We present robot games, and we give the simplest definitions
for which decidability is not known.
1. Definition.
Let U,V ⊆ ℤ2 be two finite sets of two-dimensional integer vectors.
A robot game is played in rounds from an initial configuration x0 ∈ ℤ2 as follows.
In each round, player 2 chooses a vector v ∈ V, then player 1 chooses a vector u ∈ U,
and the configuration in the next round is x+v+u where x is the configuration
in the current round. The objective of player 1 is to reach the configuration (0,0).
r50mm
(55,40)(0,0)
(5,-10)(10,0)5 [AHnb=0, linegray=.8](0,0)(0,50)
(0,-5)(0,10)5 [AHnb=0, linegray=.8](0,0)(50,0)
(35,-10)(35,40) (0,25)(50,25)
Nh=2, Nw=2 [Nmarks=n, Nmr=0, ExtNL=y, NLdist=1, NLangle=270](q)(5,-5)(−3,−3) [Nmarks=n, Nmr=2](r)(15,15) [Nmarks=n, Nmr=2](s)(25,-5) [Nmarks=n, Nh=1, Nw=1, Nmr=2, fillgray=0, NLdist=5, NLangle=32](t)(35,25)(0,0)
[ELpos=50, ELside=l, ELdist=.5, curvedepth=0](q,r) [ELpos=50, ELside=l, ELdist=.5, curvedepth=0](q,s) [ELpos=50, ELside=l, ELdist=.5, curvedepth=0](r,t) [ELpos=50, ELside=l, ELdist=.5, curvedepth=0](s,t)
A strategy for player 1 is a function σ: ℤ2 → U and a strategy for player 2 is a function π: ℤ2 → V. The play according to σ and π from initial configuration x0 is the infinite sequence x0 x1 … such that for all i≥ 0, we have xi+1 = xi + v + u where v = π(xi) and u = σ(xi + v).
A configuration x0 is winning for player 1 if there exists a strategy σ such that for all strategies π, in the resulting play from x0 there exists i ≥ 0 such that xi = (0,0).
2. Example. Let U = {(1,3),(2,1)} and V={(2,0),(1,2)}.
The initial configuration (−3,−3) is winning for player 1.
The set of winning configurations for player 1 is {(−3k,−3k) ∣ k ≥ 0}.
Note that the set of winning configurations is closed under sum (i.e., x+y is
a winning configuration if x and y are winning configurations).
3. Decision problem. Given an initial configuration x0 ∈ ℤ2
and two finite sets U,V ⊆ ℤ2, the problem is to decide whether x0
is a winning configuration in the robot game defined by U,V.
Whether this problem is decidable and what is its complexity are open questions.
The problem is undecidable if the game is played on a graph with states of player 1 and states of player 2, with ℤ2 or ℕ2 as the vector space (as in games on VASS, vector-addition systems with states) [1, 2].
The one-player version of robot games (i.e., where V={(0,0)}) is decidable in polynomial time by a reduction to linear programming. The robot games defined in one dimension (with U,V ⊆ ℤ and x0 ∈ ℤ) are also decidable.
4. Extension.
Extensions can be considered in several directions:
5. Partial results.
| Laurent Doyen |
| CNRS Researcher - LSV, ENS Cachan |
| 61, avenue du President Wilson |
| 94235 Cachan |
| Tel: +33 (0)1 47 40 22 74 |
| Email: doyen@lsv.ens-cachan.fr |
This document was translated from LATEX by HEVEA.