Phase transition of a nonlinear opinion dynamics with noisy interactions

Abstract

In several real Multi-Agent Systems, it has been observed that only weaker forms of metastable consensus are achieved, in which a large majority of agents agree on some opinion while other opinions continue to be supported by a (small) minority of agents. In this work, we take a step towards the investigation of metastable consensus for complex (nonlinear) opinion dynamics by considering the popular Undecided-State dynamics in the binary setting, which is known to reach consensus exponentially faster than the Voter dynamics. We propose a simple form of uniform noise in which each message can change to another one with probability p and we prove that the persistence of a metastable consensus undergoes a phase transition for $p=\frac{1}{6}$. In detail, below this threshold, we prove the system reaches with high probability a metastable regime where a large majority of agents keeps supporting the same opinion for polynomial time. Moreover, this opinion turns out to be the initial majority opinion, whenever the initial bias is slightly larger than its standard deviation. On the contrary, above the threshold, we show that the information about the initial majority opinion is “lost” within logarithmic time even when the initial bias is maximum. Interestingly, we show our results have explicit connections to two different concrete frameworks. The first one concerns a specific setting of a well-studied value-sensitive decision mechanism inspired by cross-inhibition in house-hunting honeybee swarms. The second framework consists of a consensus process where a subset of agents behave in a stubborn way.

Appendices

Appendix 1: Useful tools

We recall the concentration results that we use in the analysis. For an overview on the forms of Chernoff bounds see the works by Dubhashi and Panconesi (2009) or Doerr (2020).

Lemma 16

(Multiplicative forms of Chernoff bounds) Let $X_1, X_2, \dots , X_n$ be independent $\{0,1\}$ random variables. Let $X = \sum _{i=1}^n X_i$ and $\mu =\mathbb {E}[X]$. Then:

• (i) for any $\delta \in (0,1)$ and $\mu \le \mu _+ \le n$, it holds that

  $$\begin{aligned} P\big (X\ge (1+\delta )\mu _+\big )\le e^{-\frac{1}{3}\delta ^2\mu _+}, \end{aligned}$$

  (A1)

• (ii) for any $\delta \in (0,1)$ and $0 \le \mu _- \le \mu$, it holds that

  $$\begin{aligned} P\big (X\le (1-\delta )\mu _-\big )\le e^{-\frac{1}{2}\delta ^2\mu _-}. \end{aligned}$$

  (A2)

Lemma 17

(Additive forms of Chernoff bounds) Let $X_1, X_2, \dots , X_n$ be independent $\{0,1\}$ random variables. Let $X = \sum _{i=1}^n X_i$ and $\mu =\mathbb {E}[X]$. Then:

• (i) for any $0< \lambda < n$ and $\mu \le \mu _+ \le n$, it holds that

  $$\begin{aligned} P\big (X\ge \mu _+ +\lambda \big )\le e^{-\frac{2}{n}\lambda ^2}, \end{aligned}$$

  (A3)

• (ii) for any $0< \lambda < \mu _-$ and $0 \le \mu _- \le \mu$, it holds that

  $$\begin{aligned} P\big (X\le \mu _- - \lambda \big )\le e^{-\frac{2}{n}\lambda ^2}. \end{aligned}$$

  (A4)

We also make use of the Hoeffding bounds (Mitzenmacher & Upfal, 2005).

Lemma 18

(Hoeffding bounds) Let $0< a < b$ be two constants. Let $X_1, X_2, \dots , X_n$ be independent random variables such that $\mathbb {P}\left[ a \le X_i \le b\right] = 1$ and ${\mathbb {E}}\left[ X_i \right] = \mu /n$ for all $i \le n$, and let $X = \sum _{i=1}^n X_i$. Then:

• (i) for any $\lambda > 0$ and $\mu \le \mu _+$, it holds that

  $$\begin{aligned} \mathbb {P}\left[ X \ge \mu _+ + \lambda n\right] \le \exp \left( - \frac{2\lambda ^2 n }{(b-a)^2}\right) ; \end{aligned}$$

  (A5)

• (ii) for any $\lambda > 0$ and $0 \le \mu _- \le \mu$, it holds that

  $$\begin{aligned} \mathbb {P}\left[ X \le \mu _- - \lambda n\right] \le 2\exp \left( - \frac{2\lambda ^2 n }{(b-a)^2}\right) . \end{aligned}$$

  (A6)

The Berry–Esseen theorem is well treated by Korolev and Shevtsova (2010), and it gives an estimation on “how far” is the distribution of the normalized sum of i.i.d. random variables to the standard normal distribution.

Lemma 19

(Berry-Esseen) Let $X_1, \dots , X_n$ be n i.i.d. (either discrete or continuous) random variables with zero mean, variance $\sigma ^2>0$, and finite third moment. Let Z the standard normal random variable, with zero mean and variance equal to 1. Let $F_n(x)$ be the cumulative function of $\frac{S_n}{\sigma \sqrt{n}}$, where $S_n = \sum _{i=1}^n X_i$, and $\Phi (x)$ that of Z. Then, there exists a positive constant $C>0$ such that

$$\begin{aligned} \sup _{x\in \mathbb {R}} \left|F_n(x) - \Phi (x) \right| \le \frac{C}{\sqrt{n}} \end{aligned}$$

for all $n\ge 1$.

Our analysis makes use of the following probabilistic results, which hold for n large enough.

Lemma 20

Let $\eta> \lambda > 0$ be two constants, and $0 < p \le 1$ be a probability. Consider any family of events $\{\xi _i\}_{i \le M}$ with $M > 1$ being some integer. Suppose $\mathbb {P}\left[ \xi _1\right] \ge p$, and, for $i \ge 2$, that $\mathbb {P}\left[ \xi _i \big |\ \xi _1, \dots , \xi _{i-1}\right] = \mathbb {P}\left[ \xi _i \big |\ \xi _{i-1}\right] \ge p$. The following holds.

• 1. If $p = 1 - n^{-\eta }$ and $M \le n^{\lambda }$, then $\cap _{i \le M} \xi _i$ holds with probability $1 - \mathcal {O}\left( n^{-(\eta - \lambda )}\right)$.

• 2. If $p = 1 - \exp \left( -\eta n \right)$ and $M \le e^{\lambda }$. Then $\cap _{i \le M} \xi _i$ holds with probability $1 - \mathcal {O}\left( e^{-(\eta - \lambda )n}\right)$.

Proof of Lemma 20

We have that

$$\begin{aligned} \mathbb {P}\left[ \cap _{i \le M} \xi _i\right]&= \mathbb {P}\left[ \xi _M \big |\ \cap _{i \le M-1} \xi _i\right] \cdot \mathbb {P}\left[ \cap _{i \le M-1} \xi _i\right] \\&= \mathbb {P}\left[ \xi _M \big |\ \xi _{M-1}\right] \cdot \mathbb {P}\left[ \cap _{i \le M-1} \xi _i\right] \\&= \prod _{i = 2}^{M} \mathbb {P}\left[ \xi _i \big |\ \xi _{i-1}\right] \cdot \mathbb {P}\left[ \xi _1\right] \\&\ge p^M \end{aligned}$$

Now, let $f(n) = 1 - p = o(1)$. Notice that $1 - x \ge e^{-x/(1-x)}$ for $\left|x \right| < 1$. Then,

$$\begin{aligned} p^M&= \left[ 1 - f(n)\right] ^M \\&\ge \exp \left( - \frac{M f(n)}{1 - f(n)}\right) \\&\ge \exp \left( - 2 M f(n)\right) \\&{\mathop {\ge }\limits ^{(a)}} 1 - 4Mf(n), \end{aligned}$$

where (a) holds of the exponential function Taylor’s expansion, since $Mf(n) = o(1)$ by the hypotheses. As for (i), we get

$$\begin{aligned} p^M \ge 1 - \mathcal {O}\left( \frac{n^\lambda }{n^{\eta }}\right) = 1 - \mathcal {O}\left( \frac{1}{n^{\eta - \lambda }}\right) . \end{aligned}$$

As for (ii), we get

$$\begin{aligned} p^M \ge 1 - \mathcal {O}\left( \frac{\exp (\lambda n )}{\exp (\eta n)}\right) = 1 - \mathcal {O}\left( e^{-(\eta - \lambda )n}\right) . \end{aligned}$$

$\square$

More easily, the union bound implies the following.

Lemma 21

Let $\eta> \lambda > 0$ be two constants, and $0 < p \le 1$ be a probability. Consider any family of events $\{\xi _i\}_{i \le M}$ with $M > 1$ being some integer. Suppose $\mathbb {P}\left[ \xi _i\right] \ge p$ for all i. The following holds.

• 1. If $p = 1 - n^{-\eta }$ and $M \le n^{\lambda }$, then $\cap _{i \le M} \xi _i$ holds with probability $1 - n^{-(\eta - \lambda )}$.

• 2. If $p = 1 - \exp \left( -\eta n \right)$ and $M \le e^{\lambda }$. Then $\cap _{i \le M} \xi _i$ holds with probability $1 - e^{-(\eta - \lambda )n}$.

Appendix 2: Omitted proofs

1.1 Proofs: preliminaries

Proof of Equations 3–6

Conditional on any configuration at time t, the probability an agent pulls opinion Alpha at the next round is

$$\begin{aligned} (1 - 3p)\frac{a}{n} + p, \end{aligned}$$

and symmetrical expressions hold for opinion Beta and the undecided state. An agent updates its opinion to Alpha at time $t+1$ if it is undecided at time t and pulls opinion Alpha, or if it supports opinion Alpha at time t and pulls either opinion Alpha or the undecided state. If $V_q$ and $V_a$ denote the sets of agents supporting the undecided state and opinion Alpha, respectively, at time t, we have

$$\begin{aligned} {\mathbb {E}}\left[ A \big |\ \mathbf {X}_t = \mathbf {x} \right]&= \sum _{u \in V_a} \left( (1 - 3p)\frac{a}{n} + p + (1 - 3p)\frac{q}{n} + p\right) + \sum _{u \in V_q}\left( (1 - 3p)\frac{a}{n} + p\right) \\&= a\left( (1 - 3p)\frac{a}{n} + p + (1 - 3p)\frac{q}{n} + p\right) + q \left( (1 - 3p)\frac{a}{n} + p\right) \\&= (1-3p)\frac{a}{n}(a + 2q) + p(2a + q). \end{aligned}$$

Similarly, we get the conditional expectation of B. Then

$$\begin{aligned} {\mathbb {E}}\left[ S\big |\ \mathbf {X}_t = \mathbf {x} \right] = &{\mathbb {E}}\left[ A\big |\ \mathbf {X}_t = \mathbf {x} \right] - {\mathbb {E}}\left[ B\big |\ \mathbf {X}_t = \mathbf {x} \right] \\ = &\frac{1-2p}{n}\left[ s(a+b)+2qs\right] + \frac{p}{n}\left[ s(a+b)\right] \\ = &s\frac{1-2p}{n}\left[ n+q\right] + s\frac{p}{n}\left[ n-q\right] = s\left( 1-p+(1-3p)\frac{q}{n}\right) ,\\ {\mathbb {E}}\left[ Q\big |\ \mathbf {X}_t = \mathbf {x} \right] = &+ \frac{a}{n}\left[ p(a+q)+(1-2p)b\right] + \frac{b}{n}\left[ p(b+q)+(1-2p)a\right] \\&+ \frac{q}{n}\left[ p(a+b)+(1-2p)q\right] \\ = &\frac{p}{n}\left[ a^2+b^2+2q(a+b)\right] + \frac{1-2p}{n}\left[ 2ab+q^2\right] \\ = &pn + \frac{1-3p}{n}\left[ 2ab+q^2\right] = pn+\frac{1-3p}{2n}\left[ 2q^2+(n-q)^2-s^2\right] . \end{aligned}$$

$\square$

1.2 Proofs: victory of noise

Proof of Lemma 12

Define f(q) equal to ${\mathbb {E}}\left[ Q \big |\ \mathbf {x} \right] = \frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) q^2 - \frac{1-6\epsilon }{2}q + \frac{5-6\epsilon }{12}n - \frac{1-6\epsilon }{n}\left( \frac{s}{2}\right) ^2$. We are going to evaluate f(q) in $\bar{q}_{i+1}$ and in $\bar{q} = \frac{n(1+3\epsilon )}{3(1-6\epsilon )}$. We take care of different cases: first, we assume $i=-1$, with the condition that $s> 0$. Thus

$$\begin{aligned} f(\bar{q}_0) \le \&\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) n^2 \bigg [\frac{1}{9} + \frac{4\epsilon ^2}{(1+6\epsilon )^2}\left( \frac{3}{2}\right) ^{6} - \frac{4\epsilon }{3(1+6\epsilon )}\left( \frac{3}{2}\right) ^{3}\bigg ]\\&- \frac{1-6\epsilon }{2}n \bigg [\frac{1}{3} - \frac{2\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{3} \bigg ] + \frac{5-6\epsilon }{12}n \\ = \&\frac{n}{3} + n\frac{3\epsilon ^2(1-6\epsilon )}{(1+6\epsilon )^2}\left( \frac{3}{2}\right) ^{6} = \frac{n}{3}\left[ 1+\frac{3^8\epsilon ^2(1-6\epsilon )}{2^6(1+6\epsilon )^2}\right] ; \end{aligned}$$

now, we observe that

$$\begin{aligned}&1+\frac{3^8\epsilon ^2(1-6\epsilon )}{2^6(1+6\epsilon )^2} - \frac{1}{1-6\epsilon }< \frac{-6\epsilon }{1-6\epsilon }+ \frac{3^8\epsilon ^2}{2^6(1+6\epsilon )^2} \\ = \&\frac{-2^6\cdot 6\epsilon (1+6\epsilon )^2+3^8\epsilon ^2(1-6\epsilon )}{2^6(1-6\epsilon )(1+6\epsilon )^2} \\ = \&\frac{3\epsilon \left( -2^7-3\cdot 2^9\epsilon -3^2\cdot 2^9\epsilon ^2+3^7\epsilon -2\cdot 3^8\epsilon ^2\right) }{2^6(1-6\epsilon )(1+6\epsilon )^2} \\ = \&\frac{3\epsilon \left( -2^7+3\epsilon (3^6-2^9)-2\cdot 3^2\epsilon ^2(2^8+3^6)\right) }{2^6(1-6\epsilon )(1+6\epsilon )^2} <0 \end{aligned}$$

where in the last inequality we have used that $-2^7+3\epsilon (3^6-2^9)<0$ for $\epsilon \le \frac{1}{12}$. Thus, $f(\bar{q}_0) < \frac{n}{3(1-6\epsilon )}$.

Second, we assume $0\le i\le k-3$, with the condition that $s > \left( \frac{3}{2}\right) ^i\beta n$.

$$\begin{aligned} f(\bar{q}_{i+1}) \le \&\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) n^2 \bigg [\frac{1}{9} + \frac{4\epsilon ^2}{(1+6\epsilon )^2}\left( \frac{3}{2}\right) ^{4i+10} - \frac{4\epsilon }{3(1+6\epsilon )}\left( \frac{3}{2}\right) ^{2i+5}\bigg ]\\&- \frac{1-6\epsilon }{2}n \bigg [\frac{1}{3} - \frac{2\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2i+5} \bigg ] + \frac{5-6\epsilon }{12}n - \frac{2\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2i}n \\ = \&\frac{n}{3} + n\frac{3\epsilon ^2(1-6\epsilon )}{(1+6\epsilon )^2}\left( \frac{3}{2}\right) ^{4i+10} - n\frac{2\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2i} \\ = \&\frac{n}{3}\bigg \{1+\frac{3\epsilon }{(1+6\epsilon ^2)}\left( \frac{3}{2}\right) ^{2i}\bigg [3\epsilon (1 -6\epsilon )\left( \frac{3}{2}\right) ^{2(i+5)} -2(1+6\epsilon ) \bigg ]\bigg \}; \end{aligned}$$

for the evaluation of $f(\bar{q}_{i+1})$ we observe that $\beta$ is a constant in (0, 1) and that

$$\begin{aligned}&3\epsilon (1-6\epsilon )\left( \frac{3}{2}\right) ^{2(i+5)}-2(1+6\epsilon )< 3\epsilon \left( \frac{2}{3}\frac{1}{\beta }\left( \frac{3}{2}\right) ^3\right) ^2 - 2< \frac{3^5}{2^7} - 2 < 0 \end{aligned}$$

because $\left( \frac{3}{2}\right) ^{i+2}< \frac{2}{3\beta }$ for $i+2 \le k$; thus $f(\bar{q}_{i+1}) \le \frac{n}{3}$.

Let now $i=k-2$; thus $s> \left( \frac{3}{2}\right) ^{k-2}\beta n$. We now evaluate $f(\bar{q}_{k-1})$.

$$\begin{aligned} f(\bar{q}_{k-1}) \le \frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) n^2 \left[ \frac{4}{81}+\frac{\epsilon ^2}{9}+\frac{4\epsilon }{27}\right] - \frac{1-6\epsilon }{2}n \left[ \frac{2}{9} + \frac{\epsilon }{3}\right] + \frac{5-6\epsilon }{12}n \\&- \frac{2\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2k-4}n \\ = \frac{-8(1-6\epsilon )+45-54\epsilon }{108}n-\frac{\epsilon (1-6\epsilon )}{18}n - \frac{2\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2k-4}n \\ = \frac{37 - 12\epsilon + 36\epsilon ^2}{108}n - \frac{2\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2k-4}n \\ = \&\frac{n}{3}\left[ 1+\frac{(1-6\epsilon )^2}{36}-\frac{6\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2k-4}\right] \\ = \frac{n}{3}\biggl [1+\left( \frac{1-6\epsilon }{6}-\sqrt{\frac{6\epsilon }{1+6\epsilon }}\cdot \frac{3^{k-2}}{2^{k-2}}\right) \cdot \left( \frac{1-6\epsilon }{6}+\sqrt{\frac{6\epsilon }{1+6\epsilon }}\cdot \frac{3^{k-2}}{2^{k-2}}\right) \biggr ]. \end{aligned}$$

Observe that, by definition of k, we have

$$\begin{aligned}&\frac{1-6\epsilon }{6}-\sqrt{\frac{6\epsilon }{1+6\epsilon }}\cdot \frac{3^{k-2}}{2^{k-2}} \le \frac{1-6\epsilon }{6}-\sqrt{\frac{6\epsilon }{1+6\epsilon }}\cdot \frac{2^3}{3^3\beta } \\ = \&\frac{1-6\epsilon }{6} - \frac{\sqrt{3(1-6\epsilon )}\cdot 2^2}{3^3} = \frac{9-54\epsilon -8\sqrt{3(1-6\epsilon )}}{54} \\< \&\frac{9-8\sqrt{3(1-6\epsilon )}}{54} < 0 \end{aligned}$$

for $\epsilon \le \frac{1}{12}$. Thus, $f(\bar{q}_{k-1})\le \frac{n}{3}$.

Let now $i=k-1$, which implies that $s > \left( \frac{3}{2}\right) ^{k-1}\beta n$. We evaluate $f(\bar{q}_k)$:

$$\begin{aligned} f(\bar{q}_k) \le &\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) n^2 \left[ \frac{1}{144}+\epsilon ^2+\frac{\epsilon }{6}\right] - \frac{1-6\epsilon }{2}n \left[ \frac{1}{12} + \epsilon \right] + \frac{5-6\epsilon }{12}n \\&- \frac{2\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2k-2}n \\ = &\frac{-7(1-6\epsilon )+80-96\epsilon }{192}n-\frac{3\epsilon (1-6\epsilon )}{8}n - \frac{2\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2k-2}n \\ = &\frac{n}{3}\biggl [\frac{219-162\epsilon - 216\epsilon +1296\epsilon ^2}{192}-\frac{6\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2k-2}\biggr ] \\ = &\frac{n}{3}\biggl [1 + \frac{27-378\epsilon +1296\epsilon ^2}{192} -\frac{6\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2k-2}\biggr ]. \end{aligned}$$

By definition of k, we have that

$$\begin{aligned}&\frac{27-378\epsilon +1296\epsilon ^2}{192} -\frac{6\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2k-2} \le \frac{27-378\epsilon +1296\epsilon ^2}{192} - \frac{1-6\epsilon }{2}\left( \frac{2}{3}\right) ^3 < 0, \end{aligned}$$

where the first and the second inequalities hold for $\epsilon \le \frac{1}{12}$. Thus, $f(\bar{q}_k)\le \frac{n}{3}$.

We finally evaluate $f(\bar{q})$:

$$\begin{aligned} f\left( \bar{q}\right) \le &\frac{n}{12}\left( \frac{(1+3\epsilon )^2}{(1-6\epsilon )}\right) - \frac{n(1+3\epsilon )}{6} + \frac{n(5-6\epsilon )}{12} \\ = &\frac{n}{12}\left( \frac{1-6\epsilon +9\epsilon ^2}{1-6\epsilon }-2-6\epsilon +5-6\epsilon \right) \\ = &\frac{n}{12(1-6\epsilon )}\biggl (1-6\epsilon +9\epsilon ^2-2+6\epsilon +36\epsilon ^2 + 5 -36\epsilon +36\epsilon ^2\biggr ) \\ = &\frac{n}{12}\left( \frac{4-36\epsilon +81\epsilon ^2}{1-6\epsilon }\right) = \frac{n}{12}\left( \frac{(2-9\epsilon )^2}{1-6\epsilon }\right) . \end{aligned}$$

It holds that $f(\bar{q}) \le \frac{n}{3(1-6\epsilon )}$ (remember that $\epsilon \le 1/12$); in all cases, f is no more than $\frac{n}{3(1-6\epsilon )}$, and, from an immediate application of the additive Chernoff bound (Lemma 17) with $\lambda = \frac{\epsilon n}{1-6\epsilon }$, and by observing that $\bar{q}_i \ge \bar{q}_{i+1}$, we get that

$$\begin{aligned} \bar{q}_i \le Q \le \check{q}_{\epsilon /2} = \frac{n(1+3\epsilon )}{3(1-6\epsilon )} \end{aligned}$$

with probability $1 - \exp \left( - 2 \epsilon ^2 n /(1-6\epsilon )^2\right)$. $\square$

Proof of Lemma 14

Consider the first claim. We define $f(q)=\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) q^2 - \frac{1-6\epsilon }{2}q + \frac{5-6\epsilon }{12}n$. We now show that $f(q)\le q\left( 1-2\epsilon \right)$. Indeed, $f(q) - q\left( 1-2\epsilon \right)$ is equal to

$$\begin{aligned}&\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) q^2 - \left( \frac{1-6\epsilon }{2}+1-2\epsilon \right) q + \frac{5-6\epsilon }{12}n \\ = &\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) q^2 - \left( \frac{3-10\epsilon }{2}\right) q + \frac{5-6\epsilon }{12}n. \end{aligned}$$

This expression is a convex parabola which has its maximum in either $\bar{q}_1=\frac{n}{3(1-6\epsilon )}$ or $\bar{q}_2=n$. We calculate $f(q)-q\left( 1-2\epsilon \right)$ in these two points

$$\begin{aligned}&\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) q_1^2 - \left( \frac{3-10\epsilon }{2}\right) q_1+ \frac{5-6\epsilon }{12}n \\ = &\frac{n^2}{12(1-6\epsilon )}\left( 1-6+20\epsilon +5-36\epsilon +36\epsilon ^2\right) = \frac{n^2}{12(1-6\epsilon )}\left[ -4\epsilon (4-9\epsilon )\right] <0 \end{aligned}$$

for all $0<\epsilon \le \frac{1}{12}$. At the same time it holds that

$$\begin{aligned}&\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) q_2^2 - \left( \frac{3-10\epsilon }{2}\right) q_2+ \frac{5-6\epsilon }{12}n = \frac{n}{12}\left[ 9-54\epsilon -18+60\epsilon +5-6\epsilon \right] \\ = &-\frac{-n}{3}<0. \end{aligned}$$

Thus, ${\mathbb {E}}\left[ Q \big |\ \mathbf {x} \right] \le q(1-2\epsilon )$. The additive form of Chernoff bound (Lemma 17) with $\lambda = \frac{\epsilon n}{3(1-6\epsilon )}$ implies that $Q < q(1-\epsilon )$ with probability $1 - \exp \left( -2\epsilon ^2 (n/3)/(1-6\epsilon )^2\right)$.

As for the second claim, we consider two cases. First, assume $i<k-1$, thus $s\le \left( \frac{3}{2}\right) ^{i+1}\beta n$. From Eq. (6), we observe that

$$\begin{aligned} {\mathbb {E}}\left[ Q \big |\ \mathbf {x} \right] \ge &\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) \frac{n^2}{9(1-6\epsilon )^2} +\frac{5-6\epsilon }{12}n -\left( \frac{1-6\epsilon }{2}\right) \frac{n}{3(1-6\epsilon )} - \frac{1-6\epsilon }{n}\left( \frac{s}{2}\right) ^2 \\ \ge &\frac{n}{12}\left( \frac{4-24\epsilon +36\epsilon ^2}{1-6\epsilon }\right) - \frac{2n\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2i+2} \\ \ge &\frac{n}{3(1-6\epsilon )}\left( 1-3\epsilon \right) ^2 - \frac{2n\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2i+2} \\ \ge &\frac{n}{3}-\frac{2n\epsilon }{1+6\epsilon }\left( \frac{3}{2}\right) ^{2i+5}+\frac{19n\epsilon }{4}\left( \frac{3}{2}\right) ^{2i+2} = \bar{q}_{i+1}+\frac{19n\epsilon }{4}\left( \frac{3}{2}\right) ^{2i+2} . \end{aligned}$$

We conclude applying the additive Chernoff bound with $\lambda = \frac{19\epsilon n}{4}\left( \frac{3}{2}\right) ^{2i+2}$, obtaining $Q \ge \bar{q}_{i+1}$ with probability $1 - \exp \left( -19^2 \epsilon ^2 (n/8) (3/2)^{4i+4} \right)$.

Second, let $i=k-1$; then $s\le \frac{2}{3}n$. As before

$$\begin{aligned} {\mathbb {E}}\left[ Q \big |\ \mathbf {x} \right] \ge &\frac{3}{4}\left( \frac{1-6\epsilon }{n}\right) \frac{n^2}{9(1-6\epsilon )^2}+\frac{5-6\epsilon }{12}n -\left( \frac{1-6\epsilon }{2}\right) \frac{n}{3(1-6\epsilon )} - \frac{1-6\epsilon }{n}\left( \frac{s}{2}\right) ^2 \\ \ge &\frac{n}{12(1-6\epsilon )}\left( 4-24\epsilon +36\epsilon ^2\right) - \frac{n(1-6\epsilon )}{9} \ge \frac{n\left( 1-3\epsilon \right) ^2}{3(1-6\epsilon )} - \frac{n(1-6\epsilon )}{9} \\ = &\frac{n}{9(1-6\epsilon )}\left( 2-6\epsilon -9\epsilon ^2\right) = \frac{n}{12}\left( \frac{4(2-6\epsilon -9\epsilon ^2)}{3(1-6\epsilon )}\right) \\ \ge &\frac{n}{12}(1+14\epsilon ) = \bar{q}_k+\frac{\epsilon n}{6} \end{aligned}$$

since

$$\begin{aligned} \frac{4(2-6\epsilon -9\epsilon ^2)}{3(1-6\epsilon )} -(1+14\epsilon )>\frac{5-48\epsilon }{3(1-6\epsilon )}>0 \end{aligned}$$

for $\epsilon \le \frac{1}{12}$. Thus, we conclude applying the additive form of Chernoff bound with $\lambda = \frac{\epsilon }{6}n$, obtaining $Q \ge \bar{q}_k$ with probability $1 - \exp \left( -\epsilon ^2 n / 18\right)$. Hence, the second claim holds with probability at least $1 - \exp \left( -\epsilon ^2 n / 18\right)$, which is less than $1 - \exp \left( -19^2 \epsilon ^2 (n/8) (3/2)^{4i+4} \right)$. $\square$

1.3 Proofs: symmetry breaking

The proof of Theorem 3 essentially relies on the following lemma which has been proved by Clementi et al. (2018), for which we report a proof that corrects some minor mistakes.

Lemma 22

Let $\{X_{t}\}_{t\in \mathbb {N}}$ be a Markov Chain with finite-state space $\Omega$ and let $f:\Omega \mapsto [0,n]$ be a function that maps states to integer values. Let $c_3$ be any positive constant and let $m = c_3\sqrt{n}\log n$ be a target value. Assume the following properties hold:

• (1) for any positive constant h, a positive constant $c_1 < 1$ exists such that for any $x \in \Omega : f(x) < m$,

  $$\begin{aligned} \mathbb {P}\left[ f(X_{t+1})< h\sqrt{n} \big |\ X_{t} = x\right] < c_1; \end{aligned}$$

• (2) there exist two positive constants $\delta$ and $c_2$ such that for any $x \in \Omega : h\sqrt{n} \le f(x) < m$,

  $$\begin{aligned} \mathbb {P}\left[ f(X_{t+1})< (1+\delta )f(X_{t}) \big |\ X_{t} = x\right] < e^{-c_2f(x)^2/n}. \end{aligned}$$

Then the process reaches a state x such that $f(x) \ge m$ within $\mathcal {O}_{c_1,c_3}(\log n)$ rounds with probability at least $1 - 2/n$.

Proof

Define a set of hitting times $T :=\left\{ \tau (i)\right\} _{i \in \mathbb {N}}$, where

$$\begin{aligned} \tau (i) = \inf _{i \in \mathbb {N}}\left\{ t: t > \tau (i-1), f(X_t) \ge h\sqrt{n}\right\} , \end{aligned}$$

setting $\tau (0) = 0$. By the first hypothesis, for every $i\in \mathbb {N}$, the expectation of $\tau (i)$ is finite. Now, define the following stochastic process which is a subsequence of $\{X_t\}_{t\in \mathbb {N}}$:

$$\begin{aligned} \{R_i\}_{i \in \mathbb {N}} = \{X_{\tau (i)}\}_{i\in \mathbb {N}}. \end{aligned}$$

Observe that $\{R_i\}_{i \in \mathbb {N}}$ is still a Markov chain. Indeed, if $\{x_1, \dots , X_{i-1}\}$ is a set of states in $\Omega$, then

$$\begin{aligned}&\ \mathbb {P}\left[ R_i = x \big |\ R_{i-1} = x_{i-1}, \dots , R_1 = x_1\right] \\ =&\ \mathbb {P}\left[ X_{\tau (i)} = x \big |\ X_{\tau (i-1)} = x_{i-1}, \dots , X_{\tau (1)} = x_1\right] \\ =&\ \sum _{t(i)> \dots > t(1) \in \mathbb {N}} \mathbb {P}\left[ X_{t(i)} = x \big |\ X_{t(i-1)}=x_{i-1}, \dots , X_{t(1)}=x_1\right] \\&\cdot \mathbb {P}\left[ \tau (i)=t(i), \dots , \tau (1)=t(1)\right] \\ =&\ \mathbb {P}\left[ X_{\tau (i)}=x \big |\ X_{\tau (i-1)=x_{i-1}}\right] \\ =&\ \mathbb {P}\left[ R_i = x \big |\ R_{i-1} = x_{i-1}\right] . \end{aligned}$$

By definition, the state space of R is $\{x \in \Omega : f(x) \ge h\sqrt{n}\}$. Moreover, the second hypothesis still holds for this new Markov chain. Indeed:

$$\begin{aligned}&\ \mathbb {P}\left[ f(R_{i+1}< (1+\epsilon )f(R_i) \big |\ R_i = x \right] \\ =&\ 1- \mathbb {P}\left[ f(R_{i+1} \ge (1+\epsilon )f(R_i) \big |\ R_i = x\right] \\ =&\ 1- \mathbb {P}\left[ f(X_{\tau (i+1)} \ge (1+\epsilon )f(X_{\tau (i)}) \big |\ X_{\tau (i)} = x\right] \\ \le&\ 1- \mathbb {P}\left[ f(X_{\tau (i+1)} \ge (1+\epsilon )f(X_{\tau (i)}), \tau (i+1) = \tau (i) +1 \big |\ X_{\tau (i)} = x\right] \\ =&\ 1- \mathbb {P}\left[ f(X_{\tau (i)+1} \ge (1+\epsilon )f(X_{\tau (i)}) \big |\ X_{\tau (i)} = x\right] \\ =&\ 1- \mathbb {P}\left[ f(X_{t+1} \ge (1+\epsilon )f(X_{t})\big |\ X_{t} = x\right] \\ <&\ e^{-c_2f(x)^2/n}. \end{aligned}$$

These two properties are sufficient to study the number of rounds required by the new Markov chain $\{R_i\}_{i\in \mathbb {N}}$ to reach the target value m. Indeed, by defining the random variable $Z_i = \frac{f(R_i)}{\sqrt{n}}$, and considering the following “potential” function, $Y_i = exp\left( \frac{m}{\sqrt{n}}-Z_i\right)$, we can compute its expectation at the next round as follows. Let us fix any state $x\in \Omega$ such that $h\sqrt{n} \le f(x) < m$, and define $z = \frac{f(x)}{\sqrt{n}}$, $y=exp\left( \frac{m}{\sqrt{n}}-z\right)$. We have

$$\begin{aligned} {\mathbb {E}}\left[ Y_{i+1} \vert R_i = x \right]&\le \mathbb {P}\left[ f(R_{i+1})< (1+\epsilon )f(x)\right] e^{m/\sqrt{n}}\nonumber \\&\quad + \mathbb {P}\left[ f(R_{i+1}) \ge (1+\epsilon )f(x)\right] e^{m/\sqrt{n} - (1 + \epsilon )z}\nonumber \\ \text{(from } \text{ Hypothesis } \text{(2)) }&\le e^{-c_2 z^2} \cdot e^{m/\sqrt{n}} + 1 \cdot e^{m/\sqrt{n} - (1 + \epsilon )z}\nonumber \\&= e^{m/\sqrt{n} - c_2 z^2} + e^{m/\sqrt{n} - z - \epsilon z}\nonumber \\&= e^{m/\sqrt{n} - z} (e^{z - c_2 z^2} + e^{-\epsilon z})\nonumber \\&\le e^{m/\sqrt{n} - z}(e^{-2} + e^{-2})\\&<\frac{e^{m/\sqrt{n} - z}}{e}\nonumber \\&=\frac{y}{e}\nonumber , \end{aligned}$$

(B7)

where in (B7) we used that z is always at least h and thanks to Hypothesis (1) we can choose a sufficiently large h.

By applying the Markov inequality and iterating the above bound, we get

$$\begin{aligned} \mathbb {P}\left[ Y_{i} > 1\right] \le \frac{{\mathbb {E}}\left[ Y_{i} \right] }{1} \le \frac{{\mathbb {E}}\left[ Y_{{i}-1} \right] }{e}\le \cdots \le \frac{{\mathbb {E}}\left[ Y_0 \right] }{e^{\tau _R}} \le \frac{e^{m/\sqrt{n}}}{e^{i}}. \end{aligned}$$

We observe that if $Y_{i} \le 1$ then $R_{i} \ge m$, thus by setting ${i} = m/\sqrt{n} + \log n = (c_3 + 1)\log n$, we get:

$$\begin{aligned} \mathbb {P}\left[ R_{(c_3 + 1)\log n}< m\right] = \mathbb {P}\left[ Y_{(c_3 + 1)\log n} > 1\right] < \frac{1}{n}. \end{aligned}$$

(B8)

Our next goal is to give an upper bound on the hitting time $\tau _{(c_3 + 1)\log n}$. Note that the event “$\tau _{(c_3 + 1)\log n} > c_4\log n$” holds if and only if the number of rounds such that $f(X_t) \ge h\sqrt{n}$ (before round $c_4\log n$) is less than $(c_3 + 1)\log n$. Thanks to Hypothesis (1), at each round t there is at least probability $1-c_1$ that $f(X_{t}) \ge h\sqrt{n}$. This implies that, for any positive constant $c_4$, the probability $\mathbb {P}\left[ \tau _{(c_3 + 1)\log n} > c_4\log n\right]$ is bounded by the probability that, within $c_4\log n$ independent Bernoulli trials, we get less than $(c_3 + 1)\log n$ successes, where the success probability is at least $1-c_1$. We can thus choose a sufficiently large $c_4 = c_4(c_1,c_3)$ and apply the multiplicative form of the Chernoff bound (Lemma 16), obtaining

$$\begin{aligned} \mathbb {P}\left[ \tau _{(c_3 + 1)\log n} > c_4\log n\right] < \frac{1}{n}. \end{aligned}$$

(B9)

We are now ready to prove the Lemma using (B8) and (B9), indeed

$$\begin{aligned} \mathbb {P}\left[ X_{c_4 \log n} \ge m\right]&> \mathbb {P}\left[ R_{(c_3 + 1)\log n} \ge m \wedge \tau _{(c_3 + 1)\log n} \le c_4 \log n\right] \\&= 1 - \mathbb {P}\left[ R_{(c_3 + 1)\log n}< m \vee \tau _{(c_3 + 1)\log n}> c_4 \log n\right] \\&\ge 1 - \mathbb {P}\left[ R_{(c_3 + 1)\log n} < m\right] + \mathbb {P}\left[ \tau _{(c_3 + 1)\log n}> c_4 \log n\right] \\&> 1 - \frac{2}{n}. \end{aligned}$$

Hence, choosing a suitable large $c_4$, we have shown that in $c_4 \log n$ rounds the process reaches the target value m, w.h.p. $\square$

Our goal is to apply the above lemma to the Undecided-State process (which defines a finite-state Markov chain) starting with bias of size $o(\sqrt{n\log n})$ where we set $f(\mathbf {X}_t)=s(\mathbf {X}_t)$, $c_3 = \gamma >0$ for some constant $\gamma > 0$, and $m=\gamma \sqrt{n}\log n$: this would imply the upper bound $\mathcal {O}(\log n)$ on the number of rounds needed to reach a configuration having bias $\Omega (\sqrt{n\log n})$, w.h.p., breaking the symmetry because Theorem 1 then holds. To this aim, with the next two lemmas we show that the Undecided-State process satisfies the hypotheses of Lemma 22 in this setting, w.h.p.

Lemma 23

Let $\epsilon$ be any constant such that $(1-6p)^2/2 \le \epsilon < (1-6p)(1-2p)$. Let $\mathbf {x}$ be any configuration in which $s \le \check{s}_{\epsilon }$, and $q \le \frac{n}{2}$. Then, in the next round, it holds that $\hat{q}_{\epsilon /12} \le Q \le \frac{n}{2}$ with probability $1 - \exp \left( - \epsilon ^2 p^2 n / (2^43^2)\right)$.

Proof of Lemma 23

Lemma 3 implies that $Q \ge \hat{q}_{\epsilon /12}$ with probability $1-\exp (-\epsilon ^2 n / \left( 2^33^2(1-3p)^2\right) )$. At the same time, by Eq. (6) it holds that

$$\begin{aligned} {\mathbb {E}}\left[ Q \big |\ \mathbf {x} \right] \le pn + \frac{1-3p}{2n}\left[ 2q^2 + (n - q)^2\right] = f(q). \end{aligned}$$

For $q \le \frac{n}{2}$, the maximum of f is obtained either at $q_1=0$ or at $q_2=\frac{n}{2}$. Then,

$$\begin{aligned} f(q_1)&= \left( p + \frac{1 - 3p}{2}\right) n = \frac{(1-p)n}{2}; \\ f(q_2)&= pn + \frac{1-3p}{2n}\left[ \frac{n^2}{2} + n^2 - n^2 + \frac{n^2}{4} \right] = \left( p + \frac{3(1-3p)}{8}\right) n \\&= \frac{(3- p)n}{8}. \end{aligned}$$

Therefore, we have $f(q) \le (1-p)n/2$ for $q \le \frac{n}{2}$, since $(1-p)/2 > (3-p)/8$ for $p < 1/2$. By Chernoff bound (Lemma 17),

$$\begin{aligned} \mathbb {P}\left[ Q \ge \frac{n}{2} \big |\ \mathbf {x}\right]&= \mathbb {P}\left[ Q \ge \frac{(1-p)n}{2} + \frac{pn}{2} \big |\ \mathbf {x}\right] \le \mathbb {P}\left[ Q \ge {\mathbb {E}}\left[ Q \big |\ \mathbf {x} \right] + \frac{pn}{2} \big |\ \mathbf {x}\right] \\&\le \exp \left( - p^2 n / 2 \right) . \end{aligned}$$

Hence, the joint probability that $Q \le n/2$ and $Q \ge \hat{q}_{\epsilon /12}$ is at least $1 - \exp \left( - \epsilon ^2 p^2 n / (2^43^2)\right)$, since $\epsilon <1$ and $p < 1$. $\square$

Lemma 24

Let $\epsilon$ be any constant such that $(1-6p)^2/2 \le \epsilon < (1-6p)(1-2p)$. Let $\mathbf {x}$ be any configuration such that $q(\mathbf {x})\in \left[ \hat{q}_{\epsilon /12}, n/2\right]$. Then, it holds that

• (1) for any constant $h>0$ there exists a constant $c_1>0$ (depending only on p) such that

  $$\begin{aligned} \mathbb {P}\left[ |S |< h\sqrt{n}) \big |\ \mathbf {X}_t = \mathbf {x}\right] < c_1; \end{aligned}$$

• (2) there exist two positive constants $\delta$ and $c_2$ (depending only on p) such that

  $$\begin{aligned} \mathbb {P}\left[ |S |\ge (1+\delta )s \big |\ \mathbf {X}_t = \mathbf {x}\right] \ge 1-e^{-c_2\frac{s^2}{n}}. \end{aligned}$$

Proof of Lemma 24

As for the first item, let $\mathbf {x}$ and $\mathbf {x}_0$ be two states such that $|s(\mathbf {x})|< h \sqrt{n} \log n$, $|s(\mathbf {x}_0)|=0$, $q(\mathbf {x})=q(\mathbf {x}_0)$. A simple domination argument implies that

$$\begin{aligned} \mathbb {P}\left[ |S |< h\sqrt{n}\big |\ \mathbf {X}_t = \mathbf {x}\right] \le \mathbb {P}\left[ |S |< h\sqrt{n}\big |\ \mathbf {X}_t = \mathbf {x}_0\right] . \end{aligned}$$

Thus, we can bound just the second probability, where the initial bias is zero, which implies that $a=b$.

Define $A^q$, $B^q$, $Q^q$ the random variables counting the nodes that were undecided in the configuration $\mathbf {x}_0$ and that, in the next round, get the opinion Alpha, Beta, and undecided, respectively. Similarly, $A^a$ ($B^b$) counts the nodes that support opinion Alpha (Beta) in the configuration $\mathbf {x}_0$ and that, in the next round, still support the same opinion. Trivially, $A=A^q+A^a$ and $B=B^q+B^b$. Moreover, observe that, among these random variables, only $A^q$ and $B^q$ are mutually dependent. Thus, conditioned to the event $\{\mathbf {X}_t=\mathbf {x}_0\}$, if $\alpha = {\mathbb {E}}\left[ A^a\big |\ \mathbf {X}_t = \mathbf {x}_0 \right] = {\mathbb {E}}\left[ B^b\big |\ \mathbf {X}_t = \mathbf {x}_0 \right]$, it holds that

$$\begin{aligned} \mathbb {P}\left[ |S |\ge h\sqrt{n}\right] \ge&\ \mathbb {P}\left[ A\ge B + h\sqrt{n}\right] \ge \mathbb {P}\left[ A^q\ge B^q + h\sqrt{n}\right] \mathbb {P}\left[ A^a\ge \alpha \right] \mathbb {P}\left[ B^b\le \alpha \right] . \end{aligned}$$

The random variables $A^a-\alpha$ and $B^b-\alpha$ follow binomial distributions with expectation 0 (recall that $a=b$), and finite second and third moment. Thus, the Berry-Esseen theorem ( Lemma 19 in Appendix 1) allows us to approximate up to an arbitrary-small constant $\epsilon _1>0$ (as long as n is large enough) both the random variables with a normal distribution that has expectation 0. Thus,

$$\begin{aligned} \mathbb {P}\left[ A^a\ge \alpha \right] =\mathbb {P}\left[ B^b\le \alpha \right] \ge \left( \frac{1}{2}-\epsilon _1\right) . \end{aligned}$$

As for the random variable $A^q-B^q$, notice that conditioned to the event $\{q-Q^q = k\}$, it is the sum of k Rademacher random variables. The hypothesis $q\le \frac{n}{2}$ allows us to use the Chernoff bound on $Q^q$ and show that $Q^q\le \frac{3}{4}q$ w.h.p. Thus, since $q\ge \hat{q}_{\epsilon /12}$, it holds that $q-Q^q = \Theta (n)$ w.h.p. It follows that the conditional variance of $A^q-B^q$ given $q-Q^q$ yields $\Theta (n)$ w.h.p., and $A^q-B^q$ conditional on the event $E = \{q-Q^q = \Theta (n)\}$ can be approximated by a normal distribution up to an arbitrary-small constant $\epsilon _2>0$. Then, we have that

$$\begin{aligned} \mathbb {P}\left[ A^q\ge B^q + h\sqrt{n}\right] \ge \mathbb {P}\left[ A^q\ge B^q + h\sqrt{n} \big |\ E\right] \mathbb {P}\left[ E\right] \ge \epsilon _2. \end{aligned}$$

Setting $c_1 = \epsilon _1 \cdot \epsilon _2$, we get property (1). We can choose $\epsilon _1,\epsilon _2$ to be equal to p, so that $c_1$ depends only on p. As for property (2), by Eq. 5 and the hypothesis on q we have that

$$\begin{aligned} {\mathbb {E}}\left[ S \big |\ \mathbf {X}_t = \mathbf {x} \right] \ge s\left( 1+\epsilon /12\right) . \end{aligned}$$

We can get the property applying the Hoeffding bound (Lemma 18), getting that

$$\begin{aligned} \mathbb {P}\left[ S \le s\left( 1+\frac{\epsilon }{24}\right) \big |\ \mathbf {x}\right]&= \mathbb {P}\left[ S \le s\left( 1 + \frac{\epsilon }{12}\right) - \frac{\epsilon s }{24} \big |\ \mathbf {x}\right] \le \exp \left( - \frac{\epsilon ^2 s^2}{2^73^2n}\right) , \end{aligned}$$

which is the thesis. $\square$

The reader may notice that Lemma 23 requires the number of undecided nodes to be inside the interval $\left[ \hat{q}_{\epsilon /12}, n/2\right]$. We will later take care of this issue with Lemma 27, showing that whenever this number is not within the above interval, in at most $\mathcal {O}(\log n)$ rounds it will. Furthermore, Lemma 23 guarantees that the condition on the undecided nodes holds “only” w.h.p., while Lemma 22 requires this condition to hold with probability 1. We show this issue can be solved using a coupling argument similar to that by Clementi et al. (2018). The key point is that, starting from any configuration $\mathbf {x}$ with $q(\mathbf {x}) \in \left[ \hat{q}_{\epsilon /12}, n/2\right]$, the probability that the process goes in one of those “bad” configurations with q outside the above interval is negligible. Intuitively speaking, the configurations actually visited by the process before breaking symmetry do satisfy the hypothesis of Lemma 22. In order to make this argument rigorous, we define a pruned process, by removing all the unwanted transitions.

Let $\bar{s}\in \{0,1,\dots , n\}$, and $\mathbf {z}(\bar{s})$ the configuration such that $s(\mathbf {z}(\bar{s})) = \bar{s}$, and $q(\mathbf {z}(\bar{s})) = n/2$. Let $p_{\mathbf {x},\mathbf {y}}$ be the probability of a transition from the configuration $\mathbf {x}$ to the configuration $\mathbf {y}$ in the Undecided-State process. The Pruned process behaves exactly as the original process but every transition from a configuration $\mathbf {x}$ such that $q(\mathbf {x}) \in \left[ \hat{q}_{\epsilon /12}, n/2\right]$ and $s(\mathbf {x}) = \mathcal {O}(\sqrt{n\log n})$ to a configuration $\mathbf {y}$ such that $q(\mathbf {y}) < \hat{q}_{\epsilon /12}$ or $q(\mathbf {y}) > n/2$ has probability $p_{\mathbf {x},\mathbf {y}}'=0$. Moreover, for any $\bar{s}\in [n]$, starting from the configuration $\mathbf {x}$, the probability of reaching the configuration $\mathbf {z}(\bar{s})$ is

$$\begin{aligned} p_{\mathbf {x},\mathbf {z}(\bar{s})}'=p_{\mathbf {x},\mathbf {z}(\bar{s})}+\sum _{\begin{array}{c} \mathbf {y}:\ s(\mathbf {y}) = \bar{s} \text { and} \\ q(\mathbf {y}) \notin \left[ \hat{q}_{\epsilon /12}, \frac{n}{2}\right] \end{array}}p_{\mathbf {x},\mathbf {y}}. \end{aligned}$$

All the other transition probabilities remain the same. Observe that the Pruned process is defined in such a way that it has exactly the same marginal probability of the original process w.r.t. the random variable $s(\mathbf {X}_t)$; thus, Lemma 24 holds for the Pruned process as well and we can apply Lemma 22. Then, the Pruned process reaches a configuration having bias $\Omega (\sqrt{n\log n})$ within $\mathcal {O}(\log n)$ rounds, w.h.p., as shown in the following lemma.

Lemma 25

Let $\epsilon$ be any constant such that $(1-6p)^2/2 \le \epsilon < (1-6p)(1-2p)$. Starting from any configuration $\mathbf {x}$ such that $q(\mathbf {x}) \in \left[ \hat{q}_{\epsilon /12}, n/2\right]$ and $s(\mathbf {x}) = \mathcal {O}(\sqrt{n\log n})$, the Pruned process reaches a configuration having bias $\Omega (\sqrt{n\log n})$ within $\mathcal {O}_{p,\gamma }(\log n)$ rounds with probability $1 - 2/n$.

Proof of Lemma 25

Let $\gamma >0$ be a constant and $m=\gamma \sqrt{n}\log n$ be the target value of the bias in Lemma 22. Since $q(\mathbf {x}) \in \left[ \hat{q}_{\epsilon /12},n/2\right]$ and $s(\mathbf {x}) = \mathcal {O}(\sqrt{n\log n})$, the Pruned process always satisfies Lemma 24, and thus we can apply Lemma 22 (setting the function $f(\mathbf {X}_t) = s(\mathbf {X}_t)$), which gives us that the Pruned process process reaches a configuration $\mathbf {y}$ having bias $s(\mathbf {y})\ge m = \Omega (\sqrt{n\log n})$ within $\mathcal {O}_{p,\gamma }(\log n)$ rounds with probability $1 - 2/n$. $\square$

We now want to go back to the original process. The definition of the Pruned process suggests a natural coupling between it and the original one. If the two process are in different states, then they act independently, while, if they are in the same state $\mathbf {x}$, they move together unless the Undecided-State process goes in a configuration $\mathbf {y}$ such that $q(\mathbf {y}) \notin \left[ \hat{q}_{\epsilon /12}, n/2\right]$. In that case, the Pruned process goes in $\mathbf {z}(s(\mathbf {y}))$. In the proof of the next lemma, we show that the time the Pruned process takes to reach bias $\Omega (\sqrt{n\log n})$ stochastically dominates the one of the original process, giving the result.

Lemma 26

Let $\gamma ,\epsilon$ be any two positive constants, with $(1-6p)^2/2 \le \epsilon < (1-6p)(1-2p)$. Starting from any configuration $\mathbf {x}$ such that $q(\mathbf {x}) \in \left[ \hat{q}_{\epsilon /12}, n/2\right]$ and $s(\mathbf {x}) \le \gamma \sqrt{n\log n}$, the Undecided-State process reaches a configuration having bias $s \ge \gamma \sqrt{n\log n}$ within $\mathcal {O}_{p,\gamma }(\log n)$ rounds with probability at least $1-3/n$.

Proof of Lemma 26

Let $\{\mathbf {X}_t\}$ and $\{\mathbf {Y}_t\}$ be the original process and the pruned one, respectively. Denote the set of possible initial configuration according to the hypothesis by H, and let $\mathbf {x}\in H$. Note that if $\mathbf {X}_t=\mathbf {Y}_t=\mathbf {x}$, then

$$\begin{aligned} \mathbf {Y}_{t+1} = {\left\{ \begin{array}{ll} \mathbf {X}_{t+1} &{} {\textbf {if }} \mathbf {X}_{t+1} \in H \\ \mathbf {z}(s(\mathbf {X}_t)) &{} \text {otherwise} \end{array}\right. }. \end{aligned}$$

Let $\tau = \inf \{t: \mathbb {N}: \left|s(\mathbf {X}_t) \right| \ge \sqrt{n\log n}\}$, and let $\tau * = \inf \{t \in \mathbb {N}: \left|s(\mathbf {Y}_t) \right| \ge \sqrt{n\log n}\}$. For any configuration $\mathbf {x}\in H$, define $\rho _\mathbf {x}^t$ the event that the two processes $\{\mathbf {X}_t\}$ and $\{\mathbf {Y}_t\}$ have separated at round $t+1$, i.e. $\rho _\mathbf {x}^t = \{\mathbf {X}_t = \mathbf {Y}_t = \mathbf {x}_t\}\cap \{\mathbf {X}_{t+1} \ne \mathbf {Y}_{t+1}\}$. Observe that, if the two couple processes in the same configuration $\mathbf {x}_0\in H$ and $\tau > c\log n$, then either $\tau * > c\log n$ or there exists a round $t\le c\log n$ such that for some $\mathbf {x}\in H$ the event $\rho ^t_{\mathbf {x}}$ has occurred. Hence, if $\mathbb {P}'_{\mathbf {x}_0, \mathbf {x_0}}\left[ \cdot \right]$ is the joint probability for the couple $(\mathbf {X}_t, \mathbf {Y_t})$ which both start at $\mathbf {x}_0$, we have

$$\begin{aligned}&\ \mathbb {P}'_{\mathbf {x}_0, \mathbf {x_0}}\left[ \tau> c\log n\right] \\ \le&\ \mathbb {P}'_{\mathbf {x}_0, \mathbf {x_0}}\left[ \{\tau *> c\log n\}\cup \{\exists t\le c\log n, \exists \mathbf {x}\in H : \rho _\mathbf {x}^t\}\right] \\ \le&\ \mathbb {P}'_{\mathbf {x}_0, \mathbf {x_0}}\left[ \tau * > c\log n\right] + \mathbb {P}'_{\mathbf {x}_0, \mathbf {x_0}}\left[ \exists t\le c\log n, \exists \mathbf {x}\in H : \rho _\mathbf {x}^t\right] . \end{aligned}$$

We choose a suitable constant c which depends on p and $\gamma$, in order to apply Lemma 22, and 23. As for the first item, since Lemma 22 holds for the Pruned process, we have that it is upper bounded by 2/n. As for the second term, we get that

$$\begin{aligned}&\ \mathbb {P}'_{\mathbf {x}_0, \mathbf {x_0}}\left[ \exists t\le c\log n, \exists \mathbf {x}\in H : \rho _\mathbf {x}^t\right] \le \sum _{t=1}^{c\log n}\mathbb {P}'_{\mathbf {x}_0, \mathbf {x_0}}\left[ \exists \mathbf {x}\in H : \rho ^t_\mathbf {x}\right] \\ =&\ \sum _{t=1}^{c\log n}\sum _{\mathbf {x}\in H}\mathbb {P}'_{\mathbf {x}_0, \mathbf {x_0}}\left[ \rho ^t_\mathbf {x}\right] {\mathop {\le }\limits ^{(a)}} \sum _{t=1}^{c\log n} \frac{n^2}{e^{-\frac{\epsilon ^2 p^2 n}{2^43^2}}} \le \frac{1}{n}, \end{aligned}$$

where in (a) the second inequality we used Lemma 23, and the fact that $\left|H \right|$ is at most all the combinations of parameters q and s. $\square$

Now, we take care of those cases in which the starting configuration is such that $q \notin \left[ \hat{q}_{\epsilon /12}, n/2\right]$. If $q < \hat{q}_{\epsilon /12}$ then, for Lemma 23, $Q \in \left[ \hat{q}_{\epsilon /12}, n/2\right]$ w.h.p. Next lemma takes care of the case $q > n/2$.

Lemma 27

Let $\mathbf {x}$ be any starting configuration such that $q(\mathbf {x}) > \frac{n}{2}$. Then, at the next round, it holds that $Q \le q\left( 1 - p/2\right)$ with probability $1 - \exp \left( -p^2 n / 2^3\right)$.

Proof of Lemma 27

From Eq. (6), we have that

$$\begin{aligned} {\mathbb {E}}\left[ Q \big |\ \mathbf {x} \right] - q(1 - 2p)&\le pn + \frac{1 - 3p}{2n}\left[ 2q^2 + (n-q)^2 \right] - q(1-p) \\&= 3q^2 \left( \frac{1-3p}{2n}\right) - q(2-4p) + n \left( \frac{1 - p}{2}\right) = f(q). \end{aligned}$$

Now, f(q) takes its maximum either in $q_1 = n/2$, or in $q_2 = n$. Then,

$$\begin{aligned} f(q_1)&= \frac{3n}{4}\left( \frac{1-3p}{2}\right) - \frac{n}{2}(2-4p) + n \left( \frac{1 - p}{2}\right) = n \cdot \frac{3 - 9p - 8 + 16 p + 4 - 4p}{8} \\&= n \cdot \frac{-1 + 3p}{8}; \\ f(q_2)&= 3n\left( \frac{1-3p}{2}\right) - n(2 - 4p) + n \left( \frac{1 - p}{2}\right) = n \cdot \frac{3 - 9p - 4 + 8p + 1 - p}{2} \\&= -pn. \end{aligned}$$

Since $p < 1/6$, we have that $f(q) < 0$ for $n/2 < q \le n$. Thus,

$$\begin{aligned} {\mathbb {E}}\left[ Q \big |\ \mathbf {x} \right] \le q ( 1- p), \end{aligned}$$

and we can use the Chernoff bound (Lemma 17).

$$\begin{aligned} \mathbb {P}\left[ Q \ge q \left( 1 - \frac{p}{2}\right) \big |\ \mathbf {x}\right]&= \mathbb {P}\left[ Q \ge q \left( 1 - p\right) + \frac{pq}{2} \big |\ \mathbf {x}\right] \le \mathbb {P}\left[ Q \ge {\mathbb {E}}\left[ Q \big |\ \mathbf {x} \right] + \frac{pn}{4} \big |\ \mathbf {x}\right] \\&\le \exp \left( - \frac{p^2 n}{2^3}\right) . \end{aligned}$$

$\square$

Finally, we are ready to prove Theorem 3.

Proof of Theorem 3: Wrap-up

Let $\gamma = p$, and let $m=\gamma \sqrt{n}\log n$ be the target value of the bias in Lemma 22. Let $\mathbf {x}$ be any initial configuration having bias $|s |< m$. Let $\epsilon$ be any constant such that $(1-6p)^2/2 \le \epsilon < (1-6p)(1-2p)$. We have two cases.

• (i) If the number of undecided nodes is such that $q(\mathbf {x}) \in \left[ \hat{q}_{\epsilon /12},n/2\right]$, then Lemma 26 implies that the Undecided-State process reaches a configuration having bias $s \ge \gamma \sqrt{n\log n}$ in $\mathcal {O}_{p}(\log n)$ rounds with probability $1-3/n$;

• (ii) else, if the starting configuration is such that $q(\mathbf {x}) \notin \left[ \hat{q}_{\epsilon /12},n/2\right]$, then, for Lemma 23, 27, the Undecided-State process reaches within $\mathcal {O}_{p}(\log n)$ rounds a configuration having the number of undecided nodes $q \in \left[ \hat{q}_{\epsilon /12},n/2\right]$, with probability $1 - \exp \left( -p^2 n / 2^4\right)$ for Lemma 21, 20. Then, either the bias is $s \ge \gamma \sqrt{n\log n}$, or we are in case (i). As Lemmas 21, and 20 in the preliminaries imply, the probability that case (ii) and then case (i) take place is a high probability (in this case, with probability at least $1 - 4/n$).

Then, Theorem 1 gives the desired result. $\square$