processing times. The approximation factor of the algorithm is O(logρ), where ρ= ttmax
min,tmin and tmax are the minimum and maximum sweep periods among the vertices.
• If speeds of the mobile sensors are different, we prove that it is impossible to design any constant factor approximation algorithm to solve the sweep coverage problem, unless P=NP.
4.2 Sweep Coverage with Different Sweep Periods
4.2.1 Proposed algorithm
Our proposed algorithm solves Problem 2 in two phases. First phase finds number of mobile sensors and second phase calculates movement strategy of the mobile sensors.
Finding number of mobile sensors
For G = (U, E, w), we construct complete graph G0 = (U, E0, w0). Define weight of an edge (ui, uj) ∈E0 as w0(ui, uj) = d(ui, uj) + v2(τi +τj) for i, j = 1 to n, where d(ui, uj) is the weight of the shortest path between ui and uj inG. Let ρ= ttmax
min, where tmin and tmax are the minimum and maximum sweep periods among the vertices. We partition U into subsets U1, U2,· · · , Udlogρe, whereUi={uj ∈U|2i−1·tmin ≤tj <2i·tmin}. Let G0i be the induced subgraph of G0 for the vertex set Ui. For i= 1 todlogρe, we apply step 1 to step 12 of the Algorithm 2 (chapter 3) on G0i to find the number of mobile sensors for (2i−1·tmin)-sweep coverage of all vertices in Ui.
Movement strategy of the mobile sensors
Now we explain how to deploy the mobile sensors to ensure sweep coverage for all the vertices of G. Deployments of the mobile sensors are explained below for Ui, i = 1 to logdρe. Let C1, C2,· · · , CJi be the connected components of G0i computed in step 13 of Algorithm 2. Find a tour Tk by doubling all edges of Ck for k = 1,2,· · · , Ji. If Tk contains only one vertex, one mobile sensor is deployed at the vertex for periodical monitoring. If there are more than one vertex in Tk, we compute a tour Tk0 by replacing vertices and edges of Tk as explained below. Let ui1, ui2,· · · , uil ∈ Ui be the vertices along Tk in the clockwise direction. Replace each vertex uih in Tk by two vertices u0ih and u00ih and introduce an edge (u0ih, u00ih) with edge weight w0(u0ih, u00ih)=vτih. The edges (ui1, ui2),(ui2, ui3),· · · ,(uil−1, uil),(uil, u1) ofTkare replaced by the edges (u00i1, u0i2), (u00i2, u0i3),· · · ,(u00il−1, u0il), (u00il, u0i1) respectively. The weight of the edge (u00ij, u0ij+1) is given by w0(u00ij, u0ij+1) =w0(uij, uij+1)− v2(τij +τij+1).
An example of a tour Tk onG0i is shown in Fig. 4.1 and corresponding construction of Tk0 from Tk is shown in Fig. 4.2.
Partition Tk0 into l w0
(Tk0) v·2i−1tmin
m
parts of weight at most v ·2i−1tmin. According to the
ui3
ui2
ui1
Figure 4.1: Tour Tk onG0i
u0i1
u00i1
u0i2 u00i2
u0i3 u00i3
Figure 4.2: Construction of Tk0 from Tk
positions of the partitioning points, mobile sensors are deployed on the original graph G as follows. If the position of a partitioning point is on the edge (u0ij, u00ij) ∈ Tk0 for some j, then one mobile sensor is deployed at the vertex uij in G. If the position of a partitioning point is on the edge (u00ip, u0iq) ∈ Tk0 for p 6= q, then one mobile sensor is deployed at the corresponding position on the edge (uip, uiq) in G.
After deployment, all mobile sensors move around the tour Tk0 in the same direction and the movement of the mobile sensors are reflected in the original graphGas explained below. If position of a mobile sensor is on the edge (u0ij, u00ij) ∈ Tk0 for some j, then it waits at the vertex uij onG for the time which is equal to the time taken by the mobile sensor to move from its current position to u00ij on the tour Tk0 with uniform speed v.
Otherwise it continues its movement with uniform speed v along the respective edge in G.
An example is shown in Fig. 4.3 to explain deployment and movement strategy of the mobile sensors. The partitioning points are shown by the crossed marks on the tour Tk0. Corresponding to the partitioning pointP onTk0, a mobile sensor is deployed at the vertex ui1 in G. The mobile sensor waits atui1 in G for the time it takes to move from
P Q
R S
u0i1
u00i1
u0i2
u00i2
u0i3 u00i3
Figure 4.3: Partitioning a tourTk0 and initial deployment of mobile sensors P tou00i1 along the tourTk0. Then it moves to the next vertexui2 along the shortest path from from ui1 toui2 inGwith uniform speedv. Corresponding to the partitioning point Q, a mobile sensor is deployed at the same position on respective edge in G. Then it moves to the next vertex in the tour along the shortest path of the two vertices with uniform speed v.
4.2.2 Analysis
Theorem 4.2.2. According to the movement strategy of the mobile sensors each vertex ui ∈U is ti-sweep covered with processing time τi.
Proof. Ifui belongs to a component with one vertex, thenti sweep coverage is trivial by the mobile sensor deployed atui. Now if ui belongs to a component with more than one vertex, then according to the proposed algorithm for Problem 2, ui ∈Uj for somej = 1 to dlogρe. By Theorem 3.3.2, ui is sweep covered with sweep period 2j−1 ·tmin ≤ ti. Therefore ui is ti sweep covered. Also, when a mobile sensor visits ui, it waits at ui for τi time which follows from the construction of the tour explained in Section 4.2.1.
Theorem 4.2.3. The approximation factor of the proposed algorithm for Problem 2 is 6dlogρe.
Proof. Let OP T be the optimal solution of the Problem 2 on the graph G and OP T1, OP T2, · · · OP Tdlogρe be the optimal solutions for G01, G02, · · · G0dlogρe respectively (ref.
Section 4.2.1). Then OP Ti ≤OP T fori= 1 to dlogρe, as the sweep coverage of all the vertices ofGensures sweep coverage for all the vertices inG0i. According to the algorithm
for Problem 2 let OU T1, OU T2,· · ·, OU Tdlogρe be the number of mobile sensors forG01, G02, · · · G0dlogρe respectively. Then OU Ti ≤ 6OP Ti, as the length of the partition in OP Ti is at most twice of the length of the partition in OU Ti and the Algorithm 2 is a 3-approximation algorithm. Hence, total number of mobile sensors needed is equals to Pdlogρe
j=1 OU Tj ≤6Pdlogρe
j=1 OP Tj ≤6dlogρeOP T. Therefore, the algorithm for Problem 2 is 6dlogρe-approximation algorithm.
Theorem 4.2.4. The running time of the algorithm for Problem 2 is O(n2lognlogρ).
Proof. The time complexity of the Algorithm 2 is O(n2logn). The algorithm for Prob- lem 2 uses the Algorithm 2 as a subroutine forO(logρ) disjoint set of of vertices. Hence the running time is O(n2lognlogρ).
Remark 4.2.5. Although it seems that the time complexity of the algorithm for Problem 2 is pseudo polynomial but according to the decomposition strategy of the vertices of the graph there can be at most n non-empty disjoint vertex sets. Hence the complexity of the algorithm is O(n3logn) which is polynomial inn.