Team Persuasion

Abstract

We consider two teams of agents engaging in a debate to persuade an audience of the acceptability of a central argument. This is modelled by a bipartite abstract argumentation framework with a distinguished topic argument, where each argument is asserted by a distinct agent. One partition defends the topic argument and the other partition attacks the topic argument. The dynamics are based on flag coordination games: in each round, each agent decides whether to assert its argument based on local knowledge. The audience can see the induced sub-framework of all asserted arguments in a given round, and thus the audience can determine whether the topic argument is acceptable, and therefore which team is winning. We derive an analytical expression for the probability of either team winning given the initially asserted arguments, where in each round, each agent probabilistically decides whether to assert or withdraw its argument given the number of attackers.

D. Kohan Marzagão—Supported by CNPq (206390/2014-9)

M. M. Gauy—Supported by CNPq (248952/2013-7).

A Lemmas and Proofs

Proof

(of Proposition 1 ). To show that $\gamma _u$ is a state stable configuration, notice that in round $i\in \mathbb {N}$, if $\gamma _u$ is attained, then for $a\in U-\left\{ {t}\right\}$, the probability (Eq. 1) a will be off in round $i+1$ is zero, because $a^-\subseteq W$ and all attackers of a are off. Therefore, a will still be on in round $i+1$. Similarly, we can show that the probability of being off for all $b\in W$ in round $i+1$ is one. Therefore, in round $i+1$, the state is still $\gamma _u$. A similar argument to this proves that if $\gamma _w$ is attained in round i, then it will also be the state for round $i+1$. By induction over i, $\gamma _u$ and $\gamma _w$ satisfy Definition 9.

We now show that both $\gamma _u$ and $\gamma _w$ are the only state stable configurations. Assuming the contrary. Then, we have a configuration different from $\gamma _u$ and $\gamma _w$ in which no argument has a positive probability of changing their state. In this case, we would have two nodes, say $u_1$ and $u_2$, in the same partition, say U, that have different colors (otherwise we have $\gamma _u$ and $\gamma _w$). Since G is weakly connected, there is a path that ignores edges’ directions from $u_1$ to $u_2$. This path has even length and, therefore, since $u_1$ to $u_2$ are different, there must be at least two consecutive nodes in this path with the same color. One it attacking the other, therefore, the attacked one has a positive probability of changing their color. We have a contradiction. Thus $\gamma _u$ and $\gamma _w$ must be the only state-stable configurations in a bipartite AF. $\blacksquare$

We now answer a more general version of Question F2 using the framework of consensus games and colors. We derive a formula for a color to win the consensus game on a strongly connected digraph, given that consensus will be achieved. We then investigate the necessary conditions for consensus to be achieved, and derive an expression for the probability of failing to achieve consensus that depends on $s'_0$. We then generalise to the case of weakly connected graphs, and answer Question F2 via our translation back into team persuasion games.

The in-matrix H of the digraph G can be seen as a transition matrix of a time homogeneous Markov chain, where the each node v represents a state and the reversed edges represent the transitions. If the Markov chain is irreducible and finite, there is a unique stationary distribution, which is a row vector $\mu \in \mathbb {R}_+^V$ that satisfies $\mu H = \mu$.

Proof

(of Theorem 1 ). The theorem follows from the following lemmas. Note that these lemmas are considering a general digraph $G=(V,E)$ and colors 0 and 1. We also denote $\mathbf 0$ and $\mathbf 1$ as the consensus on color 0 and 1 respectively.

Lemma 1

A consensus game on a strongly-connected digraph $G=\left\langle {V,E}\right\rangle$ reaches consensus with probability for all initial configurations 1 iff $\gcd C=1$, where $C\subseteq \mathbb {N}$ is the set of the lengths of all cycles in G. In the case $\gcd {C} = g > 1$, then G is g-partite with parts $V_1,\ldots ,V_g$ where all edges go from $V_i$ to $V_{i+1}$.

Proof

(of Lemma 1 ). ($\Leftarrow$) Assuming $\gcd C=1$. Then, given an initial configuration, a game has already reached consensus or it has not. If not, we can assume, WLOG, that there is at least one $v\in V$ colored 0. We note that the $\gcd {C_v}=1$, where $C_v$ is the set of the lengths of the cycles passing through v. This follows from the fact that G is strongly connected. We can then show that there is a large enough $n_0 > 0$ such that for any $n \ge n_0$, we have $\mathbb {P}(s_n(u) = 0 \mid s_0) > 0$ for all $u \in V$. For that it is enough to show that there is finite $n_0$, such that for every $n\ge n_0$ there is a directed path from v to u of length n. The existence of such $n_0$ follows from Lemma 2.1 of [12]. Thus, if the game runs long enough, it will reach consensus (either $\mathbf 0$ or $\mathbf 1$) with probability 1.

($\Rightarrow$) We now want to prove that if the game reaches consensus with probability 1, then $\gcd {C}=1$. We are going to prove this by showing that if $\gcd {C}>1$, then there is a positive chance that the game never reaches consensus. Let $\gcd {C} = g > 1$. We start by showing that the graph must be not only a g-partite graph, but also of the form that every edge from a node in partition i points to a node in partition $i +1 ({{\mathrm{\text {mod}}}}{g})$. Let $v \in V$. For all $w \in V$, we define the partition that w belongs to by taking the $x ({{\mathrm{\text {mod}}}}{g})$, where x is the length of any path from v to w.

We show that this is well defined. First, the existence of such a path is guaranteed by the strongly connectivity of G. Also, the lengths of all paths from v to w must coincide modulo g. If not, by concatenating both paths to the same returning path from w to v, we would have created two cycles from v to v that differ in length modulo g (by assumption, all cycles must be $0 ({{\mathrm{\text {mod}}}}g)$).

We now observe that, if the game reaches a configuration in which a partition is all 0 and another all 1, consensus will never be reached. Thus it can not be reaching consensus for sure from all possible initial configurations. We will show that no non-consensual initial configuration reaches consensus with probability 1. $\blacksquare$

Lemma 2

Consider a consensus game in a strongly connected and direct graph $G=\left\langle {V,E}\right\rangle$ in which $\gcd {C}=g$. Then, we know by Lemma 1 that G is g-partite¹¹ and we denote the partitions $V_1, \dots , V_g$. We further denote $\mu (U) = \sum _{v \in U} \mu (v)$ for $U \subset V$. In these conditions,

$$\begin{aligned} \mathbb {P}(\text {Colour } 1 \text { wins in }G \mid s_0) = \prod _{i = 1}^g\left( \dfrac{1}{\mu (V_i)} \sum _{v \in V_i} \mu (v)s_0(v) \right) \end{aligned}$$

(9)

Proof

(of Lemma 2 ). We use a similar approach to the one in [10] and apply Theorem 1 of [4]. Note that the state of vertices of $V_{i+1}$ in the round $n+1$, depends only in the state of vertices of $V_i$ in the round n. We can then consider g parallel consensus games on g copies of G, where in the i-th consensus game we set the initial state of the vertices in $V_i$ to their original initial state in the consensus game, but set the state of all other vertices to 1. Denote by $p_i$ the probability of the i-th consensus game reaching a 1 winning state. It is then easy to see that $\mathbb {P}($1$\text { wins in }G \mid s_0) = \prod _{i = 1}^g p_i$.

We are left to show that $p_i = \dfrac{1}{\mu (V_i)} \sum _{v \in V_i} \mu (v)s_0(v)$. For that end, over the i-th consensus game define the random variable $X_n = \sum _{v\in V_j} \mu (v)s_n(v)$, where $j=n+i-1({{\mathrm{\text {mod}}}}g)$. We show that the process $(X_n)_{n\in \mathbb {N}}$ is a martingale with respect to the sequence $s_n$. We need to show that $\mathbb {E}(X_{n+1}|s_n) = X_n$. By linearity of expectation $\mathbb {E}(X_{n+1}|s_n) = \sum _{v\in V_{j+1}} \mu (v)\mathbb {E}(s_{n+1}(v)|s_n)$. Note that $\mathbb {E}(s_{n+1}(v)|s_n) = \sum _{u\in V_j}h_{vu}s_n(u)$ and by changing the order of summation we get that: $\mathbb {E}(X_{n+1}|s_n) = \sum _{u\in V_j}s_n(u)\sum _{v\in V_{j+1}} \mu (v)h_{vu}$. Due to stationarity of $\mu$ and the fact that $h_{vu}$ is non-zero only for $v\in V_{j+1}$, we have that $\sum _{v\in V_{j+1}} \mu (v)h_{vu} = \mu (u)$, which implies that $\mathbb {E}(X_{n+1}|s_n) = X_n$.

Now, it is easy to see that $\mu (V_i)p_i = \mathbb {E}(X_{\infty }|X_0) = \mathbb {E}(X_0)$ and this proves that $p_i = \dfrac{1}{\mu (V_i)} \sum _{v \in V_i} \mu (v)s_0(v)$, which concludes the result. $\blacksquare$

Lemma 3

Consider a consensus game played in a weekly connected digraph $G=\left\langle {V,E}\right\rangle$ and let $\mathcal {K} = \{K_1, \dots , K_n\}$ be the set of strongly connected components (SCC) of G. We define $\mathbf source _{\mathcal {K}}$ as the set of SCCs that have no attack coming from the outside, i.e., if $K \in \mathcal {K}$, then $K \in \mathbf source _{\mathcal {K}}$ if for every $(a,b) \in E$ such that $b \in K$, we have $a \in K$. Then,

$$\begin{aligned} \mathbb {P}(\text {Colour }1 \text { wins in }G \mid s_0) = \prod _{K \in \mathbf source _{\mathcal {K}}} \mathbb {P}(\text {Colour }1 \text { wins in }K \mid s_0) \end{aligned}$$

(10)

Proof

(of Lemma 3 ). First note that each $K \in \mathbf source _{\mathcal {K}}$ is independent of each other, since they are independent from anything outside each of these SCCs. Then, we cannot have consensus if they reach different consensus. It remains now to observe that, in the case they reach same consensus colors, then all the other SCCs will eventually stabilise in the same color. That happens because of the influence they receive from components in $\mathbf source _{\mathcal {K}}$, so consensus cannot be achieved by any other color. Finally, for every node not in a source component, there is a path from a source node to it, therefore there is a non-zero probability that the game achieves the sources’ color. $\blacksquare$