Measuring and moderating opinion polarization in social networks

Abstract

The polarization of society over controversial social issues has been the subject of study in social sciences for decades (Isenberg in J Personal Soc Psychol 50(6):1141–1151, 1986, Sunstein in J Polit Philos 10(2):175–195, 2002). The widespread usage of online social networks and social media, and the tendency of people to connect and interact with like-minded individuals has only intensified the phenomenon of polarization (Bakshy et al. in Science 348(6239):1130–1132, 2015). In this paper, we consider the problem of measuring and reducing polarization of opinions in a social network. Using a standard opinion formation model (Friedkin and Johnsen in J Math Soc 15(3–4):193–206, 1990), we define the polarization index, which, given a network and the opinions of the individuals in the network, it quantifies the polarization observed in the network. Our measure captures the tendency of opinions to concentrate in network communities, creating echo-chambers. Given this numeric measure of polarization, we then consider the problem of reducing polarization in the network by convincing individuals (e.g., through education, exposure to diverse viewpoints, or incentives) to adopt a more neutral stand towards controversial issues. We formally define the ModerateInternal and ModerateExpressed problems, and we prove that both our problems are NP-hard. By exploiting the linear-algebraic characteristics of the opinion formation model we design polynomial-time algorithms for both problems. Our experiments with real-world datasets demonstrate the validity of our metric, and the efficiency and the effectiveness of our algorithms in practice.

Appendix A: Theorem 1 proof

Theorem 1 The ModerateInternal problem is NP-hard.

Proof

Our proof uses a reduction from the m-SubsetSum problem, where given a set of N positive integer numbers $v_1,\ldots ,v_N$, a value m, and a target value b, we ask if there is a set of numbers B of size m, such that $\sum _{v_i \in B} v_i = b$.

Given an instance of the m-SubsetSum problem, we construct an instance of ModerateInternal as follows. The graph is a star with $N+1$ nodes: we have a central node $u_0$, and a spoke node $u_i$ for each integer $v_i$. For the center of the star (node $u_0$) we have that $w_{00} = t$, for an appropriately selected value of t (we will discuss this below), and $s_0 = -1$. The weight of the edge $(u_0,u_i)$ from the center to node $u_i$ is $w_{0i} = v_i$, and the weight of node $u_i$ to its internal opinion is also $w_{ii} = v_i$. The opinion of all spoke nodes is $s_i = 1$. We set $k = N-m$, and we ask for a set of nodes $T_s$, $|T_s| = k$, such that, when setting $s_i = 0$ for $u_i \in T_s$ $\pi (\mathbf {z}\mid T_z) = \Vert \mathbf {z}\Vert ^2$ is minimized.

The intuition of the proof is that the expressed opinion of the center node $z_0$ determines $\pi (\mathbf {z})$. The value of $z_0$ is determined by the weight t of the internal opinion of $u_0$, and the weights of the edges of nodes whose opinion is not set to zero. If we select t appropriately, we can guarantee that $\Vert \mathbf {z}\Vert ^2$ is minimized when the nodes whose opinion is not set to zero sums to the value b.

Formally, assume that we have selected the set $T_s$, $|T_s| = k$. Assume that $u_0 \not \in T_s$. Also let $R = V {\setminus } T_s\cup \{u_0\}$ denote the set of spoke nodes whose opinion was not set to 0. According to the opinion formation model, the equations for the expressed opinions of the spoke nodes are as follows. For every node $u_i \in R$, $z_i = \frac{z_0}{2} + \frac{1}{2}.$ while for every node $u_i \in T_s$, $z_i = \frac{z_0}{2}.$

We can thus write:

$$\begin{aligned} \pi (\mathbf {z}\mid T_s)= & {} \Vert \mathbf {z}\Vert ^2 = z_0^2 + k\frac{1}{4} z_0^2 + (N-k)\frac{1}{4}(z_0^2 + 2z_0 + 1)\\= & {} \frac{N+4}{4}z_0^2 + \frac{N-k}{2}z_0 + \frac{N-k}{4}. \end{aligned}$$

Recall that we want to minimize $\pi (\mathbf {z}\mid T_s)$. To find the value of $z_0$ that minimizes $\pi (\mathbf {z}\mid T_s)$, we take the derivative of the expression above, we set it zero, and solve for $z_0$. We get that the value of $z_0$ that minimizes $\pi (\mathbf {z})$ is:

$$\begin{aligned} z_0^*= \frac{k-N}{N+4} . \end{aligned}$$

It follows that the minimum value of $\pi (\mathbf {z}\mid T_s)$ is

$$\begin{aligned} \pi ^*= \frac{(N-k)(k+4)}{4(N+4)}. \end{aligned}$$

We now set the value of t such that if the set of numbers in R sums to the value of b, then $z_0$ achieves the $z_0^*$ value. First we compute the value of $z_0$ as a function of t. In the following we set $W = \sum _{i=1}^N v_i$. We have that:

$$\begin{aligned} z_0= & {} \sum _{i = 1}^N \frac{v_iz_i}{W+t} - \frac{t}{W+t} = \sum _{u_i \in T_s} \frac{v_i z_0}{2(W+t)} + \sum _{u_i \in R} \frac{v_i(z_0 + 1)}{2(W+t)} - \frac{t}{W+t} \\= & {} \frac{\sum _{i = 1}^N v_i}{2(W+t)}z_0 + \frac{\sum _{u_i \in R} v_i}{2(W+t)} - \frac{t}{W+t} = \frac{W}{2(W+t)}z_0 + \frac{\sum _{u_i \in R} v_i - 2t}{2(W+t)} \end{aligned}$$

Solving for $z_0$ we get:

$$\begin{aligned} z_0 = \frac{\sum _{u_i \in R} v_i - 2t}{W+2t}. \end{aligned}$$

We want the minimum to be achieved when $\sum _{u_i \in R} v_i = b$. Setting $z_0 = z_0^*$ we get:

$$\begin{aligned} \frac{b - 2t}{W+2t} = \frac{K-N}{N+4} \end{aligned}$$

Solving for t we get:

$$\begin{aligned} t = \frac{(N+4)b + (N-k)W}{2(k+4)}. \end{aligned}$$

Now, we want to prove the following. There is a set B of m numbers such that $\sum _{v_i \in B} v_i = b$, if and only if there is a set of nodes $T_s$ of size $k = N-m$ such that when setting their internal opinion to zero, $\pi (\mathbf {z}\mid T_s) < \pi ^*+\epsilon$ for some appropriate value of $\epsilon$.

The forward direction is easy. If there exists this set B, then there is a set $T_s$ such that when setting their opinions to zero, for the set R we have that

$$\begin{aligned} z_0 = \frac{\sum _{u_i \in R} v_i - 2t}{W+2t} = \frac{b - 2t}{W+2t} = \frac{k-N}{N+4} , \end{aligned}$$

and therefore $\pi (\mathbf {z}\mid T_s) = \pi ^*$.

For the backwards direction, if no such set of numbers exists, then it is not possible to find a set of nodes $T_s$ such the nodes in R give $z_0 = \frac{K-N}{N+4}$ that minimizes $\pi (\mathbf {z}\mid T_s)$. Therefore, there must be an $\epsilon$ such that $\pi (\mathbf {z}\mid T_s) \ge \pi ^*+ \epsilon$.

To set $\epsilon$ note that for any $z_0 \ne z_0^*$

$$\begin{aligned} \left| z_0 - z_0^*\right| = \left| \frac{\sum _{u_i \in R} v_i - b}{W+2t} \right| \ge \frac{1}{W+2t} = \frac{k+4}{(N+4)(W+b)} , \end{aligned}$$

where the inequality follows from the fact that the values $v_1,\ldots ,v_N, b$ are integers and their difference is at least one. Now, let $\mathbf {z}^*$ be the vector with $z_0^*$ that achieves the minimum value $\pi ^*$. For any other $\mathbf {z}$ we have

$$\begin{aligned} \pi (\mathbf {z}) - \pi ^*= & {} \frac{N+4}{4}\left( z_0^2- (z_0^*)^2\right) + \frac{N-k}{2}(z_0 - z_0^*)\\= & {} \left( z_0-z_0^*\right) \left( \frac{N+4}{4}z_0 + \frac{N+4}{4}z_0^*-\frac{2(N+4)}{4} \frac{k-N}{N+4}\right) \\= & {} \left( z_0-z_0^*\right) \left( \frac{N+4}{4}z_0 - \frac{N+4}{4}z_0^*\right) = \frac{N+4}{4} \left( z_0-z_0^*\right) ^2\\\ge & {} \frac{N+4}{4}\left( \frac{1}{W+2t}\right) ^2 = \frac{(k+4)^2}{4(N+4)(W+b)^2}. \end{aligned}$$

So it suffices to set $\epsilon < \frac{(k+4)^2}{4(N+4)(W+b)^2}$.

Finally, in our computations so far we have assumed that our set $T_s$ does not contain node $u_0$. This is not a restrictive assumption. Consider a solution $T_s$, where $u_0 \in T_s$, and $s_0 = 0$. Then, since $s_0$ is the only negative opinion value in our instance, it follows that $z_0 \ge 0$, and for any node $u_i\in R$ we have that $z_i = \frac{1}{2} z_0 + \frac{1}{2} \ge \frac{1}{2}$. There are $N+1-k$ nodes in R. Therefore,

$$\begin{aligned} \pi (\mathbf {z}\mid T_s) \ge \frac{N+1-k}{2}. \end{aligned}$$

Note that $\pi ^*= (N-k)(k+4)/4(N+4) \le (N-k)/4$, since $k \le N$. Therefore, $\pi (\mathbf {z}) \ge 2\pi ^*+1/4$. Selecting $\epsilon < \pi ^*+ \frac{1}{4}$ guarantees that $\pi (\mathbf {z}| T_s) > \pi ^*+ \epsilon$. Thus, if there is a set $T_s$ such that $\pi (\mathbf {z}| T_s)$ is minimized, it cannot contain $u_0$. $\square$