The Game Inside the Game Karl Lieberherr based

The Game Inside the Game Karl Lieberherr based on Master Thesis of Anna Hoepli at ETH Zurich in 2007 (communicated by Emo Welzl)

Classical Game Theory • As pointed out in http: //www. ccs. neu. edu/research/demeter/ biblio/vavasis. html the SDG break-even prices can also be determined through Linear Programming which has a close connection to classical game theory.

John von Neumann 1929 • solving a game and solving a feasible pair of dual linear programs are essentially the same. • Players I and II choose independently a row and column. II pays I chosen entry. • A two-person, zero sum game. 2 3 7 6 5 4

minimax theorem • • 2 x 1+6 x 2 ≥ w 3 x 1+5 x 2 ≥ w 7 x 1+4 x 2 ≥ w x 1+x 2=1 column inner product x = (x 1, x 2) is a mixed strategy by which player I achieves an expected gain of at least w. 2 3 7 6 5 4

minimax theorem • 2 y 3+3 y 4+7 y 5≤ w • 6 y 3+5 y 4+4 y 5≤ w • y 3+y 4+y 5=1 row inner product y = (y 1, y 2, y 3) is a mixed strategy by which player II insures itself against an expected loss of more than w. 2 3 7 6 5 4

Solving game • Find x 1≥ 0, x 2 ≥ 0, , y 3 ≥ 0, y 4 ≥ 0, y 5 ≥ 0 satisfying both (primal and dual) sets of inequalities. • Unique answer. • (1/5, 4/5) and (0, 3/5, 2/5) are optimal mixed strategies for player I and II, respectively, and 23/5 is the value of the game.

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.

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.

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. m 1 and a constraint of type 2 with probability m 2=1 -m 1.

Game matrix sik for symmetric F variables set 1 #1/ R 0 1 A 0 !A or 1 !B relation R 2 k n-1 1/n 2/n k/n 1 -1/n 1 1 1 b(k, 2)/b(n, 2) 2/n 11/b(n, 2) b(n, k) = binomial(n, k) n 0

Game matrix sik for symmetric F: General Case variables set 1 #1/ R i relation R 0 1 2 k sik = probability that relation i is true if k variables set 1 b(n, k) = binomial(n, k) n-1 n

General Formula • http: //www. ccs. neu. edu/research/demeter/ papers/evergreen/cp 07 -submission. pdf • page 6

Linear program (Alice maximizes t) • • • max t (t = satisfaction ratio) s 10*l 0+s 11*l 1+ … +s 1 n*ln ≥ t s 20*l 0+s 21*l 1+ … +s 2 n*ln ≥ t l 0+l 1+ … +ln = 1 li ≥ 0 for all i in [0, 1, … , n]

Linear program (Alice maximizes t) • • • max t (t = satisfaction ratio) -s 10*l 0 -s 11*l 1 - … -s 1 n*ln+t ≤ 0 -s 20*l 0 -s 21*l 1 - … -s 2 n*ln+t ≤ 0 l 0+l 1+ … +ln = 1 li ≥ 0 for all i in [0, 1, … , n]

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 1 n*m 1+s 2 n*m 2 ≤ t m 1+m 2 = 1 mi ≥ 0 for all i in [1, 2]

Visualizing dual linear program μ corresponds to m

Dual Linear Program for {(1, 1) (2, 0)}

Dual Linear Program for {(1, 1) (2, 0)}

Mechanical way of finding best price and worst raw material • Given derivative d = ((R 1, … ), p? , seller) • Choose n, e. g. n = 20. • Generate matrix sik. It has one row per relation and n columns. • Create input to LP solver using dual program, because we need the mi for the raw materials. The minimum is the break-even price.

SDG classic tpred = 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

Spec for RM and FP tpred = 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)

Contention Resolution page 709, chapter 13. 1 • Problem: – n processes P 1, P 2, … , Pn, each competing for access to a single shared database. (distributed SDG: n SDG robots trying to access the store. ) – Time divided into discrete rounds. DB can be accessed by at most one process in a single round. If two or more access: all locked out for the duration of that round. – All processes get through to the data base on a regular basis. – Processes cannot communicate.

Max bias again! What is the relation R(n)? • A(i, t) = event that Pi attempts to access DB in round t. Pr(A(i, t)) = p. • S(i, t) = event that Pi successfully accesses the DB in round t. Pr(S(i, t))=p*(1 -p)^(n-1). • What is the max bias? ? ? • f(p) = p*(1 -p)^(n-1) • f’(p) has a single zero at p=1/n where the maximum is achieved. • Intuitive choice!

substitute • (1/n)*(1 -1/n)^(n-1) • The function (1 -1/n)^(n-1) converges monotonically from ½ to 1/e as n increases from 2. Price for derivative will be between ½ and 1/2. 718. • R(n) is of arity n and is true for row 00000… 1 (n-1 zeros) and false for all other truth table rows. Relation number 1!

Waiting for a process to succeed • • 1/(e*n) <= Pr(S(i, t)) <= 1/(2*n) Pr(S(i, t))=Theta(1/n) e = base of natural logarithm = 2. 7… Failure event F(i, t)= process Pi does not succeed in any of the rounds 1 through t. Pr(F(i, t)) <= 1/e.