franck petit inria, lip lab. univ. / ens of lyon france optimal probabilistic ring exploration by...
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Franck PetitINRIA, LIP Lab.
Univ. / ENS of LyonFrance
Optimal Probabilistic Ring Exploration by Semi-Synchronous Oblivious Robots
Joint work withStéphane Devismes, VERIMAG, Grenoble, France
Sébastien Tixeuil, Univ. Pierre et Marie Curie - Paris 6, France
Context
o A team of k “weak” robots evolving into a ring of n nodes
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o Autonomous
: No central authorityo Anonymous: Undistinguishable o Oblivious : No mean to know the past o Disoriented : No mean to agree on a
common direction or orientation
Context
o A team of k “weak” robots evolving into a ring of n nodes
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o Atomicity
: In every configuration, each robot is located at exactly one node o Multiplicit
y: In every configuration, each node contains zero, one, or more than one robot (every robot is able to detect it)
Context
o A team of k “weak” robots evolving into a ring of n nodes
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o SSM : In every configuration, k’ robots are activated (0 < k’ ≤ k)
1. Look : Instantaneous snapshot with multiplicity detection
o The k’ activated robots execute the cycle:
2. Compute
: Based on this observation, decides to either stay idle or move to one of the neighboring nodes
3. Move : Move toward its destination
Problem
o Exploration:Each node must be visited by at least one robot
o Termination:Eventually, every robot stays idle
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o Performance: Number of robots (k<n)
Starting from a configuration where no two robots are located at the same node:
Related works (Deterministic)
o Tree networks Ω(n) robots are necessary in generalA deterministic algorithm with O(log n/log log n) robots, assuming that Δ ≤ 3[Flocchini, Ilcinkas, Pelc, Santoro, SIROCCO 08]
o Ring networks Θ(log n) robots are necessary and sufficient, provided that n and k are coprimeA deterministic algorithm for k ≥ 17[Flocchini, Ilcinkas, Pelc, Santoro, OPODIS 07] 6F. Petit – SIROCCO 2009
Contribution
o n and k are not required to be coprime
1. Exploration impossible with less than 4 robots
2. An algorithm working with 4 probabilistic robots (n > 8)
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Theorem. 4 probabilistic robots are necessary and sufficient, provided that n > 8
Oblivious Robots
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At least one configuration that cannot be an initial configuration
Remark. If n > k, any terminal configuration of any protocol contains at least one tower.
Termination Exploration
Implicit memory
Tower
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Definition. A node with at least two robots.
k ≥ 2
Tower Building
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Cannot be a terminal configuration
Enabling Exploration
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k ≥ 3
Lemma. Every execution must contain a suffix of at least n–k+1 configurations containing a tower of less than k robots and any two of them are distinguishable.
Enabling Exploration
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Two undistinguishable
configurationsTwo other
undistinguishable configurations
Lemma. With 3 robots and a fixed tower of 2 robots, the maximum number of distinguishable configurations is equal to .
€
n
2
⎢ ⎣ ⎢
⎥ ⎦ ⎥
Enabling Exploration
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Lemma. For every n > 4, there exists no exploration protocol (even probabilistic) of a n-size ring with 3 robots.
Proof :
€
n
2
⎢ ⎣ ⎢
⎥ ⎦ ⎥≥ n − k +1⇒ n ≤ 4
Negative result
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Theorem. For every n ≥ 4, there exists no exploration protocol (even probabilistic) of a n-size ring with three robots.
Proof :There exists no protocol with 3 robots in a 4-size ring with a distributed scheduler.
Contribution
o n and k are not required to be coprime
1. Exploration impossible with less than 4 robots
2. Give an algorithm working with 4 probabilistic robots (n > 8)
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Theorem. 4 probabilistic robots are necessary and sufficient, provided that n > 8
Definitions
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Segment. A maximal non-empty elementary path of occupied nodes.
2 segments of length 1
a 2-segment
Definitions
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Hole. A maximal non-empty elementary path of free nodes.
1 hole of length 4
a 2-hole
Definitions
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Arrow. A 1-segment, followed by a non-empty elementary path of free nodes, a tower, and a 1-segment.
1 arrow
Head
of length 4
Tail
Definitions
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Arrow. A 1-segment, followed by a non-empty elementary path of free nodes, a tower, and a 1-segment.
final arrow
Definitions
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Arrow. A 1-segment, followed by a non-empty elementary path of free nodes, a tower, and a 1-segment.
Primary arrow
Algorithm
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o Initially, there is no tower
1. Converge toward a 4-segment2. Build a tower3. Visit the ring and terminate
0
0
If I am an internal node, then I try to move on the other internal node.
1
Algorithm
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o Initially, there is no tower
1. Converge toward a 4-segment2. Build a tower3. Visit the ring and terminate
Primary arrow
Algorithm
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o Initially, there is no tower
1. Converge toward a 4-segment2. Build a tower3. Visit the ring and terminate Final arrow
Primary arrow
Algorithm
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o Initially, there is no tower
1. Converge toward a 4-segment2. Build a tower3. Visit the ring and terminate
a) 3-segment
Final arrow
Primary arrow
If I am the isolated node, then I move through a shortest hole.
Algorithm
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o Initially, there is no tower
1. Converge toward a 4-segment2. Build a tower3. Visit the ring and terminate
a) 3-segmentb) a unique 2-segment
Final arrow
Primary arrow
If I am at the closest distance from the 2-segment, then I move toward the closest extremity.
Algorithm
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o Initially, there is no tower
1. Converge toward a 4-segment2. Build a tower3. Visit the ring and terminate
a) 3-segmentb) a unique 2-segmentc) two 2-segments
Final arrow
Primary arrow
If I am a neighbor of the longest hole, then I try to move toward the other 2-segment.
10
Algorithm
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o Initially, there is no tower
1. Converge toward a 4-segment2. Build a tower3. Visit the ring and terminate
a) 3-segmentb) a unique 2-segmentc) two 2-segmentsd) four isolated nodes
Final arrow
Primary arrow
L: length of the longest hole
If 4 robots are neighbors of an L-hole, then I try to move through my longest neighboring hole.
1
0
Algorithm
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o Initially, there is no tower
1. Converge toward a 4-segment2. Build a tower3. Visit the ring and terminate
a) 3-segmentb) a unique 2-segmentc) two 2-segmentsd) four isolated nodes
Final arrow
Primary arrow
L: length of the longest holeIf 3 robots are
neighbors of an L-hole, then if I am one of this 3 robots and a neighbor of a smaller hole h, then I move through h.
Algorithm
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o Initially, there is no tower
1. Converge toward a 4-segment2. Build a tower3. Visit the ring and terminate
a) 3-segmentb) a unique 2-segmentc) two 2-segmentsd) four isolated nodes
Final arrow
Primary arrow
L: length of the longest holeIf 2 robots are
neighbors of an L-hole, then if I am neighbor of the L-hole, then I move through the other neighboring hole.
Phase 1, Summary
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Proof
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Lemma. No tower is created during Phase 1 in a n-ring with n > 8.
Proof Base:With n > 8 and 4 robots, there always exists a hole of length greater than 1.
Proof
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Lemma. No tower is created during Phase 1 in a n-ring with n > 8.
Lemma. Starting from any initial configuration, the system reaches in finite expected time a configuration containing a 4-segment.
Theorem. The algorithm (Phases 1 to 3) is a probabilistic exploration protocol for 4 robots in a ring of n > 8 nodes.
Conclusion
o 4 probabilistic robots are necessary and sufficient, provided that n > 8
o Future works: Ad hoc solutions for n ≤ 8 (done) Convergence time Full asynchronous model
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Conclusion
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Thank you.