Games
Miheer Dewaskar
Chennai Mathematical Institute April 27, 2016
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Games Miheer Dewaskar Chennai Mathematical Institute April 27, - - PowerPoint PPT Presentation
Games Miheer Dewaskar Chennai Mathematical Institute April 27, 2016 1 / 19 Outline Finite Duration Games Win-Lose Games Payoff Games Infinite Duration Games Parity Games Mean Payoff Games Simple Stochastic Games 2 / 19 Outline Finite
Miheer Dewaskar
Chennai Mathematical Institute April 27, 2016
1 / 19
Finite Duration Games Win-Lose Games Payoff Games Infinite Duration Games Parity Games Mean Payoff Games Simple Stochastic Games
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Finite Duration Games Win-Lose Games Payoff Games Infinite Duration Games Simple Stochastic Games
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Win-Lose game
Circle Wins Box Wins
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Win-Lose game
Circle Wins Box Wins
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Win-Lose game
Circle Wins Box Wins
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Win-Lose game
Circle Wins Box Wins
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Win-Lose game
Box wins
Circle Wins Box Wins
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Win-Lose game
Circle Wins Box Wins
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Win-Lose game
Circle wins
Circle Wins Box Wins
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Win-Lose game
Circle Wins Box Wins
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Win-Lose game
Algorithm for optimal play
Circle Wins Box Wins
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Win-Lose game
Algorithm for optimal play
Circle Wins Box Wins
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Win-Lose game
Algorithm for optimal play
Box can always win
Circle Wins Box Wins
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Payoff game
4
4 1
Maximizer Minimizer
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Payoff game
4
4 1
Maximizer Minimizer
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Payoff game
4
4 1
Maximizer Minimizer
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Payoff game
Payoff
Min pays 4 units to Max
4
4 1
Maximizer Minimizer
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Payoff game
4
4 1
Maximizer Minimizer
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Payoff game
Payoff
Min pays -1 units to Max
4
4 1
Maximizer Minimizer
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Payoff game
MinMax algorithm
4
4 1 1
Maximizer Minimizer
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Payoff game
MinMax algorithm
4 4
1
4 1 1
Maximizer Minimizer
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Payoff game
MinMax algorithm
Value = 1 Min can ensure a payoff ≤ 1 Max can ensure a payoff ≥ 1
1 4 4
1
4 1 1
Maximizer Minimizer
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Payoff game
MinMax algorithm
Value = 1 Min can ensure a payoff ≤ 1 Max can ensure a payoff ≥ 1 When both play optimally the payoff is exactly 1.
1 4 4
1
4 1 1
Maximizer Minimizer
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Finite Duration Games Infinite Duration Games Parity Games Mean Payoff Games Simple Stochastic Games
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Winning conditions
5 6 1 2 3 Odd Even
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Winning conditions
π1 =
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 =
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins π =
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins π = 1
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins π = 1 2
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins π = 1 2 3
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins π = 1 2 3 3
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins π = 1 2 3 3 6
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins π = 1 2 3 3 6 5
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins π = 1 2 3 3 6 5 2
5 6 1 2 3 Odd Even
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Winning conditions
π1 = 1 5 2 1 2 1 2 . . . inf(π1) = {1, 2} max Inf(π1) = 2 Even wins π2 = 1 5 2 1 5 2 . . . inf(π2) = {1, 2, 5} max Inf(π2) = 5 Odd wins π = 1 2 3 3 6 5 2 1 . . . Parity(max Inf(π)) wins
5 6 1 2 3 Odd Even
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Questions
Does either Even or Odd have a strategy to always win? If so, then how to compute the winning strategy?
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Questions
Does either Even or Odd have a strategy to always win? Yes If so, then how to compute the winning strategy? By reduction to finite duration games
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5 6 1 2 3 Odd Even 1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
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5 6 1 2 3 Odd Even 1 5 6 5 2 3 6 5 3 1 2 1 3 3 6 5 2 6 Odd Wins Even Wins
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5 6 1 2 3 Odd Even 1 5 6 5 2 3 6 5 3 1 2 1 3 3 6 5 2 6 Odd Wins Even Wins
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5 6 1 2 3 Odd Even 1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
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5 6 1 2 3 Odd Even
Finite game
Even has a winning strategy
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 Stack = 1
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 Stack = 1 2
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 Stack = 1 2 1
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 Stack = 1
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 Stack = 1 5
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 6 Stack = 1 5 6
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 6 5 Stack = 1 5 6 5
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 6 5 Stack = 1 5
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 6 5 2 Stack = 1 5 2
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 6 5 2 3 Stack = 1 5 2 3
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 6 5 2 3 6 Stack = 1 5 2 3 6
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 6 5 2 3 6 5 Stack = 1 5 2 3 6 5 Every eliminated cycle has max priority even
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 6 5 2 3 6 5 Stack = 1 5 . . . Every eliminated cycle has max priority even
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5 6 1 2 3 Odd Even
Finite game
Every loop has max priority even
1 5 6 2 3 6 2 3 6 5 Odd Wins Even Wins
Extension to infinite plays
π = 1 2 1 5 6 5 2 3 6 5 Stack = 1 5 . . . Every eliminated cycle has max priority even Hence max Inf priority in π is Even
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Better Algorithms
Marcin Jurdzinski and Jens Vöge. “A discrete strategy improvement algorithm for solving parity games”. In: Computer Aided Verification. Springer, 2000, pp. 202–215 Upper bound1 : O
1see also Friedmann, “Exponential Lower Bounds for Solving Infinitary Payoff Games
and Linear Programs”.
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Better Algorithms
Marcin Jurdzinski and Jens Vöge. “A discrete strategy improvement algorithm for solving parity games”. In: Computer Aided Verification. Springer, 2000, pp. 202–215 Upper bound1 : O
Marcin Jurdzinski, Mike Paterson, and Uri Zwick. “A Deterministic Subexponential Algorithm for Solving Parity Games”. In: SIAM Journal on Computing 38.4 (Jan. 2008), pp. 1519–1532 nO(√n)
1see also Friedmann, “Exponential Lower Bounds for Solving Infinitary Payoff Games
and Linear Programs”.
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Finite Duration Games Infinite Duration Games Parity Games Mean Payoff Games Simple Stochastic Games
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Payoffs
a b c
2
3 Min Max
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Payoffs (ab)ω
a b c
2
3 Min Max
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Payoffs (ab)ω
a b c
2
3 Min Max
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Payoffs (ab)ω
a b c
2
3 Min Max
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Payoffs (ab)ω
a b c
2
3 Min Max
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Payoffs (ab)ω
a b c
2
3 Min Max
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Payoffs (ab)ω
a b c
2
3 Min Max
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Payoffs (ab)ω
a
a b c
2
3 Min Max
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Payoffs (ab)ω
a −2 − − →b − 2 1
a b c
2
3 Min Max
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Payoffs (ab)ω
a −2 − − →b +3 − − →a − 2 + 3 2
a b c
2
3 Min Max
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Payoffs (ab)ω
a −2 − − →b +3 − − →a −2 − − →b − 2 + 3 − 2 3
a b c
2
3 Min Max
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Payoffs (ab)ω
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a − 2 + 3 − 2 + 3 4
a b c
2
3 Min Max
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Payoffs (ab)ω
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b − 2 + 3 − 2 + 3 − 2 5
a b c
2
3 Min Max
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Payoffs (ab)ω
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a − 2 + 3 − 2 + 3 − 2 + 3 6
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a −1 − − →c − 1 1
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a −1 − − →c −2 − − →b − 1 − 2 2
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a −1 − − →c −2 − − →b +3 − − →a − 1 − 2 + 3 3
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a −1 − − →c −2 − − →b +3 − − →a −1 − − →c − 1 − 2 + 3 − 1 4
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a −1 − − →c −2 − − →b +3 − − →a −1 − − →c −2 − − →b − 1 − 2 + 3 − 1 − 2 5
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω
a −1 − − →c −2 − − →b +3 − − →a −1 − − →c −2 − − →b +3 − − →a − 1 − 2 + 3 − 1 − 2 + 3 6
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω Min pays −1
3 units to Max
a −1 − − →c −2 − − →b +3 − − →a −1 − − →c −2 − − →b +3 − − →a . . . − 1 − 2 + 3 − 1 − 2 + 3 6 ∼ n(−1−2+3)
3n
→ 0
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω Min pays −1
3 units to Max
a −1 − − →c −2 − − →b +3 − − →a −1 − − →c −2 − − →b +3 − − →a . . . − 1 − 2 + 3 − 1 − 2 + 3 6 ∼ n(−1−2+3)
3n
→ 0 Min tries to minimize lim Max tries to maximize lim
a b c
2
3 Min Max
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Payoffs (ab)ω Min pays 1
2 units to Max
a −2 − − →b +3 − − →a −2 − − →b +3 − − →a −2 − − →b +3 − − →a . . . − 2 + 3 − 2 + 3 − 2 + 3 6 ∼ n(−2+3)
2n
→ 1 2
(acb)ω Min pays −1
3 units to Max
a −1 − − →c −2 − − →b +3 − − →a −1 − − →c −2 − − →b +3 − − →a . . . − 1 − 2 + 3 − 1 − 2 + 3 6 ∼ n(−1−2+3)
3n
→ 0 Generally Min tries to minimize lim sup Max tries to maximize lim inf
a b c
2
3 Min Max
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Questions
Does the game have a value? i.e. is there a v so that
Max can ensure lim inf ≥ v Min can ensure lim sup ≤ v
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Questions
Does the game have a value? i.e. is there a v so that
Max can ensure lim inf ≥ v Min can ensure lim sup ≤ v
Yes How to compute the optimal strategies?
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Questions
Does the game have a value? i.e. is there a v so that
Max can ensure lim inf ≥ v Min can ensure lim sup ≤ v
Yes How to compute the optimal strategies? Solution using the finite game
13 / 19
Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
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Finite Game
a b c
3 1 2
c
b
3 1 2
a b c
2
3 Min Max
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Finite Game
a b c
3 1 2
c
b
3 1 2
a b c
2
3 Min Max
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Max can ensure ≥ 0 in the finite game
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a Stack = a
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b Stack = a b
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c Stack = a b c
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b Stack = a b c b
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b Stack = a b
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b c Stack = a b c
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b c a Stack = a b c a
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b c a Stack = a
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b c a b Stack = a b
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b c a b c Stack = a b c
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b c a b c a Stack = a b c a Every time a cycle with average value ≤ 0 is eliminated
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b c a b c a Stack = a Every time a cycle with average value ≤ 0 is eliminated
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Min can ensure ≤ 0 in the mean payoff game too π = a b c b c a b c a Stack = a Hence limsup of averages of π is ≤ 0
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Finite Game
a b c c b
3 1 2
a b c
2
3 Min Max
Max can ensure ≥ 0 Similarly Max can ensure liminf
Hence the value of Mean payoff game is 0
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Finite Duration Games Infinite Duration Games Simple Stochastic Games
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1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
Circle Wins
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
Box Wins
1 2 1 2 1 2 1 2 16 / 19
Or
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
1 2 1 2 1 2 1 2 16 / 19
Circle can win from with probability 1
1 2 1 2 1 2 1 2 16 / 19
Values
1 1
1 2
1
1 2 1 2 1 2 1 2 17 / 19
Values
1 1
1 2 1 2
1
1 2 1 2 1 2 1 2 17 / 19
Values
v( ) = 1 v( ) = 0 v( ) = 1 2(v( ) + v( )) v( ) = 1 2(v( ) + v( )) v( ) = max{v( ), v( )} v( ) = min{v( ), v( )}
1 1
1 2 1 2
1
1 2 1 2 1 2 1 2 17 / 19
Values
v( ) = 1 v( ) = 0 v( ) = 1 2(v( ) + v( )) v( ) = 1 2(v( ) + v( )) v( ) = max{v( ), v( )} v( ) = min{v( ), v( )}
1 1
1 2 1 2
1
1 2 1 2 1 2 1 2
These equations have a unique solution.
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Values
v( ) = 1 v( ) = 0 v( ) = 1 2(v( ) + v( )) v( ) = 1 2(v( ) + v( )) v( ) = max{v( ), v( )} v( ) = min{v( ), v( )}
1 1
1 2 1 2
1
1 2 1 2 1 2 1 2
These equations have a unique solution. From state s - has a strategy to reach with probability ≥ v(s) has a strategy to reach with probability ≥ 1 − v(s)
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Does Even win the Parity Game? Is the value of the Mean Payoff Game ≥ 0 Is the value in the Simple Stochasic Game ≥ 1
2
NP∩coNP
2
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Does Even win the Parity Game? Is the value of the Mean Payoff Game ≥ 0 Is the value in the Simple Stochasic Game ≥ 1
2
NP∩coNP
2
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Does Even win the Parity Game? Is the value of the Mean Payoff Game ≥ 0 Is the value in the Simple Stochasic Game ≥ 1
2
NP∩coNP
2
2Chatterjee and Fijalkow, “A reduction from parity games to simple stochastic games”. 18 / 19
Does Even win the Parity Game? Is the value of the Mean Payoff Game ≥ 0 Is the value in the Simple Stochasic Game ≥ 1
2
NP∩coNP
2
Open Problem
Is there a polynomial time algorithm for any of them?
2Chatterjee and Fijalkow, “A reduction from parity games to simple stochastic games”. 18 / 19
Lloyd S. Shapley. “Stochastic games”. In: Proceedings of the National Academy of Sciences 39.10 (1953), pp. 1095–1100 E.A. Emerson and C.S. Jutla. “Tree automata, mu-calculus and determinacy”. In: IEEE Comput. Soc. Press, 1991, pp. 368–377 Anne Condon. “The complexity of stochastic games”. In: Information and Computation 96.2 (Feb. 1992), pp. 203–224 Uri Zwick and Mike Paterson. “The complexity of mean payoff games
In: Theoretical Computer Science 158.1 (May 1996),
Marcin Jurdziski. “Deciding the winner in parity games is in UP co-UP”. . In: Information Processing Letters 68.3 (1998), pp. 119–124
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Lloyd S. Shapley. “Stochastic games”. In: Proceedings of the National Academy of Sciences 39.10 (1953), pp. 1095–1100 E.A. Emerson and C.S. Jutla. “Tree automata, mu-calculus and determinacy”. In: IEEE Comput. Soc. Press, 1991, pp. 368–377 Anne Condon. “The complexity of stochastic games”. In: Information and Computation 96.2 (Feb. 1992), pp. 203–224 Uri Zwick and Mike Paterson. “The complexity of mean payoff games
In: Theoretical Computer Science 158.1 (May 1996),
Marcin Jurdziski. “Deciding the winner in parity games is in UP co-UP”. . In: Information Processing Letters 68.3 (1998), pp. 119–124
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