Average Case - Worst Case Tradeoffs for Evacuating 2 Robots from the Disk in the Face-to-Face Model

Abstract

The problem of evacuating two robots from the disk in the face-to-face model was first introduced in [16], and extensively studied (along with many variations) ever since with respect to worst case analysis. We initiate the study of the same problem with respect to average case analysis, which is also equivalent to designing randomized algorithms for the problem. First we observe that algorithm $\mathscr {B}_{2}$ of [16] with worst case cost $\mathrm {Wrs}\left( \mathscr {B}_{2}\right) :=5.73906$ has average case cost $\mathrm {Avg}\left( \mathscr {B}_{2}\right) :=5.1172$. Then we verify that none of the algorithms that induced worst case cost improvements in subsequent publications has better average case cost, hence concluding that our problem requires the invention of new algorithms. Then, we observe that a remarkable simple algorithm, $\mathscr {B}_{1}$, has very small average case cost $\mathrm {Avg}\left( \mathscr {B}_{1}\right) :=1+\pi$, but very high worst case cost $\mathrm {Wrs}\left( \mathscr {B}_{1}\right) :=1+2\pi$. Motivated by the above, we introduce constrained optimization problem $_2{\textsc {Evac}}_{F2F}^w$, in which one is trying to minimize the average case cost of the evacuation algorithm given that the worst case cost does not exceed w. The problem is of special interest with respect to practical applications, since a common objective in search-and-rescue operations is to minimize the average completion time, given that a certain worst case threshold is not exceeded, e.g. for safety or limited energy reasons.

Our main contribution is the design and analysis of families of new evacuation parameterized algorithms $\mathscr {A}(p)$ which can solve $_2{\textsc {Evac}}_{F2F}^w$, for every $w \in [\mathrm {Wrs}\left( \mathscr {B}_{1}\right) ,\mathrm {Wrs}\left( \mathscr {B}_{2}\right) ]$. In particular, by letting parameter(s) p vary, we obtain parametric curve $\left( \mathrm {Avg}\left( \mathscr {A}(p)\right) , \mathrm {Wrs}\left( \mathscr {A}(p)\right) \right)$ that induces a continuous and strictly decreasing function in the mean-worst case space, and whose endpoints are $\left( \mathrm {Avg}\left( \mathscr {B}_{1}\right) , \mathrm {Wrs}\left( \mathscr {B}_{1}\right) \right)$, $\left( \mathrm {Avg}\left( \mathscr {B}_{2}\right) , \mathrm {Wrs}\left( \mathscr {B}_{2}\right) \right)$. Notably, the worst case analysis of the problem, since it’s introduction, has been relying on technical numerical, computer-assisted, calculations, following tedious robots trajectories’ analysis. Part of our contribution is a novel systematic procedure, which, given any evacuation algorithm, can derive it’s worst and average case performance in a clean and unified way.

K. Georgiou—Research supported in part by NSERC Discovery Grant.

A Appendix

1.1 A.1 Observation 3

Proof

(Observation 3). Note that it takes time $\pi$ to search the entire circle, and that the two trajectories are symmetric with respect to horizontal axis. Therefore, we may assume that the instance $\mathrm {cycle}(x)$ satisfies $x\in [0,\pi ]$.

Clearly, for any such x, we have that $\mathcal {S}(x)=x$. By Lemma 2, we have that $\mathcal {C}(x)=1+\mathcal {S}(x)+2\mathcal {E}(x) =1+x+2\delta _x$. Numerical calculations (software assisted) show that

$$\begin{aligned}&\mathrm {Wrs}\left( \mathscr {B}_{1}\right) = \sup _{x\in [0,\pi ]} \{ \mathcal {C}(x) \} = \sup _{x\in [0,\pi ]} \{ 1+x+2\delta _x \} \approx 5.73906, \\&\mathrm {Avg}\left( \mathscr {B}_{1}\right) = \mathrm {E}_{x\in [0,\pi ]}[\mathcal {C}(x)] =\frac{1}{\pi } \int _{x=0}^{\pi }\left( 1+x+2\delta _x \right) dx \approx 5.1172. \end{aligned}$$

   $\square$

1.2 A.2 Observation 4

Proof

(Observation 4). It is easy to see that for all $x\in [0,2\pi )$ we have $\bar{t}(x)=\mathcal {S}(x)=x$ and $\mathcal {E}(x)=0$. Therefore $\mathcal {C}(x)=1+x$, and hence

$$\begin{aligned}&\mathrm {Wrs}\left( \mathscr {B}_{2}\right) = \sup _{x\in [0,2\pi )} \{ \mathcal {C}(x) \} = 1+2\pi , \\&\mathrm {Avg}\left( \mathscr {B}_{2}\right) = \mathrm {E}_{x\in [0,2\pi )}[\mathcal {C}(x)] = \int _{x=0}^{2\pi }\left( 1+x\right) dx = 1+\pi . \end{aligned}$$

   $\square$

1.3 A.3 Lemma 4

Proof

(Lemma 4). First it is easy to show that the worst case evacuation time is induced either when $\mathscr {R}_{1}$ finds the exit while moving from $\mathrm {cycle}(0)$ to $\mathrm {cycle}(\alpha )$, or while $\mathscr {R}_{1}, \mathscr {R}_{2}$ are exploring the circle together (after having met). By Lemma 2, the cost in the first case would be

$$ \max _{0\le x \le \alpha }\{1+x+2\delta _x\} = \left\{ \begin{array}{ll} 1+\alpha +2\delta _\alpha , &{} \text {if}~\alpha \le \alpha _0 \\ \mathrm {Wrs}\left( \mathscr {B}_{1}\right) , &{} \text {otherwise} \end{array} \right. $$

where the values of the piecewise function above follow from Lemma 3. In the other case, the worst placement of exit is obtained using instances $\mathrm {cycle}(\alpha +\epsilon )$ for arbitrary small values of $\epsilon >0$ in which case the evacuation cost becomes $1+2\pi -\alpha$.

Overall, is easy to see that $1+\alpha _0 +2\delta _{\alpha _0} \le 1+2\pi -\alpha _0$ showing that the dominant evacuation cost when $\alpha \le \bar{\alpha }$ is $1+2\pi -\alpha$. For $\alpha >\bar{\alpha }$ the evacuation cost becomes equal to $w_{1}$.    $\square$

1.4 A.4 Lemma 5

Proof

(Lemma 5). We distinguish three cases as to where the exit is. If $x\in [0,\alpha )$, then the worst instance $\mathrm {cycle}(x)$ is when $x=\alpha -\epsilon$ for arbitrarily small $\epsilon >0$, and the cost is $1+\alpha +2\delta _{\alpha /2}$. In the second case $x \in [\alpha , 2\alpha +\delta _{\alpha /2})$ and it is not difficult to see that the worst case induced cost in this case is not more than that of the first case. Finally, in the third case $x\in [ 2\alpha +\delta _{\alpha /2}, 2\pi )$, and the two robots move together, so the total cost, in the worst case, is $1+2\pi -\alpha$, when $x=2\pi -\epsilon$ for arbitrarily small $\epsilon >0$. It is not difficult to see that the dominant case is actually the third one, and in fact the two cases induce the same cost when $\pi =\alpha +\delta _{\alpha /2}$. By the definition of $\delta _{\alpha /2}$ we know that $\delta _{\alpha /2}=2\sin \left( {\frac{\alpha +\delta _{\alpha /2}}{2}}\right) =2\sin \left( {\pi /2}\right) =2$. Hence the costs become equal when $\alpha =\pi -2$.    $\square$

1.5 A.5 Lemma 6

Proof

(Lemma 6). Let $w(\beta ) = 1+\pi - \beta /2+2\cos \left( {\beta /4}\right)$. First we show that $w(\beta )$ is the worst case performance of $\mathscr {A}_{2}'(\alpha _\beta , \beta )$ for two specific placements of the exit.

We proceed by describing evacuation cost $\mathcal {C}(x)$ assuming two arbitrary $\alpha ,\beta$ for two different instances $\mathrm {cycle}(x)$. Using Lemma 2, we see that

$$\begin{aligned} \lim _{\epsilon \rightarrow 0^+} \mathcal {C}(\alpha -\epsilon ) = 1+ \lim _{\epsilon \rightarrow 0^+} \mathcal {S}(\alpha -\epsilon ) +2\lim _{\epsilon \rightarrow 0^+} \mathcal {E}(\alpha -\epsilon ) = 1+\alpha +2\delta _{\alpha /2}. \end{aligned}$$

(3)

Since the total search time is $2\pi -\alpha -\beta$, we also see that

$$\begin{aligned} \lim _{\epsilon \rightarrow 0^+} \mathcal {C}(2\pi -\epsilon )=1+2\pi -\alpha -\beta . \end{aligned}$$

(4)

Now we claim that (3), (4) are equal when $\alpha =\alpha _\beta$. Indeed, equating (3), (4) gives

$$\begin{aligned} a+\delta _{\alpha /2}=\pi -\beta /2. \end{aligned}$$

(5)

But then, using (2), we see that

$$\begin{aligned} \delta _{\alpha /2}=2\sin \left( {\frac{\alpha +\delta _{\alpha /2}}{2}}\right) =2\sin \left( {\frac{\pi -\beta /2}{2}}\right) =2\cos \left( {\beta /4}\right) . \end{aligned}$$

(6)

Substituting (6) into (5), we see that the value of $\alpha$ for which (3), (4) are equal satisfies $\alpha = \pi - \beta /2 - 2\cos \left( {\beta /4}\right)$, as promised. Substituting this special value of $\alpha =\alpha _\beta$ either in (3) or in (4) induces evacuation cost $w(\beta )=1+\pi - \beta /2+2\cos \left( {\beta /4}\right)$.

Next we show that as long as $\beta$ is not too big, $w(\beta )$ is indeed the worst case evacuation cost. We consider the following cases $x\in I_i$, $i=1,\ldots ,4$ for possible instances $\mathrm {cycle}(x)$; $I_1:=[0,\alpha ), I_2:=[\alpha , 2\alpha +\beta +\delta _{(\alpha +\beta )/2}), I_3:=[2\alpha +\beta +\delta _{(\alpha +\beta )/2}, 2\alpha +2\beta +\delta _{(\alpha +\beta )/2}+\delta _{\beta /2}), I_4:=[2\alpha +2\beta +\delta _{(\alpha +\beta )/2}+\delta _{\beta /2}, 2\pi ).$ Clearly, (3), (4) demonstrate the worst case evacuation costs for instances in $I_1, I_4$, respectively, and the cost in both cases, for $\alpha =\alpha _\beta$ is equal to $w(\beta )$.

If $x\in I_2$ then $\mathcal {C}(x)=1+\mathcal {S}(x)+2\mathcal {E}(x)$. It is easy to see that both $\mathcal {S}(x), \mathcal {E}(x)$ are monotone in $I_2$, so the worst case evacuation in this case is

$$\begin{aligned} \lim _{\epsilon \rightarrow 0^+} \mathcal {C}(2\alpha _\beta +\beta +\delta _{(\alpha _\beta +\beta )/2}-\epsilon )= 1+\alpha _\beta +\beta +\delta _{(\alpha _\beta +\beta )/2}+2\delta _{\beta /2}. \end{aligned}$$

(7)

Denote $\delta _{\beta /2}$ satisfying (2) by $\delta _\beta '$. Using (2) and the definition of $\alpha _\beta$, we see that

$$ \delta _{(\alpha _\beta +\beta )/2} =2\sin \left( {\frac{\alpha _\beta +\beta +\delta _{(\alpha _\beta +\beta )/2}}{2}}\right) = 2\cos \left( {\cos \left( {\beta /4}\right) - \beta /4- \delta _{(\alpha _\beta +\beta )/2}}\right) $$

For simplicity, we denote $\delta _{(\alpha _\beta +\beta )/2}$ that satisfies the equation above by $\delta _\beta ''$. Then, continuing from (7), the worst case evacuation cost when $x\in I_2$ becomes $1+\pi +\beta /2-2\cos \left( {\beta /4}\right) +\delta _\beta ''+2\delta _\beta '$, an expression that depends exclusively on $\beta$. The latter cost is no more than $w(\beta )$ if and only if $4\cos \left( {\beta /4}\right) -\beta -\delta _\beta ''-2\delta _\beta ' \ge 0$, and numerically we verify that this is satisfied as long as $\beta \le \beta _0$ (see also Fig. 5).

Fig. 5.

The behavior of expression $4\cos \left( {\beta /4}\right) -\beta -\delta _\beta ''-2\delta _\beta '$, for $\beta =0,\ldots , 0.8$.

Finally, it is easy to verify that $\delta _{\beta /2}$ and $|I_4|$ are increasing and decreasing respectively for $\beta \le \beta _0$ and that $\delta _{\beta _0/2}=0.977997\le 1.01099 = |I_4|$ (for $\beta =\beta _0$). As a result, the worst case evacuation cost of case $x\in I_3$ cannot exceed that of case $x\in I_4$, and hence the lemma follows.    $\square$

1.6 A.6 Theorem 6

Proof

(Theorem 6). The claims for the worst case performances of $\mathscr {A}_1, \mathscr {A}_2', \mathscr {A}_2$ follow directly from Lemmata 4, 6 and 5, respectively. Next we argue that as the parameters vary in their specified range, we obtain the entire spectrum of $w\in [w_1,w_2]$, and this for unique values of the parameters. For this, we will rely on that for all evacuation algorithm families, the worst case cost is monotone in the parameters.

First, we argue about $\mathscr {A}_1$. We observe that by the definition of $\bar{\alpha }$, $\mathrm {Wrs}\left( \mathscr {A}_1(\bar{\alpha })\right) =w_1$, and $\mathrm {Wrs}\left( \mathscr {A}_1(1)\right) =1+2\pi -1=2\pi$. Together with the fact that $v(\alpha )$ is strictly decreasing, we see that $\mathrm {Wrs}\left( \mathscr {A}_1(\alpha )\right)$ is 1-1 and onto to $[w_1,2\pi ]$ as $\alpha$ ranges in $[1, \bar{\alpha }]$.

Second, we study $\mathscr {A}_2'$ whose worst case cost $v_2(\beta )$ is strictly decreasing in $\beta$. Moreover, by definition of $\beta _0$, we have $\mathrm {Wrs}\left( \mathscr {A}_2(\alpha _{\beta _0},\beta _0)\right) =w_0$. Then we note that for $\beta = 0$, $\mathscr {A}_2(\alpha _\beta ,\beta )$ coincides with $\mathscr {A}_2(\pi -2)$, and in particular the induced worst case cost becomes $3+\pi$. Therefore $\mathrm {Wrs}\left( \mathscr {A}_2'(\alpha _\beta ,\beta )\right)$ is 1-1 and onto to $[w_0,3+\pi ]$ as $\beta$ ranges in $[0, \beta _0]$.

Third, we study $\mathscr {A}_2$, for which we know that $\mathrm {Wrs}\left( \mathscr {A}_2(\pi -2)\right) =3+\pi$. Again, the worst case cost is monotone in $\alpha$ and $\mathscr {A}_2(0)$ coincides with benchmark algorithm $\mathscr {B}_{2}$, that is $\mathrm {Wrs}\left( \mathscr {A}_2(0)\right) =w_2$. Hence, $\mathrm {Wrs}\left( \mathscr {A}_2(\alpha )\right)$ is 1-1 and onto to $[3+\pi ,w_2]$ as $\alpha$ ranges in $[0,\pi -2]$.

Finally, we argue that

$$\begin{aligned} \mathrm {Avg}\left( \mathscr {A}_{1}(\alpha )\right) \le u_1(\alpha ),&\forall \alpha \in [1, \bar{\alpha }]\\ \mathrm {Avg}\left( \mathscr {A}_{2}'(\alpha _\beta ,\beta )\right) \le u_2'(\beta ),&\forall \beta \in [0, \beta _0]\\ \mathrm {Avg}\left( \mathscr {A}_{2}(\alpha )\right) \le u_2(\alpha ),&\forall \alpha \in [0, \pi -2] \end{aligned}$$

For this, we numerically compute $\mathrm {Avg}\left( \mathscr {A}_{1}(\alpha )\right) , \mathrm {Avg}\left( \mathscr {A}_{2}'(\alpha _\beta ,\beta )\right) , \mathrm {Avg}\left( \mathscr {A}_{2}(\alpha )\right)$ for various values of parameters $\alpha ,\beta$, and we heuristically choose $u_1, u_2',u_2$ so as to upper bound the average case performance of $\mathscr {A}_1, \mathscr {A}_2', \mathscr {A}_2$, effectively verifying our claim numerically. For each evacuation algorithm, we utilize Corollary 1, which together with the analytic description of our evacuation algorithms (see Definitions 4, 6, and 5) allow us to compute their average case performance using computer-assisted calculations. Our numerical calculations are depicted in Fig. 6.

Fig. 6.

On the right $u_1(\alpha )-\mathrm {Avg}\left( \mathscr {A}_{1}(\alpha )\right)$, for $\alpha '\le \alpha \le \bar{\alpha }$. In the middle, $u_2'(\beta )-\mathrm {Avg}\left( \mathscr {A}_{2}'(\alpha _\beta , \beta )\right)$, for $0\le \beta \le \beta _0$. On the right $u_2(\alpha )-\mathrm {Avg}\left( \mathscr {A}_{2}(\alpha )\right)$, for $0\le \alpha \le \pi -2$.

   $\square$