A 2-Approximation Algorithm for Barrier Coverage by Weighted Non-uniform Sensors on a Line

Abstract

Barrier coverage is an approach to the intruder detection problem that relies on monitoring a perimeter, or barrier, of an area of interest using sensors placed around it. In this paper, we propose a weighted generalization of the unweighted line segment barrier coverage problem studied in [5] for which the authors demonstrate an FPTAS. We develop a fast, simple 2-approximation algorithm for the weighted case likely to be of interest to practitioners, and show that the FPTAS developed in [5] can be adapted to the weighted problem.

Wireless & Geometry Track

R. Benkoczi—Acknowledges the support for this research received from an NSERC Discovery Grant.

M. Thom—Acknowledges the support for this research received from the University of Lethbridge School of Graduate Studies.

Appendices

A Appendix

A.1 Proofs from Sect. 3

Proof

(of Proposition 4 ). We charge $s \in \text {opt}$ twice, noting

$$ 2w_s(d_s + p_{s, \text {opt}}) \ge w_s(d_s + 2p_{s, \text {opt}}) \ge w_s(d_s + x) $$

   $\square$

Proof

(of Proposition 5 ). We break the proof into cases centered around the total size of the sensors in the set $\mathcal {I} = \{ t \in \text {app} : t \in [x - p_{s,\text {opt}}, x] \cap \text {opt} \}$.

If $|\mathcal {I}|_{size} = \sum _{t \in \mathcal {I}} l_t \ge l_s$, we select $l_s$ in total length of $t \in \mathcal {I}$ and in order of descending $p_{t, \text {opt}}$. Let $\mathcal {I}'$ be this subset of $\mathcal {I}$, so that $|\mathcal {I}'|_{size} = l_s$. Since $2p_{s, \text {opt}} < x$ and $t \in \mathcal {I}$ implies $t \in [x - p_{s, \text {opt}}, x]$, we have $p_{s, \text {opt}} < p_{t, \text {opt}}$ for all $t \in \mathcal {I}$. Therefore, $ratio(s) \ge ratio(t)$ for all $t \in \mathcal {I}$ by Lemma 1.

From Fact 2, we have

$$ \frac{w_s}{l_s} \ge \frac{\sum _{t \in \mathcal {I}'} w_t}{\sum _{t \in \mathcal {I}'} l_t} \ge \frac{\sum _{t \in \mathcal {I}'} w_t}{l_s} $$

so that $w_s \ge \sum _{t \in \mathcal {I}'} w_t$.

Since $|x - p_{t, \text {opt}}| \le p_{s, \text {opt}}$ for each $t \in \mathcal {I}'$, we can use a single charge of s to move each $t \in \mathcal {I}'$ to right-align with x. We notice that each $t \in \mathcal {I}$ is viable with respect to s at x, since t precedes s in $\text {app}$. Since $|\mathcal {I}'|_{size} = l_s$, we can pay for $s \in \text {app}$ using a single charge of $s \in \text {opt}$ and a single charge of each $t \in \mathcal {I}'$. The process is concluded by substituting a gap for $s \in \text {opt}$ and for each $t \in \mathcal {I}'$.

We note that this is an adaptation of the shifting trick originally developed for use in Proposition 3. If instead $|\mathcal {I}|_{size} < l_s$, we use the same shifting trick to move each $t \in \mathcal {I}$ to right-align with x, charging the additional charge to $s \in \text {opt}$. This leaves s with $l_s - |\mathcal {I}|_{size}$ of a single charge remaining, which we use to shift $l_s - |\mathcal {I}|_{size}$ in total of sensors (any sensors) in $[x - p_{s, \text {opt}}, x] - \mathcal {I}$.

The sensors we select to shift, whether in $\mathcal {I}$ or not, are viable with respect to $s \in \text {app}$ at x. The selected sensors not in $\mathcal {I}$ aren’t found in $\text {app}$, and so, are elements of $S_x$ and therefore viable with respect to s at x. Once more, the cost of s is paid and the selected sensors moved by s and $s \in \text {opt}$ itself are replaced by gaps once their charges are spent.    $\square$

Proof

(of Proposition 6 ). By induction on the number of sensors whose cost remains to be bounded in $\text {app}_C$.

Suppose $n=1$. A single sensor is in $\text {app}_C$, with the possibility of sensors punctuated by gaps “below” it in $\text {opt}_C$, as depicted in Fig. 1g.

By Invariants 1 and 3, we can pay for exactly the proportion of s equal to the total length of the gaps using the cost of sensors drawn from the priority queue. If this does not cover the cost of s entirely, the configuration is made to resemble Fig. 1h, where s is shrunk down to $s'$, the remaining unpaid length of s, with the sensors in $\text {opt}_C$ compacted to align with the new cursor $p_x$.

Since each $t \in \text {opt}_C$ is distinct from s (except possibly the t containing 0, but this makes no difference to the argument), we have $t \in S_x$. By Invariant 1, the sensors $t \in \text {opt}_C$ satisfy

$$ \frac{w_s(d_s + x)}{\min (x, l_s)} \le \frac{w_t(d_t + x)}{\min (x, l_t)} $$

and therefore, with $p = p_x / x$ the unpaid portion of s,

$$\begin{aligned} p \cdot \frac{w_s(d_s + x)}{\min (x, l_s)}&\le p \cdot \frac{w_t(d_t + x)}{\min (x, l_t)} \\&\le \frac{w_t(d_t + p \cdot x)}{\min (x, l_t)} \\&\le \frac{w_t(d_t + p_x)}{\min (x, l_t)} \end{aligned}$$

With the conditions of Propositions 1, 2 and 3 all satisfied, we conclude the case with whichever of the three is applicable.

Now suppose $n = k+1$ and that the statement of the Proposition holds if $\text {app}_C$ has k or fewer sensors. We consider the sensor s at cursor x by case.

Case 1: $s \in \text {app}_C \cap \text {opt}_C$

If $2p_{s, \text {opt}_C} \ge x$, we apply Proposition 4 to pay for $s \in \text {app}_C$, closing the gap left in place of s by shifting the optimal sensors to the right of s leftward. We note that Invariants 1, 3 and 4 are maintained after the shift.

If $2p_{s, \text {opt}_C} < x$, we apply Proposition 5 with a modification to its argument made to maintain Invariant 4. Let $u_1, \ldots u_n \in \text {opt}_C$ be the first $l_s$ in total length of sensors u satisfying $2p_{u, \text {opt}_C} < p_{u, \text {app}_C}$ leftward from x. We use the sensors $u_i$ instead of s in providing the charge to shift the selected sensors in the range $[x - p_{s, \text {opt}_C}, x]$ forward. Gaps $g_{u_1}, \ldots , g_{u_n}$ are introduced in their place, and s is carved into contiguous fragments of length $l_{u_1}, \ldots , l_{u_n}$, which are then identified with sensors $u_1, \ldots , u_n \in \text {app}_C$ respectively.

Since $2p_{s, \text {opt}_C} \le 2p_{u_i, \text {opt}_c} < p_{u_i, \text {app}_c}$ for each $u_i$, we can consider that of all the other optimal sensors of its type, s was nearest to x all along, although that may not have been true. Under this assumption, it’s clear that Invariant 4 is maintained, since any gaps following s in $\text {opt}_C$ correspond to larger gap regions, and the secondary sensors t selected in those cases therefore have larger $p_{t, \text {opt}_C}$ values.

Secondary copies of $u_1, \ldots , u_n$ are added to the priority queue, and as we’ve just opened gaps of those sizes, Invariant 3 is preserved. It is also clear that Invariant 1 is preserved.

Case 2: $s \not \in \text {app}_C \cap \text {opt}_C$.

If the priority queue is not empty, we use Invariant 3 to draw viable sensors from the priority queue to pay for s.

Suppose we can entirely pay for s using auxiliary sensors drawn from the queue. Since $p_{t, \text {opt}_C} \le x$ for all $t \in PQ$, and t is ordered maximally by $p_{t, \text {opt}_C}$, we have that the sensors $t'$ remaining in the queue after s is paid must satisfy $p_{t', \text {opt}_C} \le x - l_s$, since every positive $p_{t, \text {opt}_c}$ is unique. If the auxiliary sensors in the priority queue can only pay for a partial length of s (or no length, which is to say the queue is empty), we fall back on the argument given in case $n = 1$, which does not rely on the assumption that s is the only sensor left in $\text {app}_C$.

In either case, Invariant 3 is preserved and the gaps tagged by the secondary sensors are compacted against by the length of those sensors. The remaining regions and gaps are unaffected except as a result of compaction. Because of the queue ordering, the only effect this has is to reduce or close the rightmost gaps, leaving Invariant 4 intact.    $\square$

B The WeightedDisjointMinSum FPTAS

The FPTAS of [5] approximates the LeftDisjointMinSum problem, and is derived from a discretization of the continuous recurrence given by

$$ f^*(i+1,z) = \max \{ \min \{ f^*(i,z), g^*(i+1,z) \}, 0 \} $$

where

$$ g^*(i+1,z) = \min _{x \in [0,z]} \{ f^*(i, z-x) - 2r_{i+1} : |f^*(i,z-x) - r_{i+1} - x_{i+1}| \le x \} $$

and

$$ f^*(0,z) = 0 \text { for all } z \in [0,+\infty )\ $$

OPT is computed as

$$ OPT = \min _{z \ge 0} \{ z : f^*(n, z) \le 0 \} $$

Aside from the use of z to denote the budget, the notation used here matches our own. First, the sensors are ordered by Lemma 1, each of them of unit weight. $f^*(i,z)$ is the length of the largest subcovering $[f^*(i, z), L]$ formed from the first i sensors whose total movement cost is no greater than z.

We see that the recurrence can use sensor $i+1$ to augment the best possible coverage of i sensors using budget $z-x$ for any $0 \le x \le z$, where x is the part of the budget funding the movement of sensor $(i+1)$. The length of the old covering is extended by $2r_{i+1}$ units if $|f^*(i,z-x) - r_{i+1} - x_{i+1}| \le x$, meaning that x is enough to pay for the movement of sensor $(i+1)$. If sensor $(i+1)$ is not used, the recurrence reverts to the optimal covering under z restricted to the first i sensors, of cost $f^*(i, z)$.

Sorting the sensors as prescribed by Lemma 1, we rewrite the continuous recurrence for WeightedLeftDisjointMinSum as

$$ f^*(i+1,z) = \max \{ \min \{ f^*(i,z), g^*(i+1,z) \}, 0 \} $$

where

$$ g^*(i+1,z) = \min _{x \in [0,z]} \{ f^*(i, z-x) - 2r_{i+1} : w_{i+1} \cdot |f^*(i,z-x) - r_{i+1} - x_{i+1}| \le x \} $$

and

$$ f^*(0,z) = 0 \text { for all } z \in [0,+\infty ) $$

The second dimension of the continuous dynamic programming table is discretized in units of $\zeta = \epsilon Z/(n(n+1))$ over the interval [0, Z], where Z is the upper bound on budgets. Z is taken to be the cost of the solution produced by the naive $n \cdot OPT$ approximation algorithm described in [5], meaning $Z \le n \cdot OPT$.

Substituting our 2-approximation algorithm for the naive $n \cdot OPT$ algorithm we revise the bound to $Z \le 2 \cdot OPT$. $\zeta$ becomes $\zeta = \epsilon Z/2(n+1)$, so that the units of $\zeta$ that form the second dimension of the discretized table are between 0 and $Z / \zeta + n + 1 = 2(n+1) / \epsilon + n + 1$. With these change in place, the derivation and proof of correctness of the FPTAS for WeightedDisjointMinSum proceeds as it does for DisjointMinSum in [5].

We gain a quadratic factor improvement in run time, since the work of computing g drops from quadratic to linear time, resulting in time complexity $O(n^5/\epsilon ^3)$ for the whole FPTAS.