1. Amnon Ta-Shma Uri Zwick Tel Aviv University Deterministic Rendezvous, Treasure Hunts and Strongly Universal Exploration Sequences 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 unknown environment 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 Vertices are undistinguishable Exits are numbers. No consistently assumed 1 3 2 2 1 2 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 = σ 1 σ 2 σ 3…  {0,1,2,3}*. In the i-th step, if σ i =0, stay in place, otherwise, take exit no. σ i. 1 3 2 2 1 2 2 σ = 3202… 2

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

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) Kowalski, Malinowski (2006) Rendezvous after O * (n 5 (τl) 1/2 +n 10 l) steps Rendezvous after O * (n 15+ l 3 ) steps when backtracking allowed Our result Rendezvous after O * (n 5 l) 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 steps and polynomial time solution in the backtracking model

9. Symmetry breaking 1 2 2 1 1 2 2 2 1 1 1 2 If the two robots are identical, no deterministic solution is possible To break the symmetry, the robots are assigned distinct labels L 1 and L 2. 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(n 3 ) Coppersmith, Tetali, Winkler (1993) Alternatively, one of the robots performs a random walk while the other stays put.

11. Universal Traversal Sequences (UTS) A sequence σ  {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 σ covers the graph. Aleliunas, Karp, Lipton, Lovasz, Rackoff (1979) A random sequence of length O(d 2 n 3 log n) is, with high probability, a UTS for d-regular graphs of size n. No efficient construction known!

12. A natural solution - That doesn’t quite work… Robot 0 stays put Robot 1 executes U 1 U 2 U 4 …U 2 k … where U n is a UTS for graphs of size n Fails as robot 0 may be activated when robot 1 is executing U k, where k  >> n. Robot 0 activated

13. Strongly Universal Traversal Sequences (SUTS) An infinite sequence σ  {1,2,3} ω is a SUTS for (cubic) graphs with cover time p(n), if for any n≥1, any contiguous subsequence of σ of 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. In the memoryless model, equivalent to the existence of SUTS. Treasure may be activated after the seeking robot. We obtain an efficient solution when backtrackings are allowed.

15. Solution of Rendezvous problem A robot with label uses the sequence: where U n is a UTS for graphs of size n

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

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

18. Strongly Universal Exploration Sequences (SUES) Suppose that U n =u 1 u 2 …u n is a UES for graphs of size n α, for 0<α<1. Suppose that U 2 i is a prefix of U 2 j for i<j. Theorem:  S n is a SUES

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

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

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