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Published byAmanda Bailey Modified over 9 years ago
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The Game Inside the Game Karl Lieberherr based on Master Thesis of Anna Hoepli at ETH Zurich in 2007 (communicated by Emo Welzl)
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SDG Partial Satisfaction Game Theoretic View 2 person game, Bob and Alice. Unsatisfiable CSP Formula F is the “board” of the game. –Bob chooses a constraint C. –Alice chooses an assignment A. –If A satisfies C, Alice wins; otherwise Bob.
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Traditional Game F must be unsatisfiable, otherwise Bob would not have a chance if Alice knows a satisfying assignment. The unsatisfiable constraint is not allowed to be in F. (Relation 0). Both play simultaneously. We play with symmetric formulas.
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T = {(1,1) (2,0)} Two constraint types: 1. A / 2. !A or !B Assume F is symmetric. Bob chooses a constraint of type 1 with prob. m1 and a constraint of type 2 with probability m2=1-m1.
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Game matrix s ik for symmetric F #1/ R 012kn-1n A01/n2/nk/n1-1/n1 !A or !B 111- 1/b(n,2) 1- b(k,2)/b(n,2) 2/n0 b(n,k) = binomial(n,k) relation R variables set 1
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Game matrix s ik for symmetric F: General Case #1/ R 012kn-1n is ik = probability that constraint i is true if k variables set 1 b(n,k) = binomial(n,k) relation R variables set 1
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General Formula http://www.ccs.neu.edu/research/demeter/ papers/evergreen/cp07-submission.pdfhttp://www.ccs.neu.edu/research/demeter/ papers/evergreen/cp07-submission.pdf page 6
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Linear program (Alice maximizes t) max t (t = satisfaction ratio) s 10 *l 0 +s 11 *l 1 + … +s 1n *l n ≥ t s 20 *l 0 +s 21 *l 1 + … +s 2n *l n ≥ t l 0 +l 1 + … +l n = 1 l i ≥ 0 for all i in [0,1, …,n]
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Linear program (Alice maximizes t) max t (t = satisfaction ratio) -s 10 *l 0 -s 11 *l 1 - … -s 1n *l n +t ≤ 0 -s 20 *l 0 -s 21 *l 1 - … -s 2n *l n +t ≤ 0 l 0 +l 1 + … +l n = 1 l i ≥ 0 for all i in [0,1, …,n]
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Dual linear program (Bob minimizes t) min t (t = satisfaction ratio) s 10 *m 1 +s 20 *m 2 ≤ t s 11 *m 1 +s 21 *m 2 ≤ t … s 1n *m 1 +s 2n *m 2 ≤ t m 1 +m 2 = 1 m i ≥ 0 for all i in [1,2]
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Mechanical way of finding best price and worst raw material Given derivative d = ((R1, … ), p?,seller) Choose n, e.g. n = 20. Generate matrix s ik. It has one row per relation and n columns. Create input to LP solver using dual program, because we need the the m i for the raw materials. The minimum is the break-even price.
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SDG classic t pred = lim n -> ∞ min all raw materials rm of size n satisfying predicate pred max all finished products fp produced for rm q(fp) Seller approximates minimum efficiently Buyer approximates Max efficiently
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Spec for RM and FP t pred = lim n -> ∞ min all raw materials rm of size n satisfying predicate pred and having property WORST(rm) max small subset of all finished products fp produced for rm q(fp)
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