Chargeable Sweep Coverage Problem

The network model of the CSC problem is showed below. Assuming that the charging stations and targets are distributed in the metric space, the network model can be represented by an undirected complete graph $G(\mathcal{T}\cup\mathcal{C},\mathcal{E})$, where $\mathcal{T}=\{t_{1},t_{2},..,t_{N}\}$ represents the set of targets, $\mathcal{C}=\{c_{1},c_{2},..,c_{K}\}$ represents $K$ charging stations, $d:\mathcal{E}\rightarrow R^{+}$ is the weight of the edge $e\in\mathcal{E}$, which conforms to triangular inequality constraint, that is, for any three points $i,j,k\in\mathcal{T}\cup\mathcal{C}$, all satisfy $d(i,j)=d(j,i)$ and $d(i,j)\leq d(i,k)+d(k,j)$. Given $M$ sensors $\mathcal{S}=\{s_{1},s_{2},..,s_{M}\}$ with rechargeable batteries, each of them can sweep at the velocity $V$ for $T_{c}$ time when the battery is fully charged. be charged. Each target in $\mathcal{T}$ needs to be covered by sensor every $T_{t}$ time, referred to as ”$T_{t}$-sweep covered” or ”covered” for simplication. Since the sensors’ moving distances are much larger than the sensing radius of the sensors, this study simplified the coverage model, thinking that the sensor cover the target only when it moves to the position of the target. Also, for simplication, the analysis in this paper assumes that the charging time $T_{r}$ can be ignored. It is also assumed that the relationship between $T_{c}$ and $T_{t}$ is an integer multiple, that is, when $T_{c}\geq T_{t}$, then $T_{c}/T_{t}=Q$, where $Q\geq 1$ is a positive integer; when $T_{c}<T_{t}$, then $T_{t}/T_{c}=\hat{Q}$, where $\hat{Q}\geq 2$ is also a positive integer.

Under this network model, the definition of the CSC problem is as follows:

Next, we study the CSC problem under several special situations. These CSC problems under some special situations can be transformed into some problems that have been studied, including the orienteering problem, the team orienteering problem, and the distance constrained vehicle routing (DCVR) problem. These problems are all NP-hard problems. We first give the definitions below.

When $s=t$, The orienteering problem is also called the rooted orienteering problem, The team orienteering problem is called the rooted team orienteering problem .

It can be seen from the above definition that when the number of charging stations $K=1$, the number of sensors $M=1$, and $T_{c}/T_{t}=1$, the CSC problem can be transformed into the rooted orienteering problem. When the number of charging stations $K=1$, the number of sensors $M\geq 1$, and $T_{t}/T_{c}=\hat{Q}$, the CSC problem can be transformed to the rooted team orienteering problem with the number of loops $M\times\hat{Q}$. When the number of charging stations $K=1$, to find the minimum number of sensors covering all targets, then the CSC problem can be transformed into the DCVR problem. The table I lists the best approximation algorithms for these problems.

Form the table I, it is known that the CSC problem is so a trivial problem. In this paper, we consider the CSC problem under two special constraints. One is that each sensor must return to its original charging station for charging. This situation occurs when the charging stations are far away, that is, the length of the shortest path between each charging station is $SP(c_{i},c_{j})=d(c_{i},c_{j})>L_{c}$, where $c_{i},c_{j}\in\mathcal{C}$, and $i\neq j$; or when the sensors and charging stations have matching models. We define the CSC problem in this case as a Restricted Chargeable Sweep Coverage (RCSC) problem. The other is that the sensors can go to any charging stations for charging, but the number of charging stations is two. We define this problem as a 2-chargeable sweep coverage (2-CSC) problem. In the RCSC problem, when the number of charging stations $K=1$, the number of sensors $M=1$, and the charging period and sweep period are the same, i.e, $L_{c}=L_{t}$, the RCSC problem can be transformed into a rooted orienteering problem. It is NP-hard. In the 2-CSC problem, if the distance between the charging stations is far, greater than $VT_{c}$, the number of sensors $M=1$, and the charging period and sweep period are the same, i.e., $L_{c}=L_{t}$, the 2-CSC problem can be transformed into the problem of choosing the larger one between two rooted orienteering problems. It is also NP-hard. Therefore, we get the theorem 5.

For the convenience of description, we call a loop that starts from any charging station $c\in\mathcal{C}$ with a length of $L$ and returns to $c$ without passing through other charging stations as $c$-loop $r_{c}(L)$. A loop starts from any charging station $c_{i}\in\mathcal{C}$ and pass at least one other charging station $c_{j}\in\mathcal{C}$ ($j\neq i$) is called an inter-loop. If a inter-loop passes through two charging stations $a,b\in\mathcal{C}$, the inter-loop with length $L$ is expressed as $r_{{\{a,b\}}}(L)$. The symbols used in this paper are listed in the table II.

When $T_{c}\geq T_{t}$, $T_{c}/T_{t}=Q$ is a positive integer. Considering that the sensors travel periodically, the travel path $r\in\mathcal{R}$ of each sensor must be a loop. When the number of sensors required for a path is the same, the longer the path is, the more targets it can cover. Therefore, the set of paths $\mathcal{R}$ can be reduced to include only paths whose length is an integer multiple of $L_{t}$. Therefore, $r\in\mathcal{R}$ is loop $r_{c_{i}}(kL_{t})$ which starts from any charging station $c_{i}\in\mathcal{C}$ and has a length of $k\times L_{t}$, where $k=1,2,...,min\{Q,M\}$. That is, $\mathcal{R}=\{r_{c_{i}}(kL_{t})\mid k\in[1,min\{Q,M\}],c_{i}\in\mathcal{C}\}$. And because the targets need to be $T_{t}$-sweep covered, the sensors need to be placed on loop $r_{c_{i}}(kL_{t})$ at a distance of $VT_{t}$ to ensure the coverage period of the targets, so a loop $r_{i}$ with a length of $kL_{t}$ requires $v_{i}=k$ sensors. The sweeping path of a sensor is shown in figure 1, the number of charging stations in the figure is 2, the number of sensors is 8, and $Q=3$.

Algorithm 1 is implemented using a greedy strategy, gradually selecting the path $r\in\mathcal{R}$ with the most unity gain, and calling the sensors to cover it. Since the path is always found in the induced subgraph of uncovered targets, unity gain is the average number of targets covered by the sensors on the path. The steps of the algorithm are as follows:

1. 1.

   Let $\hat{\mathcal{T}}\subseteq\mathcal{T}$ be the set of targets that have not yet been covered, and construct a induced subgraph of graph $G$, $G^{\prime}=(\hat{\mathcal{T}}\cup\mathcal{C},\mathcal{E}^{\prime})$, where $\mathcal{E}^{\prime}=\{(i,j)|i,j\in\hat{\mathcal{T}}\cup\mathcal{C},(i,j)\in\mathcal{E}\}$;

2. 2.

   Find the $c_{i}$ loop $r_{c_{i}}(kL_{t})$ whose length is $kL_{t}$ and sensors on the loop cover the largetest number o targets in the graph $G^{\prime}$, where $k=1,2,...,min\{M^{\prime},Q\}$, $M^{\prime}$ is the number of sensors that have not yet been allocated;

3. 3.

   In the $K\times min\{M^{\prime},Q\}$ loops $r_{c_{i}}(kL_{t})$, to select the path $r_{i}$ of maximum unity gain. If the cost is $m_{i}$, distribute $m_{i}$ sensors on $r_{i}$ separated by $L_{t}$ to $T_{t}$-sweep cover targets on $r_{i}$;

4. 4.

   Repeat steps 1, 2, 3 until $M^{\prime}=0$.

From the III-B section, we know that the problem of finding the path with a limited length and the largest number of vertices is also called the orienteering problem, which is NP-hard. Algorithm 1 calls the rooted-orienteering approximation algorithm as a subroutine, with the approximation ratio is $\alpha$ ($0<\alpha<1$). The input is the induced subgraph $G^{\prime}$ of the current uncovered targets and the the charging stations in the graph $G$, rooted node $c_{i}\in\mathcal{C}$ and the limited length $L$. The final output of Algorithm 1 is the set of covered target points $\mathcal{CT}$. We call the greedy algorithm that calls an approximation algorithm like Algorithm 1 as the approximate greedy algorithm.

Suppose that the approximation solution obtained by this approximate greedy algorithm is $APP$, and $OPT$ is the optimal solution, the number of covered targets is $\omega(APP)$, $\omega(OPT)$. $r_{i}^{*}$ is the path with the most unity gain selected in the $i^{th}$ step of the greedy algorithm, and its cost is $v_{i}$. $r_{i}$ is the path selected in step $i$ of the approximation algorithm, with the same cost, i.e., $\omega(r_{i})\geq\alpha\omega(r_{i}^{*})$. Here is a lemma 6:

The theorem 7 can be easily derivated from the lemma 6.

In each interation of Algorithm 1, it is needed to find the optimal path in the $K\times min\{Q,M^{\prime}\}$ $c_{i}$-loops, then the-rooted-orienteering algorithm is called $K\times min\{Q,M^{\prime}\}$ times. And there is $M$ interations at most. The-rooted-orienteering algorithm is a polynomial approximation algorithm, and the time complexity is set to $O(L1)$, so the time complexity of Algorithm 1 is $O(MK\times min\{Q,M\}\times L1)$, it is polynomial-time solvable.

Since the best approximation ratio for the rooted orienteering problem in metric space is $\alpha=\frac{1}{2}$[20], we get the theorem LABEL:thm:CSC-RCSC-Lt_-appr as below.

When $T_{t}>T_{c}$, $T_{t}/T_{c}=\hat{Q}$ is a positive integer. A sensor can start from one charging station $c_{i}\in\mathcal{C}$ and go through a set of $c_{i}$-loops with a length not exceeding $L_{c}$ and finally return to $c_{i}$ within a time period of $T_{t}$. The number of paths is exponential. Therefore, in Algorithm 2, only the path including $\hat{Q}$ loops starting from charging station $c_{i}\in\mathcal{C}$ and whose length is $L_{c}$ are considered, recorded as $r_{c_{i}}^{\hat{Q}}(Lc)$. Therefore, $\mathcal{R}=\{r_{c_{i}}^{\hat{Q}}(Lc)|c_{i}\in\mathcal{C}\}$. The impression on the approximation ratio will be analysized later. The sweeping path of a sensor is shown in Figure 2, where the number of charging stations i is 2, the number of sensors is 2, and $\hat{Q}=3$.

Algorithm 2 uses greedy method to gradually select the path $r\in\mathcal{R}$ with the most unity gain, and call a sensor to cover. The steps are as follows:

1. 1.

   Let $\hat{\mathcal{T}}\subseteq\mathcal{T}$ be the set of targets that have not yet been covered, and construct a induced subgraph of graph $G$, $G^{\prime}=(\hat{\mathcal{T}}\cup\mathcal{C},\mathcal{E}^{\prime})$, where $\mathcal{E}^{\prime}=\{(i,j)|i,j\in\hat{\mathcal{T}}\cup\mathcal{C},(i,j)\in\mathcal{E}\}$;

2. 2.

   Find $\hat{Q}$ $c_{i}$-loops $r_{c_{i}}^{\hat{Q}}(L_{c})$ with the largest number of targets in the graph $G^{\prime}$. and deploy one sensor to cover;

3. 3.

   Repeat steps 1 and 2 until the number of sensors is 0.

Finding the $r_{c_{i}}^{\hat{Q}}(Lc)$ that covers the most number of targets is called the rooted team orienteering problem [26], which is NP-hard. therefore, Algorithm 2 calls the rooted-team-orienteering approximation algorithm for the rooted team orienteering problem as a subroutine, whose approximation ratio is $\beta$($0<\beta<1$). The input of Algorithm rooted-team-orienteering is the induced subgraph $G^{\prime}$, the root $c_{i}\in\mathcal{C}$, the path length $L$, and the number of paths $\hat{Q}$.

Since the best approximation ratio of the rooted team orienteering problem in metric space is $\beta\approx 0.394$[26], therefore, we get the theorem 9 as follows.

When $T_{c}\geq T_{t}$, $T_{c}/T_{t}=Q$ is a positive integer. Similar to the RCSC problem, the travelling paths of a sensor $r\in\mathcal{R}$ must be a loop. When the number of sensors $v_{i}$ required for the path is the same, the longer the path, the more the number of targets can be covered. Therefore, the set of paths $\mathcal{R}$ can be reduced to only include paths with the length $k\times L_{t}$. Supposed $\bar{Q}=\lceil\frac{d(a,b)}{L_{t}}\rceil$, the path set includes self-loop $\{r_{a}(i\times L_{t}),r_{b}(i\times L_{t})\mid i\in[1,min\{Q,M\}]\}$, mutual-loop $\{r_{\{a,b\}}(j\times L_{t})\mid j\in[2\bar{Q},min\{M,2Q\}]\}$. When $M<2\bar{Q}$, the mutual-loop does not exist. The cost of self-loop $r_{c}(i\times L_{t})$ is $i$, $c\in\{a,b\}$; the cost of mutual-loop $r_{\{a,b\}}(j\times L_{t})$ is $j$. Suppose $j_{1}+j_{2}=j$, if $j_{1}$ and $j_{2}$ are not integers, then $j_{1}$ and $j_{2}$ can have countless combinations. Therefore, Algorithm 4 only considers the mutual loops composed of two paths with length that are integer multiples of $L_{t}$, denoted as $r_{\{a,b\}}((j_{1}+j_{2})\times L_{t})$. The set of candidate paths is $\mathcal{R}=\{r_{a}(i\times L_{t}),r_{b}(i\times L_{t})\mid i\in[1,min\{Q,M\}]\}\cup\{r_{\{a,b\}}((j_{1}+j_{2})\times L_{t})\mid j_{1},j_{2}\in Z^{+},j_{1}+j_{2}=j\in[2\bar{Q},min\{M,2Q\}]\}$. The sweeping path of the sensor is shown in Figure 3, the number of charging stations in the figure is 2, the number of sensors is 8, and $Q=2$.

Algorithm 4 uses a greedy strategy to gradually select the path $r\in\mathcal{R}$ with the most unity gain, and call the sensor to cover it. The specific steps of the algorithm are as follows:

1. 1.

   Let $\hat{\mathcal{T}}\subseteq\mathcal{T}$ be the set of targets that have not yet been covered, and construct a induced subgraph of graph $G$, $G^{\prime}=(\hat{\mathcal{T}}\cup\mathcal{C},\mathcal{E}^{\prime})$, where $\mathcal{E}^{\prime}=\{(i,j)|i,j\in\hat{\mathcal{T}}\cup\mathcal{C},(i,j)\in\mathcal{E}\}$;

2. 2.

   In the graph $G^{\prime}$, find an mutual loop $r_{\{a,b\}}((j_{1}+j_{2})L_{t})$ with the average largest number of targets covered by the sensors, where $j_{1},j_{2}\in Z^{+}$, $j_{1}+j_{2}=j$, $j=2\bar{Q},,...,min\{M^{\prime},2Q\}$, $M^{\prime}$ is the number of sensors that have not yet been allocated;

3. 3.

   In the graph $G^{\prime}$, find the $c_{i}$ self loop $r_{c_{i}}(kL_{t})$ whose length is $kL_{t}$ and cover the largetest average number of targets, where $c_{i}\in\{a,b\}$$k=1,2,...,min\{M^{\prime},Q\}$, $M^{\prime}$ is the number of sensors that have not yet been allocated;

4. 4.

   Compare the unity gain of the path obtained in steps 2 and 3, choose the path with the larger unity gain, if the cost is $m_{i}$, then call $m_{i}$ sensors to distribute on $r_{i}$ eparated by $L_{t}$, $T_{t}$-sweep coverage of the targets on $r_{i}$;

5. 5.

   Repeat steps 1-4 until $M^{\prime}=0$.

In the step 2, it is NP-hard to find the mutual-loop $r_{\{a,b\}}((j_{1}+j_{2})L_{t})$ whose length is $(j_{1}+j_{2})L_{t}$ and covers the largest number of targets. We have known a s-t orienteering problem, aiming to find a path with the length not greater than $L$ and coveloop the most number of targets, is an NP-hard problem. Its approximation algorithm, s-t-orienteering, whose input is the current induced subgraph $G^{\prime}$ , two root nodes $\{a,b\}$ and limited length $L$. Assuming that the approximation ratio of the approximation algorithm is $\gamma$, we design an approximation algorithm to find the mutual loop $r_{\{a,b\}}((j_{1}+j_{2})L_{t})$ covering the largest number of targets ( Algorithm 3).

This is also a greedy approximation algorithm. From the proof of the theorem 7: $\omega(APP)\geq(1-(1-\frac{\gamma}{m})^{m})\omega(OPT)$, when $m=2$, the approximation ratio of Algorithm 3 is $1-(1-\frac{\gamma}{2})^{2}$. The lemma 10 is as follows:

Supposed that the number of targets covered by Algorithm 4 is $\omega(APP)$, and the number of targets covered by the optimal solution is $\omega(OPT)$. $r_{i}^{*}$ is the path with the most unity gain selected in the $i^{th}$ interation, whose cost is $v_{i}$. $r_{i}$ is the path chosen by Algorithm 4 in the $i^{th}$ interation with the same cost. From the lemma 6, we know that if the path chosen for the $l$th time is a self-loop,