## 4 Problem Formulation and Algorithm Proposed
### Algorithm Proposed
This work was supported by the National Natural Science Foundation of China (Shandong United Foundation) (Grant No. U1806201), and the National Natural Science Foundation of China (Grant No. 11774074). The authors declare no conflict of interest. ## Appendix A

An equilateral triangular widget can connect adjacent edges in a hexagon grid graph. From Figure 7a, it can be seen that there are totally 6 variants of equilateral triangular widget, whose orientations are one of the sectors set as \(\mathcal{S}=[\)[2,6], [3,1], [4,2], [5,3], [6,4], [5,1]\(]\) in a counter-clockwise direction. Obviously, their orientations cover the whole Euclidean plane. In addition, the degree of a vertex is intuitive, at most three in a hexagon grid graph, which is also the most sophisticated condition of how a widget can be linked. As illustrated in Figure A1, the construction also includes the situation of a node degree of two. Without loss of generality, the structure depicted above is used to represent connections in a hexagon grid graph, where the black solid lines indicate an equilateral triangular widget that can only connect to at most two other widgets, because it has only two _Contacting Nodes_. The side's length is \(r\), while the dashed line indicates that the vertex is incapable of connecting to other widgets at present but this is not always so in other variants. The conclusions in [18] show that there is no need to make a widget connected to more than two adjacent widgets to construct a Hamiltonian path, but a capacity to orientate to six directions is necessary, which is satisfied by the structure above. Thus, we make the proof. A HPHGG for TOP of nodes set \(V\) with side length \(r\),its total inner cost is \([(5n-6)(5K\log r+3 Ar-2K)-2(Ar+K\log 2)]/5\), where \(n=\|V\|\). Proof. : From Lemma A1, it is known that a widget can only connect to at most two other widgets, and from Figure 6, we notice an _Emitting Node_ can only connect to one of the _Contacting Nodes_ directly, so the distance a signal travels is \(r\). When another connection (the connection between the _Emitting Node_ and another _Contacting Node_) is required, a signal need to relay three times to reach, so the total distance is \(4*\frac{r}{2}\); they cannot be multiplied directly, as they are not of a linear relationship. Another observation lies in the fact that the Hamiltonian path is a one-way directed path, so a widget should receive from its previous transmission and transmit to its next hop. For the reason explained above, the link distance between two _Contacting Node_ is the sum of their distances to the _Emitting Node_, so the distance of the inner widget is \(r+4*\frac{r}{2}\). Then, the according cost is obtained as follows:

\[ \begin{split} c_{in}&=cost(r)+4*cost(\frac{r}{2})\\ &=5K\log r+3 Ar-2K\log 2\end{split}  \tag{1}\]

Considering the particularity of head and tail nodes of the path, the total inner cost of HPHGG for TOP is as follows:

\[ \begin{split} C_{t}&=(n-2)*c_{in}+4*cost(\frac{r}{2 })\\ &=(n-2)(5K\log r+3 Ar-2K\log 2)+4K\log r+2(Ar-K\log 2)\\ &=(n-\frac{6}{5})(5K\log r+3 Ar-2K)-\frac{2}{5}(Ar+K\log 2)\\ &=\frac{(5n-6)(5K\log r+3 Ar-2K)-2(Ar+K\log 2)}{5}\end{split}  \tag{2}\]

Thus, the lemma is proved. ## References

* (1) [PERSON]; [PERSON]; [PERSON] Analysis of a High-Performance MAC Protocol for Underwater Acoustic Networks. _IEEE J. Ocean. Eng._**2014**, _39_, 74-89. [CrossRef]
* (2) [PERSON]; [PERSON]; [PERSON]; [PERSON] A Voronoi-Based Optimized Depth Adjustment Deployment Scheme for Underwater Acoustic Sensor Networks. _IEEE Sens. J._**2020**, _20_, 13849-13860. [CrossRef]
* (3) [PERSON]; [PERSON]; [PERSON] Robust adaptive acoustic vector sensor beamforming using automated diagonal loading. _Appl. Acoust._**2009**, _70_, 1029-1033. [CrossRef]
* (4) [PERSON]; [PERSON]; [PERSON]; [PERSON]; [PERSON] Sparse Bayesian Learning Based Direction-of-Arrival Estimation under Spatially Colored Noise Using Acoustic Hydrophone Arrays. _J. Mar. Sci. Eng._**2021**, \(9\), 127.