The Social Medium Selection Game

Abstract

We consider in this paper competition of content creators in routing their content through various media. The routing decisions may correspond to the selection of a social network (e.g., Twitter versus Facebook or Linkedin) or of a group within a given social network. The utility for a player to send its content to some medium is given as the difference between the dissemination utility at this medium and some transmission cost. We model this game as a congestion game and compute the pure potential of the game. In contrast to the continuous case, we show that there may be various equilibria. We show that the potential is M-concave which allows us to characterize the equilibria and to propose an algorithm for computing it. We then give a learning mechanism which allow us to give an efficient algorithm to determine an equilibrium. We finally determine the asymptotic form of the equilibrium and discuss the implications on the social medium selection problem.

Appendices

Appendix A. Proof of Proposition 1

1.1 A.1. Proof of the Upper Bound $|\mathcal {E}_\varGamma | \le \left( {J}\atop {\left\lfloor \frac{J}{2} \right\rfloor }\right)$

Lemma 1

Let $\varvec{x}$ be a load vector at a Nash equilibrium, $\varvec{x}\in \mathcal {E}_\varGamma$, and u a social medium, $u \in \mathbb {J}$. Then

$$ \left( \exists \varvec{y}\in \mathcal {E}_\varGamma ,\ x_u > y_u \right) \Rightarrow \left( \forall \varvec{z}\in \mathcal {E}_\varGamma ,\ x_u \ge z_u \right) . $$

Proof

Assume that there exists $\varvec{y}$ and $\varvec{z}$ in $\mathcal {E}_\varGamma$ such that $x_u > y_u$ and $x_u < z_u$. Then, by Theorem 3, $y_u = x_u - 1$ and $z_u = x_u + 1$. Hence, $z_u - y_u = 2$ which contradicts Theorem 3.

Lemma 2

Let $\alpha \in \mathbb {J}$. Then

$$ \sum _{k=0}^{\min (\alpha ,J-\alpha )} \left( {\begin{array}{c}\alpha \\ k\end{array}}\right) \left( {\begin{array}{c}J-\alpha \\ k\end{array}}\right) = \left( {\begin{array}{c}J\\ \alpha \end{array}}\right) .$$

Proof

We show this result using a combinatorial argument. First, note that since $\displaystyle \left( {\begin{array}{c}J\\ \alpha \end{array}}\right) = \left( {\begin{array}{c}J\\ J-\alpha \end{array}}\right)$, then one can restrict the analysis to the case where $\alpha \le J-\alpha$.

We want to select $\alpha$ elements in $\mathbb {J}$. To do that, we partition the set $\mathbb {J}$ into two subsets A and B such that $|A| = \alpha$ (so $|B|=J-\alpha$).

Selecting $\alpha$ elements in $\mathbb {J}$ amounts to choosing k the number of elements we select in B, then select these k elements and finally select $\alpha -k$ elements in A. Therefore,

$$ \displaystyle \qquad \qquad \left( {\begin{array}{c}J\\ \alpha \end{array}}\right) = \displaystyle \sum _{k=0}^{|A|} \left( {\begin{array}{c}|B|\\ k\end{array}}\right) \left( {\begin{array}{c}|A|\\ \alpha -k\end{array}}\right) = \sum _{k=0}^{\alpha } \left( {\begin{array}{c}\alpha \\ \alpha -k\end{array}}\right) \left( {\begin{array}{c}J-\alpha \\ k\end{array}}\right) = \sum _{k=0}^{\min (\alpha ,J-\alpha )} \left( {\begin{array}{c}\alpha \\ k\end{array}}\right) \left( {\begin{array}{c}J-\alpha \\ k\end{array}}\right) . \blacksquare $$

We can now proceed to the proof of Theorem 1:

Proof

Let $\varvec{x}\in \mathcal {E}_\varGamma$ and:

By Lemma 1, we have $\mathcal {U} \cap \mathcal {V} = \emptyset$ and $|\mathcal {U}|+|\mathcal {V}| \le J$. We then define the set $\mathcal {A}=$

We know by Proposition 1 and Lemma 1 that all vectors in $\mathcal {E}_\varGamma$ are of the form given in the previous expression; hence $\mathcal {E}_\varGamma \subset \mathcal {A}$.

Let $\alpha = |\mathcal {U}|$. We have $|\mathcal {V}| \le J - \alpha$. Then

$$\begin{aligned} |\mathcal {E}_\varGamma |&\le |\mathcal {A}| = 1 + \sum _{k=1}^{\min (|\mathcal {U}|,|\mathcal {V}|)} \left( {\begin{array}{c}|\mathcal {U}|\\ k\end{array}}\right) \left( {\begin{array}{c}|\mathcal {V}|\\ k\end{array}}\right) \\&\le 1 + \sum _{k=1}^{\min (\alpha ,J-\alpha )} \left( {\begin{array}{c}\alpha \\ k\end{array}}\right) \left( {\begin{array}{c}J-\alpha \\ k\end{array}}\right) . \end{aligned}$$

We conclude the proof by applying Lemma 2, using the increasing property of function $\left( {\begin{array}{c}J\\ .\end{array}}\right)$ over $\{0,...,\lfloor J/2 \rfloor \}$ and the fact that $\left( {\begin{array}{c}J\\ p\end{array}}\right) = \left( {\begin{array}{c}J\\ J-p\end{array}}\right)$ for all p:

$$\begin{aligned} |\mathcal {E}_\varGamma | \le \left( {\begin{array}{c}J\\ \alpha \end{array}}\right) \le \left( {\begin{array}{c}J\\ \left\lfloor \frac{J}{2} \right\rfloor \end{array}}\right) . \blacksquare \end{aligned}$$

1.2 A.2. A Tight Class of Settings

Let $J \ge 2$, $m \in \mathbb {N}^*$ and $\gamma \in \mathbb {R}^+$. We define the game $\varGamma$ by $K = \left\lfloor \frac{J}{2} \right\rfloor \text { and } \forall j \in \mathbb {J}, N_j = m,\ \gamma _j = \gamma$.

Lemma 3

The Nash equilibria of game $\varGamma$ satisfy the property:

$$ \varvec{\ell }\in \mathcal {E}_\varGamma \Rightarrow \exists A \subset \mathbb {J},\ |A| = \left\lfloor \frac{J}{2} \right\rfloor \text { and } \varvec{\ell }= \sum _{u \in A} \varvec{e}_u. $$

Proof

Assume that there exists $\varvec{x} \in \mathcal {E}_\varGamma$ and $u \in \mathbb {J}$ such that $x_u > 1$. Since $K < J$, there exists $v \in \mathbb {J}$ such that $x_v = 0$. Consider the vector $\varvec{y}= \varvec{x}- x_u \varvec{e}_u + x_u \varvec{e}_v.$

Since all the $N_j$ and $\gamma _j$ are equal, the potential of $\varvec{y}$ is equal to the potential of $\varvec{x}$. Therefore, $\varvec{y}\in \mathcal {E}_\varGamma$. But we have $y_v - x_v = x_u > 1$ which contradicts Theorem 3 and concludes the proof.

Since $\mathcal {E}_\varGamma \ne \emptyset$, let $\varvec{x}\in \mathcal {E}_\varGamma$. By Lemma 3, we can note $\varvec{x}= \sum _{u \in A} \varvec{e}_u$ for some $A \subset \mathbb {J}$. Let $B \subset \mathbb {J}$ verifying $|B| = \left\lfloor \frac{J}{2} \right\rfloor$ and $\varvec{y}= \sum _{v \in B} \varvec{e}_v$. Then, we have

$$\begin{aligned} {\text {Pot}}(\varvec{x}) = \sum _{u \in A} (m - \gamma ) = \left\lfloor \frac{J}{2} \right\rfloor (m - \gamma ) = \sum _{v \in B} (m - \gamma ) = {\text {Pot}}(\varvec{y}). \end{aligned}$$

$$ \text {Therefore, } |\mathcal {E}_\varGamma | = \left| \left\{ A \subset \mathbb {J}\ |\ |A| = \left\lfloor \frac{J}{2} \right\rfloor \right\} \right| = \left( {\begin{array}{c}J\\ \left\lfloor \frac{J}{2} \right\rfloor \end{array}}\right) .$$

Appendix B. A Polynomial Algorithm to Find a Maximum of f

A polynomial algorithm to determine the minimum of an M-convex function is given in [15, Sect. 4.2]. We can adapt it for maximizing our M-concave function f. In fact, this algorithm does not have a polynomial complexity in general. We proceed to show the property required which is that the M-concavity of f is respected by a scaling operation.

Proposition 4

Let $\alpha \in \mathbb {N}^*$ and $\varvec{x}\in \mathcal {D}$. We define the function $\widetilde{f}$ as

$$ \widetilde{f}(\varvec{y}) = {\left\{ \begin{array}{ll} f(\varvec{x}+\alpha \varvec{y}) &{}\text { if } \varvec{x}+\alpha \varvec{y}\in \mathcal {D} \\ - \infty &{}\text { otherwise.} \end{array}\right. }$$

Then $\widetilde{f}$ is an M-concave function.

Proof

Let $\varvec{y},\varvec{z}\in \mathbb {Z}^J$ such that $\varvec{x}+\alpha \varvec{y}\in \mathcal {D}$ and $\varvec{x}+\alpha \varvec{z}\in \mathcal {D}$. Let $u \in {\text {supp}}^+(\varvec{y}-\varvec{z})$. Since $\alpha > 0$, the same argument as in the proof of the M-concavity of f gives us that there exists some $v \in {\text {supp}}^+(\varvec{z}-\varvec{y})$. Then we calculate

$$\begin{aligned} \widetilde{f}(\varvec{y}-\varvec{e}_u+\varvec{e}_v)-\widetilde{f}(\varvec{y})&= f(\varvec{x}+\alpha (\varvec{y}-\varvec{e}_u+\varvec{e}_v)) - f(\varvec{x}+\alpha \varvec{y}) \\&= \sum _{i=x_v+\alpha y_v+1}^{x_v+\alpha (y_v+1)} \frac{1}{i} - \sum _{i=x_u+\alpha (y_u-1)+1}^{x_u+\alpha y_u} \frac{1}{i}\\&= \sum _{i=1}^{\alpha } \frac{1}{x_v+\alpha y_v+i} - \sum _{i=1}^{\alpha } \frac{1}{x_u+\alpha (y_u-1)+i}. \end{aligned}$$

We have a similar expression for $\widetilde{f}(\varvec{z}-\varvec{e}_v+\varvec{e}_u)-\widetilde{f}(\varvec{z})$. Then

$$\begin{aligned} \widetilde{f}(\varvec{y}-\varvec{e}_u+\varvec{e}_v)-\widetilde{f}(\varvec{y})+\widetilde{f}(\varvec{z}-\varvec{e}_v+\varvec{e}_u)-\widetilde{f}(\varvec{z})&=\sum _{i=1}^\alpha \frac{1}{x_v+\alpha y_v+i} - \sum _{i=1}^{\alpha } \frac{1}{x_v+\alpha (z_v-1)+i}\\&+\sum _{i=1}^\alpha \frac{1}{x_u+\alpha z_u+i} - \sum _{i=1}^{\alpha } \frac{1}{y_u+\alpha (y_u-1)+i}. \end{aligned}$$

Since $y_v \le z_v-1$ and $z_u \le y_u-1$, then the last quantity is positive. Hence, $\widetilde{f}$ is M-concave.

In order to implement Algorithm SCALING_MODIFIED_STEEPEST_DESCENT given in [15], we represent the active domain of the search of a maximizer B using two vectors $\varvec{m}$ and $\varvec{M}$ such that

$$ B = \left\{ \varvec{x}\ |\ \sum _j x_j = K \text { and } \forall j,\ m_j \le x_j \le M_j \right\} .$$

We choose the origin point to be $\varvec{x}= K\varvec{e}_1$. Then the only difficulty that remains is to compute some $\varvec{y}$ such that $\varvec{x}+\alpha \varvec{y}\in B$ in order to search for a maximum of the scaled auxiliary function. In fact, a solution is also given in [15, Sect. 5.1] with Algorithm FIND_VECTOR_IN_$N_B$ which find a vector $\varvec{y}$ whose components are within the constraints $\frac{\varvec{x}-\varvec{m}}{\alpha }$ and $\frac{\varvec{M}-\varvec{x}}{\alpha }$ and for which $\sum _j y_j = 0$. This algorithm has a time complexity in $\mathcal {O}\!\left( J\right)$.

Hence, we have an algorithm to compute some $\varvec{\ell }\in \mathcal {E}_\varGamma$ in $\mathcal {O}\!\left( J^3 \log (K/J)\right)$, which is better than $\mathcal {O}\!\left( KJ\right)$ if we study the case when there are a lot more seeds than social media.

Appendix C. Proof of the Asymptotic Behavior

Note that, by definition of the learning mechanism implemented in Algorithm 2, for all $K \in \mathbb {N}$ and $j \in \mathbb {J}$ we have

$$\begin{aligned} \ell ^{(K+1)}_j = \ell ^{(K)}_j+1 \Rightarrow j \in arg\,max_{t \in \mathbb {J}} \left( \frac{N_t}{\ell _t+1} - \gamma _t \right) . \end{aligned}$$

(5)

1.1 C.1. Social Media with Non-minimal Cost

We want to prove that

$$\begin{aligned} \forall j \in \mathbb {J}\setminus G,\ \ell ^{(K)}_j \underset{K \rightarrow \infty }{\longrightarrow } \left\lceil \frac{N_j}{\gamma _j-\gamma _m} \right\rceil -1. \end{aligned}$$

(6)

We begin by proving the following two lemmas.

Lemma 4

The quantity

$$M^{(K)} = \max _{t \in \mathbb {J}} \left( \frac{N_t}{\ell ^{(K)}_t+1}-\gamma _t \right) $$

is arbitrarily close to $\gamma _m$ for K large enough.²

Proof

First, this quantity is decreasing. Moreover, by definition of the $\varvec{\ell }^{(K)}$, we have that for all K, $\displaystyle \sum _j \ell ^{(K)}_j = K \underset{K \rightarrow \infty }{\longrightarrow } \infty .$ Therefore, there exists some $u \in \mathbb {J}$ such that $\ell ^{(K)}_u \underset{K \rightarrow \infty }{\longrightarrow } \infty$. It means that there exists $(K_n)_{n\in \mathbb {N}}$ such that $\forall n,\ \ell ^{(K_n+1)}_u = \ell ^{(K_n)}_u + 1$, which implies by (5) that $\forall n,\ \frac{N_u}{\ell ^{(K_n)}_u+1}-\gamma _u = M^{(K_n)}.$

Hence, $M^{(K_n)}$ is arbitrarily close to $-\gamma _u$ for n large enough. We conclude by noticing that $-\gamma _u\le -\gamma _m$.

Lemma 5

Let $K>0$ and $u \in \mathbb {J}$. Then $\displaystyle \frac{N_u}{\ell ^{(K)}_u+1} - \gamma _u> -\gamma _m\Leftrightarrow \exists \, K'\!>\!K,\ \ell ^{(K')}_u > \ell ^{(K)}_u.$

Proof

First, assume that $\frac{N_u}{\ell ^{(K)}_u+1} - \gamma _u \le -\gamma _m$. Then for some $w \in G$ and for all $K'\ge K$ we have

$$ \frac{N_u}{\ell ^{(K)}_u+1} - \gamma _u < \frac{N_w}{\ell ^{(K')}_w+1} - \gamma _w $$

since $\gamma _w = \gamma _m$. This implies that $\frac{N_u}{\ell ^{(K)}_u+1} - \gamma _u < M^{(K')}$. Therefore, for all $K'>K$, (5) leads to $\ell ^{(K')}_u = \ell ^{(K)}_u$.

Then assume that $\frac{N_u}{\ell ^{(K)}_u+1} - \gamma _u > -\gamma _m$. According to Lemma 5, $M^{(K')}$ is arbitrarily close to $-\gamma _m$ for $K'$ large enough. Therefore, there exists $K'>K$ such that $\displaystyle M^{(K')} < \frac{N_u}{\ell ^{(K)}_u+1} - \gamma _u.$ Hence, $\ell ^{(K')}_u > \ell ^{(K)}_u$ which concludes the proof. $\blacksquare$

We can now proceed with the proof of (6). Let $j \in \mathbb {J} \setminus G$. We know by Lemma 5 that $\ell ^{(K)}_j$ increases with K as long as $\displaystyle \frac{N_j}{\ell ^{(K)}_j+1} - \gamma _j > -\gamma _m.$ Therefore, for K large enough we have

$$\ell ^{(K)}_j = 1 + \max \left\{ p \in \mathbb {N}\ |\ \frac{N_j}{p+1}-\gamma _j > -\gamma _m\right\} .$$

Then, let $p\in \mathbb {N}$. Since $j \in \mathbb {J}\setminus G$, we have $\gamma _j>\gamma _m$. We solve $\displaystyle \frac{N_j}{p+1}-\gamma _j > -\gamma _m\Leftrightarrow p+1 < \frac{N_j}{\gamma _j-\gamma _m}.$ Hence, $\displaystyle \ell ^{(K)}_j + 1 = \left\lceil \frac{N_j}{\gamma _j-\gamma _m} \right\rceil$ which concludes the proof.

1.2 C.2. Social Media with Minimal Cost

We can directly conclude from Lemma 5 that the load of any social medium having a minimal cost goes to infinity as K increases. Formally:

$$\begin{aligned} \forall w \in G,\ \ell ^{(K)}_w \underset{K \rightarrow \infty }{\longrightarrow } \infty . \end{aligned}$$

(7)

Now we proceed to find the values of $\ell ^{(K)}_w$ for the social media with minimal cost. Let K be large enough so that  (6) is verified. Let $\displaystyle K_G = K - \sum _{j \in \mathbb {J} \setminus G} \ell ^{(K)}_j$ be the number of seeds sharing the social media in G and

$$\mathcal {D}_G = \left\{ (x_t)_{t\in G}\ | \sum _{t \in G} x_t = K_G \text { and } \forall t\in G, x_t > 0 \right\} .$$

Consider the game $\varGamma _G = (K_G,(N_t,\gamma _m)_{t \in G})$. From (7), the loads of the social media in G can be arbitrarily high with K large enough, so we determine an approximation of a load of a Nash equilibrium for the social media in G by solving

$$ \max _{\varvec{x}\in \mathbb {R}^G} P(\varvec{x})=\sum _{t \in G} \left( N_t \ln (x_t) - \gamma _mx_t\right) \text { s.t. } \varvec{x}\in \mathcal {D}_G.$$

Since P is concave, we apply a Lagrangian maximization method. Let L be the Lagrangian for this problem:

$$ L(\varvec{x},\lambda )\! =\! P(\varvec{x}) - \lambda \left( \sum _{t \in G} x_t - K_G \right) , $$

where $\lambda$ and the $x_t$ are nonnegative.

Since P is concave, the unique maximum $\varvec{x}^*$ verifies $\displaystyle \forall t\in G,\ \frac{\partial L}{\partial x_t}(\varvec{x}^*) = 0.$ Therefore, we get that for any t: $\quad \displaystyle \frac{N_t}{x^*_t} - \gamma _m- \lambda = 0 \Leftrightarrow x^*_t = \frac{N_t}{\gamma _m+ \lambda }.$

Now we determine the value of $\lambda$:

$$\begin{aligned} \sum _{t\in G} x^*_t = K_G \quad \Rightarrow \quad \sum _{t\in G} \frac{N_t}{\gamma _m+ \lambda } = K_G \quad \Rightarrow \quad \lambda = \frac{1}{K_G} \left( \sum _{t\in G} N_t\right) - \gamma _m. \end{aligned}$$

Hence, $\forall w\in G,\ x^*_w = K_G\frac{N_w}{\sum _{t\in G} N_t}.$

Thanks to (7) and since $H_n-\mu \underset{n \rightarrow \infty }{\sim } \ln n$, we finally get that $\displaystyle \forall w\in G, \frac{\ell ^{(K)}_w}{\sum _{t \in G} \ell ^{(K)}_t} \underset{K \rightarrow \infty }{\longrightarrow } \frac{N_w}{\sum _{t \in G} N_t}.$