Barrier Coverage Problem in 2D

Abstract

This paper deals with the NP-hard problem of covering a line segment by n initially arbitrarily arranged circles on the plane by moving their centers to the segment in such a way that the sum of the Euclidean distances between the initial and final positions of the centers of the disks would be minimal. In the case of identical circles, a dynamic programming algorithm is known, which constructs a $\sqrt{2}$–approximate solution to the problem with $O(n^4)$–time complexity. In this paper, we propose a new algorithm that has the same accuracy, but the complexity of which is reduced by $n^2$ times to $O(n^2)$.

A Appendix

Example. Let it is required to cover the line segment [0, 4.5] by three identical disks with radii equal to 1, which initial positions of the centers are $p_1=(-0.5,1)$, $p_2=(2.5,2)$ and $p_3=(5.5,0)$ (Fig. 4(a)).

Fig. 4.

(a) Initial position of the disks; (b) one disk in the case when $l\le 1$; (c) one disk in the case when $1<l\le 2$; (d) two disks in the case when $l\le 1$.

Fig. 5.

(a) Two disks in the case when $1<l\le 2$; (b) two disks in the case when $2<l\le 3.5$; (c) two disks in the case when $3.5<l\le 4$; (d) the optimal OPC.

Since $x_1<0$, then

$$ S_1(l) = {\left\{ \begin{array}{ll} -x_1+y_1=1.5, &{} l\le 1\\ l-1-x_1+y_1=l+0.5, &{} l>1\\ +\infty , &{} l>2. \end{array}\right. } $$

The disk 1 moves to the point (0, 0), if $l\le 1$ (Fig. 4(b)) and it moves to the point $(l-1,0)$, if $l>1$ (Fig. 4(c)). Thus, we have the switching points 0, 1, 2 and 4.5.

Let now two circles participate in the covering. If $l\le 1$, then it is easy to see, that only disk 1 covers the segment [0,l] and $S_2(l)=1.5$ (Fig. 4(d)).

If $1<l\le 2$, then the segment [0, l] can be covered ether by one disk 1 or by one disk 2. We have that $d(p_1,\hat{p}_1)=l+0.5\le d(p_2,\hat{p}_2)$. So, in this case only disk 1 covers the segment [0, l]. Suppose that both disks 1 and 2 participate in the covering of the segment [0, l]. Let us denote by $x\in (1,3)$ the point at which the center of disk 2 moves. Then the segment $[0,x-1]$ must be covered by disk 1. If $x\le 2.5$ then $S_2(l)=2+2.5-x+S_1(x-1)=4$. If $2.5<x\le 3$ then $S_2(l)=\min \limits _{x\in [2.5,3]}\{2+x-2.5+S_1(x-1)\}=\min \limits _{x\in [2.5,3]}\{2x-1\}$. Therefore, in this case only the center of disk 1 moves to the point $(l-1,0)$ (Fig. 5(a)).

If $2<l\le 3.5$, then the both disks 1 and 2 must participate in the covering of the segment [0, l]. If x is a point where the center of disk 2 moves, then the segment $[0,x-1]$ must be covered by disk 1. For any $x\in [l-1,2.5]$, we get the same value of $S_2(l)=4$ and set $x=2.5$ (Fig. 5(b)).

If $3.5<l\le 4$, then both disks participate in the covering of the segment [0, l]. If $x\in [l-1,3]$ is a point where the center of disk 2 moves, then the segment $[0,x-1]$ must be covered by disk 1. In this case $2.5\le x\le 3$. Moreover, $x=l-1$ and $S_2(l)=l-1-2.5+2+1+l-2+0.5=2l-2$ (Fig. 5(c)).

Therefore, the following formula holds

$$ S_2(l) = {\left\{ \begin{array}{ll} 1.5, &{} 0<l\le 1\\ l+0.5, &{} 1<l\le 2\\ 4, &{} 2<l\le 3.5\\ 2l-3, &{} 3.5<l\le 4\\ +\infty , &{} l>4, \end{array}\right. } $$

where the switching points are 0, 1, 2, 3.5, 4, 4.5.

Let now all three sensors participate in the covering. The center of disk 3 can move to the point $x\in [3.5,4.5]$. Then the segment $[0,x-1]$ must be covered by disks 1 and 2 and

$$ S_3(4.5)=\min \limits _{x\in [3.5,4.5]}\{5.5-x+S_2(x-1)\}=\min \limits _{x\in [3.5,4.5]}\{9.5-x\}=5. $$

Then the center of disk 3 moves to the point (4.5, 0).

The backward recursion allows us to restore the optimal coverage, which is shown in the Fig. 5(d).