Amnon Ta-Shma Uri ZwickTel Aviv University

Deterministic Rendezvous, Treasure Hunts

andStrongly Universal Exploration Sequences

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

The unknown environment

An undirected (cubic) graph. Graph and its size uknown to robots

Vertices are undistinguishable

Exits are numbers. No consistently assumed

1

32

2

12

No pebbles allowed

Robots meet when they are in the same vertex at the same time

Robots move synchronously

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

322

12

2

σ = 3202…

2

Instructions – backtracking model

When a robot enters a vertex, it learns the number of the

edge used to enter the vertex.

1

322

12

2

σ = 3202…

2

ρ = 122…

Action taken may now depend on entrance labels

In particular, backtracking is possible

Parameters

n – size of the environment

l – length (in bits) of smaller label

τ – difference in activation time

Results

Dessmark, Fraigniaud, Pelc (2003)Dessmark, Fraigniaud, Kowalski, Pelc (2006)

Kowalski, Malinowski (2006)

Rendezvous after O*(n5(τl)1/2+n10l) steps

Rendezvous after O*(n15+l3) stepswhen backtracking allowed

Our resultRendezvous after O*(n5l) steps

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

Symmetry breaking

12

2

1

1

22

2

1

1

1

2

If the two robots are identical,

no deterministic solutionis possible

To break the symmetry, the robots are assigned distinct

labels L1 and L2.

In this talk we assume that the labels are 0 and 1.

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.

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!

A natural solution -That doesn’t quite work…

Robot 0 stays putRobot 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

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?

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.

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.

A more complicated solution guarantees a meeting after O*(|Un|) steps,

where Un are the best UTS currently known

Theorem: The two robots meet after O(|Un|4) steps, where n is the size of the environment.

Universal Exploration Sequences (UES)

Instructions are now interpreted as offsets

1

322

12

2

σ = 1202…

2

33

1

UES are analogous to UTS

Theorem: (Reingold ’05) UESs (of polynomial length)

can be constructed in polynomial time.

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

Open problems

Strongly Universal Traversal Sequences ???

Efficient construction of Universal Traversal

Sequences ???

Solution of rendezvous problem

A robot with label uses the sequence:

0 0