Proof of Theorem 2.7. Suppose $\mathcal{R}_1(X_{0}) = O\!\left(n^{4/3}\right)$ and $\mathcal{R}_1(Y_{0}) = O\!\left(n^{4/3}\right)$ . It follows from Lemmas 2.8, 2.9, 2.10 and 2.11 that there exists a coupling of the CM steps such that after $T = O({\log}\ n)$ steps, $X_{T}$ and $Y_{T}$ could have the same component structure. This coupling succeeds with probability at least

\begin{equation*}\rho = \Omega\big(\omega (n)^{-\beta_1}\big) \cdot \exp\big({-}O\big(\omega (n)^9\big)\big) \cdot \exp \big({-}O \big( ({\log} \omega (n))^2 \big) \big) \cdot \Omega \big( ({\log} \log \log n)^{-\beta_2} \big), \end{equation*}

where $\beta_1, \beta_2 > 0$ are constants. Thus, $\rho = \Omega\big( ({\log} \log n)^{-1} \big)$ , since $\omega (n) = \log \log \log \log n$ .

Proof of Theorem 1.1. By Theorem 2.4, after $t_0 = O({\log}\ n)$ steps, with probability $\Omega(1)$ , we have $\mathcal{R}_1(X_{t_0}) = O\!\left(n^{4/3}\right)$ and $\mathcal{R}_1(Y_{t_0}) = O\!\left(n^{4/3}\right)$ . If this is the case, Theorem 2.7 and Lemma 2.13 imply that there exists a coupling of the CM steps such that with probability $\Omega\!\left( ({\log} \log n)^{-1} \right)$ after an additional $t_1 = O({\log}\ n)$ steps, $X_{t_0 + t_1} = Y_{t_0 + t_1}$ . Consequently, we obtain that $\tau_{\textrm{mix}}^{\textrm{CM}} = O({\log}\ n \cdot \log \log n)$ as claimed.

Proof. When $\zeta = q < 2$ , the drift function f does not have a positive root, it is continuous in (0,1], and $f(1)>0$ ; see Lemma 2.5 in [Reference Bollobás, Grimmett and Janson9] and Fact 3.5 in [Reference Blanca4]. Since $\lim_{\theta \rightarrow 0}f(\theta) = 0$ , the result follows.

Proof of Lemma 2.2. Let $\hat{T}$ be the first time t when $\hat{L}(X_t) \leq \delta n^{1/3}$ , let T ′ be a large constant we choose later; we set $T \,:\!=\, \min\{\hat{T}, T^{\prime}\}$ . Observe that with constant probability the largest component in the configuration is activated by the CM dynamics for every $t \le T^{\prime}$ , that is, the event $\Lambda_t$ occurs for every $t \le T^{\prime}$ . Let us assume this is the case and fix $t < T$ . Suppose $\mathcal{R}_2(X_t) \leq \mathcal{R}_2(X_0) + t \cdot \frac{C}{\sqrt{\delta}} n^{\frac{7}{6}}$ where C is the positive constant from Lemma 4.2. We show that with high probability:

1. i. $\mathcal{R}_2(X_{t+1}) \leq \mathcal{R}_2(X_0) + t \cdot \frac{C}{\sqrt{\delta}} n^{\frac{7}{6}}$ ; and

2. ii. $L_1(X_{t+1}) \le L_1(X_t) - \xi n$ where $\xi$ is a positive constant independent of t and n.

In particular, it suffices to set $T^{\prime} = (1-\delta)/\xi$ for the lemma to hold.

First, we show that $A(X_t)$ is concentrated around its mean. Let $L_1(X_t) \,:\!=\, \theta_t n$ and $L_1(X_{t+1}) \,:\!=\, \theta_{t+1} n$ . Let ${\mathbb{E}}[A(X_t) \mid \Lambda_t] = \mu_t = \frac{n}{q}+ \left(1-\frac{1}{q}\right)\cdot\theta_t n$ , $\gamma\,:\!=\,n^{5/6}$ , and $J_t\,:\!=\,[\mu_t - \gamma, \mu_t + \gamma]$ . Hoeffding’s inequality implies

\begin{align*}{\mathbb{P}}\left[A(X_t) \in J_t \mid \Lambda_t\right]& \geq 1 - 2\exp\!\left(\frac{-2\gamma^2}{ \mathcal{R}_2(X_t)}\right)= 1 - e^{-\Omega\left(n^{1/3}\right)}.\end{align*}

If $A(X_t) \in J_t$ , then the random graph $G(A(X_t), p)$ is super-critical since

\begin{equation*}A(X_t) \cdot p \geq (\mu_t - \gamma)\cdot \frac{q}{n}= \left[\frac{n}{q}+ \left(1-\frac{1}{q}\right)\cdot\theta_t n - n^{5/6} \right] \cdot \frac{q}{n}= 1 + (q-1)\theta_t - o(1) > 1.\end{equation*}

Next, for a super-critical random graph, Lemma 3.3 provides a concentration bound for the size of largest new component, provided $A(X_t) \in J_t$ . To see this, we write $G(A(X_t), \zeta/n)$ as

\begin{equation*}G \!\left(\mu_t + m, k(\theta_t, q)\cdot q/\mu_t \right),\end{equation*}

where $m\,:\!=\,A(X_t)-\mu_t$ ; notice that $|m| \leq \gamma = o(n)$ . Let $H \sim G \!\left(\mu_t + m, k(\theta_t, q)\cdot q/\mu_t \right)$ . Since

\begin{equation*}k(\theta_t, q)\cdot q = 1 + (q-1) \theta_t > 1 + \delta (q-1) > 1 \end{equation*}

holds regardless of n, Lemma 3.3 implies that for $\phi(\theta_t) > 0$ defined in Section 4.1, with high probability

\begin{equation*}L_1(H) \in \left[\phi(\theta_t)n - \sqrt{n\log n} - |m|,\phi(\theta_t)n+ \sqrt{n\log n} + |m|\right].\end{equation*}

Note that $L_1(H) = \Omega(n)$ w.h.p.; hence, since $L_2(X_t) = O\big(n^{2/3}\big)$ we have $L_1(X_{t+1}) = L_1(H)$ w.h.p. We have shown that w.h.p.

\begin{equation*}\theta_{t+1} - \theta_t\leq \phi(\theta_t) + \frac{\sqrt{n\log n}}{n} -\theta_t - \frac{|m|}{n}= -f(\theta_t) + \sqrt{\frac{\log n}{n}} - \frac{|m|}{n},\end{equation*}

where f is the drift function defined in Section 4.1. By Lemma 4.1, we know $f(\theta_t) > \xi_1 > 0$ for sufficiently small constant $\xi_1$ (independent of n and t). Hence, w.h.p. for sufficiently large n

\begin{equation*}L_1(X_{t+1}) - L_1(X_t) \leq -\xi_1 n + o(n) \leq \frac{-\xi_1 n}{2};\end{equation*}

this establishes (ii) from above.

For (i), note that for $t < T$ we have $\hat{L}(X_t) > \delta n^{1/3}$ , so Lemma 4.2 implies,

\begin{equation*} {\mathbb{P}}\left[\mathcal{R}_2(X_{t+1}) < \mathcal{R}_2(X_0) + t \cdot \frac{C}{\sqrt{\delta}} n^{\frac{7}{6}} + \frac{Cn^{4/3}}{\sqrt{\delta} n^{1/6}}\right] = 1 - o(1).\end{equation*}

A union bound implies that these two events occur simultaneously w.h.p. and the result follows.

Proof of Lemma 2.3. Let B be a constant large enough so that $\forall x \geq B^6$ , we have $x \geq ({\log} x)^{48}$ . Suppose $\hat{L}(X_{0}) \leq \delta n^{1/3}$ and $\mathcal{R}_2(X_{0}) = O\!\left(n^{4/3}\right)$ for the constant $\delta = \delta(q,B)$ from Lemma 4.4. Suppose also $ \hat{L}(X_{0}) \geq B^6$ ; otherwise there is nothing to prove.

Let $g_0(n)\,:\!=\,\delta n^{1/3}$ and $g_{i+1}(n)\,:\!=\, B ({\log} g_i(n))^8$ for all $i \geq 0$ . Let K be defined as the minimum natural number such that $g_K(n) \leq B^6$ . Note that $K = O({\log}^* n)$ . Assume at time $t \ge 0$ , there exists an integer $j \ge 0$ such that $X_{t}$ satisfies:

1. 1. $g_{j+1}(n) \leq \hat{L}(X_{t}) \leq g_j(n)$ , and

2. 2. $\mathcal{R}_2(X_{t}) = O\!\left(n^{4/3}\right) + O\!\left(\sum_{k=0}^{j-1} \frac{n^{4/3}}{\log g_{k}(n)}\right)$ .

We show there exists time $t^{\prime} > t$ such that properties 1 and 2 hold for $X_{t^{\prime}}$ for a different index $j^{\prime} > j$ . The following bounds on sums and products involving the $g_i$ ’s will be useful; the proof is elementary and delayed to the end of this section.

By part (ii) of this claim, note that

\begin{equation*}O\!\left(\sum_{k=0}^{j-1} \frac{n^{4/3}}{\log g_{k}(n)}\right) =O\!\left(\frac{n^{4/3}}{\log g_{j-1}(n)}\right) = O\!\left(n^{4/3}\right).\end{equation*}

Hence, Lemma 4.4 implies that with probability $1 - O\!\left(\left({\log} g_j(n)\right)^{-1} \right)$ there exist a time $t^{\prime} \leq t + O( \log g_j(n) )$ and a large constant D such that $ \hat{L}\!\left(X_{t^{\prime}}\right) \leq \max\left\{B\!\left({\log} g_j(n)\right)^8, D\right\}$ and $\mathcal{R}_2\!\left(X_{t^{\prime}}\right) \leq \mathcal{R}_2(X_{t}) + O\!\left(\frac{n^{4/3}}{\log g_j(n)}\right).$ If $\hat{L}\!\left(X_{t^{\prime}}\right) \le \max \left\{D, B^6\right\}$ we are done. Hence, suppose otherwise that $\hat{L}\!\left(X_{t^{\prime}}\right) \in \left(B^6, \log g_{j+1}(n)\right]$ . Since the interval $(B^6, \log g_{j+1}(n)]$ is completely covered by the union of the intervals $[g_{j+2}, g_{j+1}]$ , …, $[g_{K}, g_{K-1}]$ , there must be an integer $j^{\prime}\ge j+1$ such that $g_{j^{\prime}+1}(n) \leq \hat{L}\!\left(X_{t^{\prime}}\right) \leq g_{j^{\prime}}(n)$ . Also, notice

\begin{align*} \mathcal{R}_2\!\left(X_{t^{\prime}}\right) & \le \mathcal{R}_2(X_{t}) + O\!\left(\frac{n^{4/3}}{\log g_j(n)}\right) = O\!\left(n^{4/3}\right) + O\!\left(\sum_{k=0}^{j-1} \frac{n^{4/3}}{\log g_{k}(n)}\right) + O\!\left(\frac{n^{4/3}}{\log g_j(n)}\right) \\ &= O\!\left(n^{4/3}\right) + O\!\left(\sum_{k=0}^{j} \frac{n^{4/3}}{\log g_{k}(n)}\right) = O\!\left(n^{4/3}\right) + O\!\left(\sum_{k=0}^{j^{\prime}-1} \frac{n^{4/3}}{\log g_{k}(n)}\right). \end{align*}

By taking at most K steps of induction, we obtain that there exist constants C and c such that with probability at least $\rho \,:\!=\, \prod_{i=0}^{K-1} \left(1 - \frac{c}{\log g_{i}(n)}\right),$ there exists a time

\begin{equation*}t_{K} \leq \sum_{i=0}^{K-1} C \log g_i(n)\end{equation*}

that satisfies $ \hat{L}(X_{t_{K}}) \leq g_{K}(n) \leq B^6 $ and $\mathcal{R}_2(X_{t_{K}}) = O\!\left(n^{4/3}\right)$ . Observe that $t_{K}$ is a time when our goal has been achieved, so it only remains to show that $\rho = \Omega(1)$ and $t_{K} = O({\log}\ n)$ . The lower bound on $\rho$ follows from part (i) of Claim 4.5:

\begin{align*} \prod_{i=0}^{K-1} 1 - \frac{c}{\log g_{i}(n)} & \geq 1 - \frac{1.5 c}{\log g_{K-1}(n)} > 1 - \frac{1.5 c}{\log B^6} = \Omega(1). \end{align*}

By noting that $K = O({\log}^*n)$ , we can also bound $t_{K}$ since $ \sum_{i=0}^{K-1} C \log g_i(n)$ is at most $\log g_0(n) + (K-1)\log g_1(n) = O({\log}\ n)$ .

Proof of Lemma 4.3. We start with part 1. Let $\mu_t \,:\!=\, {\mathbb{E}}[A(X_t) | \Lambda_t, X_t]$ , $\gamma_t\,:\!=\,\sqrt{\hat{L}(X_{t})} \cdot n^{2/3}$ , and $J_t\,:\!=\,[\mu_t - \gamma_t, \mu_t + \gamma_t]$ . Hoeffding’s inequality implies that

\begin{equation*} \begin{split} {\mathbb{P}}\left[A(X_t) \in J_t \mid \Lambda_t, X_t\right] & \geq 1 - 2\exp\!\left(\frac{-2\gamma_t^2}{ \mathcal{R}_2(X_t)}\right) = 1 - \exp \!\left({-} \Omega\big(\hat{L}(X_{t})\big) \right). \end{split} \end{equation*}

Let $m \,:\!=\, \mu_t +\gamma_t$ , $G \sim G\!\left(m, \frac{q}{n}\right)$ and $\hat{G} \sim G(A(X_t), p)$ . Then, the monotonicity of the largest component in a random graph implies that for any $\ell > 0$

\begin{align*} & {\mathbb{P}}\left[L_1(\hat{G}) > \ell \mid A(X_t) \in J_t\right] \\ & = \sum_{a \in J_t} {\mathbb{P}}\left[L_1(\hat{G}) > \ell\mid A(X_t) = a\right]{\mathbb{P}}\left[A(X_t) = a \mid A(X_t) \in J_t\right] \\ & \leq \sum_{a \in J_t} {\mathbb{P}}\left[L_1(\hat{G}) > \ell\mid A(X_t) = m\right] {\mathbb{P}}\left[A(X_t) = a \mid A(X_t) \in J_t\right] \\ & = {\mathbb{P}}[L_1(G) > \ell] \sum_{a \in J_t} {\mathbb{P}}[A(X_t) = a \mid A(X_t) \in J_t] \\ & = {\mathbb{P}}[L_1(G) > \ell]. \end{align*}

We bound next ${\mathbb{P}}[L_1(G) > \ell]$ . For this, we rewrite $G\!\left(m, \frac{q}{n}\right)$ as $G\!\left(m, \frac{1+\varepsilon}{m}\right)$ ; since

\begin{equation*}\mu_t = \hat{L}(X_{t}) \cdot n^{2/3} + \left(n- \hat{L}(X_{t}) n^{2/3}\right)q^{-1}\end{equation*}

we have

\begin{equation*}\varepsilon = m\cdot \frac{q}{n} - 1 = \left(q - 1 + \frac{q}{\sqrt{\hat{L}(X_{t})}}\right) \frac{\hat{L}(X_{t})}{n^{1/3}}.\end{equation*}

Thus,

\begin{align*} \varepsilon^3\cdot m &= \left(q - 1 + \frac{q}{\sqrt{\hat{L}(X_{t})}}\right)^3 \frac{ \hat{L}(X_{t})^3}{n} \left(\hat{L}(X_{t}) n^{2/3} + \frac{n- \hat{L}(X_{t}) n^{2/3}}{q} + \sqrt{\hat{L}(X_{t})}n^{2/3}\right) \\ & \geq \left(q - 1 + \frac{q}{\sqrt{\hat{L}(X_{t})}}\right)^3 \cdot \frac{\hat{L}(X_{t})^3}{n} \cdot \frac{n}{q} \\ & \geq \frac{1}{q}\cdot \left((q-1)^3 \hat{L}(X_{t})^3 + q^3\sqrt{\hat{L}(X_{t})}^3\right)\\ & \geq q^2 \hat{L}(X_{t})^{3/2} \geq 100 \hat{L}(X_{t}), \end{align*}

where the last inequality follows from the fact that $\hat{L}(X_{t}) > B$ , where $B=B(q)$ is a sufficiently large constant.

Since $\varepsilon^3\cdot m\geq1$ , Lemma 3.5 implies

\begin{equation*} {\mathbb{P}}\left[ \lvert L_1(G) - 2\varepsilon m \rvert > \sqrt{\hat{L}(X_{t})} \sqrt{\frac{m}{\varepsilon}}\right] = e^{-\Omega\left(\hat{L}(X_{t})\right) }.\end{equation*}

Let $c_1=2\sqrt{\frac{1+(q-1)\delta}{q(q-1)}}$ . The upper tail bound implies

\begin{equation*}{\mathbb{P}}\left[L_1(G) \leq 2\varepsilon m + c_1 n^{2/3}\right] \geq 1 - e^{- \Omega\left( \hat{L}(X_{t}) \right)}.\end{equation*}

We show next that $2\varepsilon m + c_1 n^{2/3} \le \alpha L_1(X_t)$ for some $\alpha \in (0,1)$ .

\begin{align*} &2\varepsilon m + c_1 n^{2/3} \\ = & 2\!\left(q - 1 + \frac{q}{\sqrt{\hat{L}(X_{t})}}\right) \frac{\hat{L}(X_{t})}{n^{1/3}} \left(\hat{L}(X_{t}) n^{2/3} + \frac{n-\hat{L}(X_{t}) n^{2/3}}{q} + \sqrt{\hat{L}(X_{t})}n^{2/3}\right) + c_1 n^{2/3}\\ = & \frac{2}{q} \left(q - 1 + \frac{q}{\sqrt{\hat{L}(X_{t})}}\right) \frac{\hat{L}(X_{t})}{n^{1/3}} \left[ n + \left( q - 1 + \frac{q}{\sqrt{\hat{L}(X_{t})}}\right) \hat{L}(X_{t}) n^{2/3} \right] + c_1 n^{2/3} \\ = & \frac{2}{q} \left(q - 1 + \frac{q}{\sqrt{\hat{L}(X_{t})}} + \frac{c_1q}{\hat{L}(X_{t})}\right) \hat{L}(X_{t}) n^{2/3} + \frac{2}{q} \left(q - 1 + \frac{q}{\sqrt{\hat{L}(X_{t})}}\right)^2 \hat{L}(X_{t})^2 n^{1/3} \\ \leq & \frac{2}{q} \left[ \delta\!\left(q - 1 + O\!\left(\hat{L}(X_{t})^{-1/2}\right) \right)^2 + \left(q - 1 + O\!\left(\hat{L}(X_{t})^{-1/2} \right)\right)\right] \hat{L}(X_{t}) n^{2/3}, \end{align*}

where in the last inequality we use the assumption that $\delta n^{1/3} \geq \hat{L}(X_{t})$ . For sufficiently small $\delta$ and sufficiently large B, $\exists \ \alpha < 1$ such that

\begin{equation*}\alpha > \frac{2}{q}\left[ \delta\!\left(q - 1 + \frac{2q}{B^{1/2}}\right)^2 + \left(q - 1 + \frac{2q}{B^{1/2}}\right)\right].\end{equation*}

Consequently, $ L_1\!\left(G\right) \leq 2\varepsilon m + c_1 n^{2/3} \leq \alpha L_1(X_t)$ with probability $1 - \exp\big({-}\Omega\big(\hat{L}(X_{t}\big)\big)$ . If that is the case, $L_1(X_{t+1}) \leq \max\left\{\alpha L_1(X_t), L_2(X_t)\right\} \,=\!:\, L^+$ . Therefore,

\begin{align*} & {\mathbb{P}}\left[L_1(X_{t+1}) \leq L^+ \mid X_t, \Lambda_t\right] \\[3pt] & \geq {\mathbb{P}}\left[L_1(X_{t+1}) \leq L^+ \mid X_t, \Lambda_t , A(X_t) \in J_t\right] \cdot {\mathbb{P}}\left[A(X_t) \in J_t \mid X_t, \Lambda_t \right] \\[3pt] & \geq 1 - \exp\big({-}\Omega\big(\hat{L}(X_{t})\big)\big), \end{align*}

which concludes the proof of part 1.

For part 2, note first that when the largest component is inactive, we have $L_1(X_{t+1}) \geq L_1(X_t)$ ; hence, it is sufficient to show that $L_1(X_{t+1}) \leq L_1(X_t)$ with the desired probability.

Let $\mu^{\prime}_t \,:\!=\, {\mathbb{E}}\left[A(X_t) \mid \neg \Lambda_t, X_t\right] = \big(n- \hat{L}(X_{t}) n^{2/3} \big)q^{-1}$ , $\gamma^{\prime}_t\,:\!=\,\sqrt{\hat{L}(X_{t})} \cdot n^{2/3}$ , and $J^{\prime}_t \,:\!=\, [\mu^{\prime}_t - \gamma^{\prime}_t, \mu^{\prime}_t + \gamma^{\prime}_t]$ . By Hoeffding’s inequality,

\begin{equation*}{\mathbb{P}}\left[A(X_t) \in J^{\prime}_t \mid \neg\Lambda_t, X_t\right] \geq 1 - \exp\big({-} \Omega\big( \hat{L}(X_{t}) \big)\big).\end{equation*}

Let $G\sim G(A(X_t), p)$ , $m = \mu^{\prime}_t + \gamma^{\prime}_t$ and let $G^+ \sim G\!\left(\mu^{\prime}_t + \gamma^{\prime}_t, p\right) $ , By monotonicity of the largest component in a random graph,

\begin{equation*}{\mathbb{P}}\left[L_1(G) > L_1(X_t) \mid A(X_t) \in J^{\prime}_t\right] \leq {\mathbb{P}}\left[L_1(G^+) > L_1(X_t)\right].\end{equation*}

Rewrite $G\!\left(\mu^{\prime}_t + \gamma^{\prime}_t, p\right)$ as $G\!\left(m, \frac{1 + \varepsilon}{m}\right)$ , where

\begin{equation*}\varepsilon = \left(\frac{n - \hat{L}(X_{t}) n^{2/3}}{q} + \sqrt{\hat{L}(X_{t})} n^{2/3} \right) \cdot \frac{q}{n} - 1 = \left(\sqrt{\hat{L}(X_{t})}q -\hat{L}(X_{t}) \right) n^{-1/3}. \end{equation*}

From this bound, applying Lemma 3.7 to $G^+$ , we obtain

\begin{equation*}{\mathbb{E}}\left[\mathcal{R}_1(G^+)\right] = O\!\left( \frac{m}{ \varepsilon } \right) = O\!\left( \frac{n^{4/3}}{\hat{L}(X_{t})} \right). \end{equation*}

Hence, ${\mathbb{E}}\left[L_1(G^+)^2\right] = O\!\left(n^{4/3}/\hat{L}(X_{t})\right)$ and by Markov’s inequality

\begin{equation*}{\mathbb{P}}\left[L_1(G^+) > \hat{L}(X_{t}) n^{2/3}\right] = {\mathbb{P}}\left[L_1(G^+)^2 > \hat{L}(X_{t})^2 n^{4/3}\right] \leq \frac{{\mathbb{E}}[L_1(G^+)^2]}{\hat{L}(X_{t})^2 n^{4/3}} = O\!\left(\frac{1}{\hat{L}(X_{t})^3} \right).\end{equation*}

To conclude, we observe that

\begin{align*} & {\mathbb{P}}[L_1(X_{t+1}) \leq L_1(X_t) \mid X_t, \neg\Lambda_t] \\ & \geq {\mathbb{P}}\left[L_1(G) \leq L_1(X_t) \mid X_t, \neg\Lambda_t, A(X_t) \in J^{\prime}_t\right] {\mathbb{P}}\left[A(X_t) \in J^{\prime}_t \mid X_t, \neg\Lambda_t\right] \\ & \geq \left(1 - e^{- \Omega\left( \hat{L}(X_{t}) \right) }\right) \left(1 - O\!\left(\frac{1}{\hat{L}(X_{t})^3} \right)\right) = 1 - O\!\left(\frac{1}{\hat{L}(X_{t})^3} \right), \end{align*}

as desired.

Proof of Lemma 4.4. Suppose $\mathcal{R}_2(X_0) \le D_1^2n^{4/3}$ for a constant $D_1$ . Let $T^{\prime}\,:\!=\,B^{\prime}\log g(n)$ , where B ′ is a constant such that $B^{\prime} \log g(n)= 2q \log_{1/\alpha}\left(\frac{ g(n)}{B \big({\log} g(n)\big)^8}\right)$ and $\alpha \,:\!=\, \alpha(B, q, \delta)$ is the constant from Lemma 4.3. Let $\hat{T}$ be the first time

\begin{equation*}\hat{L}(X_{t}) \leq \max\big\{ B \big({\log} g(n)\big)^8, D\big\},\end{equation*}

where D is a large constant we choose later. Let $T\,:\!=\,T^{\prime}\wedge \hat{T}$ , where the operator $\wedge$ takes the minimum of the two numbers. Define e(t) as the number of steps up to time t in which the largest component of the configuration is activated.

To facilitate the notation, we define the following events. (The constants C and $\alpha$ are those from Lemmas 4.3 and 4.2, respectively).

1. 1. Let $H_i $ denote $\hat{L}(X_{i}) > \max\big\{ B \big({\log} g(n)\big)^8, D\big\}$ ;

2. 2. Let $F_i $ denote $\mathcal{R}_2(X_{i}) \leq \mathcal{R}_2(X_{i-1}) + C n^{4/3}\hat{L}(X_{i-1})^{-1/2}$ ; let us assume $F_0$ occurs;

3. 3. Let $F^{\prime}_i$ denote $\mathcal{R}_2(X_{i}) \leq \mathcal{R}_2(X_{i-1}) + C n^{4/3} \big({\log} g(n)\big)^{-4} B^{-1/2}$ ; again, we assume $F^{\prime}_0$ occurs;

4. 4. Let $Q_i$ denote $\hat{L}(X_{i}) \leq \max\big\{\alpha^{e(i)}\hat{L}(X_{0}), D \big\}$ ;

5. 5. Let $Base_i$ be the intersection of $\big\{F^{\prime}_0, Q_0, H_0\big\}, ..., \big\{F^{\prime}_{i-1}, Q_{i-1}, H_{i-1}\big\},$ and $ \big\{F^{\prime}_i, Q_i\big\}$ .

By induction, we find a lower bound for the probability of $Base_T$ . For the base case, note that ${\mathbb{P}}[Base_0] $ $= 1$ by assumption. Next we show

\begin{equation*}{\mathbb{P}}\left[Base_{i+1\wedge T} \mid Base_{i\wedge T}\right] = 1 - O\!\left( \big({\log} g(n)\big)^{-4} \right).\end{equation*}

If $T \leq i$ , then $Base_{i\wedge T} = Base_{ T} = Base_{i+1\wedge T}$ , so the induction holds. If $T > i$ , then we have $H_i$ .

By the induction hypothesis $F^{\prime}_1, F^{\prime}_2, ..., F^{\prime}_{i-1}$ ,

\begin{equation*} \mathcal{R}_2(X_{i}) \leq \mathcal{R}_2(X_{0}) + i \cdot C n^{4/3} \big({\log} g(n)\big)^{-4} B^{-1/2}.\end{equation*}

Moreover, since $i < T \leq T^{\prime} = B^{\prime} \log g(n)$ and $\mathcal{R}_2(X_0) \le D_1^2n^{4/3}$ , we have

\begin{equation*} \mathcal{R}_2(X_{i}) \leq D_1^2n^{4/3} + CB^{\prime} n^{4/3} \big({\log} g(n)\big)^{-3} B^{-1/2}.\end{equation*}

Given $\mathcal{R}_2(X_{i}) = O\!\left(n^{4/3}\right)$ and $H_i$ , Lemma 4.2 implies that $F_{i+1}$ occurs with probability

\begin{equation*}1 - O\Big(\hat{L}(X_i)^{-1/2}\Big) = 1 - O\!\left( \big({\log} g(n)\big)^{-4} \right).\end{equation*}

In addition, note that $F_{i+1} \cup H_i $ leads to $F^{\prime}_{i+1}$ . Let $\unicode{x1D7D9}(\Lambda_t)$ be the indicator function for the event $\Lambda_t$ . Given $H_i, Q_i$ and $\mathcal{R}_2(X_{i}) = O\!\left(n^{4/3}\right)$ , Lemma 4.3 implies

(5) \begin{equation} L_1(X_{i+1}) \leq \max \{\alpha^{\unicode{x1D7D9}(\Lambda_t)} L_1(X_i) ,L_2(X_i) \}\end{equation}

with probability at least $1 - O\big(\hat{L}(X_{t})^{-3}\big) = 1 - O\!\left( \big({\log} g(n)\big)^{-24} \right)$ .

Dividing equation (5) by $n^{2/3}$ , we obtain $Q_{i+1}$ for large enough D. In particular, we can choose D to be $D_1 + 2$ . A union bound then implies

\begin{equation*}{\mathbb{P}}[Base_{i+1\wedge T} \mid Base_{i\wedge T}] \geq {\mathbb{P}}\left[Base_{i+1 \wedge T} \mid Base_i, H_i\right] = 1 - O\!\left( \big({\log} g(n)\big)^{-4} \right).\end{equation*}

The probability for $Base_{T}$ can then be bounded as follows:

\begin{equation*}{\mathbb{P}}[Base_{T}] \geq \prod_{i=0}^{T-1} {\mathbb{P}}\big[Base_{i+1 \wedge T} \mid Base_{i \wedge T}\big]= \prod_{i=0}^{T-1} 1 - O\!\left( \big({\log} g(n)\big)^{-4} \right)= 1 - O\Big( \big({\log} g(n)\big)^{-3} \Big).\end{equation*}

Next, let us assume $Base_T$ . Then we have

\begin{equation*}\mathcal{R}_2(X_{T}) \leq \mathcal{R}_2(X_{0}) + T^{\prime} \cdot C n^{4/3} \big({\log} g(n)\big)^{-4} B^{-1/2} = \mathcal{R}_2(X_{0}) + O\!\left( n^{4/3} \big({\log} g(n)\big)^{-3} \right).\end{equation*}

Notice that if $T = \hat{T}$ then the proof is complete. Consequently, it suffices to show $\hat{T} \le T^{\prime}$ with probability at least $1 - g(n)^{-\Omega(1)}$ .

Observe that $K \,:\!=\, e(T^{\prime})$ is a binomial random variable $Bin\!\left(T^{\prime}, 1/q\right)$ , whose expectation is $\frac{T^{\prime}}{q} = \frac{B^{\prime}}{q} \log g(n)$ . By Chernoff bound

\begin{equation*}{\mathbb{P}}\left[K < \frac{B^{\prime}}{2q} \log g(n)\right] \leq \exp\!\left({-}\frac{B^{\prime}}{16q} \log g(n)\right) = g(n)^{-\Omega(1)}.\end{equation*}

If indeed $T = T^{\prime}$ and $ K \geq \frac{B^{\prime}}{2q} \log g(n)$ , then the event $Q_T$ implies

\begin{equation*}\hat{L}(X_{T}) < \alpha^{e(T)} \hat{L}(X_{0}) \leq \alpha^{\log_{\alpha}\left(\frac{B \big({\log} g(n)\big)^8}{g(n)}\right)} \hat{L}(X_{0})= \frac{B \big({\log} g(n)\big)^8}{g(n)} \hat{L}(X_{0}) \leq B \big({\log} g(n)\big)^8,\end{equation*}

which leads to $\hat{T} \le T$ . Therefore,

\begin{equation*}{\mathbb{P}} \left[\hat{T} > T^{\prime} \mid Base_T\right] \le {\mathbb{P}} \left[K < \frac{B^{\prime}}{2q} \log g(n) \right] = g(n)^{-\Omega(1)},\end{equation*}

as desired.

Proof of Claim 4.5. We first show the following inequality:

(6) \begin{equation} \frac{1.5}{\log g_j(n)} + \frac{1}{\log g_{j+1}(n)} \leq \frac{1.5}{\log g_{j+1}(n)}. \end{equation}

Note that by direction computation

\begin{equation*} \begin{split} \frac{1.5}{\log g_j(n)} + \frac{1}{\log g_{j+1}(n)} & = \frac{1.5 \big({\log} B + \log \big({\log} g_j(n)\big)^8\big) + \log g_j(n)}{\log g_j(n) \log g_{j+1}(n)}. \\ \end{split} \end{equation*}

From the definition of K, we know that $g_j(n) > B^6$ for all $j < K$ . Hence, $\log B < \log g_j(n)^{1/6}$ . In addition, recall that B is such that $\forall\,x \geq B^6$ , we have $x \geq ({\log} x)^{48}$ ; therefore, $g_j(n) \geq \big({\log} g_j(n)\big)^{48}$ . Then, $\log \big({\log} g_j(n)\big)^8 \le \log g_j(n)^{1/6}$ . Putting all these together,

\begin{equation*} \begin{split} \frac{1.5 \big({\log} B + \log \big({\log} g_j(n)\big)^8\big) + \log g_j(n)}{\log g_j(n) \log g_{j+1}(n)} & \leq \frac{1.5 \big(\frac{1}{6}\log g_j(n) + \frac{1}{6} \log g_j(n)\big) + \log g_j(n)}{\log g_j(n) \log g_{j+1}(n)} \\ & = \frac{ 1.5 \log g_j(n)}{\log g_j(n) \log g_{j+1}(n)} = \frac{ 1.5 }{ \log g_{j+1}(n)} \end{split} \end{equation*}

The proof of part (i) is inductive. The base case ( $i=0$ ) holds trivially. For the inductive step note that

\begin{equation*} \begin{split} \prod_{i=0}^{j+1} \left(1 - \frac{c}{\log g_{i}(n)}\right) & = \left(1 - \frac{c}{\log g_{j+1}(n)}\right) \prod_{i=0}^{j} \left(1 - \frac{c}{\log g_{i}(n)}\right) \\[3pt] & \geq \left(1 - \frac{c}{\log g_{j+1}(n)}\right) \left(1 - \frac{1.5 c}{\log g_j(n)} \right) \\[3pt] & \geq 1 - c\!\left(\frac{1.5}{\log g_j(n)} + \frac{1}{\log g_{j+1}(n)}\right) \\[3pt] & \geq 1 - \frac{ 1.5 c }{ \log g_{j+1}(n)}, \end{split} \end{equation*}

where the last inequality follows from (6).

For part (ii) we also use induction. The base case ( $i=0$ ) can be checked straightforwardly. For the inductive step,

\begin{equation*} \begin{split} \sum_{i=0}^{j+1} \frac{1}{\log g_i(n)} & \leq \frac{1}{\log g_{j+1}(n)} + \sum_{i=0}^{j} \frac{1}{\log g_i(n)} \leq \frac{1}{\log g_{j+1}(n)} + \frac{1.5}{\log g_j(n)} \leq \frac{1.5}{\log g_i(n)}, \end{split} \end{equation*}

where the last inequality follows from (6).

Proof of Lemma 2.8. We show that there exist suitable constants C, $D > 0$ and $\alpha \in (0,1)$ such that if $\mathcal{R}_1(X_t)\le C n^{4/3}$ and $\widetilde{\mathcal{R}}_\omega(X_t) > Dn^{4/3}\omega (n)^{-1/2}$ , then

(7) \begin{align} \mathcal{R}_1(X_{t+1}) &\le C n^{4/3},\ \text{and} \end{align}

(8) \begin{align} \widetilde{\mathcal{R}}_\omega(X_{t+1}) &\le (1-\alpha) \widetilde{\mathcal{R}}_\omega(X_t) \end{align}

with probability $\rho = \Omega(1)$ . This implies that we can maintain (7)–(8) for T steps with probability $\rho^T$ . Precisely, if we let

\begin{align*} \tau_1 &= \min \left\{t > 0\,:\, \mathcal{R}_1(X_{t}) > C n^{4/3}\right\}, \\ \tau_2 &= \min \left\{t > 0\,:\,\widetilde{\mathcal{R}}_\omega(X_{t}) > (1-\alpha) \widetilde{\mathcal{R}}_\omega(X_{t-1})\right\}, \\ T &= \min \left\{ \tau_1,\tau_2,c\log \omega (n)\right\}, \end{align*}

where the constant $c>0$ is chosen such that $(1-\alpha)^{c\log \omega (n)} = O(\omega (n)^{-1/2})$ , then $T = c\log \omega (n)$ with probability $\rho^{c \log \omega (n)}$ . (Note that $\rho^{c \log \omega (n)}= {\omega (n)}^{-\beta}$ for a suitable constant $\beta > 0$ .) Hence, $\mathcal{R}_1(X_T) = O\!\left(n^{4/3}\right)$ and

\begin{equation*}\widetilde{\mathcal{R}}_\omega(X_T) \le \widetilde{\mathcal{R}}_\omega(X_0) \cdot O\!\left(\omega (n)^{-1/2}\right) \le \mathcal{R}_1(X_0) \cdot O\!\left(\omega (n)^{-1/2}\right) = O\!\left(n^{4/3}\omega (n)^{-1/2}\right).\end{equation*}

The lemma then follows from the fact that $I(X_T) = \Omega(n)$ with probability $1-o(1)$ by Lemma 3.2 and a union bound.

To establish (7)–(8), let $\mathcal{H}^1_t$ be the event that $A(X_t) \in \left[n/q - \delta n^{2/3},n/q + \delta n^{2/3}\right]$ , where $\delta > 0$ is a constant. By Hoeffding’s inequality, for a suitable $\delta > 0$ , ${\mathbb{P}}[\mathcal{H}^1_t] \ge 1 - \frac{1}{8q^2}$ since $\mathcal{R}_1(X_t) = O\!\left(n^{4/3}\right)$ . Let $K_t$ denote the subgraph induced on the inactivated vertices at the step t. Observe that ${\mathbb{E}}\big[\widetilde{\mathcal{R}}_\omega(K_t)\big] = \left(1-\frac{1}{q}\right)\widetilde{\mathcal{R}}_\omega(X_{t})$ . Similarly, ${\mathbb{E}}\big[\mathcal{R}_1(K_t) - \widetilde{\mathcal{R}}_\omega(K_t)\big] = \left(1-\frac{1}{q}\right) \big(\mathcal{R}_1(X_{t}) - \widetilde{\mathcal{R}}_\omega(X_{t+1})\big)$ . Hence, by Markov’s inequality and independence between activation of each component, with probability at least $1/4q^2$ , the activation sub-step is such that $G_u$ satisfies

\begin{equation*}\widetilde{\mathcal{R}}_\omega(K_t) \le \left(1-\frac{1}{2q}\right)\widetilde{\mathcal{R}}_\omega(X_{t}),\end{equation*}

and

\begin{equation*}\mathcal{R}_1(K_t) - \widetilde{\mathcal{R}}_\omega(K_t) \le \left(1-\frac{1}{2q}\right)\left( \mathcal{R}_1(X_{t}) - \widetilde{\mathcal{R}}_\omega(X_{t+1}) \right).\end{equation*}

We denote this event by $\mathcal{H}^2_t$ . It follows by a union bound that $\mathcal{H}^1_t$ and $\mathcal{H}^2_t$ happen simultaneously with probability at least $1/8q^2$ . We assume that this is indeed the case and proceed to discuss the percolation sub-step.

Lemma 3.10 implies that there exists $C_1 > 0$ such that with probability $99/100$ ,

\begin{equation*}\widetilde{\mathcal{R}}_\omega\!\left(G\!\left(A(X_t), \frac{q}{n}\right)\right) \le C_1\frac{n^{4/3}}{{\omega (n)}^{1/2}}.\end{equation*}

Hence,

\begin{align*} \widetilde{\mathcal{R}}_\omega(X_{t+1}) =\widetilde{\mathcal{R}}_\omega(K_t) + \widetilde{\mathcal{R}}_\omega\!\left(G\!\left(A(X_t), \frac{q}{n}\right)\right) \le \left(1-\frac{1}{2q}\right)\widetilde{\mathcal{R}}_\omega(X_{t}) + C_1\frac{n^{4/3}}{{\omega (n)}^{1/2}} \le (1-\alpha) \widetilde{\mathcal{R}}_\omega(X_{t}), \end{align*}

where the last inequality holds for a suitable constant $\alpha \in (0,1)$ and a sufficiently large D since $\widetilde{\mathcal{R}}_\omega(X_t) > Dn^{4/3}\omega (n)^{-1/2}$ .

On the other hand, Lemma 3.9 implies ${\mathbb{E}}\left[\mathcal{R}_1\left(G\!\left(A(X_t), \frac{q}{n}\right)\right)\right] = O\!\left(n^{4/3}\right)$ . By Markov’s inequality, there exists $C_2$ such that, with probability $99/100$ ,

\begin{equation*} \mathcal{R}_1\!\left(G\!\left(A(X_t), \frac{q}{n}\right)\right) \le C_2n^{4/3}. \end{equation*}

For large enough C,

\begin{align*} \mathcal{R}_1(X_{t+1}) &\le \mathcal{R}_1(K_t) + \mathcal{R}_1\left(G\!\left(A(X_t), \frac{q}{n}\right)\right) \le \left( 1- \frac{1}{2q} \right)\mathcal{R}_1(X_{t}) + \mathcal{R}_1\left(G\!\left(A(X_t), \frac{q}{n}\right)\right) \\ &\le \left( 1- \frac{1}{2q} \right) Cn^{4/3} + C_2n^{4/3} \le Cn^{4/3} \end{align*}

Finally, it follows from a union bound that (7) and (8) hold simultaneously with probability at least $\frac{98}{100\cdot 8q^2}$ .

Proof of Lemma 2.9. First, both $\{X_t\}$ , $\{Y_t\}$ perform one independent CM step from the initial configurations $X_0$ , $Y_0$ . We start by establishing that $X_1$ and $Y_1$ preserve the structural properties assumed for $X_0$ and $Y_0$ .

By assumption $\mathcal{R}_1(X_{0}) = O\!\left(n^{4/3}\right)$ , so Hoeffding’s inequality implies that the number of activated vertices from $X_0$ is such that

\begin{equation*} A(X_0) \in I \,:\!=\, \left[n/q - O( n^{2/3}),n/q + O\big(n^{2/3}\big)\right] \end{equation*}

with probability $\Omega(1)$ . Then, the percolation step is distributed as a

\begin{equation*}G\!\left(A(X_0), \frac{1 + \lambda A(X_0)^{-1/3}}{A(X_0)}\right)\end{equation*}

random graph, with $|\lambda| = O(1)$ with probability $\Omega(1)$ . Conditioning on this event, from Lemma 3.2 we obtain that $I(X_1)=\Omega(n)$ w.h.p. Moreover, from Lemma 3.9 and Markov’s inequality we obtain that $\mathcal{R}_1(X_1) = O\!\left(n^{4/3}\right)$ with probability at least $99/100$ and from Lemma 3.10 that $\widetilde{\mathcal{R}}_\omega(X_1) = O\big(n^{4/3}{\omega (n)}^{-1/2}\big)$ also with probability at least $99/100$ .

We show next that $X_1$ and $Y_1$ , in addition to preserving the structural properties of $X_0$ and $Y_0$ , also have many connected components with sizes in certain carefully chosen intervals. This fact will be crucial in the design of our coupling. When $A(X_0) \in I$ , by Lemmas 3.11 and 3.12 and a union bound, for all integer $k \ge 0$ such that $n^{2/3}\omega (n)^{-2^k} \rightarrow \infty$ , $N_{k}(X_1, \omega) = \Omega(\omega (n)^{3 \cdot 2^{k-1}})$ w.h.p. (Recall, that $N_{k}(X_1, \omega)$ denotes the number of connected components of $X_1$ with sizes in the interval $\mathcal I_k(\omega)$ .) We will also require a bound for the number of components with sizes in the interval

\begin{equation*}J = \left[\frac{cn^{2/3}}{\omega (n)^{6}},\frac{2cn^{2/3}}{\omega (n)^{6}}\right], \end{equation*}

where $c > 0$ is a constant such that J does not intersect any of the $\mathcal I_k(\omega)$ ’s intervals. Let $W_X$ (resp., $W_Y$ ) be the set of components of $X_1$ (resp., $Y_1$ ) with sizes in the interval J. Lemma 3.11 then implies that for some positive constants $\delta_1, \delta_2$ independent of n,

\begin{equation*} {\mathbb{P}}\left[|W_X|\ge \delta_1n \left(\frac{\omega (n)^{6}}{cn^{2/3}}\right)^{3/2} \right] \ge 1- \frac{\delta_2}{n} \left(\frac{cn^{2/3}}{\omega (n)^{6}}\right)^{3/2} = 1 - O\!\left(\omega (n)^{-9}\right). \end{equation*}

All the bounds above apply also to the analogous quantities for $Y_1$ with the same respective probabilities. Therefore, by a union bound, all these properties hold simultaneously for both $X_1$ and $Y_1$ with probability $\Omega(1)$ . We assume that this is indeed the case and proceed to describe the second step of the coupling, in which we shall use each of the established properties for $X_1$ and $Y_1$ .

Recall $\mathcal{S}_{\omega}(X_1)$ and $\mathcal{S}_{\omega}(Y_1)$ denote the sets of connected components in $X_1$ and $Y_1$ , respectively, with sizes larger than $B_\omega$ . (Recall that $B_\omega = n^{2/3} \omega (n)^{-1}$ , where $\omega (n) = \log \log \log \log n$ .) Since $\mathcal{R}_1(X_1) = O\!\left(n^{4/3}\right)$ , the total number of components in $\mathcal{S}_{\omega}(X_1)$ is $O\big(\omega (n)^2\big)$ ; moreover, it follows from the Cauchy–Schwarz inequality that the total number of vertices in the components in $\mathcal{S}_{\omega}(X_1)$ , denoted $\|\mathcal{S}_{\omega}(X_1)\|$ , is $O\big(n^{2/3}\omega (n)\big)$ ; the same holds for $\mathcal{S}_{\omega}(Y_1)$ . Without loss of generality, let us assume that $\|\mathcal{S}_{\omega}(X_1)\| \ge \|\mathcal{S}_{\omega}(Y_1)\|$ . Let

\begin{equation*} \varGamma = \{C \subset W_Y\,:\, \|\mathcal{S}_{\omega}(Y_1) \cup C\| \ge \|\mathcal{S}_{\omega}(X_1)\| \}, \end{equation*}

and let $C_{\textrm{min}} = \arg \min_{C \in \varGamma} \|\mathcal{S}_{\omega}(Y_1) \cup C\|$ . In words, $C_{\textrm{min}}$ is the smallest subset C of components of $W_Y$ so that the number of vertices in the union of $\mathcal{S}_{\omega}(Y_1)$ and C is greater than that in $\mathcal{S}_{\omega}(X_1)$ . Since every component in $W_Y$ has size at least $cn^{2/3}\omega (n)^{-6}$ and $|W_Y| = \Omega\big(\omega (n)^9\big)$ , the number of vertices in $W_Y$ is $\Omega\big(n^{2/3}\omega (n)^3\big)$ and so $\varGamma \neq \emptyset$ . In addition, the number of components in $C_{\textrm{min}}$ is $O\big(\omega (n)^9\big)$ . Let $\mathcal{S}^{\prime}_{\omega}(Y_1) = \mathcal{S}_{\omega}(Y_1) \cup C_{\textrm{min}}$ and observe that the number of components in $\mathcal{S}^{\prime}_{\omega}(Y_1)$ is also $O\big(\omega (n)^9\big)$ and that

\begin{equation*}0 \le \|\mathcal{S}^{\prime}_{\omega}(Y_1)\| - \|\mathcal{S}_{\omega}(X_1)\| \le 2cn^{2/3}\omega (n)^{-6}. \end{equation*}

Note that $\|\mathcal{S}_{\omega}(X_1)\| - \|\mathcal{S}_{\omega}(Y_1)\|$ may be $\Omega\big(n^{2/3}\omega (n)\big)$ (i.e., much larger than $ \|\mathcal{S}^{\prime}_{\omega}(Y_1)\| - \|\mathcal{S}_{\omega}(X_1)\| $ ). Hence, if all the components from $\mathcal{S}_{\omega}(Y_1)$ and $\mathcal{S}_{\omega}(X_1)$ were activated, the difference in the number of active vertices could be $\Omega\big(n^{2/3}\omega (n)\big)$ . This difference cannot be corrected by our coupling for the activation of the small components. We shall require instead that all the components from $\mathcal{S}^{\prime}_{\omega}(Y_1)$ and $\mathcal{S}_{\omega}(X_1)$ are activated so that the difference is $O\big(n^{2/3}\omega (n)^{-6}\big)$ instead.

We now describe a coupling of the activation sub-step for the second step of the CM dynamics. As mentioned, our goal is to design a coupling in which the same number of vertices are activated from each copy. If indeed $A(X_1) = A(Y_1)$ , then we can choose an arbitrary bijective map $\varphi$ between the activated vertices of $X_1$ and the activated vertices of $Y_1$ and use $\varphi$ to couple the percolation sub-step. Specifically, if u and v were activated in $X_1$ , the state of the edges $\{u,v\}$ in $X_2$ and $\{\varphi(u),\varphi(v)\}$ in $Y_2$ would be the same. This yields a coupling of the percolation sub-step such that $X_2$ and $Y_2$ agree on the subgraph update at time 1.

Suppose then that in the second CM step all the components in $\mathcal{S}_{\omega}(X_1)$ and $\mathcal{S}^{\prime}_{\omega}(Y_1)$ are activated simultaneously. If this is the case, then the difference in the number of activated vertices is $d \le 2c n^{2/3}\omega (n)^{-6}$ . We will use a local limit theorem (i.e., Theorem 5.1) to argue that there is a coupling of the activation of the remaining components in $X_1$ and $Y_1$ such that the total number of active vertices in both copies is the same with probability $\Omega(1)$ . Since all the components in $\mathcal{S}_{\omega}(X_1)$ and $\mathcal{S}^{\prime}_{\omega}(Y_1)$ are activated with probability $\exp\big({-}O\big(\omega (n)^9\big)\big)$ , the overall success probability of the coupling will be $\exp\big({-}O\big(\omega (n)^9\big)\big)$ .

Now, let $x_1,x_2,\dots,x_m$ be the sizes of the components of $X_1$ that are not in $\mathcal{S}_{\omega}(X_1)$ (in increasing order). Let $\hat{A}(X_1)$ be the random variable corresponding to the number of active vertices from these components. Observe that $\hat{A}(X_1)$ is the sum of m independent random variables, where the j-th variable in the sum is equal to $x_j$ with probability $1/q$ , and it is 0 otherwise. We claim that sequence $x_1,x_2,\dots,x_m$ satisfies all the conditions in Theorem 5.1.

First, note that since the number of isolated vertices in $X_1$ is $\Omega(n)$ , $m = \Theta(n)$ and consequently $x_m = O\big(m^{2/3}\omega(m)^{-1}\big)$ , $\sum_{i=1}^m x_i^2 = \tilde{R}_\omega(X_1) = O\big(m^{4/3}\omega(m)^{-1/2}\big)$ and $x_i=1$ for all $i \le \rho m$ , where $\rho \in (0,1)$ is independent of m. Moreover, since $N_{k}(X_1, \omega) = \Omega\Big(\omega (n)^{3 \cdot 2^{k-1}}\Big)$ for all $k \ge 1$ such that $n^{2/3}\omega (n)^{-2^k} \rightarrow \infty$ ,

\begin{equation*}|\{i\,:\,x_i \in \mathcal{I}_k(\omega )\}| = \Omega\!\left(\omega(m)^{3\cdot2^{k-1}}\right).\end{equation*}

Since $N_0(X_1,\omega) = \Omega\!\left( \omega (n)^{3/2}\right) $ , we also have

\begin{equation*} \sum\nolimits_{i=1}^m x_i^2 \ge N_0(X_1,\omega) \cdot \frac{\vartheta^2 n^{4/3}}{4\omega (n)^2} = \Omega \!\left({m^{4/3}\omega(m)^{-1/2}} \right). \end{equation*}

Let $\mu_X = {\mathbb{E}}\big[\hat{A}(X_1)\big] = q^{-1}\sum_{i=1}^m x_i$ and let

\begin{equation*}\sigma_X^2 = {\textrm{Var}}\big(\hat{A}(X_1)\big) = q^{-1}\big(1-q^{-1}\big) \sum_{i=1}^m x_i^2 = \Theta\big(m^{4/3}\omega(m)^{-1/2}\big). \end{equation*}

Hence, Theorem 5.1 implies that ${\mathbb{P}} \big[\hat{A}(X_1) = a\big] = \Omega\!\left(\sigma_X^{-1}\right)$ for any $a \in [\mu_X-\sigma_X,\mu_X+\sigma_X]$ . Similarly, we get ${\mathbb{P}} \big[\hat{A}(Y_1) = a \big] = \Omega\big(\sigma_Y^{-1}\big)$ for any $a \in [\mu_Y-\sigma_Y,\mu_Y+\sigma_Y]$ , with $\hat{A}(Y_1)$ , $\mu_Y$ and $\sigma_Y$ defined analogously for $Y_1$ . Note that $\mu_X - \mu_Y = O\big(n^{2/3}\omega (n)^{-6}\big)$ and $\sigma_X , \sigma_Y= \Theta\big(n^{2/3} \omega (n)^{-1/4}\big)$ . Without loss of generality, suppose $\sigma_X < \sigma_Y$ . Then for any $a \in [\mu_X-\sigma_X /2,\mu_Y+\sigma_X /2]$ and $d = O\big(n^{2/3}\omega (n)^{-6}\big)$ , we have

\begin{equation*} \min \left\{{\mathbb{P}} \big[\hat{A}(X_1) = a \big], {\mathbb{P}} \big[\hat{A}(Y_1) = a - d \big]\right\} = \min \left\{\Omega \!\left( \sigma_X^{-1}\right), \Omega \!\left( \sigma_Y^{-1}\right) \right\} = \Omega \!\left(\sigma_Y^{-1}\right). \end{equation*}

Hence, there exists a coupling $\mathbb P$ of $\hat{A}(X_1)$ and $\hat{A}(Y_1)$ so that $\mathbb P\big[\hat{A}(X_1) = a, \hat{A}(Y_1) = a - d\big] = \Omega\big(\sigma_Y^{-1}\big)$ for all $a \in \left[\mu_X-\sigma_X /2,\mu_Y+\sigma_X /2\right]$ . Therefore, there is a coupling of $\hat{A}(X_1)$ and $\hat{A}(Y_1)$ such that

\begin{equation*} {\mathbb{P}}\big[\hat{A}(X_1) - \hat{A}(Y_1) = d \big] = \Omega \!\left( {\sigma_X}/{\sigma_Y} \right) = \Omega(1). \end{equation*}

Putting all these together, we deduce that $A(X_1) = A(Y_1)$ with probability $\exp\big({-}O\big(\omega (n)^{9}\big)\big)$ . If this is the case, the edge re-sampling step is coupled bijectively (as described above) so that $\mathcal{S}_{\omega}(X_2) = \mathcal{S}_{\omega}(Y_2)$ .

It remains for us to guarantee the additional desired structural properties of $X_2$ and $Y_2$ , which follow straightforwardly from the random graph estimates stated in Section 3. First note that by Hoeffding’s inequality, with probability $\Omega(1)$ ,

\begin{equation*} \left| A(X_1) - \frac{n}{q} - \frac{(q-1)\|\mathcal{S}_{\omega}(X_1)\|}{q} \right| = O\big(n^{2/3}\big).\end{equation*}

Hence, in the percolation sub-step the active subgraph is replaced by $F \sim G\!\left(A(X_1), \frac{1 + \lambda A(X_1)^{-1/3}}{A(X_1)}\right)$ , where $|\lambda| = O(\omega (n))$ with probability $\Omega(1)$ since $\|\mathcal{S}_{\omega}(X_1)\| = O\big(n^{2/3}\omega (n)\big)$ . Conditioning on this event, since the components of F contribute to both $X_2$ and $Y_2$ , Corollary 3.12 implies that w.h.p. $\hat{N}_k(2, \omega (n)) $ $ = \Omega\big( \omega (n)^{3 \cdot 2^{k-1}} \big)$ for all $k \ge 1$ such that $n^{2/3}\omega (n)^{-2^k}\rightarrow \infty$ . Moreover, from Lemma 3.2 we obtain that $I(X_2)=\Omega(n)$ w.h.p. From Lemma 3.4 and Markov’s inequality, we obtain that $\mathcal{R}_2(X_2) = O\!\left(n^{4/3}\right)$ with probability at least $99/100$ and from Lemma 3.10 that $\widetilde{\mathcal{R}}_\omega(X_2) = O\big(n^{4/3}{\omega (n)}^{-1/2}\big)$ also with probability at least $99/100$ . All these bounds apply also to the analogous quantities for $Y_2$ with the same respective probabilities.

Finally, we derive the bound for $L_1(X_2)$ and $L_1(Y_2)$ . First, notice $L_1(F)$ is stochastically dominated by $L_1(F^{\prime})$ , where $F^{\prime}\sim G\Big(A(X_1), \frac{1 + |\lambda| A(X_1)^{-1/3}}{A(X_1)}\Big)$ . Under the assumption that $|\lambda| = O(\omega (n))$ , if $|\lambda| \rightarrow \infty$ , then Corollary 3.6 implies that $L_1(F^{\prime}) = O(|\lambda| A(X_1)^{2/3}) = O\big(n^{2/3} \omega (n)\big)$ w.h.p.; otherwise, $|\lambda| = O(1)$ and by Lemma 3.9 and Markov’s inequality, $L_1(F^{\prime}) = O\big(n^{2/3}\big)$ with probability at least $99/100$ . Thus, $L_1(F) = O\big(n^{2/3}\omega (n)\big)$ with probability at least $99/100$ . We also know that the largest inactivated component in $X_1$ has size less than $n^{2/3}\omega (n)^{-1}$ , so $L_1(X_2) = O\big(n^{2/3} \omega (n)\big)$ with probability at least $99/100$ . The same holds for $Y_2$ . Therefore, by a union bound, all these properties hold simultaneously for both $X_2$ and $Y_2$ with probability $\Omega(1)$ , as claimed.

Proof. The activation coupling has two parts. First we use the maximal matching $W_t$ to couple the activation of a subset of the components in $M(X_t)$ and $M(Y_t)$ . Specifically, let $\ell$ be defined as in Theorem 5.1; for all $k \in [1, \ell]$ , we exclude $\Theta\big(\omega (n) ^ {3 \cdot 2 ^{k-1}}\big)$ pairs of components of size in the interval $\mathcal{I}_k(\omega)$ and we exclude $\Theta(n)$ pairs of matched isolated vertices. (These components exist by Assumptions 3 and 4.) All other pairs of components matched by $W_t$ are jointly activated (or not). Hence, the number of vertices activated from $X_t$ in this first part of the coupling is the same as that from $Y_t$ .

Let $\mathcal{C}(X_t)$ and $\mathcal{C}(Y_t)$ denote the sets containing the components in $X_t$ and components in $Y_t$ not considered to be activated in the first step of the coupling. This includes all the components from $D(X_t)$ and $D(Y_t)$ , and all the components from $M(X_t)$ and $M(Y_t)$ excluded in the first part of the coupling. Let $A^{\prime}(X_t)$ and $A^{\prime}(Y_t)$ denote the number of activated vertices from $\mathcal{C}(X_t)$ and $\mathcal{C}(Y_t)$ respectively. The second part is a coupling of the activation sub-step in a way such that

\begin{equation*}{\mathbb{P}} \left[ A^{\prime}(X_t) = A^{\prime}(Y_t) \right] = \Omega\big(\omega (n)^{-1}\big).\end{equation*}

Let $m_x \,:\!=\, |\mathcal{C}(X_t)| = \Theta(n)$ , and similarly for $m_y\,:\!=\,|\mathcal{C}(Y_t)|$ . Let $\mathcal{C}_1 \le \dots \le \mathcal{C}_{m_x}$ (resp., ${\mathcal{C}^{\prime}}_1 \le \dots \le {\mathcal{C}^{\prime}}_{m_y}$ ) be sizes of components in $\mathcal{C}(X_t)$ (resp., $\mathcal{C}(Y_t)$ ) in ascending order. For all $i\le m_x$ , let $\mathcal{X}_i$ be a random variable that equals to $\mathcal{C}_i$ with probability $1/q$ and 0 otherwise, which corresponds to the number of activated vertices from ith component in $\mathcal{C}(X_t)$ . Note that $\mathcal{X}_1, \dots, \mathcal{X}_{m_x}$ are independent. We check that $\mathcal{X}_1, \dots, \mathcal{X}_{m_x}$ satisfy all other conditions of Theorem 5.1.

Assumption $\mathcal{S}_{\omega}(X_t) = \mathcal{S}_{\omega}(Y_t)$ and the first part of the activation ensure that

\begin{equation*}\mathcal{C}_{m_x} \le B_\omega = O\!\left( n^{2/3}\omega (n)^{-1} \right) = O\!\left(m_x^{2/3}\omega(m_x)^{-1}\right).\end{equation*}

Observe also that there exists a constant $\rho$ such that $\mathcal{C}_i = 1$ for $i \le \rho m_x$ and $\lvert\{i \,:\, \mathcal{C}_i \in \mathcal{I}_k(\omega)\}\rvert = \Theta\!\left(\omega (n) ^ {3 \cdot 2 ^{k-1}}\right)$ for $ 1 \le k \le \ell$ ; lastly, from assumption $Z_t = O\!\left(\frac{n^{4/3}}{\omega (n)^{1/2}}\right)$ , we obtain

(10) \begin{align} \sum_{i = 1}^{m_x} \mathcal{C}_i^2 & \le Z_t + O(\rho m_x) + \sum_{k=1}^{\ell} \frac{\vartheta n^{4/3}}{\omega(n)^{2^{k+1}}} \cdot O\!\left(\omega (n)^{3\cdot 2^{k-1}} \right) \nonumber \\ & = O\!\left(\frac{m_x^{4/3}}{\sqrt{\omega(m_x)}}\right) + O\!\left( \sum_{k=1}^{\ell} \frac{m_x^{4/3}}{\omega(m_x)^{2^{k-1}}} \right) \nonumber \\ & = O\!\left(\frac{m_x^{4/3}}{\sqrt{\omega(m_x)}}\right) + O\!\left( \sum_{k=1}^{\ell} \frac{m_x^{4/3}}{\omega(m_x)^{k}} \right) \nonumber \\ & = O\!\left(\frac{m_x^{4/3}}{\sqrt{\omega(m_x)}}\right) + O\!\left(\frac{m_x^{4/3}}{\omega(m_x)}\right) = O\!\left(\frac{m_x^{4/3}}{\sqrt{\omega(m_x)}}\right) . \end{align}

Therefore, if $\mu_x = {\mathbb{E}} \left[ \sum_{i=1}^{m_x} \mathcal{X}_i \right] $ and $\sigma_x^2 = Var\!\left( \sum_{i=1}^{m_x} \mathcal{X}_i \right)$ , Theorem 5.1 implies that for any $x \in [\mu_x - \sigma_x, \mu_x + \sigma_x]$ ,

\begin{equation*}{\mathbb{P}} \left[A^{\prime}(X_t) = x\right] = {\mathbb{P}} \left[\sum_{i=1}^{m_x} \mathcal{X}_i = x\right] = \frac{1}{\sqrt{2\pi} \sigma_x} \exp\!\left({-}\frac{(x-\mu_x)^2}{2\sigma_x^2}\right) + o\!\left(\frac{1}{\sigma_x}\right) = \Omega \!\left( \frac{1}{\sigma_x}\right) .\end{equation*}

Similarly, we get that ${\mathbb{P}} \left[A^{\prime}(Y_t) = y\right] = \Omega (\sigma_y^{-1})$ for any $y \in [\mu_y - \sigma_y, \mu_y + \sigma_y]$ , with $\mu_y$ and $\sigma_y$ defined analogously. Without loss of generality, suppose $\sigma_y \le \sigma_x$ . Since $\mu_x = \mu_y$ , for $x \in \left[\mu_x - \sigma_y, \mu_x + \sigma_y\right]$ , we obtain

\begin{equation*} \min \left\{{\mathbb{P}} \left[A^{\prime}(X_t) = x \right], {\mathbb{P}} \left[A^{\prime}(Y_t) = x \right]\right\} = \Omega \!\left( \frac{1}{\sigma_x}\right). \end{equation*}

Hence, we can couple $(A^{\prime}(X_t), A^{\prime}(Y_t))$ so that ${\mathbb{P}} [A^{\prime}(X_t) = A^{\prime}(Y_t) = x] = \Omega(\sigma_x^{-1})$ for all $x \in [\mu_x - \sigma_y, \mu_x + \sigma_y]$ . Consequently, under this coupling,

\begin{equation*} {\mathbb{P}} \left[ A^{\prime}(X_t) = A^{\prime}(Y_t) \right] = \Omega\!\left(\frac{\sigma_y}{\sigma_x}\right). \end{equation*}

Since $\mathcal{X}_1, \dots, \mathcal{X}_{m_x}$ are independent, $\sigma_x^2 = \Theta \!\left(\sum_{i = 1}^{m_x} \mathcal{C}_i^2\right)$ , and similarly $\sigma_y^2 = \Theta \!\left(\sum_{i = 1}^{m_y} {\mathcal{C}^{\prime}}_i^2\right)$ . Hence, inequality (10) gives an upper bound for $\sigma_x^2$ ; meanwhile, a lower bound for $\sigma_y^2$ can be obtained by counting components in the largest interval:

\begin{equation*} \sum_{i = 1}^{m_y} {\mathcal{C}^{\prime}}_i^2 \ge \sum_{i\,:\, {\mathcal{C}^{\prime}}_i \in \mathcal{I}_1(\omega)} {\mathcal{C}^{\prime}}_i^2 \ge \frac{Bn^{4/3}}{\omega(n)^{4}} \cdot \Theta\!\left(\omega (n)^{3} \right) = \Omega\!\left(\frac{n^{4/3}}{\omega(n)} \right). \end{equation*}

Therefore,

\begin{equation*} {\mathbb{P}} \left[ A^{\prime}(X_t) = A^{\prime}(Y_t) \right] = \Omega\!\left(\frac{n^{2/3}}{\omega (n)^{1/2}} \cdot \frac{\omega(m_x)^{1/4} }{m_x^{2/3}}\right) = \Omega \!\left( \frac{1}{\omega (n)}\right), \end{equation*}

as desired.

Proof of Lemma 2.10. Let $C_1$ be a suitable constant that we choose later. We wish to maintain the following properties for all $t \le T \,:\!=\, C_1 \log \omega (n)$ :

1. 1. $\mathcal{S}_{\omega}(X_t) = \mathcal{S}_{\omega}(Y_t)$ ;

2. 2. $Z_t = O\!\left(\frac{n^{4/3}}{\omega (n)^{1/2}}\right)$ ;

3. 3. $\hat{N}_k(X_t, Y_t, \omega (n)) = \Omega\!\left(\omega (n) ^ {3 \cdot 2 ^{k-1}}\right)$ for all $k \ge 1$ such that $n^{2/3}\omega (n)^{-2^k}\rightarrow \infty$ ;

4. 4. $I(X_t), I(Y_t) = \Omega(n)$ ;

5. 5. $\mathcal{R}_2(X_t), \mathcal{R}_2(Y_t) = O\!\left(n^{4/3}\right)$ ;

6. 6. $L_1 (X_t) \le \alpha^t L_1(X_0), L_1 (Y_t) \le \alpha^t L_1(Y_0)$ for some constant $\alpha$ independent of t.

By assumption, $X_0$ and $Y_0$ satisfy these properties. Suppose that $X_t$ and $Y_t$ satisfy these properties at step $t \le T$ . We show that there exists a one-step coupling of the CM dynamics such that $X_{t+1}$ and $Y_{t+1}$ preserve all six properties with probability $\Omega \!\left(\omega (n)^{-1} \right)$ .

We provide the high-level ideas of the proof first. We will crucially exploit the coupling from Lemma 5.2. Assuming $A(X_t) = A(Y_t)$ , properties 1 and 2 hold immediately at $t + 1$ , and properties 3 and 4 can be shown by a ‘standard’ approach used throughout the paper. In addition, we reuse simple arguments from previous stages to guarantee properties 5 and 6.

Consider first the activation sub-step. By Lemma 5.2, $A(X_t) = A(Y_t)$ with probability at least $\Omega(\omega (n)^{-1})$ . If the number of vertices in the percolation is the same in both copies, we can couple the edge re-sampling so that the updated part of the configuration is identical in both copies. In other words, all new components created in this step are automatically contained in the component matching $W_{t+1}$ ; this includes all new components whose sizes are greater than $B_\omega$ . Since none of the new components contributes to $Z_{t+1}$ , we obtain $Z_{t+1} \le Z_t = O\!\left(\frac{n^{4/3}}{\omega (n)^{1/2}}\right)$ . Therefore, $A(X_t) = A(Y_t)$ immediately implies properties 1 and 2 at time $t+1$ .

With probability $1/q$ , the largest components of $X_t$ and $Y_t$ are activated simultaneously. Suppose that this is the case. By Hoeffding’s inequality, for constant $K>0$ , we have

\begin{equation*}{\mathbb{P}} \left[ \left\lvert A(X_t) -{\mathbb{E}}\left[A(X_t)\right] \right\rvert \ge Kn^{2/3} \right] \le \exp\!\left({-}\frac{K^2 n^{4/3}}{\mathcal{R}_2(X_t)}\right).\end{equation*}

Property 5 and the observation that ${\mathbb{E}}\left[A(X_t)\right] = L_1(X_t) + \frac{n - L_1(X_t)}{q}$ imply that

\begin{equation*}{\mathbb{P}} \left[ \left\lvert A(X_t) -L_1(X_t) - \frac{n - L_1(X_t)}{q} \right\rvert \ge Kn^{2/3} \right] = O(1).\end{equation*}

By noting that $L_1(Y_0), L_1(X_0) \le n^{2/3} \omega (n)$ , property 6 implies that

(11) \begin{equation} {\mathbb{P}}\left[A(X_t) \cdot \frac{q}{n} \le 1 + \frac{(q-1)\omega (n) + Kq}{n^{1/3}} \right] = \Omega(1). \end{equation}

We denote $A(X_t) = A(Y_t)$ by m. By inequality (11), with at least constant probability, the random graph for both chains is $H \sim G\!\left(m, \frac{1 + \lambda m^{-1/3}}{m}\right)$ , where $\lambda \le \omega(m)$ . Let us assume that is the case. Corollary 3.12 ensures that there exists a constant $b > 0$ such that, with probability at least $1- O\big(\omega (n)^{-3}\big)$ , $N_{k}(H,\omega (n)) \ge b \omega (n)^{3 \cdot 2^{k-1}}$ for all $k \ge 1$ such that $n^{2/3}\omega (n)^{-2^k}\rightarrow \infty$ . Since components in H are simultaneously added to both $X_{t+1}$ and $Y_{t+1}$ , property 3 is satisfied. Moreover, Lemma 3.2 implies that with high probability $\Omega(n)$ isolated vertices are added to $X_{t+1}$ and $Y_{t+1}$ , and thus property 4 is satisfied at time $t+1$ .

In addition, Lemma 3.4 and Markov’s inequality imply that there exists a constant $C_2$ such that

\begin{equation*}{\mathbb{P}} \left[\mathcal{R}_2(X_{t+1}) = C_2n^{4/3} \right] \ge \frac{99}{100};\end{equation*}

By Lemma 4.3, there exists $\alpha < 1$ such that with at least probability $99/100$

\begin{equation*}L_1(X_{t+1}) \le \max \{\alpha L_1(X_t) ,L_2(X_t) \},\end{equation*}

where $\alpha$ is independent of t and n. Potentially, property 6 may not hold when $\alpha L_1(X_t) < L_1(X_{t+1}) \le L_2(X_t) = O\big(n^{2/3}\big)$ , but then we stop at this point. (We will argue that in this case all the desired properties are also established shortly.) Hence, we suppose otherwise and establish properties 5 and 6 for $X_{t+1}$ . Similar bounds hold for $Y_{t+1}$ .

By a union bound, $X_{t+1}$ and $Y_{t+1}$ have all six properties with probability at least $92/100$ , assuming the activation sub-step satisfies all the desired properties, and thus overall with probability $\Omega \!\left(\omega (n)^{-1} \right)$ . Inductively, the probability that $X_T$ and $Y_T$ satisfy the six properties is

\begin{equation*}O\!\left(\omega (n)\right)^{-C_1 \log \omega (n)} = \exp\!\left( \log O(\omega (n))^{-C_1 \log \omega (n)}\right) = \exp \!\left({-} O\!\left( ({\log} \omega (n))^2 \right) \right).\end{equation*}

Suppose $X_T$ and $Y_T$ have the six properties. By choosing $C_1 > 1 / \log \frac{1}{\alpha}$ , properties 5 and 6 imply

\begin{equation*}\mathcal{R}_1(X_T) = L_1(X_T)^2 + \mathcal{R}_2(X_T) \le \left( \alpha^{C_1 \log \omega (n)} n^{2/3} \omega (n) \right)^2 + O\!\left(n^{4/3}\right) = O\!\left(n^{4/3}\right),\end{equation*}

and $\mathcal{R}_1(Y_T) = O\!\left(n^{4/3}\right)$ . While the lemma almost follows from these properties, notice that property 3 does not match the desired bounds on the components in the lemma statement. To fix this issue, we perform one additional step of the coupling.

Consider the activation sub-step at T. Assume again $A(X_T) = A(Y_T) \,=\!:\, m^{\prime}$ . By Hoeffding’s inequality, for some constant K ′, we obtain

(12) \begin{equation} {\mathbb{P}}\left[ \left| m^{\prime} \cdot \frac{q}{n} - 1 \right| > \frac{K^{\prime}}{n^{1/3}} \right] ={\mathbb{P}}\left[ \left| m^{\prime} - \frac{n}{q} \right| > K^{\prime}n^{2/3} \right] \le \exp\!\left( \frac{-{K^{\prime}}^2 n^{4/3}}{\mathcal{R}_1(X_T)}\right) = O(1). \end{equation}

Let $\lambda^{\prime} \,:\!=\, (m^{\prime}qn^{-1} - 1)\cdot m^{\prime 1/3}$ . Inequality (12) implies with at least constant probability the random graph in the percolation step is $H^{\prime} \sim G\!\left(m^{\prime}, \frac{1 + \lambda^{\prime} m^{\prime-1/3}}{m^{\prime}}\right)$ , where $\lambda^{\prime} \le K^{\prime}$ and $m^{\prime} \in (n/2q,n)$ . If so, Corollary 3.12 ensures with high probability $\hat{N}_k(X_{T+1}, Y_{T+1}, \omega (n)^{1/2})= \Omega\Big({\omega (n)}^{3 \cdot 2^{k-2}}\Big)$ for all $k \ge 1$ such that $n^{2/3}\omega (n)^{-2^{k-1}}\rightarrow \infty$ .

By the preceding argument, with $\Omega(\omega (n)^{-1})$ probability, the six properties are still valid at step $T+1$ , so the proof is complete. We note that if we had to stop earlier because property 6 did not hold, we perform one extra step (as above) to ensure that $\hat{N}_k\big(X_{T+1}, Y_{T+1}, \omega (n)^{1/2}\big)= \Omega\Big({\omega (n)}^{3 \cdot 2^{k-2}}\Big)$ for all $k \ge 1$ such that $n^{2/3}\omega (n)^{-2^{k-1}}\rightarrow \infty$ .

Proof of Lemma 2.11. The coupling has four phases: in phase 1 we will consider $O({\log} \log \log \log n)$ steps of the coupling, $O({\log} \log \log n)$ steps in phase 2, $O({\log} \log n)$ steps in phase 3 and phase 4 consists of $O({\log}\ n)$ steps.

We will keep track of the random variables $ \mathcal{R}_1(X_t), \mathcal{R}_1(Y_t), I(X_t), I(Y_t), Z_t$ and $\hat{N}_{k}(t, g)$ for a function g we shall carefully choose for each phase, and use these random variables to derive bounds on the probability of various events.

Phase 1. We set $g_1(n) =\omega (n)^{1/2}$ and $h_1(n) = K^2\omega (n)^{1/2}$ where $K>0$ is a constant we choose. Let $a_1 \,:\!=\, 1 - \frac{1}{2q}$ and let $T_1\,:\!=\,-12\log_{a_1} ({\log} \log \log n)$ , and we fix $t < T_1$ . Suppose we have $\mathcal{R}_1(X_t) + \mathcal{R}_1(Y_t) \le C_1 n^{4/3}$ , $I(X_t), I(Y_t) = \Omega(n)$ , and $\hat{N}_{k}(t, g_1) = \Omega\Big(g_1(n)^{3 \cdot 2^{k-1} }\Big)$ for all $k \ge 1$ such that $n^{2/3}g_1(n)^{-2^k}\rightarrow \infty$ , where $C_1 > 0$ is a large constant that we choose later.

By Lemma 5.5, for a sufficiently large constant $K>0$ , we obtain a coupling for the activation of $X_t$ and $Y_t$ such that the same number of vertices are activated in $X_t$ and $Y_t$ , with probability at least

\begin{equation*}1- 4e^{-2K}- \sqrt{\frac{K\omega (n)^{1/2}}{K^2\omega (n)^{1/2}}} - \frac{\delta}{\omega (n)^{1/2}} \ge 1 - \frac{1}{16q^2}.\end{equation*}

By Lemma 5.4, $A(X_t) \in (n/2q,n)$ and $\lambda \,:\!=\, (A(X_t)q/n - 1)\cdot A(X_t)^{1/3} = O(1)$ with probability at least $ 1 - \frac{1}{16q^2}$ . It follows a union bound that $A(X_t) = A(Y_t)$ , $A(X_t) \in (n/2q,n)$ and $\lambda = O(1)$ with probability $1 - \frac{1}{8q^2}$ . We call this event $\mathcal{H}^1_t$ .

Let $D^{\prime}_t$ denote the inactivated components in $D(X_t) \cup D(Y_t)$ at the step t, and $M^{\prime}_t$ the inactivated components in $M(X_t) \cup M(X_t)$ . Observe that

\begin{equation*}{\mathbb{E}}\left[\sum\nolimits_{\mathcal{C} \in D^{\prime}_t} |\mathcal{C}|^2\right] = \left(1 -\frac{1}{q} \right) \sum\nolimits_{\mathcal{C} \in D(X_t) \cup D(Y_t)} |\mathcal{C}|^2 = \left(1 -\frac{1}{q} \right) Z_t.\end{equation*}

Similarly,

\begin{equation*} {\mathbb{E}}\left[\sum\nolimits_{\mathcal{C} \in M^{\prime}_t} |\mathcal{C}|^2\right] = \left(1 -\frac{1}{q} \right) \sum\nolimits_{\mathcal{C} \in M(X_t) \cup M(Y_t)} |\mathcal{C}|^2 = \left(1 -\frac{1}{q} \right) \left(\mathcal{R}_1(X_t) + \mathcal{R}_1(Y_t) - Z_t \right). \end{equation*}

Hence, by Markov’s inequality and independence between activation of components in $D^{\prime}_t$ and components in $M^{\prime}_t$ , with probability at least $1/4q^2$ , the activation sub-step is such that

\begin{equation*}\sum\nolimits_{\mathcal{C} \in D^{\prime}_t} |\mathcal{C}|^2 \le \left(1-\frac{1}{2q}\right) Z_t,\end{equation*}

and

\begin{equation*}\sum\nolimits_{\mathcal{C} \in M^{\prime}_t} |\mathcal{C}|^2 \le \left(1-\frac{1}{2q}\right) \left(\mathcal{R}_1(X_t) + \mathcal{R}_1(Y_t) - Z_t \right).\end{equation*}

We denote this event by $\mathcal{H}^2_t$ . By a union bound, $\mathcal{H}^1_t$ and $\mathcal{H}^2_t$ happen simultaneously with probability $1/8q^2$ .

Suppose all these events indeed happen; then we couple the percolation step so that the components newly generated in both copies are exactly identical, and we claim that all of the following holds with at least constant probability:

1. 1. $\mathcal{R}_1(X_{t+1}) + \mathcal{R}_1(Y_{t+1}) \le C_1 n^{4/3}$ ;

2. 2. $Z_{t+1} \le a_1 Z_t$ ;

3. 3. $I(X_{t+1}), I(Y_{t+1}) = \Omega(n)$ ;

4. 4. $\hat{N}_{k}(t + 1, g_1) = \Omega\Big(g_1(n)^{3 \cdot 2^{k-1}}\Big)$ for all $k \ge 1$ such that $n^{2/3}g_1(n)^{-2^k}\rightarrow \infty$ .

First, note that $Z_{t+1}$ can not possibly increase because the matching $W_{t+1}$ can only grow under the coupling if indeed $A(X_t) = A(Y_t)$ . Observe that only the inactivated components in $X_t$ and $Y_t$ would contribute to $Z_{t+1}$ , so

\begin{equation*}Z_{t+1} = \sum\nolimits_{\mathcal{C} \in D^{\prime}_t} |\mathcal{C}|^2 \le a_1 Z_t.\end{equation*}

Next, we establish the properties 3 and 4. For this, notice that the percolation step is distributed as a $H \sim $ $G\!\left(A(X_t), \frac{1 + \lambda A(X_t)^{-1/3}}{A(X_t)}\right)$ random graph. Corollary 3.12 implies $N_{k}(H, g_1) = \Omega\Big( g_1(n)^{3 \cdot 2^{k-1}}\Big)$ for all $k \ge 1$ such that $n^{2/3}g_1(n)^{-2^k}\rightarrow \infty$ , with probability at least $1 - O\!\left(g_1(n)^{-3}\right) $ . Moreover, Lemma 3.2 implies that with high probability $I(H) = \Omega(n)$ . Since the percolation step is coupled, this implies that both $X_{t+1}$ and $Y_{t+1}$ will have all the components in H, so we have $\hat{N}_{k}(t+1, g_1) = \Omega\Big(g_1(n)^{3 \cdot 2^{k-1}}\Big)$ for all $k \ge 1$ such that $n^{2/3}g_1(n)^{-2^k}\rightarrow \infty$ , and $I(X_{t+1}), I(Y_{t+1}) = \Omega(n)$ , w.h.p.

Finally, assuming that $|\lambda| = O(1)$ , by Lemma 3.9 and Markov’s inequality, there exists $C_2 > 0$ such that ${\mathbb{E}} \left[\mathcal{R}_1(H) \right] = C_2n^{4/3}$ with probability at least $99/100$ . Then

\begin{align*} \mathcal{R}_1(X_{t+1}) + \mathcal{R}_1(Y_{t+1}) &= \sum\nolimits_{\mathcal{C} \in D^{\prime}_t} |\mathcal{C}|^2 + \sum\nolimits_{\mathcal{C} \in M^{\prime}_t} |\mathcal{C}|^2 + \mathcal{R}_1(H) \\ & \le a_1 \left(\mathcal{R}_1(X_{t}) + \mathcal{R}_1(Y_{t})\right) + C_2n^{4/3} \le a_1 C_1 n^{4/3} + C_2n^{4/3} \le C_1 n^{4/3}, \end{align*}

for large enough $C_1$ . A union bound implies that all four properties hold with at least constant probability $\varrho > 0$ .

Thus, the probability that at each step of update all four properties can be maintained throughout Phase 1 is at least

\begin{equation*}\varrho^{T_1} = \varrho^{-12\log_{a_1} ({\log} \log \log n)} = \left({\log} \log \log n\right)^{-12\log_\varrho (a_1)}.\end{equation*}

If property 2 holds at the end of Phase 1, we have

\begin{equation*}Z_{T_1} = O\!\left( \frac{n^{4/3}}{h_1(n)} \cdot a_1^{T_1} \right) = O\!\left( \frac{n^{4/3}}{h_1(n)} \cdot a_1^{ -12 \log_{a_1}({\log} \log \log n)} \right) = O\!\left( \frac{n^{4/3}}{({\log} \log \log n)^{12}} \right).\end{equation*}

To facilitate discussions in Phase 2, we show that the two copies of chains satisfy one additional property at the end of Phase 1. In particular, there exist a lower bound for the number of components in a different set of intervals. We consider the last percolation step in Phase 1. Then, Corollary 3.12 with $g_2(n) \,:\!=\, \left( \log\log\log n \cdot \log\log\log\log n \right)^2$ implies $\hat{N}_{k}(T_1, g_2) = \Omega\Big(g_2(n)^{3 \cdot 2^{k-1}}\Big)$ for all $k \ge 1$ such that $n^{2/3}g_2(n)^{-2^k}\rightarrow \infty$ , with high probability.