# CAVVPM: Challenge-Based Authentication and Verification of Vehicle Platooning at Motorway
## 4 Security Analysis
### Remote Adversary
#### 4.2.2 A Vehicle Follow Verifier
\[X_{n}=X_{0}\cdot P_{0}^{n} \tag{10}\]  
By Pr[pass], we may indicate the likelihood that the opponent will arrive \(d\) in time step \(n\), or, put another way, that the task will be successfully completed. Then:  
\[ \begin{array}{c}\Pr[\text{pass}]=\sum\limits_{i=1}^{M}\Pr[d=i]\cdot\Pr[X_{n }=i\mid d=i]\\ =\frac{1}{M}\sum_{i=1}^{M}\cdot\Pr[X_{n}=i\mid d=i]\\ =\frac{1}{M}\sum_{i=1}^{M}\sum_{j=1}^{N}\Pr[X_{0}=j]\cdot\Pr[X_{n}=i\mid X_{0 }=j]\\ =\frac{1}{NM}\sum_{i=1}^{M}\sum_{j=1}^{N}P_{j,i}^{n}\end{array}  \tag{11}\]  
All M potential checkpoint values were taken into consideration in Equation (11). Checkpoints for each task are picked randomly in Equation (11). All of the first N locations for vehicle R were examined in Equation (11), respectively, in the random process. **Passing a PoF:** A PoF challenger must pass all K challenges in set \(\Gamma=\{(d_{1},t_{1}),(d_{2},t_{2}),\ldots(d_{K},t_{K})\}\) in order to succeed. To put it another way, R must walk to each checkpoint by the given time. With random and independent selection of each checkpoint \(d_{k}\), the passing rate may be represented as follows:  
\[ \begin{array}{c}P_{M}=\Pr[\text{pass}\,\cdot\text{chal}\cdot 1]\,\ldots\,\Pr[\,\text{Pass}\,\cdot\,\text{ chal}\,\cdot\,\text{K}]\\ =\left(\frac{1}{NM}\sum_{i=1}^{M}\sum_{j=1}^{N}P_{j,i}^{n_{1}}\right)\ldots \left(\frac{1}{NM}\sum_{i=1}^{M}\sum_{j=1}^{N}P_{j,i}^{n_{1}+n_{2}+\ldots+n_{k} }\right)\\ =\left(\frac{1}{NM}\right)^{K}\prod_{k=1}^{K}\sum_{i=1}^{M}\sum_{j=1}^{N}P_{j, i}^{\sum_{k=1}^{k}n_{k}}\end{array}  \tag{12}\]  
Equation (12) is derived from the difficulties of being independent from one another. Here, we added up the time it takes to complete each task, rather than using the chance of passing a single challenge from Equation (11). For a more condensed form of Equation (12), use Equation (12). **Lemma Proof. ** The upper bound on the passage probability of the opponent is  
\[P_{M}\leq\left(\frac{1}{M}\right)^{K} \tag{13}\]  
**Proof. ** For the purpose of demonstrating Lemma 1, we look at the chance of completing a single challenge, which is given by Equation (11). The summarizing phase should be noted as \(\sum_{i=1}^{M}P_{j,i}^{n}\). In the \(j\)-th row of matrix \(P^{n}\), the \(j\)th row sums \(M\) out of \(N\) elements. Because checkpoints are a subdomain of possible positions for vehicle \(R\),  
\[\sum_{i=1}^{M}P_{j,i}^{n}\leq\sum_{i=1}^{N}P_{j,i}^{n}=1\]  
Equivalently, substituting this limit into Equation (11) results in:  
\[\mathrm{pr}[\mathrm{pass}]=\frac{1}{NM}\sum_{i=1}^{M}\sum_{j=1}^{N}P_{j,i}^{n} \leq\frac{1}{NM}\sum_{j=1}^{N}1=\frac{1}{M}\]  
The proof is completed by substituting this bound into Equation (12). \(\square\)  
With respect to the number of checkpoints, we can see that the passing probability PM diminishes exponentially with increasing checkpoint complexity, as shown by Lemma 1. It is possible to manipulate these two factors in order to achieve any desired passing probability, but this comes at a cost in the time it takes for the PoF verification process to finish. When M << N, the limit of Lemma 1 is fairly lenient, as seen in Figure 1. In reality, it is possible to demonstrate that the random walk's steady-state distribution uniformizes with time (with the exception of the two boundaries that have different transition probabilities). The chance of passing a single challenge is \(\mathrm{pr}[\mathrm{pass}]=1/N\), which is a variable of M that is under uniform distribution in most of the N states. To put it another way, the likelihood of reaching the chosen checkpoint is one in N. PoF verification has a chance of passing of \(P_{M}\sim\left(1/N\right)^{K}\) when \(K\) separate challenges are taken into consideration.