1. Amnon Ta-Shma Uri Zwick
Deterministic Rendezvous, Treasure Hunts and Strongly Universal Exploration Sequences
Amnon Ta-Shma Uri Zwick
Tel Aviv University
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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 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. n – size of the environment l – length (in bits) of smaller label
Parameters
n – size of the environment
l – length (in bits) of smaller label
τ – difference in activation time

7. Kowalski, Malinowski (2006)
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) steps when backtracking allowed
Our result
Rendezvous after O*(n5l) steps

8. But, we do not know how to compute these steps in polynomial time…
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
If the two robots are identical, no deterministic solution is possible 1 2 2 1 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 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(d2n3log n) is, with high probability, a UTS for d-regular graphs of size n.
No efficient construction known!

12. That doesn’t quite work…
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.
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.
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 2u2k steps. Stay put for u2k steps 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 after O*(|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 2 3 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 Exploration Sequences (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. Strongly Universal Traversal Sequences ???
Open problems
Strongly Universal Traversal Sequences ???
Efficient construction of Universal Traversal Sequences ???

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