that are within distance $2$ to $B$ , so that $M_{z} = A' \cup B'$ . We require the following claim, whose proof we defer to the end of this proof:

We can extend these paths to a family of $A-B$ paths of length at most $6$ (see Figure 3) and then, very crudely, since $\Delta (G)=st$ , we can greedily thin this family to a set of $\frac{|W|}{36s^6t^{8}}$ edge-disjoint $A-B$ paths of length at most $6$ .

Figure 3. An extension of a partition $A\cup B$ of $W$ to a partition $A'\cup B'$ of $M_{z}$ . Each $A'-B'$ two-path (solid) can be extended into an $A-B$ six-path (solid and dashed). Note that some vertices in these paths may lie outside of $M_{z}$ .

We can now apply Lemma 18 with $H=G_{p_1}$ , $c=\frac{1}{3}$ , $X=W$ , $p=p_2=\frac{\epsilon }{3t}=O(t^{-1})$ , $k=t^{15}$ , $r=36s^6t^{8}$ and $m=6$ . Indeed, by our assumptions on $G_{p_1}$ and by Lemma 19, we have that whp both conditions in Lemma 18 hold, and

\begin{align*} rp^{-m} &=36s^6t^{8}p_2^{-6} =O\left (t^{14}\right ) = o\left (t^{15}\right ) = o(k). \end{align*}

As a consequence of Lemma 18, we have that whp $G_p=G_{p_1}\cup G_{p_2}$ contains a component which contains at least $ \frac{2}{3}|W|$ vertices of $W$ . Combining this with the bound $|W|\ge |G|^{1-s^{-\frac{1}{5}}}$ from (11), we have that whp the largest component of $G_p$ is of order at least $\frac{2}{3}|G|^{1-s^{-\frac{1}{5}}}\ge |G|^{1-s^{-\frac{1}{6}}}$ .

Proof of Claim 21. Let $z=\frac{t}{\sqrt{s}}$ be the number of coordinates of $v$ which correspond to a centre of a star. Recall that, by Lemma 20, there is a constant $\rho \gt 0$ such that whp $|W|\ge \rho |M_{z}|$ . The following two constants will be useful throughout the proof:

\begin{align*} \alpha \;:\!=\; \frac{\rho }{4}, \quad \beta \;:\!=\; \frac{\rho }{3(4-\rho )}. \end{align*}

Note that $\beta +(1-\beta )\alpha =\frac{\rho }{3(4-\rho )}+\frac{\rho }{4}\left (1-\frac{\rho }{3(4-\rho )}\right )=\frac{\rho }{3}$ .

We are interested in the structure of the graph $G^2[M_{z}]$ . Suppose that $v,w\in M_{z}$ , and denote by $I,K\in [t]^{z}$ (respectively) the set of coordinates corresponding to centres of stars. Then $v$ and $w$ are at distance $2$ in $G$ if either

1. (T1) $I=K$ and they disagree on a single coordinate of a leaf outside $I$ ; or

2. (T2) $|I\triangle K|=2$ and they agree on all the (leaf) coordinates outside $I\cup K$ .

Hence, we can see that $G^2[M_{z}]$ can be built in the following manner (see also Figure 4) – we start with the Johnson graph $J\;:\!=\; J(t,{z})$ , which represents the $z$ -set of coordinates which are centres of stars, and we replace each vertex with a copy of $H\;:\!=\; H(t-{z},s)$ , which represents the set of leaf coordinates. There is a natural map $f: M_{z} \to V(J) \times V(H)$ which maps a vertex $v \in M_{z}$ to the pair $(I,x)$ where $I$ is the $z$ -set of coordinates of $v$ which are centres of stars and $x$ is the restriction of $v$ to the coordinates which are leaves.

Figure 4. A visualisation of the graph $G^2[M_{z}]$ in the case $t=4,z=2$ and $s=2$ . Here, each circle is a copy of $H(t-z,s)$ and the black tubes represent the edges of $J(t,z)$ . A pair of copies of $H(t-z,s)$ have been displayed, where the vertices of $M_{z}$ inside each copy have been coloured according to whether they lie in $A'$ (blue) or $B'$ (orange). Inside the black tube corresponding to the edge of $J(t,z)$ between this pair of vertices there is some matching, represented by the dashed (purple) edges. Note that this matching may not respect the structure of the $H$ copies in each circle. Each vertex $I$ of $J(t,z)$ is then coloured according to whether it is $A'$ -dominated (blue), $B'$ -dominated (orange) or evenly balanced (green), corresponding to $\chi (I)=1,2,$ and $3$ , respectively.

The edges of $G^2[M_{z}]$ then come in two types. Those of type (T1) give rise to a copy of $H$ on $I \times V(H)$ for each $I \in V(J)$ , whereas those of type (T2) give rise to a matching between $I \times V(H)$ and $K \times V(H)$ for each $(I,K) \in E(J)$ , where a vertex $(I,x)$ is matched to a vertex $(K,y)$ if and only if $f^{-1}(I,x)$ and $f^{-1}(K,y)$ agree on all coordinates outside $I \cup J$ . Note that this is similar to, but not quite the same as, the Cartesian product of $J$ and $H$ .

We use the partition $M_{z}=A'\cup B'$ to define a colouring $\chi \;:\; V(J) \to [3]$ in the following manner. We say that a copy of $H$ given by $I \times V(H)$ is evenly-split if

\begin{equation*} \alpha |H| \leq |f^{-1}(I \times V(H)) \cap A'| \leq \left (1-\alpha \right )|H|. \end{equation*}

Note that, since $M_{z}=A'\cup B'$ is a partition, if $I \times V(H)$ is evenly-split then it also satisfies

\begin{equation*} \alpha |H|\le |f^{-1}\left (I\times V(H)\right )\cap B'|\le \left (1-\alpha \right )|H|. \end{equation*}

If $I \times V(H)$ is not evenly-split then we say it is either $A'$ -dominated or $B'$ -dominated if

\begin{equation*} |f^{-1}(I \times V(H)) \cap A'| \geq \left (1-\alpha \right )|H| \qquad \text { or } \qquad |f^{-1}(I \times V(H)) \cap B'| \geq \left (1-\alpha \right )|H|, \end{equation*}

respectively. We then let $\chi (I)= 1$ if $I \times V(H)$ is $A'$ -dominated, $\chi (I)= 2$ if $I \times V(H)$ is $B'$ -dominated, and $\chi (I)= 3$ if $I \times V(H)$ is evenly-split (see Figure 4).