# FollowMe: One Social Importance-Based Collaborative Scheme in MONs
## 4 FollowMe
### Game Strategy Update  
**Income of a node in a round:** Based on the prisoner's dilemma game and the aforementioned game tree, nodes have the following four possibilities to earn income. Here, we use two nodes \(x\) and \(y\) to illustrate these possibilities and suppose \(x\) is a signal sender and \(y\) is a receiver. 1. Both of them are willing to forward packets, thus each of them has an income \(R\). Let \(p_{x}^{1}\) and \(p_{y}^{1}\) denote the forwarding probability, we have: \[p_{x}^{1}p_{y}^{1}=\theta^{2}*\left(2-\theta\right).\]  
Proof. : From the game tree, we know that there exist three paths (step 1, 2, and 3) in which both nodes take the forwarding strategy. In step 1, node \(x\) belongs to the cooperative type with a probability \(\theta\). In addition, both of them are social nodes with a signal \(m_{1}\), hence the receiver is also cooperative and forwards the packet with a probability \(\theta\). The probability of the path thus is \(\theta\times\theta\times\theta=\theta^{3}\). In step 2, the node \(y\) is non-cooperative with a probability \(1-\theta\); this is mainly because node \(y\) itself is a non-social node. The probability of this path is \(\theta\times\left(1-\theta\right)\times\theta=\theta^{2}\left(1-\theta\right)\). In step 3, the node \(y\) remains non-cooperative, but this time the reason is that node \(x\) is a social node. The probability of this path is \(\theta\times\left(1-\theta\right)\times\theta=\theta^{2}\left(1-\theta\right)\). The combination of these steps concludes the proof. 2. Node \(x\) forwards packets with an income \(S\), but \(y\) is declined with an income \(T\). This situation happens along path 4. Similarly, we obtain the probability: \[p_{x}^{1}p_{y}^{0}=\theta*\left(1-\theta\right)^{2},\] where \(p_{y}^{0}\) denotes the probability of rejection. 3. Node \(x\) is declined with an income \(T\), but \(y\) forwards packets with an income \(S\). This situation happens along paths 5, 6, and 7 since the sender belongs to \(t_{2}\). The probability is \[p_{x}^{0}p_{y}^{1}=\theta*\left(1-\theta\right)*\left(2-\theta\right),\] where \(p_{x}^{0}\) denotes the probability of rejection of node \(x\). The proof process is similar to that of (1), and we omit it here for a concise presentation. 4. Both of the nodes refuse to forward the packet and each of them has an income \(P\). The probability is \[p_{x}^{0}p_{y}^{0}=\left(1-\theta\right)^{3}.\]  
**Updates of the game strategy**: After a round of game with all neighboring nodes, the cumulative gain of node \(x\), i.e., \(g(x)\) can be calculated as follows:  
\[\mathrm{g}\left(x\right)=c_{1}\times R\times p_{x}^{1}p_{y}^{1}+c_{2}\times S \times p_{x}^{1}p_{y}^{0}+c_{3}\times T\times p_{x}^{0}p_{y}^{1}+c_{4}\times P \times p_{x}^{0}p_{y}^{0}, \tag{2}\]  
Figure 1: The signal game paths between two nodes. where \(c_{1}\), \(c_{2}\), \(c_{3}\), and \(c_{4}\) are the number of times that \(x\) gains \(R\), \(T\), \(S\), and \(P\) in a round, and \(c_{1}\ +\ c_{2}\ +\ c_{3}\ +\ c_{4}=K_{x}\). Simultaneously, node \(y\) can obtain its gains by using a similar method. After calculating the accumulated income of each node, the Fermi function is deployed to determine whether node \(x\) follows the opponent's game strategy:  
\[W\left(a_{x}\gets a_{y}\right)=\frac{1}{1+\exp\big{[}\big{(}g_{x}-g_{y} \big{)}/n_{0}\big{]}}, \tag{3}\]  
where \(a_{x}\) and \(a_{y}\) represent the strategy adopted by nodes \(x\) and node \(y\) for the current round, respectively; \(g_{x}\) and \(g_{y}\) are the corresponding cumulative earnings in the current round, and \(n_{0}\) is the noise factor. If \(W\geq\alpha\) (\(\alpha\) is a random number homogeneously distributed within the interval [0, 1]), node \(x\) follows the strategy of node \(y\); otherwise, the current strategy is retained. Algorithm 1 expresses the aforementioned process.