3.5 Shifting Strategy and its Application to the GMDS
R
c1 c2 c3 c4 c5 · · · cr−1 cr
d
X Y
Figure 3.5: Demonstration of shifting strategy.
Theorem 3.5.2. For a given point set P in R2, if there exists an algorithm A that can produce an α-factor approximation result for the GMDS problem defined for the points inside a monotone strip of width `d, where each pair of consecutive monotone chains are at distance d ≥ 2 and ` is an integer (shifting parameter), then there exists an α(1 + 1`)-factor approximation algorithm for the GMDS problem for the point set P. Proof. The algorithm is similar to the algorithm proposed by Hochbaum and Maass [55].
Since the monotone strips are of width `d and the distance between a pair of monotone strips is d≥ 2, the solution of one strip does not affect the solution of any other strip.
Thus the approximation factor of the algorithm follows.
3.5.2 A
52-factor approximation algorithm
Here we propose a 52-factor approximation algorithm for the GMDS problem for a given set P of n points in R2 using shifting strategy discussed in Subsection 3.5.1.
Definition 3.5.3. A duper-cell E is a combination of 30 cells (regular hexagons of side length 12) as shown in Figure 3.6.
The basic idea of the proposed 52-factor approximation algorithm is as follows: first optimally solve the sub-problem for a duper-cell E, we refer this problem as a single
u
v w
x
Figure 3.6: Demonstration of a duper cell and its partition.
duper-cell GMDS problem; then apply shifting strategy in both horizontal and vertical directions separately for the entire region R, and use the optimal algorithm for solving the generated sub-problems consisting of duper-cells.
3.5.2.1 Computing an optimum solution for a single duper-cell GMDS prob- lem
We divide the duper-cell E into three regions as in the case of super-cell in Section 3.4.
Let us name these regions unshaded, light-shaded, and dark-shaded regions, respectively (see Figure 3.6). The algorithm for computing an optimum solution for a single duper- cell is similar to the algorithm for a single super-cell (refer Section 3.4). Thus, we have the following lemma.
Lemma 3.5.4. A GMDS for the set of points inside a duper-cell E can be computed optimally in O(m20logm) time and using O(1) extra space, where m is the number of points inside the duper-cell E.
Now, we discuss the method of applying two-level shifting strategy to the GMDS problem. Split the entire regionRinto strips usingx-monotone chains, sayc1, c2, . . . , cr, such that (i) each segment is of length 12 unit, (ii) each pair of consecutive segments make an angle 2π3 , (iii) the distance between any two consecutive chains is 12 unit, and (iv) every chainc2i starts at exactly one unit away fromc2i−1in the opposite direction (either up or down) of c2i−1. Next, split each strip into hexagonal regions of side length 12.
In the first level of the shifting strategy, we take 3 consecutive rows of hexagons as a single horizontal strip and in the second level we take 10 columns of hexagons as a vertical strip inside that horizontal strip, i.e., we partition the horizontal strip into duper-cells. Each dupe-cell in the partition is bounded by four monotone chains, uv, vw, wx and xu, whereuv and wxare monotone with respect to x-axis, and vw and xu are monotone with respect to y-axis (see Figure 3.6). We name these chains as left, bottom, right, and top, respectively. Thus, the region R is split into duper-cells. By applying Theorem 3.5.2 in two-levels, we have the following result.
Theorem 3.5.5. A 52-factor approximation result for the GMDS problem can be obtained in O(n20logn) time, wheren is the input size.
Proof. In our GMDS problem the disks are of diameter 2. Let `1 and `2 be two shifting parameters in the first and second level shifting strategies, respectively. In the first level of partitioning we consider 3 consecutive rows of hexagons as a single horizontal strip, hence the distance between the monotone chains bottom and top bounding the strip is 2. Implies, `1 = 1 as d = 2 and `1d = 2. Again, inside a monotone strip of the first level, the distance between the monotone chains left and right corresponding to a duper-cell is greater than 8 i.e., `2d > 8, implies, `2 > 4. At the end of the second level, the region R is partitioned into duper-cells. We have an algorithm to compute an optimum solution for a duper-cell. Thus, the application of Theorem 3.5.2 in both the levels yields the approximation factor and is less than 1(1 + 11)(1 + 14) = 52. The time complexity O(n20logn) follows from Lemma 3.5.4.
3.5.3 A PTAS for the GMDS problem
We apply two-level shifting strategy (as in the previous subsection) to obtain a PTAS for the GMDS problem by solving the GMDS problem optimally for the points inside a region (sayF) bounded by two pairs of monotone chains such that the distance between left and right (resp. bottom and top) monotone chains isk (see Figure 3.7), wherek
is an integer given as input. In the first level, we partition R into horizontal strips of width k usingx monotone chains. In the second level, we further divide each horizontal strip into regions of size k ×k using y monotone chains. We solve each sub-problem independently and get an optimum solution; this yields a (1 +k1)2-factor approximation algorithm for the GMDS problem for the input P.
L1
L2
Figure 3.7: Demonstration of the PTAS.
3.5.3.1 Computing an optimum solution for P ∩ F
In order to solve the GMDS problem for the points in P ∩ F optimally, we further decompose F into four quadrants using the monotone chains L1 and L2 as shown in Figure 3.7. The number of unit disks in an optimum solution that pierce all the unit disks which intersecting the chainL1with centersleft(resp. right) side ofL1 is at most d2×2× √2k3e, which is less than 5k, and the number of disks in an optimum solution that pierce all the unit disks which intersecting the chainL2 with centersbottom (resp.
top) side of L2 is at most d2×2× √2k3e, which is less than 5k. That is, in any optimum solution the set, sayS(⊆ P ∩F), of points which cover all the points that are of distance at most 1 from the chains L1 and L2 is at most 10k. Note that|S|is independent of the number of points in P ∩ F. We consider all possible combinations of points inS of size at most 10k, and for every combination we do the following: consider the combination as part of the solution and delete the points in the quadrants that are within unit distance
from the chosen points. Compute an optimum solution for the rest of the points in the quadrants separately using the same procedure (note that any unit disk centered in the optimum solution of any quadrant does not intersect L1 and L2, and hence we can solve them independently). Next, we check whether the solutions obtained in the quadrants together with the chosen combination covers the points in P ∩ F. We consider a best possible solution once the process ends. If T(n, k) is the running time of the recursive algorithm for the GMDS problem for P ∩ F, then we have the following recurrence relation: T(n, k) = 4×T(n,k2)×n10k. This leads to the following theorem.
Theorem 3.5.6. For a given set P of n points in R2, the proposed algorithm produces a solution in nO(k) time, whose size is at most (1 + 1k)2× |OP T|, where k is a positive integer depends on ε and OP T is an optimum solution.
Though the running time of the proposed PTAS is same as the running time of the PTAS proposed by De et al. [34] in terms of O notation, but the constant involved in O is smaller than [34].