Reducing polarization and increasing diverse navigability in graphs by inserting edges and swapping edge weights

Abstract

The sets of hyperlinks in web pages, relationship ties in social networks, or sets of recommendations in recommender systems, have a major impact on the diversity of content accessed by the user in a browsing session. Bias induced by the graph structure may trap a reader in a polarized bubble with no access to other opinions. It is widely accepted that exposure to diverse opinions creates more informed citizens and consumers. We introduce the concept of the polarized bubble radius of a node, as the expected length of a random walk from it to a node of different opinion. Using the bubble radius, we define the measures of structural bias and diverse navigability to quantify the effect of links and recommendations on the diversity of content visited in a browsing session. We then propose algorithmic techniques to reduce the structural bias of the graph or improve the diverse navigability of the system through minimal modifications, such as edge insertions or flipping the order of existing links or recommendations, corresponding to switching the edge traversal probabilities. Under mild conditions, our techniques obtain a constant factor-approximation of their respective tasks. In our extensive experimental evaluation, we show that our algorithms reduce the structural bias or improve the diverse navigability faster than appropriate baselines, including some designed with the goal of reducing the polarization of a graph.

A Postponed Proofs

Lemma 3

Let X be a random variable satisfying $0 \le X \le t$. We have:

$$\begin{aligned} \mathbb {P}(X \le k) \le \frac{t - \mathbb {E}_{} \left[ X \right] }{t - k} . \end{aligned}$$

Proof

It holds

$$\begin{aligned} \mathbb {E}_{} \left[ X \right] = \int _{0}^{k} x p(x) dx +\int _{k}^{t} x p(x) dx \le k \left( 1 - \mathbb {P}(X \ge k) \right) + t \mathbb {P}(X \ge k) . \end{aligned}$$

Thus,

$$\begin{aligned} \mathbb {P}(X \ge k) \ge \frac{\mathbb {E}_{} \left[ X \right] - k}{t - k}, \end{aligned}$$

and

$$\begin{aligned} \mathbb {P}(X \le k) = 1 - \mathbb {P}(X \ge k) \le 1 - \frac{\mathbb {E}_{} \left[ X \right] - k}{t - k} = \frac{t - \mathbb {E}_{} \left[ X \right] }{t - k} . \end{aligned}$$

$\square$

Lemma 4

Let $v \in \mathcal {P}_{C_v}(G)$, then, for any $t'\le t$, it holds $\mathsf {B}^{t'}_{G}\left( v \right) \ge r\frac{t'}{t}$.

Proof

From the hypothesis, using the definition of BR, it holds

$$\begin{aligned} r&\le \mathsf {B}^{t}_{G}\left( v \right) = t \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{\ge t}_{G} \bar{C}_v \right) + \sum _{i=1}^{t-1} i \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{= i}_{G} \bar{C}_v \right) \\&\le t \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{\ge t'}_{G} \bar{C}_v \right) + \sum _{i=1}^{t'-1} i \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{= i}_{G} \bar{C}_v \right) . \end{aligned}$$

By rearranging the terms, we obtain

$$\begin{aligned} \left( r- \sum _{i=1}^{t'-1} i \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{= i}_{G} \bar{C}_v \right) \right) \frac{1}{t} \le \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{\ge t'}_{G} \bar{C}_v \right) . \end{aligned}$$

Thus, for $t'\le t$, it holds

$$\begin{aligned} \mathsf {B}^{t'}_{G}\left( v \right)&= t' \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{\ge t'}_{G} \bar{C}_v \right) + \sum _{i=1}^{t'-1} i \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{= i}_{G} \bar{C}_v \right) \\&\ge rj \frac{t'}{t} - \sum _{i=1}^{t'-1}i\left( 1-\frac{t'}{t} \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{\ge t'}_{G} \bar{C}_v \right) \right) \ge \frac{t'}{t} r. \end{aligned}$$

$\square$

Lemma 14

Let $C \in \{R,B\}$, $v\in C$, $t'\le t$, and $(e,e')\in \Re _{C}$ with v the source vertex and w the radicalizing end point. Let $\delta = M(e) - M(e')$. It holds

$$\begin{aligned} \delta \mathsf {B}^{t'-1}_{G}\left( w \right) \le \varGamma (G,v,(e,e'),t') \le \mathcal {F}_{t'-1}\left( v\right) \delta \mathsf {B}^{t'}_{G}\left( w \right) . \end{aligned}$$

Proof

Let $G_{e,e'}$ be the graph obtained after swapping the probabilities of e and $e'$. Consider the probability space of all random walks starting from v in $G_{e,e'}$ and G. We introduce a coupling between these two probability spaces as follows: consider a walk in G, for any step that is not traversing e couple it to an identical step in G. If a step traverses e, then, with probability $1-\delta$, couple it to the same step in $G_{e,e'}$, else couple it to $e'$ in $G_{e,e'}$.

Let $\mathcal {E}_i$, $1\le i\le t'$ be the event that the coupling diverges at the i-th step, which is equivalent to being at v at step $i-1$ and the first r.w. taking e, the second taking $e'$.

When $\mathcal {E}_i$ happens, then the walk in $G_{e,e'}$ has reached the other color by taking $e'$ while the walk in G still needs, in expectation, $\mathsf {B}^{t'-i}_{G}\left( w \right)$ steps to reach the other color. Using the law of total expectation and summing over all $1\le i\le t'$, we can write

$$\begin{aligned} \varGamma (G,v,(e,e'),t') = \sum _{i=1}^{t'} \mathsf {B}^{t'-i}_{G}\left( w \right) \mathbb {P}(\mathcal {E}_i) = \sum _{i=1}^{t'-1} \mathsf {B}^{t'-i}_{G}\left( w \right) \mathbb {P}(\mathcal {E}_i) . \end{aligned}$$

It holds

$$\begin{aligned} {\mathbb {P}(\mathcal {E}_i)= \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{i-1}_{G} v \right) \delta .} \end{aligned}$$

Since $\mathsf {B}^{t'-i}_{G}\left( w \right) \ge 1$ for every $1 \le i < t'$, and clearly $\mathbb {P}(\mathcal {E}_1) = \delta$, we obtain the left hand side of the thesis.

The right hand side is concluded as follows. It holds $\mathsf {B}^{t'-i}_{G}\left( w \right) \le \mathsf {B}^{t'-1}_{G}\left( w \right)$, for every $1 \le i < t'$, and we can write, using (3),

$$\begin{aligned} {\sum _{i=1}^{t'-1} \mathbb {P}({\mathcal {E}}_i) = \sum _{i=1}^{t'-1} \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{i-1}_{G} v \right) \delta = \sum _{i=0}^{t'-2} \mathbb {P}\left( v \mathop {\rightsquigarrow }\limits ^{i}_{G} v \right) \delta = \mathcal {F}_{t'-1}\left( v\right) \delta .} \end{aligned}$$

$\square$