Fibonacci helps to evacuate from a convex region in a grid network

Abstract

This study considers an evacuation problem where the evacuees try to escape to the boundary of an affected area, which is convex, and a grid network is embedded in the area. The boundary is unknown to the evacuees and we propose an online evacuation strategy based on the Fibonacci sequence. This strategy is proved to have a competitive ratio of 19.5, which is better than the best previously reported result of 21.

Appendix

1.1 The proof of Theorem 3.1

Situation 2: The point P is on the sides of the square $g_{4n+1}$.

Similar to situation 1, draw the points $A_1$, $B_1$, $C_1$, $D_1$, $E_1$, $F_1$, $G_1$, $H_1$, $P_1'$, $E_1'$ in the corresponding positions. The case when P is at the first point $A_1$ on $g_{4n+1}$ was considered in case 3 for situation 1. In addition, the case when P is on line segment $F_1G_1$ and $G_1B_1$ may be similar to the case where P is at the end point $B_1$ on $g_{4n+1}$. Therefore, we consider two cases in the following.

Case 1 P is on the line segment $A_1F_1$.

Similar to case 2 in situation 1, we consider the ratio of the increment ${{T({A_1}{F_1}')} \over {D({H_1}{E_1}')}}$. Thus, it follows that

$$\begin{aligned} {{{h_{4n}} + {h_{4n - 2}}} \over {{h_{4n - 2}}}}= & {} 1 + {{{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^4} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^4} \cdot \left( 1 - {{\beta }^{4n}}\right) } \over {1 - {{\beta }^4}}} + 1} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^2} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^2} \cdot \left( 1 - {{\beta }^{4n}}\right) } \over {1 - {{\beta }^4}}}}}\\< & {} 1 + {{{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^4} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} + 1} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^2} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^2}} \over {1 - {{\beta }^4}}}}} \\= & {} 1 + {\alpha ^2} + {{{{{{\alpha }^2} \cdot {{\beta }^2}} \over {1 - {{\beta }^4}}} + \sqrt{5} } \over {{{{{\alpha }^2} \cdot \left( {{\alpha }^{4n}} - 1\right) } \over {{{\alpha }^4} - 1}} - {{{{\beta }^2}} \over {1 - {{\beta }^4}}}}}\\\le & {} 1 + {\alpha ^2} + {{{{{{\alpha }^2} \cdot {{\beta }^2}} \over {1 - {{\beta }^4}}} + \sqrt{5} } \over {{{{{\alpha }^2} \cdot \left( {{\alpha }^4} - 1\right) } \over {{{\alpha }^4} - 1}} - {{{{\beta }^2}} \over {1 - {{\beta }^4}}}}}\\< & {} 19.5. \end{aligned}$$

Therefore, $R(FSS)<19.5$ in this case by Lemma 3.

Case 2 P is at the end point $B_1$ on $g_{4n+1}$.

Similarly, in this case, the competitive ratio for the FSS is

$$\begin{aligned} R(FSS) = {{T(O{B_1})} \over {D(O{C_1})}} = {{2{S_{4n + 1}} - 1} \over {{h_{4n - 2}}}}. \end{aligned}$$

Note that R(FSS) is equal to ${{175} \over 9}$ and less than 19.5 when $n=2$. Since

$$\begin{aligned} - {{2\left( 2 - \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {\beta ^{4n}} - 3 \le - {{2\left( 2 - \sqrt{5} \right) } \over {\sqrt{5} }}{\beta ^{12}} - 3< 0, \end{aligned}$$

for $n \ge 3$, then we have

$$\begin{aligned} R(FSS)= & {} {{{{2\left( 2 + \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {{\alpha }^{4n}} - {{2\left( 2 - \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {{\beta }^{4n}} - 3} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^2} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^2} \cdot \left( 1 - {{\beta }^{4n}}\right) } \over {1 - {{\beta }^4}}}}}\\\le & {} {{2\left( 2 + \sqrt{5} \right) \cdot {{\alpha }^{4n}}} \over {{{{{\alpha }^2} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {{{{\beta }^2} \cdot \left( 1 - {{\beta }^{4n}}\right) } \over {1 - {{\beta }^4}}}}}\\< & {} {{2\left( 2 + \sqrt{5} \right) \cdot {{\alpha }^{4n}}} \over {{{{{\alpha }^2} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {{{{\beta }^2}} \over {1 - {{\beta }^4}}}}}\\= & {} {{2\left( 2 + \sqrt{5} \right) } \over {{{{{\alpha }^2}} \over {1 - {{\alpha }^4}}} \cdot \left( {1 \over {{{\alpha }^{4n}}}} - 1\right) - {{{{\beta }^2}} \over {1 - {{\beta }^4}}} \cdot \left( {1 \over {{{\alpha }^{4n}}}} - 1\right) }}\\\le & {} {{2\left( 2 + \sqrt{5} \right) } \over {{{{{\alpha }^2}} \over {1 - {{\alpha }^4}}} \cdot \left( {1 \over {{{\alpha }^{12}}}} - 1\right) - {{{{\beta }^2}} \over {1 - {{\beta }^4}}} \cdot \left( {1 \over {{{\alpha }^{12}}}} - 1\right) }} \\< & {} 19.5. \end{aligned}$$

In conclusion, R(FSS) is less than 19.5 when $n \ge 2$.

Situation 3 The point P is on the sides of the square $g_{4n+2}$.

Similarly, we draw the points $A_2$, $B_2$, $C_2$, $D_2$, $E_2$, $F_2$, $G_2$, $H_2$, $P_2'$, $E_2'$ and we consider two cases for this situation in the following.

Case 1 P is on the line segment $A_2F_2$.

We consider the ratio of increment ${{T({A_2}{F_2}')} \over {D({H_2}{E_2}')}}$, which is equal to

$$\begin{aligned} {{{h_{4n + 1}} + {h_{4n - 1}}} \over {{h_{4n - 1}}}}= & {} 1 + {{{1 \over {\sqrt{5} }} \cdot {{\alpha \cdot (1 - {{\alpha }^{4n + 1}})} \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{\beta \cdot \left( 1 - {{\beta }^{4n + 1}}\right) } \over {1 - {{\beta }^4}}}} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^3} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^3} \cdot \left( 1 - {{\beta }^{4n}}\right) } \over {1 - {{\beta }^4}}}}}\\< & {} 1 + {{\left( {{\alpha }^{4n + 1}} - 1\right) - {{\beta \cdot \left( {{\alpha }^4} - 1\right) } \over {\alpha \cdot \left[ 1 - {{\beta }^4}\right] }}} \over {{{\alpha }^2} \cdot \left( {{\alpha }^{4n}} - 1\right) }}\\= & {} 1 + {\alpha ^{ - 1}} + {{\alpha - 1 - {{\beta \cdot \left( {{\alpha }^4} - 1\right) } \over {\alpha \cdot \left( 1 - {{\beta }^4}\right) }}} \over {{{\alpha }^2} \cdot \left( {{\alpha }^{4n}} - 1\right) }}\\\le & {} 1 + {\alpha ^{ - 1}} + {{(\alpha - 1) - {{\beta \cdot \left( {{\alpha }^4} - 1\right) } \over {\alpha \cdot \left( 1 - {{\beta }^4}\right) }}} \over {{{\alpha }^2} \cdot \left( {{\alpha }^4} - 1\right) }}\\< & {} 19.5. \end{aligned}$$

Thus, R(FSS) is less than 19.5 by Lemma 3.

Case 2 P is at the end point $B_2$ on $g_{4n+2}$.

Note that

$$\begin{aligned} R(FSS) = {{39} \over 2} = 19.5, \end{aligned}$$

when $n=1$. For $n \ge 2$, we obtain

$$\begin{aligned} R(FSS)= & {} {{2{S_{4n + 2}} - 1} \over {{h_{4n - 1}}}}\\= & {} {{{{2\left( 2 + \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {{\alpha }^{4n + 1}} - {{2(2 - \sqrt{5} )} \over {\sqrt{5} }} \cdot {{\beta }^{4n + 1}} - 3} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^3} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^3} \cdot \left( 1 - {{\beta }^{4n}}\right) } \over {1 - {{\beta }^4}}}}}\\< & {} {{{{2\left( 2 + \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {{\alpha }^{4n + 1}}} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^3} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}}}}\\= & {} 2\left( 2 + \sqrt{5} \right) \cdot {\alpha ^{ - 2}} \cdot \left( {\alpha ^4} - 1\right) \cdot \left( 1 + {1 \over {{{\alpha }^{4n}} - 1}}\right) \\\le & {} 2\left( 2 + \sqrt{5} \right) \cdot {\alpha ^{ - 2}} \cdot \left( {\alpha ^4} - 1\right) \cdot \left( 1 + {1 \over {{{\alpha }^8} - 1}}\right) \\< & {} 19.5. \end{aligned}$$

Therefore, R(FSS) is no more than 19.5 when $n \ge 1$.

Situation 4 The point P is on the sides of the square $g_{4n+3}$.

Again, we draw the points $A_3$, $B_3$, $C_3$, $D_3$, $E_3$, $F_3$, $G_3$, $H_3$, $P_3'$, $E_3'$ and we consider two cases in the following.

Case 1 P is on the line segment $A_3F_3$.

Similar to case 2 in situation 1, we consider the ratio of increment

$$\begin{aligned} {{T({A_3}{F_3}')} \over {D({H_3}{E_3}')}}= & {} {{{h_{4n}} + {h_{4n - 2}}} \over {{h_{4n - 2}}}}\\= & {} 1 + {{{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^2} \cdot \left( 1 - {{\alpha }^{4n + 2}}\right) } \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^2} \cdot \left( 1 - {{\beta }^{4n + 2}}\right) } \over {1 - {{\beta }^4}}}} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^4} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^4} \cdot \left( 1 - {{\beta }^{4n}}\right) } \over {1 - {{\beta }^4}}} + 1}}. \end{aligned}$$

Note that ${ - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^4} \cdot (1 - {{\beta }^{4n}})} \over {1 - {{\beta }^4}}} + 1}$ is a positive value, as proved above, and the formula is

$$\begin{aligned} R(increment)< & {} 1 + {{{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^2} \cdot \left( 1 - {{\alpha }^{4n + 2}}\right) } \over {1 - {{\alpha }^4}}}} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^4} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}}}}\\= & {} 2 + {{{{\alpha }^2} - 1} \over {{{\alpha }^2} \cdot \left( {{\alpha }^{4n}} - 1\right) }} \le 2 + {{{{\alpha }^2} - 1} \over {{{\alpha }^2} \cdot \left( {{\alpha }^4} - 1\right) }}\\< & {} 19.5. \end{aligned}$$

Therefore, we find that R(FSS) is less than 19.5 by Lemma 3.

Case 2 P is at the end point $B_3$ on $g_{4n+3}$.

In this case, R(FFS) is equal to 16.25 when $n=1$. When $n \ge 2$, the ratio is

$$\begin{aligned} R(FSS)= & {} {{2{S_{4n + 3}} - 1} \over {{h_{4n}}}} \\= & {} {{{{2\left( 2 + \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {{\alpha }^{4n + 2}} - {{2\left( 2 - \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {{\beta }^{4n + 2}} - 3} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^4} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^4} \cdot \left( 1 - {{\beta }^{4n}}\right) } \over {1 - {{\beta }^4}}} + 1}}. \end{aligned}$$

Note that for $n \ge 2$,

$$\begin{aligned} - {{2\left( 2 - \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {\beta ^{4n + 2}} - 3 \le - {{2\left( 2 - \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {\beta ^{10}} - 3 < 0, \end{aligned}$$

and

$$\begin{aligned} - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^4} \cdot \left( 1 - {{\beta }^{4n}}\right) } \over {1 - {{\beta }^4}}} + 1 > - {1 \over {\sqrt{5} }} \cdot {{{{\beta }^4}} \over {1 - {{\beta }^4}}} + 1 = 0.9236 > 0. \end{aligned}$$

Thus, the solution to the formula above is less than

$$\begin{aligned} R(FSS)= & {} {{{{2\left( 2 + \sqrt{5} \right) } \over {\sqrt{5} }} \cdot {{\alpha }^{4n + 2}}} \over {{1 \over {\sqrt{5} }} \cdot {{{{\alpha }^4} \cdot \left( 1 - {{\alpha }^{4n}}\right) } \over {1 - {{\alpha }^4}}}}}\\= & {} 2\left( 2 + \sqrt{5} \right) \cdot {\alpha ^{ - 2}} \cdot \left( {\alpha ^4} - 1\right) \cdot \left( 1 + {1 \over {{{\alpha }^{4n}} - 1}}\right) \\< & {} 19.5 \end{aligned}$$

Therefore, R(FSS) is less than 19.5 when $n \ge 1$.

In summary, the competitive ratio for FSS is no more than 19.5 in this situation.

The Theorem follows.