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Deterministic Rendezvous and Strongly Universal Exploration Sequences

This paper discusses the rendezvous problem in an unknown environment and presents a polynomial time solution using deterministic sequences of instructions and strongly universal exploration sequences.

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Deterministic Rendezvous and Strongly Universal Exploration Sequences

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  1. Deterministic Rendezvous, Treasure HuntsandStrongly Universal Exploration Sequences Amnon Ta-Shma Uri Zwick Tel Aviv University TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AA

  2. The Rendezvous problem Two robots are activated, at different times and in different locations, in an unknownenvironment Should both follow deterministic sequences of instructions The instructions should guarantee that the two robots meet in a polynomial number of steps, after the activation of the second robot

  3. The unknown environment An undirected (cubic) graph. Graph and its size uknown to robots 1 2 2 Vertices are undistinguishable 3 2 Exits are numbers. No consistently assumed 1 No pebbles allowed Robots meet when they are in the same vertex at the same time Robots move synchronously

  4. Instructions – memoryless model Each robot gets a sequence σ = σ1 σ2 σ3… {0,1,2,3}*. In the i-th step, if σi=0, stay in place, otherwise, take exit no. σi. 1 2 2 3 2 1 2 2 σ = 3202…

  5. Instructions – backtracking model When a robot enters a vertex, it learns the number of the edge used to enter the vertex. 1 2 2 3 2 1 Action taken may now depend on entrance labels 2 2 In particular, backtracking is possible σ = 3202… ρ = 122…

  6. Parameters n – size of the environment l – length (in bits) of smaller label τ – difference in activation time

  7. Results Dessmark, Fraigniaud, Pelc (2003)Dessmark, Fraigniaud, Kowalski, Pelc (2006) Rendezvous after O*(n5(τl)1/2+n10l) steps Kowalski, Malinowski (2006) Rendezvous after O*(n15+l3) stepswhen backtracking allowed Our result Rendezvous after O*(n5l) steps

  8. Additional results The previous results guarantee a rendezvous after a polynomial number of steps. But, we do not know how to compute these steps in polynomial time… We give the first polynomial stepsand polynomial time solution in the backtracking model

  9. Symmetry breaking If the two robots are identical, no deterministic solutionis possible 1 2 1 2 1 2 To break the symmetry, the robots are assigned distinct labels L1 and L2. 2 1 1 2 2 1 In this talk we assume that the labels are 0 and 1.

  10. Randomized rendezvous If randomization is allowed, then achieving a rendezvous is easy: Each robot simply performs a random walk. Expected number of steps before the two robots meet is O(n3)Coppersmith, Tetali, Winkler (1993) Alternatively, one of the robots performs a random walk while the other stays put.

  11. Universal Traversal Sequences (UTS) A sequenceσ {1,2,3}* is a UTS for (cubic) graphs of size n if for every connected (cubic) graph of size at most n, every labeling and every starting point, the walk defined by σcoversthe graph. Aleliunas, Karp, Lipton, Lovasz, Rackoff (1979)A random sequence of length O(d2n3log n) is, with high probability, a UTS for d-regular graphs of size n. No efficient construction known!

  12. Robot 0 activated A natural solution - That doesn’t quite work… Robot 0 stays put Robot 1 executes U1U2U4…U2k…where Un is a UTS for graphs of size n Fails as robot 0 may be activated when robot 1 is executing Uk, where k>> n.

  13. Strongly Universal Traversal Sequences (SUTS) An infinite sequenceσ {1,2,3}ωis a SUTS for (cubic) graphs with cover time p(n), if for any n≥1, any contiguous subsequence of σof length p(n) is a UTS for (cubic) graphs of size n. Main open problem: Do SUTS exist?

  14. Treasure hunt The version of the rendezvous problem in which one of the robots is static. Treasure may be activated after the seeking robot. In the memoryless model, equivalent to the existence of SUTS. We obtain an efficient solution when backtrackings are allowed.

  15. Solution of rendezvous problem A robot with label uses the sequence: The k-th block of robot 1:Run U2k twice, then stay put for 2u2ksteps.Stay put for u2ksteps between any two steps above. Theorem: The two robots meet after O(|Un|4) steps, where n is the size of the environment. A more complicated solution guarantees a meeting afterO*(|Un|) steps, where Un are the best UTS currently known

  16. Universal Exploration Sequences (UES) Instructions are now interpreted as offsets 1 2 1 2 3 3 3 2 1 UES are analogous to UTS 2 Theorem: (Reingold ’05)UESs (of polynomial length) can be constructed in polynomial time. 2 σ = 1202…

  17. Strongly Universal ExplorationSequences (SUES) Suppose that Un=u1u2…un is a UES for graphs of size nα, for 0<α<1. Suppose that U2i is a prefix of U2j for i<j. Theorem:Sn is a SUES

  18. Open problems Strongly Universal Traversal Sequences??? Efficient construction of Universal Traversal Sequences ???

  19. Solution of rendezvous problem A robot with label uses the sequence:

  20. 0 0

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