# Information Fusion in Autonomous Vehicle Using Artificial Neural Group Key Synchronization
## 3 Proposed Methodology
### Synchronization Using B-Tree Framework  
The terminal neurons of the B-Tree are represented by the main \(USER_{j}\) (\(j=1\ to\ m\)) in the group in Figure 7. The neighboring principles (\(USER_{j}\ USER_{j+1}\)) form a couple and begin with a shared starting input. There will be \(m/2\) pairings if the number of nodes is even. In Figure 7, we examine eight users who pair up as \( \begin{array}{l}(USER\ 1,\ USER\ 2)(USER\ 3,\ USER\ 4)\\ (USER\ 5,\ USER\ 6)(USER\ 7,\ USER\ 8)\end{array} \). Each pair forms a TLTPM to create the synchronized weights \( \begin{array}{l}(USER\ 12)(USER\ 34)\\ (USER\ 56)(USER\ 78)\end{array} \). These synchronized weights are now the pair's initial weights (\(USER\ 12\)). Similarly, (\(USER\ 34)(USER\ 56)(USER\ 78) have initial synchronized weights. In the following round, (\(USER\ 12,\ USER\ 34\)) and (\(USER\ 56,\ USER\ 78\)) create consecutive pairs for the TLTPM and produce keys (\(USER\ 1234,\ USER\ 5678\)). The above mentioned merging and  
Figure 6: Ring framework. synchronization procedure is repeated until all users have merged to create a single group. Every member are linked to an identical weights that forms the neural group key. For odd number of members, as illustrated in Figure 8, say \((2^{m}+1)\), the \(2^{m}\) users will merge to an identical neural key as previously explained, while the final member will then coordinate the TLTPM with the TLTPM of \(USER\) 12345678 to get a mutually agreed key. The amount of layers necessary for key exchange is determined by the amount of users in the cluster. That is, for \(2^{m-1}<\)_number of users \(\leq 2^{m}\)_ then \(m\) levels are required. For example, for \(m=3\), we may have 5, 6, 7, or 8 members, and the amount of levels necessary for key creation stays the same for all of these users. Figure 8: Tree framework considering m as odd. Figure 7: Tree framework considering m as even. ```
Input : The random vector weight \(D^{q}\) and the identical input \(G^{q}\). Output : With the identical key pair, both sender and recipient have synchronized TLTPM. Step 1: Assign the weight matrix \(D^{q}\) to any random vector value. Where, \(\delta^{q}_{v,w}\in\{-\lambda,-\lambda+1,\ \ldots,\ +\lambda\}\). Repeat steps 2 through 5 until you have achieved ideal synchronization. Step 2: Every hidden node's output is determined by a weight dependent on the current state of the inputs. The result of the first hidden layer is determined by Equation (2). \[\kappa^{q}_{v}=\frac{1}{\sqrt{N}}\gamma^{q}_{v}\delta^{q}_{v}=\frac{1}{\sqrt{N }}\sum_{w=1}^{N}\gamma^{q}_{v,w}\delta^{q}_{v,w}\] (2) The result \(\xi^{q}_{v}\) of the \(v\)-th concealed node is denoted by \(signum(\kappa_{v})\) in Equation (3). \[\xi^{q}_{v}=signum(\kappa_{v})\] (3) If \(\kappa_{v}=0\) is true, \(\xi^{q}_{v}\) is mapped to \(-1\), then binary outcome is produced. \(\xi^{q}_{v}\) is assigned to \(+1\) if \(\kappa_{v}>0\) is true, indicating that the concealed neuron is functioning. The concealed node is deactivated if the magnitude is \(\xi^{q}_{v}=-1\). Equation (4) shows how to do this. \[signum(\kappa_{v})=\left\{ \begin{array}{ll}-1&if\ \kappa_{v}\leq 0\\ +1&if\ \kappa_{v}>0\end{array} \right.\] (4) Step 3: Calculate the final result of TLTPM. The ultimate result of TLTPM is calculated by multiplying the hidden neurons in the last layer. This is represented by \(\zeta\) (Equation (5)). \[\zeta=\prod_{v=1}^{\tau_{3}^{3}}\prod_{v=1}^{\tau_{3}^{3}}\vdots\prod_{v=1}^{ \tau_{3}^{3}}\zeta^{v}_{v}\] (5) Equation (6) shows how the magnitude of \(\zeta\) is represented. \[\zeta=\left\{ \begin{array}{ll}-1&if\ \xi_{v}=-1,\ is\ odd\\ +1&if\ \xi_{v}=-1,\ is\ even\end{array} \right.\] (6) If TLTPM has one concealed neuron, \(\zeta=\xi^{q}_{1}\). For \(2^{u-1}\) distinct (\(\xi^{q}_{1},\ \ \xi^{q}_{2},\ \ldots,\ \xi^{q}_{u}\)) options, the \(\zeta\)value is the same.