## 4 Problem Formulation and Algorithm Proposed
### Algorithm Proposed  
A Greedy Heuristic Algorithm for MNCTO  
In order to expound ideas of the algorithm, we analyze and construct a greedy heuristic algorithm to solve the problem of MNCTO step by step in this section. Given a 2D underwater Euclidean plane randomly deployed of sensors with \(\pi/3\) beamwidth directional transducer, an intuitive thought is that there must be more than 6 nodes that the whole plane can cover, although the locations and orientations of these nodes should meet some certain conditions. Fortunately, the conditions are not very strict, i.e., even though they need to occupy all the sectors of a disk, the satisfactory orientations are independent of their accurate positions, which just depend on the sector they are located in as Figure 8 shows. Each node in different sectors can choose its orientation from the directions displayed in Figure 8 to cover the disk center; in this way, its associating node laying in the opposite sector is also connected, and vice versa. The three pairs of nodes are sure to construct three symmetric connections and cover the entire plane. Obviously, this not enough to build a fully connected network for the reason that we cannot guarantee that the three pair of nodes are interconnected. In fact, this is impossible, according to the lemma bellows:  
**Lemma 1**. : _The 6 nodes located in different sectors of a disk cannot interconnect, even if their transmission ranges are unlimited, when they cover the entire plane._  
Proof. : It is known from the above analyses that the 6 nodes can form 3 pairs of symmetric connections, and each of them is composed of 2 nodes laying oppositely. Without losing generality, one of the symmetric links \(a\leftrightarrow b\) and other link nodes, \(e\) and \(f\), are taken for illustration. As can be seen from Figure 8, node \(b\) covers node \(f\) but not vice versa, because one of their coverage boundaries is parallel. It is the same with node \(e\), and thus, a fact can be deduced that nodes in different symmetric connections cannot reach each other simultaneously. In this way, the lemma is proved. Figure 8: Three pairs of symmetrically connected nodes covering the entire plane. Therefore, some more work should be done to construct a core group that guarantees connectivity. Considering the ideas that [PERSON] proposed in [22], a method to construct the structure is proposed, but first of all, another lemma is introduced as an auxiliary. Lemma 2: _Any three nodes with directional transducer of \(\pi/3\) beamwidth can form an interconnection if their orientations are unrestricted._  
Proof: Assuming that the three nodes \(x,y,z\) compose a triangle whose interior angles are \(\angle yzx<\angle xyz<\angle zxy\), as Figure 9 shows, despite the special case that it is equilateral (in this case, the three nodes are symmetrically connected, simultaneously), the three angles have the following relationship. \[\angle xyz+\angle yzx+\angle zxy=180^{\circ}\] \[\implies 3\angle yzx<180^{\circ}<3\angle zxy \tag{23}\] \[\implies \angle yzx<60^{\circ}<\angle zxy\]  
So, there must be an interior angle no more than \(60^{\circ}\), assuming it is \(\angle yzx\), which means that the node \(z\) can cover another two nodes with its \(\pi/3\) beam, so the symmetric connections between them are easily built only if \(x,y\) orientate themselves to \(z\). Combining Lemma 2 and Figure 8, it is easy to find a target group composed of at least 9 nodes that always interconnect and radiate omni-directionally, which is called a _target group_. The method is illustrated in Lemma 3 as follows.