Optimal searching path to find a hidden target in a bounded region

Abstract

This paper addresses the problem of searching for a randomly located target in a bounded, known region by using two unit-speed searchers. The searching process starts at $(0,0)$. The area is divided into a finite set of cells. The position of the target has a symmetric probability distribution. It is desired to search in an optimal manner to minimize the expected time for detecting the target. The search strategy is derived using a dynamic programming algorithm. We present some special cases in which the cells are symmetric and asymmetric, and the algorithm depends on the dimensions of the search area. We give an illustrative example to demonstrate the applicability of this technique.

Appendix

Proof of Theorem 1

If the target lies in any cell of $\beta_{1}$, then $t_{11} = 2\left( {b + \sqrt {a^{2} + b^{2} } } \right)$. In addition, when the target lies in any cell of $\beta_{2}$, then $t_{21} = 4\left( {b + \sqrt {a^{2} + b^{2} } } \right)$. And, if the target lies in any cell of $\beta_{3}$, then $t_{31} = 6\left( {b + \sqrt {a^{2} + b^{2} } } \right)$ and etc., until the searcher reaches the last path $\beta_{m}$, then

$$\begin{aligned} t_{{m1}} & = ( h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + \sqrt {a^{2} + b^{2} } + ... + \sqrt {a^{2} + b^{2} } \\ & \quad+ \sqrt {b^{2} + (r - (m - 1)a)^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } \\ & \quad+ \sqrt {a^{2} + b^{2} } + ... + \sqrt {a^{2} + b^{2} } + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + h ) \\ & = 2\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } } \right). \\ \end{aligned}$$

If the target lies in any cell of $\Gamma_{1}$, then $t_{12} = 2\left( {b + \sqrt {a^{2} + b^{2} } } \right)$. And, if the target lies in any cell of $\Gamma_{2}$, then $t_{22} = 4\left( {b + \sqrt {a^{2} + b^{2} } } \right)$. Also, if it lies in any cell of $\Gamma_{3}$, then $t_{32} = 6\left( {b + \sqrt {a^{2} + b^{2} } } \right)$ and etc., until the searcher reaches the last path $\Gamma_{m}$, then

$$ t_{m2} = 2\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } } \right). $$

Since the target has symmetric uniformly distribution in the known region which has area $A$. Thus, the probability of detecting the target on any cell is $\frac{ab}{A}$, where any rectangle cell has area $ab$. Therefore,

$$ \begin{gathered} E(t(\varphi )) = 2\left( {b + \sqrt {a^{2} + b^{2} } } \right)\frac{2ab}{A} + 4\left( {b + \sqrt {a^{2} + b^{2} } } \right)\frac{4ab}{A} + 6\left( {b + \sqrt {a^{2} + b^{2} } } \right)\frac{6ab}{A} \hfill \\\qquad + 2\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } } \right)\hfill \\ \qquad\times\left[ {\frac{a(h - (m - 1)b) + b(r - (m - 1)a) + (m - 2)ba}{A}} \right] \hfill \\ \qquad+ 2\left( {b + \sqrt {a^{2} + b^{2} } } \right)\frac{2ab}{A} + 4\left( {b + \sqrt {a^{2} + b^{2} } } \right)\frac{4ab}{A} + 6\left( {b + \sqrt {a^{2} + b^{2} } } \right)\frac{6ab}{A} \hfill \\ \qquad+ 2\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } } \right)\hfill \\ \qquad\times\left[ {\frac{a(h - (m - 1)b) + b(r - (m - 1)a) + (m - 2)ba}{A}} \right] \hfill \\ \quad= \frac{8ab}{A}\left( {b + \sqrt {a^{2} + b^{2} } } \right) + \frac{32ab}{A}\left( {b + \sqrt {a^{2} + b^{2} } } \right) + \frac{72ab}{A}\left( {b + \sqrt {a^{2} + b^{2} } } \right) \hfill \\ \qquad+ \frac{128ab}{A}\left( {b + \sqrt {a^{2} + b^{2} } } \right) + 4\left[ {\frac{a(h - (m - 1)b) + b(r - (m - 1)a) + (m - 2)ba}{A}} \right]\hfill \\ \qquad\times\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } } \right) \hfill \\ \end{gathered} $$

Consequently,

$$ \begin{gathered} E(t(\varphi )) = \frac{{8ab\left( {b + \sqrt {a^{2} + b^{2} } } \right)}}{A}\left[ {1 + 4 + 9 + 16 + 25 + ...} \right] \hfill \\ \qquad+ 4\left[ {\frac{a(h - (m - 1)b) + b(r - (m - 1)a) + (m - 2)ba}{A}} \right]\hfill \\ \qquad \times\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } } \right) \hfill \\ \quad = \frac{1}{A}\Big( {8ab(b + \sqrt {a^{2} + b^{2} } )\sum\limits_{i = 1}^{m - 1} {i^{2} } } + 4[a(h - (m - 1)b) + b(r - (m - 1)a) + (m - 2)ba].\hfill \\ \qquad\times\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } } \right). \hfill \\ \end{gathered} $$

Proof of Theorem 2

From (1),

$E(t(\varphi )) = \frac{1}{A}\left( {8ab(b + \sqrt {a^{2} + b^{2} } )\sum\limits_{i = 1}^{m - 1} {i^{2} } } \right. + 4[a(h - (m - 1)b) + b(r - (m - 1)a) + (m - 2)ba].\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } } \right)$ By differentiation with respect to $a$,

$$ \begin{gathered} \frac{\partial E(t(\varphi ))}{{\partial a}} = \frac{1}{A}\left( \left( {8b(b + \sqrt {a^{2} + b^{2} } ) + \frac{{8a^{2} b}}{{\sqrt {a^{2} + b^{2} } }}} \right)\sum\limits_{i = 1}^{m - 1} {i^{2} } + 4[h - mb)].\right.\hfill \\ \qquad\times\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } } \right) \hfill \\ \qquad+ 4[h - mb)].\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } } \right) \hfill \\ \end{gathered} $$

(9.a)

By differentiation with respect to $b$,

$$ \begin{gathered} \frac{\partial E(t(\varphi ))}{{\partial b}} = \frac{1}{A}\left( \left( {8a(b + \sqrt {a^{2} + b^{2} } ) + 8ab\left( {1 + \frac{b}{{\sqrt {a^{2} + b^{2} } }}} \right)} \right)\sum\limits_{i = 1}^{m - 1} {i^{2} } + 4[r - ma)].\right. \hfill \\ \quad\times\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } } \right) \hfill \\ \quad+ 4[r - ma)].\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } } \right) \hfill \\ \end{gathered} $$

(9.b)

From (9.a) and (9.b) we have,

$$ \begin{gathered} \frac{{b\left( {a^{2} + b^{2} \sqrt {a^{2} + b^{2} } + b\left( {a^{2} + b^{2} } \right)} \right)m(m - 1)(2m - 1)}}{{3\sqrt {a^{2} + b^{2} } }} \hfill \\ \quad+ \left[ {a(h - (m - 1)b) + b( - a(m - 1) + r) + (m - 2)ba} \right].\hfill \\ \quad\times\left( {\frac{a}{{\sqrt {a^{2} + (h - (m - 1)b)^{2} } }} + \frac{(m - 2)a}{{\sqrt {a^{2} + b^{2} } }} + \frac{(r - (m - 1)a)(1 - m)}{{\sqrt {b^{2} + (r - (m - 1)a)^{2} } }}} \right) \hfill \\ \quad+ [h - mb)].\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } } \right) \hfill \\ \quad- \frac{{(ab^{2} + 2ab\sqrt {a^{2} + b^{2} } + a(a^{2} + b^{2} ))m(m - 1)(2m - 1)}}{{3\sqrt {a^{2} + b^{2} } }} \hfill \\\quad - [a(h - (m - 1)b) + b(r - (m - 1)a) + (m - 2)ba].\hfill \\ \quad\times\left( {\frac{(h - (m - 1)b)(1 - m)}{{\sqrt {a^{2} + (h - (m - 1)b)^{2} } }} + \frac{(m - 2)b}{{\sqrt {a^{2} + b^{2} } }} + \frac{b}{{\sqrt {b^{2} + (r - (m - 1)a)^{2} } }}} \right) \hfill \\ \quad+ [r - ma)].\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } } \right) = 0. \hfill \\ \end{gathered} $$

This leads to,

$$ \begin{gathered} \frac{{m(m - 1)(2m - 1)\left( {\left( {ab(a - b} \right) + (1 - 2a)b\sqrt {a^{2} + b^{2} } + (b^{2} - a)(a^{2} + b^{2} ) + b^{3} } \right)}}{{3\sqrt {a^{2} + b^{2} } }} \hfill \\ + [r - ma)].\left( {h + \sqrt {a^{2} + (h - (m - 1)b)^{2} } + (m - 2)\sqrt {a^{2} + b^{2} } + \sqrt {b^{2} + (r - (m - 1)a)^{2} } } \right) = 0. \hfill \\ + (h - mb)(r - ma)\left( {h + \sqrt {a^{2} + (h - b(m - 1))^{2} } + (m - 2)\left( {\sqrt {a^{2} + b^{2} } } \right) + \sqrt {b^{2} + (r - a(m - 1))^{2} } } \right) = 0. \hfill \\ \end{gathered} $$