## 3 System Model
### Trajectories of AUV Mobility
#### 3.2.1 Formation of the HC: An Example of AUV Trajectory Formation  
Let random weights be assigned in terms of the Euclidean distance for each edge \(\mathbf{e}\in E\) of \(G\). Let the Euclidean distance between nodes be represented by \(d\); then, D\({}_{ij}\) = {d(8,9) = 4, d(7,9) = 7, d(4,7) = 5, d(9,13) = 3, d(13,14) = 8, d(8,14) = 6}, as shown in Figure 2. Initially FTL, RTL and TL have the following status: FTL = {\(l\left({4,7}\right),l\left({7,9}\right),l\left({9,13}\right),l\left({8,9} \right),l\left({8,14}\right)\)}, RTL = \(\emptyset\) and TL = \(\emptyset\). We start from the node with the highest ID in V\({}_{CDS}\). In our case, the highest ID node is 14, \(e\left({14,8}\right)\), which is added to TL = {\(e\left({14,8}\right)\)} and \(l\left({8,14}\right)\) is moved to the RTL. At Node 8, \(e\left({8,9}\right)\) is added to TL = {\(e\left({14,8}\right),e\left({8,9}\right)\)}, and similarly, \(l\left({9,8}\right)\) is moved to the RTL. At Node 9, we have two choices \(e\left({9,13}\right)\) and \(e\left({9,7}\right).\) The Euclidean length of \(e\left({9,13}\right)\) is shorter than that of \(e\left({9,7}\right)\), so \(e\left({9,13}\right)\) is included in TL = {\(e\left({14,8}\right),e\left({8,9}\right),e\left({9,13}\right)\)}, and \(e\left({13,9}\right)\) is added to the RTL. At Node 13, there is no more outgoing edge of CDS, so \(e\left({13,9}\right)\) is added to TL = {\(e\left({14,8}\right),e\left({8,9}\right),e\left({9,13}\right),e\left({13,9} \right)\)}. At Node 9, we have two options, one from the FTL and the other from the RTL; \(e\left({9,8}\right)\) is in the RTL, and \(e\left({9,7}\right)\) is in the FOL. \(e\left({9,7}\right)\) is chosen because it lies in the FTL. Therefore, TL = {\(e\left({14,8}\right),e\left({8,9}\right),e\left({9,13}\right),e\left({13,9} \right),e\left({9,7}\right)\)}. At Node 7, we have only one option, so \(e\left({7,4}\right)\) is added in the TL, and \(e\left({4,7}\right)\) is added to the RTL. Node 4 is a leaf node and, hence, traversed back, and \(e\left({4,7}\right)\) is added to the TL. At Node 7, we have one option, so \(e\left({7,9}\right)\) is added to the TL. At Node 9, we have two options: either go to 13 or 8. As \(e\left({9,13}\right)\) is already added in the TL, so HOL is scanned, and the next destination is Node 8. Similarly, the TL is constructed by continuing on this line, and we have,  
\[TL=\left\{{\begin{array}{*{20}c}{e\left({14,8}\right),e\left({8,9}\right),e \left({9,13}\right),e\left({13,9}\right),e\left({9,7}\right),e\left({7,4} \right),e\left({4,7}\right),}\\ {e\left({7,9}\right),e\left({9,8}\right),e\left({8,14}\right),}\end{array}} \right\} \tag{1}\]  
The length of the AUV path can be computed easily by adding all links of the CDS-MST as follows. The HC is shown in Figure 4.  
\[Lp=\sum\limits_{\forall\ e\left({i,j}\right)\in EMST}{length(i,j)}+\alpha \tag{2}\]  
where \(length(i,j)\in EMST\) is the distance between node \(i\) and \(j\). \(\alpha\) is an offset factor, if the trajectory deviates from the CDS nodes. \(\alpha\) is used to compensate the distance margin. \(length(i,j)\) is computed by using the Euclidean distance formula:  
\[d\left({i,j}\right)=\sqrt{\left({x_{i}-x_{j}}\right)^{2}+\left({y_{i}-y_{j}} \right)^{2}} \tag{3}\]  
where \(d\left({i,j}\right)\) is the Euclidean distance between \(i\) and \(j\).