On Binary Networked Public Goods Game with Altruism

For each player $v\in\mathcal{V}$, let $d_{v}$ denote the degree of $v$. At each node $v$ with parent $u$, we maintain a table of tuples $(x_{u},n_{u},x_{v},n_{v})$ of valid configurations. A tuple $(x_{u},n_{u},x_{v},n_{v})$ is said to be a valid configuration if there exists a strategy profile $\textbf{x}^{\prime}=(x_{v}^{\prime})_{v\in\mathcal{V}}$ such that the following holds true:

• $\vartriangleright$

  $x_{u}^{\prime}=x_{u}$, $x_{v}^{\prime}=x_{v}$

• $\vartriangleright$

  The number of neighbours of $u$ and $v$ playing $1$ in $\textbf{x}^{\prime}$ is $n_{u}$ and $n_{v}$ respectively

• $\vartriangleright$

  None of the players in the sub-tree rooted at $v$ deviate from their strategy in $\textbf{x}^{\prime}$

Note that the root node $r$ doesn’t have a parent. Hence, we consider an imaginary parent $p$ with $x_{p}=0$ and $g_{p}(x)=0$ for all $x\geqslant 0$. Hence if there is a tuple in the table of root node $r$ then we can conclude that there is a PSNE otherwise we can conclude that there is no PSNE.

Leaf nodes: We add a tuple $(x_{u},n_{u},x_{v},n_{v})$ to the table if $n_{v}=x_{u}$, $v$ does not deviate if it plays $x_{v}$ and $x_{v}\leqslant n_{u}\leqslant d_{u}+x_{v}-1$. Table for the leaf node can be clearly constructed in polynomial time.

Non-leaf nodes: For each tuple $(x_{u},n_{u},x_{v},n_{v})$ we do the following. If there is no child $u^{\prime}$ of $v$ having a tuple of type $(x_{v},n_{v},x_{u^{\prime}},n_{u^{\prime}})$ in its table, then we don’t add $(x_{u},n_{u},x_{v},n_{v})$ to the table of $v$. Similarly if $n_{u}>d_{u}+x_{v}-1$ or $n_{u}<x_{v}$, then we don’t add $(x_{u},n_{u},x_{v},n_{v})$ to the table of $v$. Otherwise we do the following. Let $U_{1}$ be the set of children $u^{\prime}$ of $v$ which have tuples of the type $(x_{v},n_{v},1,n_{u^{\prime}})$ in their table but don’t have tuples of the type $(x_{v},n_{v},0,n_{u^{\prime}})$. Let $U_{0}$ be the set of children $u^{\prime}$ of $v$ which have tuples of the type $(x_{v},n_{v},0,n_{u^{\prime}})$ in their table but don’t have tuples of the type $(x_{v},n_{v},1,n_{u^{\prime}})$. Let $U$ be the set of children $u^{\prime}$ of $v$ which have tuples of the type $(x_{v},n_{v},1,n_{u^{\prime}})$ and $(x_{v},n_{v},0,n_{u^{\prime}})$ in their table. First let us consider the case when $x_{v}=1$. Now for each $u^{\prime}\in(U_{1}\cup U)\cap N_{v}$, we find the tuple $(1,n_{v},1,n_{u^{\prime}})$ in its table so that $a\cdot\Delta g_{u^{\prime}}(n_{u^{\prime}})$ is maximized and let this value be $y_{u^{\prime}}$. Similarly for each $u^{\prime}\in(U_{0}\cup U)\cap N_{v}$, we find the tuple $(1,n_{v},0,n_{u^{\prime}})$ in the table so that $a\cdot\Delta g_{u^{\prime}}(n_{u^{\prime}}-1)$ is maximized and let this value be $z_{u^{\prime}}$. Also $\forall u^{\prime}\in(U_{1}\cup U_{0}\cup U)\setminus N_{v}$, $y_{u^{\prime}}=z_{u^{\prime}}=0$. If $u\in N_{v}$ then $y_{u}=a\cdot\Delta g_{u}(x_{u}+n_{u}-1)$ otherwise $y_{u}=0$. Now we include the tuple $(x_{u},n_{u},x_{v}=1,n_{v})$ in the table if the optimal value of the following ILP is at least $c_{v}-\Delta g_{v}(n_{v})-y_{u}-\sum_{u^{\prime}\in U_{1}}y_{u^{\prime}}-\sum_{u^{\prime}\in U_{0}}z_{u^{\prime}}$.

The above ILP can be solved in polynomial time as follows. First sort the values $a_{u^{\prime}}=|y_{u^{\prime}}-z_{u^{\prime}}|$ in non-increasing order breaking ties arbitrarily and order the vertices in $U$ as $\{u_{1},\ldots,u_{|U|}\}$ as per this order, that is, $a_{u_{i}}\geqslant a_{u_{j}}$ if $i\leqslant j$. Then we traverse the list of values in non-increasing order and for the $u^{\prime}$ corresponding to the value, we choose $x_{1}^{u^{\prime}}=1$ if $y_{u^{\prime}}>z_{u^{\prime}}$ otherwise we choose $x_{0}^{u^{\prime}}=1$. We do this until $\sum_{u^{\prime}\in U}x_{1}^{u^{\prime}}=n_{v}^{\prime}$ or $\sum_{u^{\prime}\in U}x_{0}^{u^{\prime}}=|U|-n_{v}^{\prime}$ where $n_{v}^{\prime}=n_{v}-x_{u}-|U_{1}|$. Remaining values are chosen in a way such that $\sum_{u^{\prime}\in U}x_{1}^{u^{\prime}}=n_{v}$ is satisfied. For a more detailed description, please see the Algorithm 1.

Now we show the correctness. Consider an optimal solution $x^{*}$. Let $i$ be the smallest number such that $x_{1}^{u_{i}}=1$ as per our algorithm and in optimal solution it is $0$. Similarly let $i^{\prime}$ be the smallest number such that $x_{1}^{u_{i^{\prime}}}=0$ as per our algorithm and in optimal solution it is $1$. We now swap the values of the variables $x_{1}^{u_{i}}$ and $x_{1}^{u_{i^{\prime}}}$ (resp. $x_{0}^{u_{i}}$ and $x_{0}^{u_{i^{\prime}}}$) in the optimal solution without decreasing the value of the objective function. Let us assume that $i<i^{\prime}$. Then it must be the case that $y_{u_{i}}>z_{u_{i}}$ otherwise $\forall j>i$, we will have $x_{1}^{u_{j}}=1$ as per our algorithm. Hence by swapping in the optimal solution, the value of the objective function increases by at least $a_{u_{i}}-a_{u_{i^{\prime}}}$ which is a non-negative quantity. Similarly when $i^{\prime}<i$, it must be the case that $y_{u_{i}}\leqslant z_{u_{i}}$ otherwise $\forall j>i$, we will have $x_{1}^{u_{j}}=0$ as per our algorithm. Hence by swapping in the optimal solution, the value of the objective function increases by at least $a_{u_{i^{\prime}}}-a_{u_{i}}$ which is a non-negative quantity. By repeatedtly finding such indices $i,i^{\prime}$ and then swaping the value of $x_{1}^{u_{i}}$ and $x_{1}^{u_{i^{\prime}}}$ (resp. $x_{0}^{u_{i}}$ and $x_{0}^{u_{i^{\prime}}}$) in the optimal solution leads to our solution.

An analogous procedure exists for the case where $x_{v}=0$. For each $u^{\prime}\in(U_{1}\cup U)\cap N_{v}$, we find the tuple $(0,n_{v},1,n_{u^{\prime}})$ in the table so that $a\cdot\Delta g_{u^{\prime}}(n_{u^{\prime}}+1)$ is minimized and let this value be $y_{u^{\prime}}$. Similarly for each $u^{\prime}\in(U_{0}\cup U)\cap N_{v}$, we find the tuple $(1,n_{v},0,n_{u^{\prime}})$ in the table so that $a\cdot\Delta g_{u^{\prime}}(n_{u^{\prime}})$ is minimized and let this value be $z_{u^{\prime}}$. Also $\forall u^{\prime}\in(U_{1}\cup U_{0}\cup U)\setminus N_{v}$, $y_{u^{\prime}}=z_{u^{\prime}}=0$. If $u\in N_{v}$ then $y_{u}=a\cdot\Delta g_{u}(x_{u}+n_{u})$ otherwise $y_{u}=0$. Now we include the tuple $(x_{u},n_{u},x_{v}=0,n_{v})$ in the table if the optimal value of the following ILP is at most $c_{v}-\Delta g_{v}(n_{v})-y_{u}-\sum_{u^{\prime}\in U_{1}}y_{u^{\prime}}-\sum_{u^{\prime}\in U_{0}}z_{u^{\prime}}$.

The above ILP can be solved in polynomial time as follows. First sort the values $a_{u^{\prime}}=|y_{u^{\prime}}-z_{u^{\prime}}|$ in non-increasing order breaking ties arbitrarily and order the vertices in $U$ as $\{u_{1},\ldots,u_{|U|}\}$ as per this order, that is, $a_{u_{i}}\geqslant a_{u_{j}}$ if $i\leqslant j$. Then we traverse the list in non-increasing order and for the $u^{\prime}$ corresponding to the value, we choose $x_{0}^{u^{\prime}}=1$ if $y_{u^{\prime}}>z_{u^{\prime}}$ otherwise we choose $x_{1}^{u^{\prime}}=1$. We do this until $\sum_{u^{\prime}\in U}x_{1}^{u^{\prime}}=n_{v}^{\prime}$ or $\sum_{u^{\prime}\in U}x_{0}^{u^{\prime}}=|U|-n_{v}^{\prime}$ where $n_{v}^{\prime}=n_{v}-x_{u}-|U_{1}|$. Remaining values are chosen in a way such that $\sum_{u^{\prime}\in U}x_{1}^{u^{\prime}}=n_{v}$ is satisfied.

Now we show the correctness. Consider an optimal solution $x^{*}$. Let $i$ be the smallest number such that $x_{1}^{u_{i}}=1$ as per our algorithm and in optimal solution it is $0$. Similarly let $i^{\prime}$ be the smallest number such that $x_{1}^{u_{i^{\prime}}}=0$ as per our algorithm and in optimal solution it is $1$. We now swap the values of the variables $x_{1}^{u_{i}}$ and $x_{1}^{u_{i^{\prime}}}$ (resp. $x_{0}^{u_{i}}$ and $x_{0}^{u_{i^{\prime}}}$) in the optimal solution without increasing the value of the objective function. Let us assume that $i<i^{\prime}$. Then it must be the case that $y_{u_{i}}\leqslant z_{u_{i}}$ otherwise $\forall j>i$, we will have $x_{1}^{u_{j}}=1$ as per our algorithm. Hence by swapping in the optimal solution, the value of the objective function decreases by at least $a_{u_{i}}-a_{u_{i^{\prime}}}$ which is a non-negative quantity. Similarly when $i^{\prime}<i$, it must be the case that $y_{u_{i}}>z_{u_{i}}$ otherwise $\forall j>i$, we will have $x_{1}^{u_{j}}=0$ as per our algorithm. Hence by swapping in the optimal solution, the value of the objective function increases by at least $a_{u_{i^{\prime}}}-a_{u_{i}}$ which is a non-negative quantity. By repeatedtly finding such indices $i,i^{\prime}$ and then swapping the value of $x_{1}^{u_{i}}$ and $x_{1}^{u_{i^{\prime}}}$ (resp. $x_{0}^{u_{i}}$ and $x_{0}^{u_{i^{\prime}}}$) in the optimal solution leads to our solution.

As mentioned earlier if there is any tuple in the table of the root $r$, then we conclude that there is a PSNE otherwise we conclude that there is no such PSNE.

In the proof of Theorem 1, just discard those entries from the table of $u\in\SS$ which don’t have $x_{u}=x_{u}^{\prime}$ and $n_{u}=n_{u}^{\prime}$.

Let $(\mathcal{G}(\mathcal{V},\mathcal{E}),p)$ be an input instance of directed public goods game. Now we create an instance $(\mathcal{G}^{\prime}=(\mathcal{V}^{\prime},\mathcal{E}^{\prime}),\mathcal{H}=(\mathcal{V}^{\prime},\mathcal{E}^{\prime\prime}),(g_{v})_{v\in\mathcal{V}^{\prime}},(c_{v})_{v\in\mathcal{V}^{\prime}},a)$ of BNPG game with symmetric altruism.

Let the constant $a$ be $1$. $\forall u\in\mathcal{V}$, $c_{u_{in}}=1+2\varepsilon$ and $c_{u_{out}}=p$. Now define the functions $g_{w}(.)$ as follows:

Now we show that given any $\varepsilon$-Nash equilibrium of the BNPG game with altruism , we can find an $\varepsilon$-Nash equilibrium of the directed public goods game in polynomial time. Let $(\Delta_{u})_{u\in\mathcal{V}^{\prime}}$ be an $\varepsilon$-Nash equilibrium of the BNPG game with symmetric altruism.

For all $v\in\{u_{in}:u\in\mathcal{V}\}$, we have the following:

Hence $1$ can’t be in the support of $\Delta_{v}$. Therefore $\forall u\in\mathcal{V}$, $\Delta_{u_{in}}(0)=1$.

Now we show that $(\Delta_{u})_{u\in\mathcal{V}}$ is an $\varepsilon$-Nash equilibrium of the directed public goods game where $\Delta_{u}=\Delta_{u_{out}}$ $\forall u\in\mathcal{V}$. Now consider a strategy profile $(x_{v})_{v\in\mathcal{V}^{\prime}}$ such that $\forall u\in\mathcal{V}$, we have $x_{u_{in}}=0$. Now let $(x_{v})_{v\in\mathcal{V}}$ be a strategy profile such that $\forall u\in\mathcal{V}$ we have $x_{u}=x_{u_{out}}$. Then we have the following:

Using the above equality and the fact that $\forall u\in\mathcal{V}$, $\Delta_{u_{in}}(0)=1$, we have $\mathbb{E}_{x_{-u}\sim\Delta_{-u}}[U_{u}^{\prime}(x_{u},x_{-u})]=\mathbb{E}_{x_{-u_{out}}\sim\Delta_{-u_{out}}}[U_{u_{out}}(x_{u_{out}},x_{-u_{out}})]$ where $x_{u}=x_{u_{out}}$. For all $u\in\mathcal{V}$, for all $x^{\prime}_{u}\in\{0,1\}$, for all $x_{u}\in\text{Supp}(\Delta_{u})$, we have the following:

Hence, given any $\varepsilon$-Nash equilibrium of the BNPG game with symmetric altruism , we can find an $\varepsilon$-Nash equilibrium of the directed public goods game in polynomial time. This concludes the proof of this theorem.

[24] showed that solving an instance of asymmetric altruistic design is equivalent to solving $n$ different instances of Minimum Knapsack problem and each of these instances have at most $\Delta+1$ items. In Minimum Knapsack problem, we are give a set of items $1,\ldots,k$ with costs $p_{1},\ldots p_{k}$ and weights $w_{1},\ldots w_{k}$. The aim is to find a subset $S$ of items minimizing $\sum_{i\in S}w_{i}$ subject to the constraint that $\sum_{i\in S}p_{i}\geqslant P$. We assume that $\sum_{i\in[k]}p_{i}\geqslant P$ otherwise we don’t have any feasible solution. Let us denote the optimal value by $OPT$. Let $W:=\sum_{i\in[k]}w_{i}$. Now consider the following integer linear program which we denote by ILP_w:

The above integer linear program can be solved in time $2^{k/2}\cdot k^{O(1)}$ [10]. Now observe that for all $w\geqslant OPT$, the optimal value $OPT_{w}$ of ILP_w is at least $P$. Similarly for all $w<OPT$, the optimal value $OPT_{w}$ of the ILP_w is less than $P$. Now by performing a binary search for $w$ on the range $[0,W]$ and then solving the above ILP repeatedly, we can compute $OPT$ in time $2^{k/2}\cdot\log W\cdot k^{O(1)}$. See Algorithm 2 for more details.

As discussed earlier, an Instance of asymmetric altruistic design is equivalent to solving $n$ different instances of Minimum Knapsack problem and each of these instances have at most $\Delta+1$ items. Hence ANM with asymmetric altruism can be solved in time $2^{\Delta/2}\cdot|x|^{O(1)}$ where $x$ is the input instance of ANM with asymmetric altruism.

Let $(\mathcal{G}=(\mathcal{V},\mathcal{E}),\mathcal{H}=(\mathcal{V},\mathcal{E}^{\prime}),(g_{v})_{v\in\mathcal{V}},(c_{v})_{v\in\mathcal{V}},a)$ be an instance of BNPG game with altruism. Let the circuit rank of $\mathcal{G}$ be $d$. W.l.o.g. let the graph $\mathcal{G}$ have $1$ connected component. First, let us compute the minimum spanning tree $\mathcal{T}=(\mathcal{V},\mathcal{E}_{1})$ of $\mathcal{G}$. Let $\mathcal{E}^{\prime\prime}:=\mathcal{E}\setminus\mathcal{E}_{1}$. Note that $|\mathcal{E}^{\prime\prime}|=d$. Let $\mathcal{V}^{\prime}=\{v_{1},v_{2},\ldots,v_{\ell}\}\subseteq\mathcal{V}$ be the set of endpoints of the edges in $\mathcal{E}^{\prime\prime}$. Note that $|\mathcal{V}^{\prime}|=\ell\leqslant 2d$. Let $\mathcal{E}_{2}=\{(u,v):\{u,v\}\in\mathcal{E}_{1}\text{ and }(u,v)\in\mathcal{E}^{\prime}\}$. For all $v\in\mathcal{V}$, let $N_{v}^{\prime}$ denote the set of out-neighbours of $v$ in $\mathcal{H}^{\prime}=(\mathcal{V},\mathcal{E}_{2})$. For every pair of tuples $t=(x_{v}^{\prime})_{v\in\mathcal{V}^{\prime}}\in\{0,1\}^{\ell}$ and $s=(n_{v}^{\prime})_{v\in\mathcal{V}^{\prime}}\in\{0,1,\ldots,n\}^{\ell}$, we do the following.

1. (i)

   $\forall v\in\mathcal{V}^{\prime}$, let $n_{v}^{t}$ be the number of neighbours of $v$ in $\mathcal{G}^{\prime}=(\mathcal{V},\mathcal{E}^{\prime\prime})$ who play $1$ in the tuple $t$. Now below we define $g^{t}_{v}$ and $c^{s}_{v}$ for all $v\in\mathcal{V}$.

   $$g^{t}_{v}(x)=\begin{cases}g_{v}(x+n_{v}^{t})&\text{if }v\in\mathcal{V}^{\prime}\\ g_{v}(x)&\text{otherwise}\end{cases}$$

   $$c^{s}_{v}=\begin{cases}c_{v}-a\sum_{u\in N_{v}\setminus N_{v}^{\prime}}\Delta g_{u}(x_{u}^{\prime}+n_{u}^{\prime}-1)\\ \text{if }v\in\mathcal{V}^{\prime},x_{v}^{\prime}=1\\ c_{v}-a\sum_{u\in N_{v}\setminus N_{v}^{\prime}}\Delta g_{u}(x_{u}^{\prime}+n_{u}^{\prime})\\ \text{if }v\in\mathcal{V}^{\prime},x_{v}^{\prime}=0\\ c_{v}\text{ otherwise}\end{cases}$$

2. (ii)

   Next we decide if there exists a PSNE $(x_{v}^{\prime\prime})_{v\in\mathcal{V}}\in\{0,1\}^{\mathcal{V}}$ for the instance $(\mathcal{T},\mathcal{H}^{\prime},(g^{t}_{v})_{v\in\mathcal{V}},(c_{v}^{s})_{v\in\mathcal{V}},a)$ of BNPG game with altruism such that

   1. (a)

      $x_{v}^{\prime\prime}=x^{\prime}_{v}$ for every $v\in\mathcal{V}^{\prime}$

   2. (b)

      $\forall v\in\mathcal{V}^{\prime}$, the number of neighbours of $v$ in $\mathcal{G}$ whose strategies are $1$ in $(x_{v}^{\prime\prime})_{v\in\mathcal{V}}$ is $n_{v}^{\prime}$.

   This can be decided by a polynomial time algorithm due to Corollary 1. We return yes if such a PSNE exists.