260 likes | 284 Views
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.
E N D
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.
minimax theorem • 2x1+6x2≥ w • 3x1+5x2 ≥ w • 7x1+4x2 ≥ w • x1+x2=1 column inner product x = (x1,x2) is a mixed strategy by which player I achieves an expected gain of at least w.
minimax theorem • 2y3+3y4+7y5≤ w • 6y3+5y4+4y5≤ w • y3+y4+y5=1 row inner product y = (y1,y2,y3) is a mixed strategy by which player II insures itself against an expected loss of more than w.
Solving game • Find x1≥0, x2 ≥0,,y3 ≥0, y4 ≥0, y5 ≥ 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 SatisfactionGame 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. m1 and a constraint of type 2 with probability m2=1-m1.
Game matrix sik for symmetric F variables set 1 relation R b(n,k) = binomial(n,k)
Game matrix sik for symmetric F:General Case variables set 1 relation R b(n,k) = binomial(n,k)
General Formula • http://www.ccs.neu.edu/research/demeter/papers/evergreen/cp07-submission.pdf • page 6
Linear program (Alice maximizes t) • max t (t = satisfaction ratio) • s10*l0+s11*l1+ … +s1n*ln≥ t • s20*l0+s21*l1+ … +s2n*ln≥ t • l0+l1+ … +ln= 1 • li≥ 0 for all i in [0,1, … ,n]
Linear program (Alice maximizes t) • max t (t = satisfaction ratio) • -s10*l0-s11*l1- … -s1n*ln+t ≤ 0 • -s20*l0-s21*l1- … -s2n*ln+t ≤ 0 • l0+l1+ … +ln= 1 • li≥ 0 for all i in [0,1, … ,n]
Dual linear program(Bob minimizes t) • min t (t = satisfaction ratio) • s10*m1+s20*m2≤ t • s11*m1+s21*m2≤ t • … • s1n*m1+s2n*m2≤ t • m1+m2= 1 • mi≥ 0 for all i in [1,2]
Visualizing dual linear program μ corresponds to m
Mechanical way of finding best price and worst raw material • Given derivative d = ((R1, … ), 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 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 Resolutionpage 709, chapter 13.1 • Problem: • n processes P1, P2, … ,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.