1. Peter van Emde Boas: Games and Computer Science 1999 GAMES AND COMPUTER SCIENCE Theoretical Models 1999 Peter van Emde Boas References available at: http://turing.wins.uva.nl/~peter/teaching/thmod99.html Most Papers will be made available in Library

2. Peter van Emde Boas: Games and Computer Science 1999

3. Games in Computer Science Information & Uncertainty (Traub ea. - 80+) Pebble Game (Register Allocation, Theory) Tiling Game (Reduction Theory) Alternating Computation Model and / or trees Interactive Proofs Arthur Merlin Games Zero Knowledge

4. Peter van Emde Boas: Games and Computer Science 1999 Game Theory Theory of Strategic Interaction Attributes –Discrete vs. Continuous –Cooperative vs. Non-Cooperative –Full Information vs. Incomplete Information (Knowledge Theory)

5. Peter van Emde Boas: Games and Computer Science 1999 Discrete / Continuous Combinatorial Analysis Backward Induction Number Theory (Conway Guy Berlekamp) Equilibria theory (Nash) Stochasitic Features Optimization Other names of importance: Von Neumann & Morgenstern Aumann Shapley Harsanyi

6. Peter van Emde Boas: Games and Computer Science 1999 OTHER ASPECTS Single player - no choices Single player - random moves Single player - choices : Solitaire Two players - choices Two players - choices and random moves Two players - concurrent moves

7. Peter van Emde Boas: Games and Computer Science 1999 Computer Science Computation Theory Complexity Theory Machine Models Algorithms Knowledge Theory Information Theory

8. Peter van Emde Boas: Games and Computer Science 1999 COMPUTATION Deterministic Nondeterministic Probabilistic Alternating Interactive protocols Concurrency

9. Peter van Emde Boas: Games and Computer Science 1999 COMPUTATION Notion of Configurations: Nodes Notion of Transitions: Edges Non-uniqueness of transition: Out- degree > 1 Initial Configuration : Root Terminal Configuration : Leaf Computation : Branch Tree Acceptance Condition: Property of trees

10. Peter van Emde Boas: Games and Computer Science 1999 © Games Workshop URGATTHORGRIM Introducing the Opponents

11. Peter van Emde Boas: Games and Computer Science 1999 A Game © Donald Duck 1999 # 35 Starting with 15 matches players alternatively take 1, 2 or 3 matches away until none remain. The player ending up with an odd number of matches wins the game

12. Peter van Emde Boas: Games and Computer Science 1999 Questions about this Game What if the number of matches is even? Can any of the two players force a win by clever playing? How does the winner depend on the number of matches Is this dependency periodic? If so WHY?

13. Peter van Emde Boas: Games and Computer Science 1999 Games as Recognizers Construct a map  :  * --> Games (simply computable; Poly-time, Logspace or NC, ….) Set recognized := {w |  (w) is won (by the first player) } How does this relate to conventional ways of recognizing languages ?

14. Peter van Emde Boas: Games and Computer Science 1999 Games as Recognizers Construct a map  :  * --> Games (simply computable; Poly-time, Logspace or NC, ….)  (w) is guaranteed to be proper Set recognized := {w |  (w) is won (by the first player) } Properness conditions frequently involve probabilistic aspects

15. Peter van Emde Boas: Games and Computer Science 1999 Game Trees Root Terminal node: Thorgrim looses Thorgrim’s turn Urgat’s turn Terminal node: Urgat looses Standard Interpretation: Player unable to move looses the game

16. Peter van Emde Boas: Games and Computer Science 1999 Game Trees Root Terminal node: Thorgrim’s turn Urgat’s turn Terminal node: Free Interpretation: Winner explicitly designated at terminal node T TU U T UT

17. Peter van Emde Boas: Games and Computer Science 1999 Game Trees Root Terminal node: Thorgrim’s turn Urgat’s turn Terminal node: Non Zero-Sum Game: Payoffs explicitly designated at terminal node 2 / 02 / 0 5 / -71 / 4 -1 / 4 3 / 13 / 1 -3 / 21 / -1

18. Peter van Emde Boas: Games and Computer Science 1999 Game Trees Root Terminal node: Thorgrim’s turn Urgat’s turn Terminal node: Free Interpretation: Winner explicitly designated at terminal node T TU U T UT SUB-GAME

19. Peter van Emde Boas: Games and Computer Science 1999 Backward Induction Terminal node: Thorgrim’s turn Urgat’s turn Free Interpretation: Winner explicitly designated at terminal node Root Terminal node: T TU U T UT T T U U U

20. Peter van Emde Boas: Games and Computer Science 1999 Backward Induction Root Terminal node: Thorgrim’s turn Urgat’s turn Terminal node: Non Zero-Sum Game: Payoffs explicitly designated at terminal node 2 / 02 / 0 5 / -71 / 4 -1 / 4 3 / 13 / 1 -3 / 21 / -1 2 / 02 / 0 3 / 13 / 1 1 / 4 -3 / 2 1 / 4

21. Peter van Emde Boas: Games and Computer Science 1999 Backward Induction 2 / 02 / 0 5 / -71 / 4 -1 / 4 3 / 13 / 1 -3 / 21 / -1 2 / 02 / 0 3 / 13 / 1 1 / 4 -3 / 2 1 / 4 At terminal nodes: Pay-off as explicitly given At Thorgrim’s nodes: Pay-off inherited from Thorgrim’s optimal choice At Urgat’s nodes: Pay-off inherited from Urgat’s optimal choice For strictly competetive games this is the Max-Min rule T T U U T UT T T U U U

22. Peter van Emde Boas: Games and Computer Science 1999 Analysis of the DD game Extension used: Thorgrim wins if he has an odd number when the game terminates. This allows for even n. Four types of configurations remain: T/E : Thorgrim has to play and has an even number T/O : Thorgrim has to play and has an odd number U/E : Urgat plays, while Thorgrim has an even number U/O : Urgat plays, while Thorgrim has an odd number Relevant feature: parity of number of matches collected so far (not the number itself!)

23. Peter van Emde Boas: Games and Computer Science 1999 Backward Induction Table nU / EU / OT / ET / O 18 U U T / 1 T / 2 17 U T T / 1 U 16 U T U T / 3 15 U U T / 2 T / 3 14 U U T / 2 T / 1 13 T U U T / 1 12 T U T / 3 U 11 U U T / 3 T / 2 10 U U T / 1 T / 2 9 U T T / 1 U 8 U T U T / 3 7 U U T / 2 T / 3 6 U U T / 2 T / 1 5 T U U T / 1 4 T U T / 3 U 3 U U T / 3 T / 2 2 U U T / 1 T / 2 1 U T T / 1 U 0 U T U T

24. Peter van Emde Boas: Games and Computer Science 1999 What is the Strategy? Play to number  0 or 1 (mod4) Switch your parity on every turn Start right: to even if n mod 8  {5,6,7,0} and to odd if n mod 8  {1,2,3,4} Question: explain the correctness of this strategy, otherwise than by inspecting the table.....

25. Peter van Emde Boas: Games and Computer Science 1999 Alternating Computation +- ++- --+ + - Computation Tree Configuration Type Existential Universal Negating Accepting Rejecting

26. Peter van Emde Boas: Games and Computer Science 1999 Alternating Computation +- ++- --+ + - Evaluation Full Computation Tree This Tree Accepts Configuration Type Existential Universal Negating Accepting Rejecting + + + + + + - -- -- - -- ++ + +

27. Peter van Emde Boas: Games and Computer Science 1999 Alternating Computation Infinite Branches ? Requires third quality : Indeterminate nodes Universal node is indeterminate iff it has no rejecting son and at least one indeterminate son Existential node is indeterminate iff it has no accepting son and at least one indeterminate one Negating node is indeterminate iff its son is

28. Peter van Emde Boas: Games and Computer Science 1999 Alternating Computation Infinite Branches ? Universal node is accepting iff it has no rejecting son and no indeterminate son (all sons are accepting) Existential node is accepting iff it has one accepting Son; indeterminate and rejecting sons don’t matter Negating node is accepting iff its son is rejecting Requires Recursive Evaluation of computation tree !

29. Peter van Emde Boas: Games and Computer Science 1999 RECURSIVE EVALUATION …. Indeterminate :       +- +- +- +- +- The proper way of Recursive evaluation ???

30. Peter van Emde Boas: Games and Computer Science 1999 RECURSIVE EVALUATION ….       +- +- +- +- +- Recursive evaluation == Solving LEAST FIXED POINT EQUATION ! +- Partial order ≤ of definedness Extends to functions defined on the tree: F ≤ G iff  x[F(x) ≤ G(x)] OKNOK

31. Peter van Emde Boas: Games and Computer Science 1999 The Knaster Tarski Theorem DOMAIN := SET U with partial order ≤ and least element  Countable chains have least upper bounds x 0 ≤ x 1 ≤ x 2 ≤ ….. ≤ x n ≤ x n+1 ≤ …. ---> x  =:  i x i  i[x i ≤ x  ] and  i[x i ≤ y] ==> x  ≤ y OPERATOR := FUNCTION  which is: MONOTONE: x ≤ y ==>  (x) ≤  (y) CONTINUOUS:  ( i x i ) = i  (x i )

32. Peter van Emde Boas: Games and Computer Science 1999 The Knaster Tarski Theorem THEOREM: If  is an operator defined over domain U then the equation X =  ( X ) has a least solution . This solution is obtained as the limit of the sequence of iterates:  ≤  ≤  ≤ ….  = i  i (  ) APPLICATION: U := domain of evaluations of tree  := single application of recursive rule

33. Peter van Emde Boas: Games and Computer Science 1999 Back to Alternation For an accepting tree there exists a witness subtree for acceptance (and similar for rejection) Witness subtree contains a single accepting son for every accepting node, and a single rejecting son for every rejecting node A witness subtree is finite, even when the tree itself is infinite! Infinite branches are irrelevant!

34. Peter van Emde Boas: Games and Computer Science 1999 Negating Nodes ? Create for every node its dual node which yields the “same” transitions Dual of accepting node is rejecting Dual of rejecting node is accepting Dual of universal node is existential Dual of existential node is universal Dual of Dual is identity Replace every negating node by an existential one, dualizing the entire subtree below it (think de Morgan!)

35. Peter van Emde Boas: Games and Computer Science 1999 Eliminating Negating Nodes +- ++- --+ + + + + + + - -- -- - -- ++ + + ++ + + - -- + + + + + + - - -- - -- + + + - - - - + - - Dualized nodes