Search and Rescue on the Line

Abstract

We propose and study a problem inspired by a common task in disaster, military, and other emergency scenarios: search and rescue. Suppose an object (victim, message, target, etc.) is at some unknown location on a path. Given one or more mobile agents, also at initially arbitrary locations on the path, the goal is to find and deliver the object to a predefined destination in as little time as possible. We study the problem for the one- and two-agent cases and consider scenarios where the object and agents are arbitrarily (adversarially, even) placed along a path of either known (and finite) or unknown (and potentially infinite) length. We also consider scenarios where the destination is either at the endpoint or in the middle of the path. We provide both deterministic and randomized online algorithms for each of these scenarios and prove bounds on their (expected) competitive ratios.

Appendices

Appendix

A Proofs From Sect. 4 (A Single Agent)

Theorem 4.6

Every online algorithm for the single-agent, line model must have a competitive ratio of at least 5.

Proof

Consider a scenario where the agent is placed at some arbitrarily small distance $s > 0$ away from 0 and the object is at least a distance 2s from s. Any algorithm must involve a sequence of positive distances $x_1, x_2, x_3, \ldots$ such that the agent moves left (or right) a distance $x_1$ and back to s, then right (or left) a distance $x_2$ and back to s, and so on. Clearly any optimal online algorithm of this form must satisfy $x_{i+2} > x_i$ and any algorithm with a competitive ratio of 5 or better must satisfy $x_i \le 3 x_{i-1}$ where $x_1 \ge 2s$ (since the object at least a distance 2s from s). Thus, $x_i < 2s \cdot 3^{i-1}$ for all $i \ge 2$ and so the number of required turning points for an optimal online algorithm can be made arbitrarily large (by setting s to be arbitrarily small).

For some sequence of turning points, suppose the object is found on the $k^\text {th}$ round (k can be arbitrarily large by the argument above) and observe the competitive ratio can then be written

$$\begin{aligned} \sup _{k,y,s}&\frac{\sum _{i=1}^{k-1} 2 x_i + 2y \pm s}{2y \pm s} = \sup _{k} \frac{\sum _{i=1}^{k-1} x_i + x_{k-2}}{x_{k-2}} \\&= 1 + \sup _{k} \frac{\sum _{i=1}^{k-1} x_i}{x_{k-2}} \ge 1 + \sup _{k} \frac{\sum _{i=1}^{k-1} r^i}{r^{k-2}} = 1 + \sup _{k} \frac{r^k - r}{(r-1)r^{k-2}} \end{aligned}$$

where $r > 1$ (the expansion factor). The above inequality follows from Corollary 7.11 of Sect. 7.2 of [21] (following the method for proving the 9-competitive search on an infinite line in Sect. 8.2.1 of [21]). Then, since $\frac{r^k - r}{(r-1)r^{k-2}}$ is increasing with respect to k (its derivative $\frac{r^{3-k} \ln r}{r-1}$ is greater than 0 for any $r>1$), we can simplify the competitive ratio to

$$\begin{aligned} 1 + \sup _{k} \frac{r^k - r}{(r-1)r^{k-2}} = 1 + \lim _{k \rightarrow \infty } \frac{r^k - r}{(r-1)r^{k-2}} = 1 + \frac{r^2}{r-1} \end{aligned}$$

which has a minimum value of 5 for $r=2$.    $\square$

Theorem 4.7

The expansion rate $r=\frac{1}{W(1/e)} \approx 3.59112$ yields an expected competitive ratio of $1 + \frac{1}{2 W(1/e)} \approx 2.79556$ for Algorithm 5 where W(x) is the product logarithm (Lambert W function [14]) of x.

Proof

Just as in the proof for Theorem 4.5, we consider the alternate algorithm $A'$ such that the agent does not turn around early upon reaching an endpoint. It is clear that our original algorithm cannot perform worse than $A'$ and in cases where an endpoint is never reached, the two algorithms are identical.

Let $d = s \cdot r^{k+\delta }$ denote the position of the object where $0 \le \delta < 1$. By executing Algorithm 5, the agent moves a distance $r^{i+\epsilon }$ to the left and right in alternating rounds $i=1,2,\ldots$. Consider the round when the agent moves a distance $s \cdot r^k$ for the first time. If the agent moves $r^k$ distance for the first time in the opposite direction of the object, then it will definitely find the object in round $k+1$. The expected competitive ratio in this case can be written:

$$\begin{aligned} \mathbb {E} \left[ \frac{\sum _{i=1}^{k}s\cdot 2 r^{i+\epsilon } + 2d \pm s}{2d \pm s} \right]&= \mathbb {E} \left[ \frac{\sum _{i=1}^{k} 2 r^{i+\epsilon } + 2r^{k+\delta } \pm 1}{2r^{k+\delta } \pm 1} \right] \\&= \mathbb {E} \left[ 1 + \frac{2r^\epsilon \left( r^{k+1} - r \right) }{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) } \right] \\&= 1 + \frac{2 \left( r^{k+1} - r \right) }{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) } \cdot \mathbb {E} \left[ r^\epsilon \right] \\&= 1 + \frac{2 \left( r^{k+1} - r \right) }{\left( 2r^{k+\delta } \pm 1 \right) \ln r} \end{aligned}$$

since $\mathbb {E}[r^\epsilon ] = \int _{1}^{r} x \frac{1}{x \ln r} dx = \frac{r-1}{\ln r}$.

On the other hand, if the agent moves a distance $s \cdot r^k$ for the first time in the direction of the object, it will find it on round k if $\epsilon \ge \delta$ and on round $k+2$ otherwise. Let B be the event that $\epsilon \ge \delta$, then the expected competitive ratio can be written:

$$\begin{aligned}&\mathbb {E} \left[ Pr[B] \left[ \frac{\sum _{i=1}^{k-1}s\cdot 2 r^{i+\epsilon } + 2d \pm s}{2d \pm s} \right] + (1-Pr[B]) \left[ \frac{\sum _{i=1}^{k+1}s\cdot 2 r^{i+\epsilon } + 2d \pm s}{2d \pm s} \right] \right] \\ =&~ \mathbb {E} \left[ Pr[B] \left[ 1+\frac{2r^\epsilon (r^k-r)}{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) } \right] + (1-Pr[B]) \left[ 1+\frac{2r^\epsilon (r^{k+2}-r)}{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) } \right] \right] \\ =&~ Pr[B] \left[ 1+\frac{2(r^k-r)}{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) } \cdot \mathbb {E} \left[ r^\epsilon | B \right] \right] \\&~+ (1-Pr[B]) \left[ 1+\frac{2(r^{k+2}-r)}{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) } \cdot \mathbb {E} \left[ r^\epsilon | \overline{B} \right] \right] \\ =&~ Pr[B] \left[ 1+\frac{2 (r^k-r) (r - r^\delta )}{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) \ln r Pr[B]} \right] \\&~+ (1-Pr[B]) \left[ 1+\frac{2(r^{k+2}-r) (r^\delta - 1)}{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) \ln r Pr[\overline{B}]} \right] \end{aligned}$$

since $\mathbb {E}[r^\epsilon | B] = \int _{r^\delta }^{r} x \frac{1}{Pr[B] \cdot x \ln r} dx = \frac{r - r^\delta }{\ln r Pr[B]}$ and $\mathbb {E}[r^\epsilon | \overline{B}] = \int _{1}^{r^\delta } x \frac{1}{Pr[\overline{B}] \cdot x \ln r} dx = \frac{r^\delta -1}{\ln r Pr[\overline{B}]}$. Then the expression can be further simplified:

$$\begin{aligned} =&~ 1 + \frac{2 (r^k-r) (r - r^\delta )}{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) \ln r} + \frac{2(r^{k+2}-r) (r^\delta - 1)}{\left( 2r^{k+\delta } \pm 1 \right) \left( r-1 \right) \ln r} \\ =&~ 1 + \frac{2}{\left( 2r^{k+\delta } \pm 1 \right) (r - 1) \ln r} \left( (r^k-r)(r-r^\delta )+(r^{k+2}-r)(r^\delta -1) \right) \\ =&~ 1 + \frac{2}{\left( 2r^{k+\delta } \pm 1 \right) (r - 1) \ln r} (r - 1) (r^{\delta + k} + r^{d + k + 1} - r^{k + 1} - r) \\ =&~ 1 + \frac{2(r^{\delta + k} + r^{d + k + 1} - r^{k + 1} - r)}{\left( 2r^{k+\delta } \pm 1 \right) \ln r} \end{aligned}$$

Observe that, since the initial search direction is chosen uniformly randomly, the total expected competitive ratio is

$$\begin{aligned}&~ \frac{1}{2} \left[ 1 + \frac{2 \left( r^{k+1} - r \right) }{\left( 2r^{k+\delta } \pm 1 \right) \ln r} \right] + \frac{1}{2} \left[ 1 + \frac{2(r^{\delta + k} + r^{d + k + 1} - r^{k + 1} - r)}{\left( 2r^{k+\delta } \pm 1 \right) \ln r} \right] \\ =&~ 1 + \frac{\left( r^{k+1} - r \right) }{\left( 2r^{k+\delta } \pm 1 \right) \ln r} + \frac{(r^{\delta + k} + r^{d + k + 1} - r^{k + 1} - r)}{\left( 2r^{k+\delta } \pm 1 \right) \ln r} \\ =&~ 1 + \frac{r^{k+\delta } (1+r) - 2r}{\left( 2r^{k+\delta } \pm 1 \right) \ln r} \le 1 + \frac{r^{k+\delta } (1+r) - (1+r)}{\left( 2r^{k+\delta } - 1 \right) \ln r - \ln r} = 1 + \frac{(1+r) (r^{k+\delta } - 1)}{\ln r \left( 2r^{k+\delta } - 2 \right) } \\ =&~ 1 + \frac{1+r}{2 \ln r}. \end{aligned}$$

Finally, with an expansion rate of $r=\frac{1}{W(1/e)} \approx 3.59112$, the above upper bound becomes $1 + \frac{1}{2 W(1/e)} \approx 2.79556$.    $\square$

B Proofs From Sect. 5 (Two Agents)

Theorem 5.2

Algorithm 7 has a competitive ratio of $\min \left( 1+\sqrt{2}, \frac{3}{1+2v} \right)$.

Proof

The proof is very similar to that of Theorem 5.1. Without loss of generality, suppose the first agent has a speed of 1 and the second agent a speed of $v \le 1$. For the case where $v \le \frac{2-\sqrt{2}}{2+2\sqrt{2}}$ or $v=1$, the first agent exactly performs Algorithm 1, and so a competitive ratio of $1+\sqrt{2}$ is achieved. The interesting case, then is when $\frac{2-\sqrt{2}}{2+2\sqrt{2}}< v < 1$. First, if the fast agent still arrives at the object first, the competitive ratio is $\frac{3+v}{1+3v}$ (using the same analysis as used in the proof for Theorem 5.1). Otherwise, if the slow agent arrives at the object before the fast agent reaches 0 (i.e. $s > \frac{y-s}{v} \Rightarrow y < s(v+1)$), then the competitive ratio is

$$\begin{aligned} \frac{t + (y - (tv - (y-s)))}{2y-s}&= \frac{s(2-v)-2y}{v(s-2y)} \le \frac{3}{1+2v} \end{aligned}$$

where $t = \frac{2\left( y-\left( s-\frac{y-s}{v} \right) \right) }{1+v}$ is the time the agents meet for a handover. The first equality follows from substituting this value for t and the final inequality follows since $\frac{s(2-v)-2y}{v(s-2y)}$ is increasing with respect to y (its derivative with respect to y, $\frac{2s(1-v)}{v(s-2y)^2}$, is positive for all $v < 1$) and $y < s(v+1)$. On the other hand, if the slow agent finds the object after the fast agent reaches 0 but still before it catches up ($s< \frac{y-s}{v} < y+s$), then the competitive ratio is

$$\begin{aligned} \frac{t + (y - (tv - (y-s)))}{2y-s}&= \frac{s(1+v)-4y}{(1+v)(s-2y)} = \frac{s(1+v)-4y}{s(1+v)- 2y(1+v)} \le \frac{3}{1+2v} \end{aligned}$$

where $t = \frac{2y}{1+v}$ is the time the agents meet for a handover. The first equality follows from substituting this value for t and the final inequality follows since $\frac{s(1+v)-4y}{s(1+v)- 2y(1+v)}$ is decreasing with respect to y (its derivative with respect to y, $\frac{2s(v-1)}{(1+v)(s-2y)^2}$, is negative for all $v < 1$) and $y > s(1+v)$. Finally, observe the second and third cases dominate the competitive ratio:

$$\begin{aligned} \frac{3}{1+2v} \ge \frac{3+v}{1+3v} ~\Rightarrow ~ 3+9v \ge 3+7v+2v^2 ~\Rightarrow ~ v(1-v) \ge 0 \end{aligned}$$

Clearly this condition is always satisfied since $0 \le v \le 1$.    $\square$