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.
Three wireless robots, one Byzantine
verifiedWith one Byzantine robot among three: between \(\approx 5.948\) and \(\approx 6.921\).
Assumed (added on the last edge are outlined)
Not assumed
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: three robots
- Three robots.
- Goal: all robots
- The task ends when the last robot reaches the exit.
- 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: Byzantine
- Up to \(f\) robots may lie: stay silent when they find the target, or claim to have found it where it is not.
- Speeds: equal speeds
- All robots move at most at unit speed.
- Randomness: deterministic
- Trajectories are fixed functions of the information received.
What is known
- Theorem 3(b)
- three robots, at most one Byzantine
Results, in order of publication
Introductory and current best (Theorem 2).
Open questions
None recorded for this variant.
Notes
Checked against the full text of Czyzowicz et al. 2017c. For \(k\) wireless robots of which one may be faulty, Leonardos, Pagourtzis and Papaioannou prove that symmetric-persistent algorithms (all robots go to the boundary and explore in the same direction until told of the exit) need time at least \(1+4\pi/k+2\sin(\pi/2-\pi/k)\) (Theorem 1, proved with a crash-faulty robot), and give a symmetric-persistent algorithm with time at most \(3+4\pi/k\) for one Byzantine robot (Theorem 3), which for \(k=3\) is about \(7.19\), above \(6.921\); with two Byzantine robots the time is at most \(3+6\pi/k\) for \(k \ge 9\) (Theorem 4); they also give face-to-face upper bounds (Theorems 5–7) Leonardos et al. 2023. Behrouz et al. give upper bounds for \(f\) Byzantine robots among \(k > 2f\) wireless robots (Theorem 1) and among \(k\) face-to-face robots (Theorem 2), and no lower bounds Behrouz et al. 2025.
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
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