Who wins and how? Sasha Rubin Cornell REU 2009
Traditional Game Theory ◮ von Neumann, Morgenstern, Nash, ... ◮ focus on incomplete information eg. Poker Q. How should a player behave to maximise payoff? A. Use randomness in strategies. eg. coin-hiding
Chess Chess is not a game ... Chess is a well-defined form of computation. You may not be able to work out the answers, but in theory there must be a solution, a right move in any position. von Neumann ie. it is possible, in principle , to play a perfect game.
Chess . . . However, even at this figure there will be 10 120 variations to be calculated from the initial position. A machine operating at the rate of one variation per micro-second would require over 10 90 years to calculate the first move! Shannon ie. but we may never know the strategy, or even if the first player has a winning strategy.
Combinatorial Game Theory Chess, Go, Checkers, Tic-Tac-Toe, Hex, ... Two players move alternately Finitely many positions A sequence of moves always comes to an end (ie. finite plays) Outcome depends on the play and is one of (Win, Lose) (Lose, Win) (Draw, Draw) Complete information (about play history) No chance in the rules
Strategies Zermelo (1912): mathematical treatment of ‘best possible move’ Strategy: game history �→ next move.
Strategies Zermelo (1912): mathematical treatment of ‘best possible move’ Strategy: game history �→ next move. A strategy for player x is winning if no matter what moves the other player makes the resulting play is a win for player x .
Fundamental Algorithmic Questions Decide if a given player has a winning strategy. ◮ For combinatorial games the problem is decidable in principle. ◮ When is it practically decidable? Eg. Does either player have a winning strategy in 5 × 5 × 5 tic-tac-toe? There are 3 125 board positions.
Fundamental Algorithmic Questions Describe/compute a winning strategy. Eg. First player has a winning strategy in Hex (strategy stealing argument). Find an explicit winning strategy in Hex.
Infinite games Like combinatorial games. We will consider: ◮ plays are infinite, ◮ no draws and determined - exactly one player has a winning strategy. Careful. Not all win-lose games are determined!
Games on graphs Two players 0 and 1 and their vertices. Move a token, from a starting vertex, along the edges of a finite directed graph. Assume no dead ends. Traces out an infinite path, called a play π . Winning condition: Every play π is declared either a win for player 0 or a win for player 1. No drawing plays.
Classes of games We group games according to their winning conditions. Reachability: ◮ hit some vertex from a fixed set of vertices T .
Classes of games We group games according to their winning conditions. Reachability: ◮ hit some vertex from a fixed set of vertices T . Avoidance: ◮ do not hit any vertex from T .
Classes of games We group games according to their winning conditions. Reachability: ◮ hit some vertex from a fixed set of vertices T . Avoidance: ◮ do not hit any vertex from T . Repeated reachability: ◮ infinitely often hit some vertex from T .
Parity Games Parity games generalise repeated reachability. Players are called Even and Odd . Vertices have colours - these are natural numbers { 1 , · · · , d } . Winning condition: Even wins if the largest colour occuring infinitely often is even Odd wins if the largest colour occuring infinitely often is odd
Subcases Just one colour is trivial to analyse.
Subcases Just one colour is trivial to analyse. With two colours, { 1 , 2 } : Even wins if in the play, he hits nodes labelled 2 infinitely often. ie. Repeated Reachability
Subcases Just one colour is trivial to analyse. With two colours, { 1 , 2 } : Even wins if in the play, he hits nodes labelled 2 infinitely often. ie. Repeated Reachability With three colours, { 1 , 2 , 3 } : Even wins if 2 is hit infinitely often while 3 is hit only finitely often.
Subcases Just one colour is trivial to analyse. With two colours, { 1 , 2 } : Even wins if in the play, he hits nodes labelled 2 infinitely often. ie. Repeated Reachability With three colours, { 1 , 2 , 3 } : Even wins if 2 is hit infinitely often while 3 is hit only finitely often. With four colours, { 1 , 2 , 3 , 4 } : Even wins if 4 is hit infinitely often OR 2 is hit infinitely often while 3 is hit only finitely often.
Solving parity games Thm. ◮ Parity games are determined. ◮ The winner has a historyless strategy ie. depends only on the current node. The proof is constructive.
Solving parity games Thm. ◮ Parity games are determined. ◮ The winner has a historyless strategy ie. depends only on the current node. The proof is constructive. So there is an algorithm that given a parity game (coloured graph with starting vertex) outputs which player wins (& a winning strategy) This is called solving parity games. but . . .
Parity in P ? . . . the known algorithms are intractable/slow. A problem is tractable if there is an algorithm solving it and taking polynomially many steps (in the size of the input). The open problem: Is parity tractable? The answer is probably ‘Yes’.
Interesting? Solving parity games is equivalent to problems in ◮ logic (satisfaction of the µ -calculus). ◮ automata theory (non-emptiness of tree-automata). The problem is intruigingly in the complexity classes NP and co-NP and thus believed to be in P .
What to take away Parity games: 1. class of games 2. generalise repeated-reachability 3. infinite plays 4. determined (exactly one player has a winning strategy) 5. can be solved algorithmically. The goal is to show how to solve them fast.
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