COS 521: Advanced Algorithms Game Theory and Linear Programming

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1 COS 521: Advanced Algorthms Game Theory and Lnear Programmng Moses Charkar February 27, 2013 In these notes, we ntroduce some basc concepts n game theory and lnear programmng (LP). We show a connecton between equlbrum strateges n a certan knd of two player game and LP dualty and connect ths to the multplcatve weghts update algorthm we saw earler n the context of learnng wth experts. 1 Game Theory 1.1 Introducton A two player game (or more correctly, a two player normal-form game) s specfed by two m n payoff matrces R and C correspondng to the row and column player respectvely. Each of these matrces has m rows correspondng to the m strateges of the row player and n columns correspondng to the n strateges of the column payer. The row player pcks a row [m], and the column player pcks a column [n]. We say that the outcome of the game s (, ). The payoff to the row player s R, and the payoff to the column player s C,. The goal of each player s to maxmze ther own payoff. The strateges for the row and column player descrbed so far are referred to as pure strateges. Both the row and column player may also pck a dstrbuton over rows and columns respectvely, referred to as mxed strateges. A mxed strategy for the row player can be specfed by a vector x = (x 1,..., x m ) such that x 0 and x = 1. Here the th coordnate x specfes the probablty that the (pure) strategy [m] s pcked by the row player. The space of mxed strateges for the row player s denoted by m. Ths s ust the set of vectors x = (x 1,..., x m ) such that x 0 and x = 1. Smlarly, a mxed strategy for the column player s specfed by a vector y = (y 1,..., y n ) such that y 0 and y = 0. The space of mxed strateges for the column player s denoted by n. If the row and column player play mxed strateges x, y respectvely, then the nterpretaton s that both players pck ndependently from the probablty dstrbutons specfed by x and y,.e. the probablty that the outcome s (, ) s x y. The expected payoff to Prnceton Unversty, Sprng

2 the row player s, R,x y = x T Ry and the expected payoff to the column player s, C,x y = x T Cy. In ths vector notaton, a pure strategy for the row player can be represented by a vector e, [m], where e s a m-dmensonal vector such that the th coordnate s 1 and all other coordnates are zero. Smlarly, a pure strategy for the column player can be represented by a vector e, [m], where e s a n-dmensonal vector such that the th coordnate s 1 and all other coordnates are zero. In these notes, we wll reserve x, e and to refer to strateges of the row player and y, e and to refer to strateges of the column player. Gven a game, what strateges should we expect the players to play? Ths s a fundamental queston n game theory. Gven full nformaton about the other player s strategy, each player attempts to pck a strategy that maxmzes hs/her payoff. We say that a par of (pure) strateges and are n equlbrum f, when the row player plays and the column player plays, nether can get better payoff by unlaterally swtchng to a dfferent strategy..e. R, R, for all [m] and C, C, for all [n]. If we have a two player game wth a par of strateges and n equlbrum, then t s reasonable to expect that one possble outcome of the game s (, ). Unfortunately, not all two player games have such an equlbrum. However, f we broaden our noton of equlbrum to nclude mxed strateges, then we always have a par of mxed strateges n equlbrum. Ths s a consequence of a famous theorem of John Nash whch shows that such equlbrum strateges exst n the more general mult-player settng ths s part of the work for whch he was awarded the Nobel Prze n Economcs n We say that a par of mxed strateges x and y are n Nash equlbrum f, when the row player plays x and the column player plays y, nether can get better payoff by unlaterally swtchng to a dfferent strategy,.e. x T Ry (x ) T Ry for all x m, and (1) x T Cy x T Cy for all y n (2) An equvalent defnton s the followng: A par of mxed strateges x, y are n Nash equlbrum f nether player can unlaterally swtch to a pure strategy that gves better payoff: x T Ry e T Ry for all [m], and (3) x T Cy x T Ce for all [n] (4) Note that e T Ry s the expected payoff when the row player plays pure strategy and the column player plays mxed strategy y. Thus e T Ry = R,y. Smlarly x T Re s the expected payoff when the row player plays mxed strategy x and the column player plays pure strategy. Thus x T Re = R,x. It s a good exercse to prove that the above two defntons are equvalent we hghly recommend ths as a way to test your understandng of these defntons! A game could have multple such Nash equlbra (x, y), but at least one s guaranteed to exst by Nash s theorem. Gven a two player game, we do not know an effcent (polynomal 2

3 tme) algorthm to fnd a Nash equlbrum. Desgnng such an algorthm (or provng that t does not exst) s a sgnfcant open problem n theoretcal computer scence. We have some evdence that suggests that fndng equlbrum strateges s a hard problem, however t s not known whether the problem s NP-hard. 1.2 Zero sum games Two player zero sum games are a specal class of two player games where R + C = 0,.e. R, + C, = 0 for all,. In ths case, the payoffs can be vewed as a payment from one player to the other. The goal of the column player s to maxmze the payoff x T Cy. On the other hand, the goal of the column player s to maxmze the payoff x T Ry = x T Cy. In other words, the goal of the row player s to mnmze x T Cy. Let s thnk about the optmzaton problem faced by the row player: If the row player pcks (mxed) strategy x, the column player wll pck (mxed) strategy y that maxmzes x T Cy. In other words, for a fxed x, the value to the row player s max y x T Cy. Snce the goal of the row player s to mnmze x T Cy, the optmzaton problem that the row player should attempt to solve s mn x max x T Cy (5) y By smlar reasonng, the optmzaton problem that the column player should attempt to solve s max mn x T Cy (6) y x In fact, t turns out that the values of the two optmzaton problems are exactly the same, and ths was proved by Von Neumann: Theorem 1.1 (Von Neumann mnmax theorem). mn x max y x T Cy = max mn x T Cy (7) y x We wll prove ths theorem later by appealng to lnear programmng dualty, a concept we wll ntroduce shortly. Before we do ths, we frst dscuss the mplcaton for Nash Equlbrum n 2 player zero sum games. Corollary 1.2. Every 2 player zero sum game has a Nash equlbrum. Proof. Ths proof looks heavy on notaton, but s conceptually qute smple. The man pont s that the strateges x and y that optmze the LHS and RHS of the equalty (7) are n Nash equlbrum. 3

4 Consder a 2 player zero sum game where C s the payoff matrx for the column player. By Theorem 1.1, mn x max y x T Cy = max mn x T Cy (8) y x Let λ be the value of the optmzaton problem on the LHS and RHS above. Further, let x be the value of x that acheves max y x T Cy = λ and let y be the value of y that acheves mn x x T Cy = λ. Hence (x ) T Cy λ for all y n (9) x T Cy λ for all x m (10) Ths mples that (x )Cy = λ. We clam that the par of strateges (x, y ) are n Nash equlbrum. Snce x T Cy (x )Cy for all strateges x m, the row player has no ncentve to devate from x. Snce (x ) T Cy (x )Cy for all strateges y n, the column player has no ncentve to devate from y. Hence the par of strateges x, y are n Nash equlbrum. In order to prepare for the proof of the mnmax theorem, we restate the equalty (7) n an equvalent form. Ths s smlar to the alternate defnton of Nash equlbrum we mentoned earler. Frst, we clam that for a fxed x, x T Cy s maxmzed at y = e for [n]. In other words, max y x T Cy = max x T Ce. Thus the optmzaton for the row player can be expressed as mn x max x T Ce. Smlarly, for a fxed y, x T Cy s mnmzed at x = e for [m]. In other words, mn x x T Cy = mn e T Cy. Thus the optmzaton for the column player can be expressed as max y mn e T Cy. Hence the mnmax theorem can be rephrased n the followng equvalent form: Alternately: mn x max x T Ce = max mn e T Cy (11) y { } { } mn max C, x = max mn C, y x y (12) The mnmax theorem fnds many uses n theoretcal computer scence. In partcular, t s the bass of a technque to show lower bounds on the compettve rato for randomzed onlne algorthms, and lower bounds for randomzed algorthms n general. We next ntroduce lnear programmng (LP) and then use LP dualty to prove the mnmax theorem. 2 Lnear Programmng A lnear program (LP) s a partcular knd of optmzaton problem. Lnear programs are very useful because they arse n many dfferent applcatons and we have effcent algorthms 4

5 to solve them. A lnear program conssts of a set of varables wth lnear constrants on them (ether equaltes or nequaltes where both sdes are lnear functons of the varables). The obectve functon s a lnear functon of the varables. The goal s to assgn real values to the varables so as to optmze (ether mnmze or maxmze) the obectve functon. Lnear programmng s a very useful tool n optmzaton because many problems can be phrased as lnear programmng problems. The optmum soluton to a lnear program can be computed n tme polynomal n the sze of the nput. We wll not actually dscuss any algorthms to solve lnear programs n these notes (and ndeed such algorthms are outsde the scope of the course) but we menton that the frst polynomal tme algorthm to solve lnear programs optmally was developed by Khachyan n Here s an example of a lnear program max 3y 1 + 4y 2 + y 3 (13) subect to y 1 + y 2 5 (14) y 2 + y 3 4 (15) y 0 (16) A feasble soluton for ths lnear program s a settng of values to the varables such that all the constrants are satsfed. An optmum soluton to the lnear program s a feasble soluton that optmzes the value of the obectve functon. Note that the optmum soluton need not be unque. Gven a soluton to ths partcular lnear program, how do we prove that t s optmum? We can do that by exhbtng an upper bound on the value of the obectve functon for any feasble soluton to the LP. One way to obtan such an upper bound s to multply the frst nequalty by 3 and the second by 1 and add them up. 3 (y 1 + y 2 ) + (y 2 + y 3 ) (17) 3y 1 + 4y 2 + y 3 19 (18) Ths gves an upper bound of 19 on the obectve functon of the LP wthout actually solvng t. In fact ths bound s tght. Consder the soluton y 1 = 5, y 2 = 0, y 3 = 4. It s easy to check that ths satsfes the constrants and the value of the obectve functon for ths soluton s 19. We have ust proved that 19 s n fact the optmum value of ths LP. Consder now a slght modfcaton of the LP we started out wth where we change one coeffcent n the obectve functon: max 2y 1 + 4y 2 + y 3 (19) subect to y 1 + y 2 5 (20) y 2 + y 3 4 (21) y 0 (22) 5

6 How can we obtan an upper bound for the modfed LP? The same bound of 19 derved earler stll holds: 2y 1 + 4y 2 + y 3 3 (y 1 + y 2 ) + (y 2 + y 3 ) (23) Can we do better? Suppose we multply the frst nequalty by x 1 0 and the second nequalty by x 2 0 where x 1 and x 2 are unknowns to be determned. (We nsst that x 1, x 2 be nonnegatve to ensure that multplyng by these values does not reverse the drecton of the nequalty). What condtons are needed to ensure that we have a vald upper bound on the value of the LP? Here s what we need: 2y 1 + 4y 2 + y 3 x 1 (y 1 + y 2 ) + x 2 (y 2 + y 3 ) 5x 1 + 4x 2 (24) For ths to be correct, note that the frst nequalty should hold for every y 1, y 2, y 3 0. One way to guarantee ths s to make sure that the coeffcent of y 1 on the LHS of the frst nequalty s then coeffcent of y 1 on the RHS and smlarly for the coeffcents of y 2 and y 3. Ths gves three lnear nequaltes for x 1 and x 2. The bound that we get from ths s 5x 1 + 4x 2 and the best bound from ths method s obtaned by mnmzng ths expresson. Ths gves a (new) lnear program on the varables x 1 and x 2 : mn 5x 1 + 4x 2 (25) subect to x 1 2 (26) x 1 + x 2 4 (27) x 2 1 (28) x 0 (29) Settng x 1 = 2, x 2 = 2 gves a soluton that satsfes all the constrants of ths new LP. The value of the soluton s = 18, whch s an upper bound on the value of the LP (19)-(22). In fact, ths bound s tght. Consder the followng soluton to the LP (19)-(22): y 1 = 1, y 2 = 4, y 3 = 0. The value of ths soluton s 2y 1 + 4y 2 + y 3 = 18. Thus we have proved that 18 s the optmal value of ths LP. In fact, what we have ust seen are smple examples of LP dualty that we now proceed to dscuss. The two LPs (19)-(22) and (25)-(29) are duals of each other. More generally, any lnear program can be expressed n the followng form: max b T y (30) Ay d (31) y 0 (32) Here y = (y 1,... y n ) s a column vector of varables and b = (b 1,..., b n ) s a column vector of coeffcents (.e. real numbers). The obectve functon s mn b y. A s an m n matrx of coeffcents and b = (b 1,... b m ) s a column vector whose entres are real numbers. 6

7 The lnear program has n constrants the th constrant s gven by A,y b. All varables y are constraned to be non-negatve. We mentoned earler that lnear programs can be solved n polynomal tme. In fact, there are three possbltes for a lnear program: (1) the optmum value s bounded, (2) the optmum value s unbounded, or (3) there s no feasble soluton. In tme polynomal n the sze of a gven LP, we can decde whch of the three cases ths LP falls nto. In case the optmum s bounded, we can obtan an optmal soluton n polynomal tme. We refer to the orgnal lnear program as the prmal LP. The dual of ths lnear program s the followng: mn d T x (33) A T x b (34) x 0 (35) Here x = (x 1,... x m ) s a vector of varables. A s the same matrx and b, d are the column vectors that appear n the descrpton of the prmal LP. Note that we have a varable x n the dual correspondng to the th constrant n the prmal. Thnk of the x varables as multplers for the constrants n the prmal LP. We would lke to multply the th constrant of the prmal by x and sum over all constrants to obtan a vald upper bound on the value of the prmal. The bound that we obtan thus s d x whch s the obectve functon of the dual LP. We have a constrant n the dual correspondng to every varable n the prmal. Snce we multply the th prmal constrant by x and add up over all, the coeffcent of y n ths lnear combnaton s A,x Ths ought to be the coeffcent of y n the prmal obectve, whch s b. Ths gves the followng constrant: A,x b whch s exactly the th constrant n the dual LP. The dual conssts of all such constrants for [n]. We state two lemmas that relate the values of the prmal and dual LPs: Lemma 2.1 (Weak Dualty). For any feasble soluton x to the LP (33)-(35) and y to the LP (30)-(32), we have b T y d T x. In fact, we have already presented all the deas needed to prove ths lemma. Prove ths as an exercse to test your understandng! Lemma 2.2 (Strong Dualty). For any optmal soluton x to the LP (33)-(35) and optmal soluton y to the LP (30)-(32), we have b T y = d T x. Dualty s a very useful property of lnear programs. We have already seen how to explot dualty to prove that a soluton to a lnear program s optmal: smply exhbt an optmal soluton to the dual lnear program. We wll establsh LP dualty later. For now, we wll use t to prove the mnmax theorem stated earler. In general, the form of LP dualty we have stated here s general enough to wrte down the dual for any lnear program. However, t s convenent to remember a few more rules 7

8 that make the ob of wrtng the dual easer n some cases these can be derved from the form that we descrbed earler. Suppose the the prmal LP has some equalty constrants. In prncple, each equalty constrant A,y = d can be wrtten down as two nequaltes: A,y d and A,y d. Let x (+) and x ( ) be the two correspondng dual varables. It s easer to wrte down the dual drectly by usng the varable x = x (+) x ( ). In ths case however, the dual varable x s unconstraned,.e. can be ether postve, negatve or zero. Let s try to get some ntuton for why ths makes sense. Recall the motvaton for the dual LP. The x varable s a multpler for the th constrant n the dual. The reason why we nssted on x 0 earler s to ensure that multplyng by x does not reverse the nequalty. However, f the th constrant s an equalty, we can multply t by any value postve or negatve and we stll have an equalty. Hence the dual varable correspondng to an equalty s unconstraned. Smlarly, f we had an unconstraned varable y n the prmal LP, the correspondng th constrant n the dual LP s an equalty constrant. One way to see ths s to convert to an LP where all varables are nonnegatve by replacng y by y (+) where the two new varables y (+), y ( ) 0. If we go through the calculatons, we wll see that that the dual constrants y ( ) and y ( ) are of the form A,x b and A,x b. correspondng to y (+) Together they mply the followng constrant: A,x = b, whch s easer to wrte down drectly. Agan ntutvely, t makes sense that the dual constrant correspondng to an unconstraned varable s an equalty constrant. Recall the motvaton for the dual LP and constrants. Suppose for [n], we multply the th constrant n the prmal LP by x and add them up. The goal s to ensure that the LHS of ths lnear combnaton of constrants s an upper bound on the obectve functon of the prmal LP. Then the th constrant of the dual s obtaned by examnng the coeffcent of prmal varable y on the LHS of ths lnear combnaton and n the obectve functon of the prmal LP. If the prmal varable y s unconstraned, the only way we can guarantee that the value of the lnear combnaton s an upper bound on the obectve functon s to ensure that the coeffcents of y n the lnear combnaton and the obectve functon match exactly. In other words, the constrant correspondng to the y should be an equalty. To summarze, we state the prmal and dual LPs agan allowng for unconstraned varables and equalty constrants n addton to nequalty constrants. It s convenent to remember the prmal and dual LPs n ths form. The prmal LP s as follows: (Here 8

9 S [m], T [n]). max b y (36) S A, y d (37) [m] \ S A, y = d (38) T y 0 [n] \ T y unconstraned The dual LP s as follows: T [n] \ T mn d x (39) A, x b (40) A, x = b (41) S x 0 [m] \ S x unconstraned Note that unconstraned varables n the prmal correspond to equalty constrants n the dual, and unconstraned varables n the dual correspond to equalty constrants n the prmal. We also remark that the dual of the dual s the prmal LP. 2.1 Proof of mnmax theorem Recall the equvalent formulaton of the mnmax theorem we derved earler: { } { } mn max C, x = max mn C, y x y (42) In ths secton, we wll prove the above equalty. The LHS s the value of the optmzaton problem faced by the row player and the RHS s the value of the optmzaton problem faced by the column player. We wll formulate both these problems as lnear programs and show that they are duals of each other. Frst we clam that the followng lnear program exactly captures the optmzaton prob- 9

10 lem for the row player: [n] mn z (43) m C, x z (44) =1 x = 1 (45) [m] x 0 z unconstraned Note that the x varables are the coordnates of the mxed strategy vector x used by the row player. For any fxng of the x varables, the mnmum value of the obectve functon s acheved for z = max { C,x }, so the optmum value of the lnear program s ndeed the mnmum of max { C,x } over all choces of mxed strateges x. By smlar reasonng, we clam that the followng lnear program exactly captures the optmzaton problem for the column player: [m] max w (46) n C, y w (47) =1 y = 1 (48) [n] y 0 w unconstraned Here the y varables are the coordnates of the mxed strategy vector y used by the row player. For any fxng of the y varables, the maxmum value of the obectve functon s { acheved for w = mn C,y }, so the optmum value of the lnear program s ndeed { } the maxmum of mn C,y over all choces of mxed strateges y. Let s reformulate the lnear program (46)-(48) n the form of the LP (36)-(38): [m] max w (49) n w C, y 0 (50) =1 y = 1 (51) [n] y 0 w unconstraned 10

11 Now, let s wrte down the dual of ths LP, usng varables x, [m], for the constrants (50), and varable z for the constrant (51). Correspondng to each y, [n], we have a dual constrant whch s an nequalty. Correspondng to w, we have a dual constrant whch s an equalty. The dual LP s [n] mn z (52) m z C, x 0 (53) =1 x = 1 (54) [m] x 0 z unconstraned It s easy to see that the dual LP we derved above s exactly the same as the LP (43)- (45) we wrote down earler to capture the optmzaton problem for the row player. Thus the lnear programs we wrote down for the row and column player are duals of each other. By strong dualty, the values of ther optmum solutons are equal. Now let s go back to the reformulaton (42) of the mnmax theorem we stated at the begnnng of ths secton. The optmum value of the row player s LP s the LHS of (42) and the optmum value of the column player s LP s the RHS of (42). Hence we have proved that both sdes are equal, establshng the mnmax theorem. 3 Multplcatve Weghts To Be Completed 11

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