## 4 Proposed Scheme
### Identity, Route, and Location Privacy (IRL)
We will use the term \"adversary is in possession of a node\" to indicate that the adversary can passively listen to any communication within the radio range of that node. **Claim 1:** Suppose \(\mathcal{A}\) is in possession of \(B\). Let \(B_{b}\) be the number of backward nodes of the base station (nodes one hop away from the base station). Then for any packet \(q\) received by \(B\) and for large enough \(N\):

\[\Pr\left[\mathcal{A}\left(q,N\right)=s\right]\approx\frac{B_{b}+1}{N} \tag{7}\]

_Proof_: The adversary can always know the ID of the last forwarding node. Let \(B_{b}\) be the number of backward nodes to the base station. The packet could only have come from one of the nodes in \(N_{B}-\{B\}\) (which only contains backward nodes to \(B\)). Since the nodes are just a hop away from the BS, so they will not send the packet to another node. Hence for large \(N\) we have:

\[\Pr\left[\mathcal{A}\left(q,N\right)=s\right]=\Pr\left[\mathcal{A }\left(q,N\right)=s|s\in N_{B}-\{B\}\right]\times\Pr\left[s\in N_{B}-\{B\} \right]+\] \[\Pr\left[\mathcal{A}\left(q,N\right)=s|s\
otin N_{B}-\{B\}\right] \Pr\left[s\
otin N_{B}-\{B\}\right]\] \[\delta t=2\left(\frac{D}{d}\times p_{t}\right) \tag{6}\]0 \[\delta t=2\left(\frac{D}{d}\times p_{t}\right) \tag{6}\]1

Now let us assume that \(\mathcal{A}\) is in possession of a node \(m\) in the network. The following probability estimate gives an upper bound of the probability of success of the adversary. It is an upper bound since it does not include the possibility of a packet sent backwards. When a packet is sent backwards over one or many hops, the probability of success of the adversary decreases since there would be more possible nodes. Thus in this scenario our result would be like an upper bound on the adversary's limitations. **Claim 2:** Suppose \(\mathcal{A}\) is in possession of a node \(m\). Let \(c=|C_{\to^{2}m}|\) denote the number of backward nodes in backward set \(C_{\to^{2}m}\) of some node \(\to^{2}m\). Then,

\[\delta t=2\left(\frac{D}{d}\times p_{t}\right) \tag{6}\]2

_Proof_: Since the adversary is in possession of a node \(m\), it knows its backward and forward nodes. Furthermore, if any of these nodes including the node \(m\) itself is the sender of a packet \(q\), then the adversary will know. This is true since the adversary can see all incoming packets to the node \(m\) and to its neighbor nodes (the forward and the backward nodes). Thus it can see if the payload of \(q\) is not equal to the payload of any \(q^{\prime}\) being received by these nodes in a given interval of time. If this is the case, then the adversary will know the sender. Now if none of the nodes in \(N_{m}\) are the senders, then the packet was forwarded by a node \(i\) that is two hops away from \(m\). The adversary knows the ID of that node through the packet \(q\). Thus the adversary makes a list of all the possible backward nodes in the backward set of \(i\). Let that number be denoted by \(c\). Notice that node \(i\) could also be the possible sender. Hence the total number of possible senders would be \(c+1\). We have:

\[\delta t=2\left(\frac{D}{d}\times p_{t}\right) \tag{6}\]3 \[\delta t=2\left(\frac{D}{d}\times p_{t}\right) \tag{6}\]4 \[\delta t=2\left(\frac{D}{d}\times p_{t}\right) \tag{6}\]5

Now, suppose the adversary is in possession of two nodes at the same time; \(m_{1}\) and \(m_{2}\). We can safely assume that \(N_{m_{1}}\cap N_{m_{2}}=\varphi\), since it would be more advantageous to the adversary to cover nodes with non-overlapping radio ranges. The adversary will always know whenever any node in \(N_{m_{1}}\) or \(N_{m_{2}}\) is the sender of a packet. How about the case when they are not the senders? There could be two possible cases: without loss of generality, first assume that \(m_{2}\in C_{m_{1}}\). If the packet \(q\) was received by some node in \(N_{m_{1}}\) and was received by some node in \(N_{m_{2}}\) before, then the adversary had already checked it when the packet was sent to a node in \(N_{m_{1}}\).