Abstract
Suppose that we want to patrol a fence (line segment) using k mobile agents with given speeds v1,..., vk so that every point on the fence is visited by an agent at least once in every unit time period. A simple strategy where the ith agent moves back and forth in a segment of length vi/2 patrols the length (v1 +· · ·+vk)/2, but it has been shown recently that this is not always optimal. Thus a natural question is to determine the smallest c such that a fence of length c(v1 + · · · + vk)/2 cannot be patrolled. We give an example showing c ≥ 4/3 (and conjecture that this is the best possible). We also consider a variant of this problem where we want to patrol a circle and the agents can move only clockwise. We can patrol a circle of perimeter rvr by a simple strategy where the r fastest agents move at the same speed, but it has been shown recently that this is not always optimal. We conjecture that this is not even a constant-approximation strategy. To tackle this conjecture, we relate it to what we call constant gap families. Using this relation, we give another example where the simple strategy is not optimal. We propose another variant where we want to patrol a single point under the constraint that for each i = 1,..., k, the time between two consecutive visits of agent i should be ai or longer. This problem can be reduced to the discretized version where the ai are integers and the goal is to visit the point at every integer time. It is easy to see that this discretized patrolling is impossible if 1/a1+· · ·+1/ak < 1, and that there is a simple strategy if 1/a1+· · ·+1/ak ≥ 2. Thus we are interested in the smallest c such that patrolling is always possible if 1/a1 + · · · + 1/ak ≥ c. We prove that α ≤ c < 1.546, where α = 1.264... (we conjecture that c = α). We also discuss the computational complexity of related problems.
| Original language | English |
|---|---|
| Title of host publication | Algorithms and Complexity - 9th International Conference, CIAC 2015, Proceedings |
| Editors | Peter Widmayer, Vangelis Th. Paschos |
| Publisher | Springer Verlag |
| Pages | 261-273 |
| Number of pages | 13 |
| ISBN (Print) | 9783319181721 |
| DOIs | |
| Publication status | Published - Jan 1 2015 |
| Event | 9th International Conference on Algorithms and Complexity, CIAC 2015 - Paris, France Duration: May 20 2015 → May 22 2015 |
Publication series
| Name | Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) |
|---|---|
| Volume | 9079 |
| ISSN (Print) | 0302-9743 |
| ISSN (Electronic) | 1611-3349 |
Other
| Other | 9th International Conference on Algorithms and Complexity, CIAC 2015 |
|---|---|
| Country | France |
| City | Paris |
| Period | 5/20/15 → 5/22/15 |
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All Science Journal Classification (ASJC) codes
- Theoretical Computer Science
- Computer Science(all)
Cite this
Simple strategies versus optimal schedules in multi-agent patrolling. / Kawamura, Akitoshi; Soejima, Makoto.
Algorithms and Complexity - 9th International Conference, CIAC 2015, Proceedings. ed. / Peter Widmayer; Vangelis Th. Paschos. Springer Verlag, 2015. p. 261-273 (Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics); Vol. 9079).Research output: Chapter in Book/Report/Conference proceeding › Conference contribution
}
TY - GEN
T1 - Simple strategies versus optimal schedules in multi-agent patrolling
AU - Kawamura, Akitoshi
AU - Soejima, Makoto
PY - 2015/1/1
Y1 - 2015/1/1
N2 - Suppose that we want to patrol a fence (line segment) using k mobile agents with given speeds v1,..., vk so that every point on the fence is visited by an agent at least once in every unit time period. A simple strategy where the ith agent moves back and forth in a segment of length vi/2 patrols the length (v1 +· · ·+vk)/2, but it has been shown recently that this is not always optimal. Thus a natural question is to determine the smallest c such that a fence of length c(v1 + · · · + vk)/2 cannot be patrolled. We give an example showing c ≥ 4/3 (and conjecture that this is the best possible). We also consider a variant of this problem where we want to patrol a circle and the agents can move only clockwise. We can patrol a circle of perimeter rvr by a simple strategy where the r fastest agents move at the same speed, but it has been shown recently that this is not always optimal. We conjecture that this is not even a constant-approximation strategy. To tackle this conjecture, we relate it to what we call constant gap families. Using this relation, we give another example where the simple strategy is not optimal. We propose another variant where we want to patrol a single point under the constraint that for each i = 1,..., k, the time between two consecutive visits of agent i should be ai or longer. This problem can be reduced to the discretized version where the ai are integers and the goal is to visit the point at every integer time. It is easy to see that this discretized patrolling is impossible if 1/a1+· · ·+1/ak < 1, and that there is a simple strategy if 1/a1+· · ·+1/ak ≥ 2. Thus we are interested in the smallest c such that patrolling is always possible if 1/a1 + · · · + 1/ak ≥ c. We prove that α ≤ c < 1.546, where α = 1.264... (we conjecture that c = α). We also discuss the computational complexity of related problems.
AB - Suppose that we want to patrol a fence (line segment) using k mobile agents with given speeds v1,..., vk so that every point on the fence is visited by an agent at least once in every unit time period. A simple strategy where the ith agent moves back and forth in a segment of length vi/2 patrols the length (v1 +· · ·+vk)/2, but it has been shown recently that this is not always optimal. Thus a natural question is to determine the smallest c such that a fence of length c(v1 + · · · + vk)/2 cannot be patrolled. We give an example showing c ≥ 4/3 (and conjecture that this is the best possible). We also consider a variant of this problem where we want to patrol a circle and the agents can move only clockwise. We can patrol a circle of perimeter rvr by a simple strategy where the r fastest agents move at the same speed, but it has been shown recently that this is not always optimal. We conjecture that this is not even a constant-approximation strategy. To tackle this conjecture, we relate it to what we call constant gap families. Using this relation, we give another example where the simple strategy is not optimal. We propose another variant where we want to patrol a single point under the constraint that for each i = 1,..., k, the time between two consecutive visits of agent i should be ai or longer. This problem can be reduced to the discretized version where the ai are integers and the goal is to visit the point at every integer time. It is easy to see that this discretized patrolling is impossible if 1/a1+· · ·+1/ak < 1, and that there is a simple strategy if 1/a1+· · ·+1/ak ≥ 2. Thus we are interested in the smallest c such that patrolling is always possible if 1/a1 + · · · + 1/ak ≥ c. We prove that α ≤ c < 1.546, where α = 1.264... (we conjecture that c = α). We also discuss the computational complexity of related problems.
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UR - http://www.scopus.com/inward/citedby.url?scp=84944725820&partnerID=8YFLogxK
U2 - 10.1007/978-3-319-18173-8_19
DO - 10.1007/978-3-319-18173-8_19
M3 - Conference contribution
AN - SCOPUS:84944725820
SN - 9783319181721
T3 - Lecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics)
SP - 261
EP - 273
BT - Algorithms and Complexity - 9th International Conference, CIAC 2015, Proceedings
A2 - Widmayer, Peter
A2 - Paschos, Vangelis Th.
PB - Springer Verlag
ER -