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Evacuation

Robots start at the centre of a disk (or a polygon) and must all leave through an exit at an unknown point of the boundary, as fast as possible.

Problem statement and measuresexpand

\(k\) unit-speed robots start together at the centre of a unit disk (or the centroid of a unit equilateral triangle or square). An exit lies at an unknown point of the boundary; a robot recognises it only on reaching it. Time is the worst case, over exit positions, of the time until the last robot reaches the exit (in priority evacuation, where one robot is the queen: until the queen does).

Measures: Evacuation time, worst-case time until the last robot (or the queen) reaches the exit; unit disk, unit speed.

no assumptions›+ disk›+ two robots›+ priority (queen), + wireless, + no faults

Disk, priority evacuation, queen and one servant

verified
Evacuation time: polynomial gap

A queen and one servant evacuate the queen in about \(4.819\), strictly faster than evacuating both; the lower bound is about \(4.568\).

Assumed (added on the last edge are outlined)

+Domainunit disk, start at the centre+Number of robotstwo robots+Goalpriority: only the queen must reach the exit+Communicationwireless: instant communication at any distance+Faultsno faults

Not assumed

Exitsone exitSpeedsrobots may have different maximum speedsRandomnessdeterministic
Formal definitions of this setting
Domain: disk
The unit disk; robots start at its centre; the exit is a point of the boundary circle.
Number of robots: two robots
Two robots.
Goal: priority (queen)
One designated robot (the queen) must reach the exit; the others (servants) only help; the task ends when the queen exits.
Exits: one exit
One exit, at an unknown point of the boundary.
Communication: wireless
Wireless: every robot can send messages to every other robot instantly at any distance.
Faults: no faults
Every robot detects and reports the target.
Speeds: distinct speeds
Each robot \(i\) has its own maximum speed \(v_i\), known to all.
Randomness: deterministic
Trajectories are fixed functions of the information received.

What is known

Evacuation time · worst-case time until the last robot (or the queen) reaches the exit; unit disk, unit speed
polynomial gap
About \(4.819\) against \(4.568\).
lower bound
\(\approx 4.56798\) Georgiou & Wang 2025
bottleneckunknown exit
holds when
  • Theorem 1
  • a linear-programming relaxation over polygon instances
introductory
\(\approx 4.81854\)
current best
\(\approx 4.81854\)

Results, in order of publication

Czyzowicz et al. 2018 introductory
Evacuation time \(\approx 4.81854\)

Theorem 8; also the first lower bound, about \(4.3896\) (Theorem 17).

Georgiou & Wang 2025 current best
Evacuation time \(\approx 4.81854\)

Same algorithm, with the improved lower bound.

Open questions

  1. Close the gap between \(4.568\) and \(4.819\) Georgiou & Wang 2025.

Notes

Weighted group search, where the two robots' arrival times are weighted \(1\) and \(w\), is solved within a gap in the same paper Georgiou & Wang 2025. Checked against the full texts.

What carries over from neighbouring settings

Read off the order of assumptions, not checked case by case: results may carry extra assumptions (shown on their pages), and asynchronous time is measured in epochs.

Algorithms from weaker settings work here

Disk, two wireless robots of different speeds
Evacuation time \(\tfrac{1+2\arccos(-2/v)}{v}+\sqrt{1-4/v^2}\)

Lower bounds from stronger settings hold here

No catalogued stronger setting.

Know a better result for this setting, or found a mistake? Suggest a result · Report an error. Accepted contributions are acknowledged.