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 at the centroid, or at a given point of the boundary, 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; in weighted evacuation of two robots, a weighted average of their arrival times; in search, until a non-faulty robot has reached the exit and every non-faulty robot knows where it is).
Measures: Evacuation time, worst-case time until the last robot (or the queen) reaches the exit, or the weighted average or search time where the goal says so; unit disk, unit speed.
Line, weighted evacuation of two wireless robots
verifiedWhen the heavy robot's arrival counts \(w\) times the light one's, two wireless robots achieve exactly \(1+\sqrt{2w/(1+w)}\), between \(2\) and \(1+\sqrt2\).
Assumed (added on the last edge are outlined)
Not assumed
Formal definitions of this setting
- Domain: line
- The robots start at the origin of the real line; the exit is at an unknown point at distance at least \(1\); the measure is the competitive ratio, the evacuation time divided by that distance.
- Number of robots: two robots
- Two robots.
- Goal: weighted
- Two robots, a light one and a heavy one of known weight \(w \ge 1\), both must reach the exit and stay there; with arrival times \(t_1\) (light) and \(t_2\) (heavy), the cost is \((t_1 + w\,t_2)/(1+w)\).
- 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: equal speeds
- All robots move at most at unit speed.
- Randomness: deterministic
- Trajectories are fixed functions of the information received.
Statement for this variant
Two unit-speed robots start at the origin of the line; the exit is at unknown distance \(d \ge 1\). Both know which robot is heavy and the weight \(w \ge 1\). The cost is \((t_1 + w\,t_2)/((1+w)d)\) in the worst case, where \(t_1, t_2\) are the times from which the light and the heavy robot stay at the exit.
What is known
- Theorem 3.6
- identities and \(w\) known
Results, in order of publication
Introductory and current best (Theorems 3.1, 3.2). Without the identities, robots walking apart at unit speed achieve \(1+2w/(1+w)\) (Theorem 5.1).
Open questions
Weighted evacuation with more than two robots Georgiou & Lucier 2023.
Notes
For \(w = 1\) the cost is the average arrival time, ratio \(2\); as \(w\) grows the ratio tends to \(1+\sqrt2\), below the ratio \(3\) of evacuating both robots. Checked against the full text.
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
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.