FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS

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1 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS FEDERICO ECHENIQUE Abstract. I present a simpe and fast agorithm that finds a the purestrategy Nash equiibria in games with strategic compementarities. This is the first non-trivia agorithm for finding a pure-strategy Nash equiibria. 1. Introduction I present an agorithm that finds a the pure-strategy equiibria in n-payer games with strategic compementarities (GSC). GSC were first formaized by Topkis (1979), and were introduced by Vives (1990) to economics. GSC are important in many areas in economics. For exampe, both priceand quantity-competition oigopoy modes can be modeed as GSC; arguaby covering most static market modes one may wish to consider. See Migrom and Roberts (1990), Migrom and Shannon (1992), Topkis (1998), and Vives (1999) for economic exampes of GSC. Many modes in operations research have recenty been anaized as GSC. Exampes are Lippman and McCarde (1997), Bernstein and Federgruen (2004a), Bernstein and Federgruen (2004b), Netessine and Rudi (2003), Netessine and Shumsky (2003), Cachon and Lariviere (1999), and Cachon (2001). See Cachon and Netessine (2003) for a survey. I wish to emphasize three features of the agorithm: (1) It finds a pure-strategy equiibria, but no mixed-strategy equiibria. The omission is justified because mixed-strategy equiibria are not good predictions in GSC (Echenique and Edin, 2002). Further, purestrategy equiibria aways exist in GSC (Topkis, 1979; Vives, 1990). (2) It is generay very fast. For exampe, it needs ess than 10 seconds to find a equiibria in a game with strategy profies. Date: March 15, JEL Cassification. C63, C72. I thank Gerard Cachon, Eddie Deke, Juan Dubra, Matt Jackson, Ivana Komunjer, Andy McLennan, John Rust, Bi Sandhom, Iya Sega, Chris Shannon, and Bernhard von Stenge for comments and suggestions. I aso thank seminar audiences in a number of institutions for their comments. 1

2 2 F. ECHENIQUE (3) It is simpe. I use the agorithm by hand on some bimatrix games to show that the agorithm is very simpe to appy. There are many agorithms for finding one equiibrium, caed a sampe equiibrium (see the surveys by McKevey and McLennan (1996) and von Stenge (2002)). But there is currenty ony one method for finding a pure equiibria: the underining method one teaches undergraduates fix one payer, for each strategy-profie of the payer s opponents, find her bestresponse, and then check if some opponent wants to deviate. The method is cose to testing a the game s strategy-profies to see if they are equiibria; I sha ca this method the trivia agorithm. Not surprisingy, the trivia agorithm is very sow, and computationay infeasibe on arge games. Some agorithms find a sampe equiibrium that survives an equiibrium refinement typicay perfection (a recent exampe is von Stenge, van den Ezen, and Taman (2002); see McKevey and McLennan (1996) and von Stenge (2002) for other exampes). This is some times adequate, but it is in genera restrictive: there is normay no guarantee that ony one equiibrium survives the refinement, and the refinements do not aways have bite. (An exception is Judd, Yetekin, and Conkin (2000); their agorithm finds a perfect-equiibrium payoffs in repeated games.) The agorithm I present is based on Topkis s (1979) resuts that Robinson s (1951) method of iterating best-responses finds an equiibrium in GSC (see aso Vives (1990)), so the agorithm uses different and simper ideas than the more recent iterature on finding equiibria. I sha not appy the agorithm to economic, or operations research, exampes. The paper presents a method, and it argues that the method works we. So the appications of the agorithm either iustrate how it works or show that the agorithm is fast. Nevertheess, there are many appications in operations and economics. I give two exampes: Suppy-chain anaysis. Cachon (2001) studies inventory competition in a suppy chain with retaiers that face stochastic demands. The resuting game is a GSC. Cachon compares numericay the systemoptima soution to the Nash equiibria of the game. He uses exhaustive search (after identifying the extrema equiibria see Section 3) to find a Nash equiibria. The agorithm I introduce can be used instead of exhaustive search; it wi be more efficient. 1 Oigopoy modes. Under mid conditions, Bertrand oigopoy with differentiated products is a GSC (Migrom and Shannon, 1994). In 1 Other appications to suppy-chain anaysis incude discretized versions of Lippman and McCarde (1997).

3 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 3 turn, Bertrand oigopoy with differentiated products is a very common market structure. The agorithm has then natura appications in the empirica anaysis of markets. One important exampe is the evauation of mergers by the US Department of Justice. The Department of Justice needs to predict the consequences of mergers between firms. They postuate a mode of a market they often use Bertrand modes with differentiated products, see for exampe Werden, Froeb, and Tschantz (2001) or Crooke, Froeb, Tschantz, and Werden (1997) -and compute a Nash equiibrium before and after the merger of some firms in the market. 2 But their concusions might of course change if they coud find a equiibria before and after the merger. For exampe, the merger coud have no effect on price if you ook at some equiibria, but a arge price increase if you compare most equiibria. Besides Bertrand oigopoy, the agorithm can aso be appied to Cournot oigopoy. By expoiting Amir and Lambson s (2000) ideas for using GSC-techniques in Cournot oigopoy, one can easiy adapt the agorithm to find a symmetric equiibria in Cournot modes as we. Arguaby, the agorithm is appicabe to most static modes of a market one may wish to consider in appied work. Finay, et me mention how the paper contributes to the theory of GSCs. The semina papers on GSCs (Topkis, 1979; Vives, 1990; Migrom and Roberts, 1990) stress how the smaest and argest equiibria can be obtained as imits of certain monotone sequences. The source of the sequences is interpreted both as out-of-equiibrium adjustments (earning), and as an agorithm. But no equiibrium-seection theory says that the smaest, or argest, equiibrium are good predictions. It is important then to use GSC methods to approach other equiibria, and the resuts I present here are the first to do that. The paper is organized as foows. Section 2 presents some preiminary definitions and resuts. Section 3 shows informay how the agorithm works. Section 4 defines the agorithm and presents the main resuts of the paper. Section 5 deveops two simpe exampes. Section 6 outines the argument that the agorithm is fast, and expains the benchmark agorithm. Section 7 presents computationa resuts for simuations of GSC. Section 8 expains why the agorithm is normay fast. Section 9 discusses an agorithm for a specia cass of GSC. 2 The software they use is in

4 4 F. ECHENIQUE 2. Preiminary Definitions and Resuts 2.1. Basic Definitions and Notation. Let X R n, and x, y R n. Denote the vector (max {x i, y i }) by x y, and the vector (min {x i, y i }) by x y. Say that X is a attice if, whenever x, y X, x y, x y X. If X is a attice, a function f : X R is quasi-supermoduar if for any x, y X, f(x) f(x y) impies f(x y) f(y) and f(x) > f(x y) impies f(x y) > f(y). Quasi-supermoduarity is an ordina notion of compementarities; it was introduced by Migrom and Shannon (1994). Let T R m. A function f : X T R satisfies the singe-crossing condition in (x, t) if whenever x < x and t < t, f(x, t) f(x, t) impies that f(x, t ) f(x, t ) and f(x, t) < f(x, t) impies that f(x, t ) < f(x, t ). For two subsets A, B of X, say that A is smaer than B in the strong set order if a A, b B impies a b A, a b B. Let φ : X X be a correspondence. Say that φ is increasing in the strong set order if, whenever x y, φ(x) is smaer in the strong set order than φ(y). A detaied discussion of these concepts is in Topkis (1998). An n-payer norma-form game (a game, for short) is a coection Γ = {(S i, u i ) : i = 1,... n}, where each payer i is characterized by a set of possibe strategies, S i, and a payoff function u i : S R, where S = n j=1s j. Say that payers have strict preferences if, for a i and s i S i, the function s i u i (s i, s i ) is one-to-one. For each payer i, et β i,γ denote i s best-response correspondence in Γ the correspondence defined by β i,γ (s) = argmax si S i u i ( s i, s i ). And et β Γ (s) = n i=1β i,γ (s) denote the game s best-response correspondence. When Γ is understood I sha write β i for β i,γ and β for β Γ. A point s S is a Nash equiibrium if s β(s). Let E(Γ) be the set of a Nash equiibria of Γ. When Γ is understood, I sha write E for E(Γ) The Mode. Say that a game Γ = {(S i, u i ) : i = 1,... n} is a finite game of strategic compementarities (GSC) if, for each i, S i R d i is a finite attice, s i u i (s i, s i ) is quasi-supermoduar for a s i, and (s i, s i ) u i (s i, s i ) satisfies the singe-crossing property. The positive integer d i is the number of dimensions of payer i s strategies. I sha assume, in addition, that S i = {1, 2,... K i } d i. The assumption that S i = {1, 2,... K i } d i simpifies notation, but I shoud stress that a my resuts hod for arbitrary finite GSC. The set of Nash equiibria of a GSC is a compete attice (Zhou, 1994).

5 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 5 Remark 1. One can think of the mode as a discretized version of a game with continuous strategy spaces, where each S i is an interva in some Eucidean space of dimension d i. For an exampe, see Section Auxiiary resuts. First, GSC have monotone best-response correspondences: Lemma 2. (Migrom and Shannon, 1994) For a i, β i is increasing in the strong set order, and inf β i (s), sup β i (s) β i (s). See Migrom and Shannon (1994) for a proof. Second, I need some resuts and notation for games where we restrict the strategies that payers can choose: For each s i S i, et Si r (s i ) = { s i S i : s i s i } be the strategy space obtained by etting i choose any strategy in S i, as ong as it is arger than s i. For each strategy profie s = (s 1,... s n ) S, et S r (s) = n i=1si r (s i ). Denote by Γ r (s) the game where each payer i is constrained to choosing a strategy arger than s i. Then, Γ r (s 1,... s n ) = { (S r i (s i ), u i S r i (s i )) : i = 1,... n }. The foowing emmata are trivia. Lemma 3. If Γ is a GSC, then so is Γ r (s), for any strategy profie s S. Lemma 4. If s is a Nash equiibrium of Γ, and z s, then s is a Nash equiibrium of Γ r (z). Lemma 3 and Lemma 4 foow immediatey from the definitions of GSC and of Nash equiibrium. Third, I sha expoit some previous resuts on finding equiibria in GSC. The method of iterating β unti an equiibrium is found is normay attributed to Robinson (1951). Topkis (1979) proved that the method works in GSC. I ca this method the Robinson-Topkis agorithm. Agorithm 1. The foowing are three variants of the Robinson-Topkis agorithm. T (s): Start with s 0 = s. Given s k S, et s k+1 = inf β Γ (s k ). Stop when s k = s k+1. T (s): Start with s 0 = s. Given s k S, et s k+1 = sup β Γ (s k ). Stop when s k = s k+1. T r (s): Do agorithm T (s) in Γ r (s). Lemma 5. (Topkis, 1979) T (inf S) stops at the smaest Nash equiibrium of Γ, and T (sup S) stops at the argest Nash equiibrium of Γ. See Topkis (1979) (or Topkis (1998)) for a proof of Lemma 5.

6 6 F. ECHENIQUE Remark 6. Note that T (inf S) is faster than iterating inf β Γ (s k ) suggests. When the agorithm has to find inf β Γ (s k ), it knows that searching in the interva [s k, sup S] is enough. The sequence {s k } is monotone increasing, so each iteration of T (inf S) is faster the previous iteration. A simiar thing happens to T (s) and T r (s). A round-robin version of RT where payers take turns in best-responding instead of jointy best-responding in each iteration is faster than the version above (see Topkis (1979)). I use the version above because its notation is easier. A resuts in the paper hod if round-robin RT is substituted for RT. In fact, the resuts reported in Section 7 are from the round-robin impementation. 3. How it works In the authors experience, an important idea in organizing the anaysis of a game by hand is to find one equiibrium, then ask how other equiibria might differ from this one; there is currenty no substantiation of this wisdom in theory or computationa experience. (McKevey and McLennan, 1996, p. 28) I sha use an exampe to expain how the agorithm works. The expanation shows that the agorithm is a rudimentary substantiation of McKevey and McLennan s wisdom. Consider a two-payer GSC, Γ. Suppose that payer 1 has strategy set S 1 = {1, 2,... 15}, and payer 2 has S 2 = {1, 2,... 11} (the numbers do not matter, they just happen to give a nice picture in Figure 1). The payers joint strategy space, S 1 S 2, is in Figure 1. Suppose that we have cacuated the payers best-response functions (to make things simpe, assume bestresponses are everywhere unique), β 1 and β 2. The game s best-response function is β, where β(s 1, s 2 ) = (β 1 (s 2 ), β 2 (s 1 )). Because Γ is a GSC, β 1, β 2 and β are monotone increasing functions (Lemma 2). I do not specify the payers payoffs, they are not needed in the expanation. First we need to understand how the Robinson-Topkis (RT from now on) agorithm works. RT starts at the smaest strategy profie, (1, 1), and iterates the game s best-response function unti two iterations are the same. Since (1, 1) is smaer than β(1, 1), and β is monotone, we have that β(1, 1) is smaer than β(β(1, 1)) = β 2 (1, 1). Simiary, β 2 (1, 1) is smaer than β 3 (1, 1), and so on iterating β we get a monotone increasing sequence in S. Now, S is finite, so there must be an iteration k such that β k (1, 1) = β k 1 (1, 1). But then of course β k (1, 1) = β(β k 1 (1, 1)), so s = β k 1 (1, 1) is a Nash equiibrium.

7 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 7 s 2 s s 2 s 1 s s 1 Equiibria States of the agorithm Figure 1. The agorithm in a two-payer game. It turns out that s is the smaest Nash equiibrium in Γ: Let s be any other equiibrium, and note that (1, 1) s. Monotonicity of β impies that β(1, 1) β(s ) = s. Then, iterating β we get s = β k 1 (1, 1) β k 1 (s ) = s. In a simiar way, RT finds the game s argest Nash equiibrium s by iterating the game s best-response function starting from the argest strategy profie, (15, 11). I now describe informay the agorithm that I propose. Then I expain heuristicay why it works. I deveop these ideas in fu generaity in Section 4. The agorithm consists of the foowing steps: (1) Find the smaest (s) and argest (s) Nash equiibrium using RT note s and s in Figure 1. (2) Consider Γ r (s 1, s 2 +1), the game where payer 1 is restricted to choosing a strategy arger than s 1, and payer 2 is restricted to choosing a strategy arger than s The strategy profie (s 1, s 2 + 1) is indicated in the figure with a circe above (s 1, s 2 ), and the strategy

8 8 F. ECHENIQUE space in Γ r (s 1, s 2 + 1) is the interva [(s 1, s 2 + 1), (15, 11)] shown with non-dotted ines in the figure. Now use RT to find s 1, the smaest Nash equiibrium in Γ r (s 1, s 2 + 1). Each iteration of β is shown with an arrow in the figure, and s 1 is the back disk reached after three iterations. Simiary, consider Γ r (s 1 + 1, s 2 ), the game where payer 1 is restricted to choosing a strategy arger than s 1 + 1, and payer 2 is restricted to choosing a strategy arger than s 2. The strategy profie (s 1 + 1, s 2 ) is indicated in the figure with a circe to the right of (s 1, s 2 ). Use RT to find s 2, the smaest Nash equiibrium in Γ r (s 1 + 1, s 2 ); s 2 shoud be a back disk at this point, I expain in next step why it is gray. (3) Check if s 1 and s 2 are Nash equiibria of Γ. First consider s 1. Because s 1 is an equiibrium of Γ r (s 1, s 2 + 1), and β is monotone increasing, we ony need to check that s 2 is not a profitabe deviation for payer 2. Simiary, to check if s 2 is an equiibrium we ony need to check that s 1 is not a profitabe deviation for payer 1. I expain beow why these checks are sufficient. Let us assume that s 1 passes the check whie s 2 fais, this is indicated in the figure by drawing s 2 as a gray circe. (4) Do steps 2 and 3 for Γ r (s 1 1, s ), Γ r (s , s 1 2), Γ r (s 2 1, s ), and Γ r (s , s 2 2). (5) Continue repeating steps 2 and 3 for each Nash equiibrium s k found, uness s k is equa to s. The picture shows what the agorithm does for a seection of s k s; note that the agorithm starts at arger and arger -circes, and that it approaches s. I phrased item 3 the check -phase in terms of the first iteration of the agorithm. In genera, et s k be a candidate equiibrium obtained as the smaest equiibrium in some Γ r (ŝ). To check if s k is an equiibrium I need to take a confirmed (in Γ) equiibrium s with s ŝ and check that payer i does not want to deviate to some strategy in the interva [s i, ŝ i ). In the first iteration ŝ = (s 1, s 2 + 1) and s = s, so I ony need to check that payer 2 does not want to deviate to s 2. Why is this check sufficient? First, s i s k i, and β is monotone increasing, so s i = β i (s i) β i (s k i) and hence the best possibe deviation β i (s k i) is arger than s i. Second, s k is an equiibrium in Γ r (ŝ), so no deviations arger than ŝ i are profitabe. Thus we ony need to check for deviations in the interva [s i, ŝ i ). As the iterations progress, we get arger and arger ŝ s; but normay we aso have arger and arger confirmed equiibria, and thus the intervas [s i, ŝ i ) often do not get very arge.

9 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 9 I now expain why the agorithm finds a the Nash equiibria of Γ. Suppose that s is an equiibrium, so s s s. If s = s or s = s, then the agorithm finds s in step 1. Suppose that s < s < s, then either (s 1, s 2 + 1) s or (s 1 + 1, s 2 ) s (or both). Suppose that (s 1, s 2 + 1) s, so s is a strategy in Γ r (s 1, s 2 +1). Note that s is aso an equiibrium of Γ r (s 1, s 2 +1): if a payer i does not want to deviate from s when aowed to choose any strategy in S i, she wi not want to deviate when ony aowed to choose the subset of strategies in Γ r (s 1, s 2 +1). But s 1 is the smaest equiibrium in Γ r (s 1, s 2 +1), so s 1 s. If s 1 = s the agorithm has found s. If s 1 < s then either (s 1 1, s ) s or (s , s 1 2) s (or both). Suppose that (s 1 1, s ) s, then repeating the steps above we wi arrive at a new s k s. The sequence of stricty increasing s k s ony stops when s k reaches s, so s < s impies that there must be a s k = s. Since s is an equiibrium, s k = s passes the test in item 3; hence the agorithm finds s. 4. The Agorithm Let e d i be the -th unit vector in R d i, i.e. e d i = (0,... 1, ) R d i, where 1 is the -th eement of e d i. Agorithm 2. Find s = inf E using T (inf S), and s = sup E using T (sup S). Let Ê = {s, s}. The set of possibe states of the agorithm is the power set 2 S, the agorithm starts at state {s}. Let the state of the agorithm be M 2 S. Whie M {s}, repeat the foowing sub-routine to obtain a new state M. Subroutine Let M =. For each s M, i {1,... n} and with 1 d i, if (s i + e d i, s i ) s, then do steps 1-4: (1) Let s be a maxima eement in { } s Ê : s (s i + e d i, s i ). (2) Run T r (s i + e d i, s i ); et ŝ be the strategy profie at which it stops. (3) Check that no payer j wants to deviate from ŝ j to a strategy in the set { z Sj : s j z and (s i + e d i, s i ) j z }. If no payer wants to deviate, add ŝ to Ê. (4) Add ŝ to M (Let M = M {ŝ}). Theorem 7. The set Ê produced by Agorithm 2 coincides with the set E of Nash equiibria of Γ. Proof. First I sha prove that the agorithm stops after a finite number of iterations, and that it stops when M = {s}, not before (step we-behaved ). Then I sha prove that Ê E, and then that E Ê.

10 10 F. ECHENIQUE Step we-behaved. Let M 2 S be the coection of states visited by Agorithm 2. Let C be the set of maps z : M S such that (1) For a M M, z(m) M; (2) If the agorithm transits from M to M, and there is at east one payer i and dimension such that (z(m) i + e d i, z(m) i ) s, then z(m ) is obtained from T r (z(m) i +e d i, z(m) i ) from one such payer and dimension in step 2 of Agorithm 2. Note that, for a M, M = {z(m) : z C}. First I sha prove that the agorithm stops when it reaches state {s}, and not before. I need to prove that s s for a s {M : M M}; which impies that z(m) s for a z C. Let the state M transit to state M. Let s M, then s must have been obtained from some s M, and some i and with (s i + e d i, s i ) s, by T r (s i + e d i, s i ) in step 2 of the subroutine. By Lemma 5, s is the smaest Nash equiibrium in Γ r (s i + e d i, s i ). By Lemma 4, s is a Nash equiibrium of Γ r (s i + e d i, s i ), so s (s i + e d i, s i ) s s. This proves that s s for a s M, for a M that transit to some state, and that s s for a s M for a states M that are obtained by transit from some other state. Uness s = s, these two possibiities cover a states in M, and if s = s there is nothing to prove. Now fix a state M. For a s M, s s, so if M = {s} then there is s M such that s < s. So there are i and such that (s i + e d i, s i ) s. Thus the agorithm must transit from M to a new state whie M {s}. Second, et z C. Let M be any state in M, and be M be the state that it transits to. I sha prove that, if z(m ) s, then z(m) < z(m ). If z(m ) s then, by item 2 of the definition of C, there is some i and such that z(m ) is obtained by T r (z(m) i + e d i, z(m) i ) in Step 2 of the subroutine. So, z(m) < (z(m) i + e d i, z(m) i ) z(m ). Hence z(m) < z(m ). Now, M 2 S, and S is finite, so M and therefore C are finite sets. Each z C is stricty increasing unti z(m) = s, so the binary reation M transits to M on M is transitive. Thus, eventuay z(m) = s for every z C. But then there is an M M such that z(m) = s for a z C, as C is finite. Hence M = { z(m) : z C } = {s}, and Agorithm 1 stops at state M, after a finite number of steps.

11 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 11 Step Ê E. I sha prove that Ê E by induction. First, in the initia state, {s}, Ê E by definition of Ê. Second, suppose that, when the agorithm is in state M, Ê E, and that when the agorithm transits from state M to M ŝ is added to Ê. I sha prove that ŝ E. By induction, this impies that Ê E. Suppose we obtained ŝ by running T r (s i + e d i, s i ), for some s M, and some payer i and dimension. Fix a payer j. I sha prove that ŝ j β j,γ (ŝ j ) by first finding a strategy z j β j,γ (ŝ j ), and then showing that u j (z j, ŝ j ) u j (ŝ j, ŝ j ). Let s be the maxima eement in { } s Ê : s (s i + e d i, s i ) found in step 2 of the agorithm. Note that s (s i + e d i, s i ), so the set of s Ê such that s (s i + e d i, s i ) is non-empty; thus s is we-defined. We have s j β j,γ (s j), as s is a Nash equiibrium. Let s j β j,γ (ŝ j ). Note that s j ŝ j, so Migrom and Shannon s (1994) Theorem 4 impies that z j = s j s j β j,γ (ŝ j ). By definition of z j, s j z j. First, if (s i + e d i, s i ) j z j, then u j (z j, ŝ j ) u j (ŝ j, ŝ j ), as ŝ j β (ŝ j,γ r (s i +e d i,s i ) j) because ŝ is a Nash equiibrium of Γ r (s i + e d i, s i ). Second, if (s i + e d i, s i ) j z j then u j (z j, ŝ j ) u j (ŝ j, ŝ j ) by step 4 of the subroutine. Hence u j (z j, ŝ j ) u j (ŝ j, ŝ j ), so z j β j,γ (ŝ j ) impies that ŝ j β j,γ (ŝ j ). Payer j was arbitrary, so ŝ β Γ (ŝ) and ŝ E. Step E Ê. Let s E. Suppose, by way of contradiction, that s / Ê. Caim: Let Agorithm 2 transit from state M to state M. If there is z M with z < s then there is z M with z < s. Proof of the caim: Since z < s, there is i and such that z i < s i. Then s is a strategy profie in Γ r (z i + e d i, z i ). If ŝ is the strategy profie found by T r (z i + e d i, z i ), then Lemma 5 impies that ŝ s, as s is a Nash equiibrium of Γ r (z i + e d i, z i ). If ŝ = s then s woud pass the test of step 4 and be added to Ê, but we assumed s / Ê so it must be that ŝ < s. Set z = ŝ, then z M by step 5, and the proof of the caim is compete. Now, s / Ê impies that s s. Initiay M = {s} so there is z(= s) in M with z < s. Using the Caim above inductivey, it must be that a stages of Agorithm 2 contain a z with z < s. But the fina state of the agorithm is M = {s}; a contradiction, since s s. Remark 8. A modification of Agorithm 2 wi make it run faster: Ony do step 3 of the subroutine if there is no s Ê such that ŝ s, and ŝ S(s i + e d i, s i ), for the s, i and at which s was found. For, if there is such an s, then we know that ŝ / E, as ŝ E woud impy that ŝ is

12 12 F. ECHENIQUE an equiibrium of Γ r (s i + e d i, s i ), which contradicts that s is the smaest equiibrium of Γ r (s i + e d i, s i ). Theorem 7 says that Agorithm 2 works. In the rest of the paper I show that it is efficient. 5. Exampes I present two exampes; they show how the agorithm works, and offer a first gance into when and how it is ikey to be fast Exampe 1. Consider the two-payer game on the eft in Figure 2. Both payers have identica strategy sets, {1, 2, 3, 4}. The strategies are ordered in the natura way: a strategy s i is arger than strategy s i if it is a arger number, so 2 is arger than 1, 4 is arger than 2, and so on. With this order it is straightforward if tedious to check that Exampe 1 is a game with strategic compementarities , 3 2, 3 3, 4 5, 5 3 1, 3 3, 3 3, 4 4, 4 2 2, 3 4, 3 4, 4 4, 4 1 4, 4 3, 2 3, 1 3, 0 Exampe , 0 0, 0 0, 0 0, 0 3 1, 3 1, 2 1, 1 0, 0 2 2, 3 2, 2 2, 1 0, 0 1 3, 3 3, 2 3, 1 0, 0 Exampe 2 Figure 2. Two exampes. Agorithm 2 starts by finding s = inf E and s = sup E by RT: et us first iterate the game s best response function starting at the smaest point in the strategy space, (1, 1). Now, β(1, 1) = (1, 1) so (1, 1) = inf β(1, 1), and the RT agorithm returns s = (1, 1) as the game s smaest equiibrium. Simiary, it returns s = (4, 4) as the game s argest equiibrium. Then, the initia state of the agorithm is M = {(1, 1)}, and the initia ist of equiibria is Ê = {(1, 1), (4, 4)}, see Tabe 1. M Ê 1 {(1, 1)} {(1, 1), (4, 4)} 2 {(2, 3)} {(1, 1), (2, 3), (4, 4)} 3 {(3, 3), (4, 4)} {(1, 1), (2, 3), (4, 4)} 4 {(4, 4)} {(1, 1), (2, 3), (4, 4)} Tabe 1. Iterations in Exampe 1

13 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 13 The initia state is M = {(1, 1)}. First, (1, 1) + (1, 0) = (2, 1) s, so we do steps 1-4 in the subroutine starting at (2, 1). inf β Γ r (2,1)(2, 1) = (2, 3), and inf β Γ r (2,1)(2, 3) = (2, 3), so RT in game Γ r (2, 1) returns (2, 3) as the smaest equiibrium in Γ r (2, 1). In step 3 we need to check that payer 1 does not want to deviate to pay strategy 1, but paying strategy 1 woud yied her a payoff of 3, whie paying strategy 2 yieds her a payoff of 4. Since the deviation is not profitabe, we add (2, 3) to Ê. Second, (1, 1)+(0, 1) = (1, 2) s, so we do steps 1-4 starting at (1, 2): inf β Γ r (1,2)(1, 2) = (2, 2), inf β Γ r (1,2)(2, 2) = (2, 3), and inf β Γ r (1,2)(2, 3) = (2, 3). Graphicay, the action of inf β Γ r (1,2) is (1, 2) (2, 2) (2, 3) (2, 3). Thus RT returns (2, 3) as the smaest equiibrium in Γ r (1, 2). To sum up, the resut of steps 1-4 is that the agorithm transits to state {(2, 3)}, and the ist of equiibria is Ê = {(1, 1), (2, 3), (4, 4)}. Now the state of the agorithm is {(2, 3)}. First, (2, 3)+(1, 0) = (3, 3) s, so we do steps 1-4 in the subroutine starting at (3, 3). Note that inf β Γ r (3,3)(3, 3) = (3, 3), so RT returns (3, 3) as the smaest equiibrium in Γ r (3, 3). In step 3 we need to check that payer 1 does not want to deviate from strategy 3 to strategy 2. In fact, strategy 2 gives a higher payoff (4) than strategy 3 (3), so (3, 3) is not an equiibrium, and we do not add (3, 3) to Ê. Second, (2, 3) + (0, 1) = (2, 4) s, so we do steps 1-4 in the subroutine starting at (2, 4). Note that inf β Γ r (2,4)(2, 4) = (4, 4), so RT returns (4, 4) as the smaest equiibrium in Γ r (4, 4). We aready know that (4, 4) is an equiibrium of Γ. The resut of steps 1-4 is that the agorithm transits to state {(3, 3), (4, 4)}, and the ist of equiibria is Ê = {(1, 1), (2, 3), (4, 4)}. The state of the agorithm is now {(3, 3), (4, 4)}. Both (4, 4) + (1, 0) and (4, 4)+(0, 1) fai to be smaer than s, so we do not run the subroutine starting from (4, 4). Now, steps 1-4 in the subroutine starting from (3, 3) + (1, 0) = (4, 3) or (3, 3)+(0, 1) = (3, 4) give (4, 4) as the smaest equiibrium of Γ r (3, 4) and Γ r (4, 3). So, the fina state of the agorithm is {(4, 4)}, and the fina ist of equiibria is {(1, 1), (2, 3), (4, 4)} Exampe 2. Now consider the game on the right in Figure 2. RT yieds (1, 1) as the smaest equiibrium, and (4, 4) as the argest equiibrium in Exampe 2. The initia state of the agorithm is thus {(1, 1)}. We start the subroutine at (2, 1) = (1, 1) + (1, 0) and get back (2, 1) as the smaest equiibrium of Γ r (2, 1). But payer 1 prefers strategy 1 over strategy 2, so (2, 1) does not survive step 3. We start the subroutine at (1, 2) = (1, 1)+(0, 1) and get back (1, 2) as the smaest equiibrium of Γ r (1, 2). But payer 1 prefers strategy 1 over strategy 2, so (1, 2) does not survive step 3.

14 14 F. ECHENIQUE If one competes a iterations (shown in Tabe 2) it is cear that the agorithm stops at a strategy profies, and discards a but the argest and the smaest equiibria of the game. M Ê 1 {(1, 1)} {(1, 1), (4, 4)} 2 {(2, 1), (1, 2)} {(1, 1), (4, 4)} 3 {(3, 1), (2, 2), (1, 3)} {(1, 1), (4, 4)} 4 {(4, 1), (3, 2), (2, 3), (1, 4)} {(1, 1), (4, 4)} 5 {(4, 2), (3, 3), (2, 4)} {(1, 1), (4, 4)} 6 {(4, 3), (3, 4)} {(1, 1), (4, 4)} 7 {(4, 4)} {(1, 1), (4, 4)} Tabe 2. Iterations in Exampe 2 Exampe 2 presents a pathoogica situation; the agorithm is forced to check a strategy profies of the game. The root of the probem is that, after each iteration, it is optima for the payers to choose their smaest aowed strategies. I argue in Section 6 that the situation wi typicay not occur in the games that one encounters in appications. 6. How fast is Agorithm 2? 6.1. Outine. The rest of the paper estabishes that Agorithm 2 is generay very fast. Here is an outine: Section 7. I simuate a arge cass of games, and show that Agorithm 2 finds a equiibria very quicky. The cass of games was chosen trying to bias the test against Agorithm 2. Section 8. I show that Agorithm 2 is faster when best-responses in each iteration have arge increases I then argue that this wi occur for many natura appications of the agorithm. Section 9. I present a version of the agorithm that appies to twopayer games with strict preferences. This version is faster than Agorithm 2. In the worst case, Agorithm 2 is sow (see Exampe 2 of Section 5). But the worst case is not surprisingy irreevant for actua appications The Benchmark. I now describe the trivia agorithm, the benchmark against which Agorithm 2 is compared in terms of speed. Let Γ = {(S i, u i ) : i = 1,... n} be an n-payer game. Fix a payer, say i. First, for each s i, find the set of best-responses by i. Second, check if any payer j i wants to deviate from her strategy in s i when i chooses one of

15 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 15 her best-responses. This agorithm is essentiay the underining method for finding the Nash equiibria that one teaches first-year students. Suppose that a payers have K strategies, and that best-responses are everywhere unique. Let r be the time required to make a payoff-function evauation. Note that r is independent of K. The trivia agorithm turns out to require O(rK n ) time: The agorithm performs K payoff-function evauations to find the best-responses, K n 1 times. Thus the agorithm performs K n payoff-function evauations. By the accounting procedure in Aho, Hopcroft, and Uman (1974), the agorithm is O(rK n ). If best-responses are not unique, the agorithm needs to check if any payer j i wants to deviate from her strategy in s i, for a of i s best-responses. In the worst case, the set of best responses grows at rate K, so the agorithm is O(rK n+1 ). Note that the O(rK n ) cacuation is not an unreaistic worst-case bound. It is the time the trivia agorithm must use, as ong as best-responses are unique. For n-payer games with n > 2, Bernhard von Stenge (persona communication) has suggested a recursive procedure that improves on the trivia agorithm: for each s n S n, fix s n as the strategy payed by payer n, and find a equiibria of the resuting (n 1)-payer game. For each equiibrium found, check if payer n wishes to deviate from s n. In the worst-case cacuation, this recursive procedure does not improve on the simpe trivia agorithm but in many games of interest it may speed up the trivia agorithm by saving on cacuations of best-response by payer n. If Γ is a GSC, for any s n S n, the resuting (n 1)-payer game is aso a GSC. So von Stenge s suggestion can be appied to Agorithm 2 as we. But I do not know if foowing his suggestion improves over the speed of Agorithm 2 or not. 7. Performance I evauate the performance of Agorithm 2, using a cass of two-payer games, where each payer has the interva [0, 1] as her strategy space. The agorithm is fast; I use Agorithm 2 with different discretizations grids of [0, 1], and show that, even when the resuting grid is quite sma (the number of strategies of each payer is quite arge), the agorithm is very fast. I use the computations to compare Agorithm 2 to the trivia agorithm Cass of games. I use a cass of games that tend to have a arge number of equiibria Agorithm 2 is faster the smaer is the number of equiibria, and I want to evauate Agorithm 2 using games where it does not have

16 16 F. ECHENIQUE One game Strat. Trivia Ag. 2 20, min 2.96 sec. 40, min 10.0 sec. 60, min 10.8 sec. Tabe 3. Simuations. 2,000 games Strat. Trivia Ag. 2 20, days 1.6 hours 40, days 5.5 hours 60, days 6.0 hours an apriori advantage. The cass of games is idiosyncratic, but that is unavoidabe: the functiona forms that economists normay use give few or unique equiibria. With those functiona forms, Agorithm 2 is faster sti. The games have two-payers, each payer i has strategy set S i = [0, 1], and payoff function u i (s i, s i ) = (α i /10)(s i s i ) β i sin(100s i ) + (1/100) [(1 α i )s i (1 + s i ) (1/2 β i )s 2 i /100]. The parameters α i and β i are in [0, 1]. I arrived at the above functiona form by trying to come up with games that have a fairy arge number of equiibria. The first summand is a purecoordination term, its roe is to produce mutipe equiibria. The roe of the second summand is to provoke mutipe maxima (so that preferences are not strict, see Section 9); the second summand aso heps in getting mutipe equiibria. 3 The third and fourth summand are variants of poynomia terms that I found by tria and error often produce mutipe equiibria. Note that, for a α i and β i, ({1, 2}, {S 1, S 2 }, {u 1, u 2 }) is a GSC. I discretized the payers strategy spaces, so each payer i chooses a strategy in S i = {k/k : 0 k K}. Parameters α i and β i are chosen at pseudorandom from [0, 1] using a uniform distribution Resuts. The resuts of the simuations are in Tabe 3. I first compare the performance of Agorithm 2 and the trivia agorithm. Then I discuss what the simuations say about Agorithm 2 in genera. I simuated a arge number of games, and used the agorithms to find the equiibria of each game. In each individua game, the parameters α i and β i were generated at pseudo-random from a uniform distribution on [0, 1]. The average resuts are in the tabe on the eft. On average, when each payer has 20,000 strategies, Agorithm 2 needed 2.96 seconds to find a equiibria. The trivia agorithm needed 2.8 minutes to do the same work. The tabe reports the resuts for 40,000 and 60,000 as we. 3 Simuations without the second summand are sighty faster, and the games have fewer equiibria.

17 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS Simuation of 2000 games, grid size X Time(seconds) Number of equiibria Figure 3. Reation between time and number of equiibria. The tabe on the right reports how much each agorithm needs to find a the equiibria of 2,000 games presumaby a representative task of the estimation or caibration of an economic mode. Aready with 20,000 strategies the trivia agorithm needs 3.8 days. Agorithm 2, on the other hand, needs ony 1.6 hours. On the whoe, Agorithm 2 is remarkaby fast. It finds a equiibria of a game with strategy-profies in ony 10 seconds, and it finds the equiibria of 2,000 such games in 6 hours. The graph in Figure 3 shows that nothing is hidden in the averages; the graph pots a 2,000 games. It shows how many equiibria each game has, and how ong it takes for the agorithm to find them. Note that, with one exception, Agorithm 2 never needs more than one minute to find a equiibria. Tabe 4 compares the performance of the two agorithms. With 20,000 strategies, Agorithm 2 is 56 times faster than the Trivia agorithm. The

18 18 F. ECHENIQUE Strat. Trivia/Ag. 2 20, , , Tabe 4. Comparison of Trivia and Agorithm 2 reative performance of Agorithm 2 improves with the number of strategies, because the performance of Agorithm 2 is sensitive to the number of equiibria more than to the number of strategies Impementation. I wrote an impementation in C. The code (and the output from the simuations reported above) can be downoaded from The difficuty in impementing Agorithm 2 is that the state, M, of the agorithm is potentiay taken from a arge set of possibe states. Reserving space for the possibe vaues that M can take may sow down the agorithm consideraby. I found a rudimentary soution in my impementation of the agorithm; hopefuy a better programmer can write a more efficient impementation. The average computations on the eft in Tabe 3 are based on 100 simuations in the case of the trivia agorithm, and 2,000 simuations in the case of Agorithm 2. Obviousy, the trivia agorithm needs the same amount of time to compute the equiibria of each game. There is a very sight difference in the actua computation of some games, probaby due to how the computer organizes the task, or because some payoff functions require sighty more time than others. But a 100 games with 20,000 strategies required essentiay 2.8 minutes (the output can be downoaded from the web-page indicated above). The cacuations for 2,000 games in the case of the trivia agorithm are thus a projection. A the simuations were done on a Linux De Precision PC with a 1.8 GHz Xeon CPU and 512 MB Ram. The computer performs foatingpoint operations per second (based on output MFLOPS(1), from A Aburto s fops.c program). 8. Cacuating best-responses 8.1. The source of sowness. The exampes is Section 5 show that Agorithm 2 can be sow. I sha argue that the exampes are in some sense pathoogica, and that the agorithm is ikey to be very fast in most situations in particuar when the source of the game is the discretization of a game with continuous strategy-spaces. The agorithm is sow when best-response cacuations advance sowy. For exampe consider a game with two payers, each with K = 10 strategies:

19 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 19 the numbers 31, 32, If payer 1 s best response to 2 paying 31 is 31, then computing 1 s best-response to 2 paying 32 requires 10 computations; it requires computing the payoffs from a strategies, and finding the maximum. But if 1 s best response to 2 paying 31 is 39, then computing 1 s best response to 32 requires ony 2 computations best responses are monotone increasing, so it is enough to compare the payoffs from paying 39 and 40. Hence, if best-responses advance sowy reative to K as K grows then Agorithm 2 is reativey sow. On the other hand, if best-responses advance at rate K or faster, the agorithm wi be fast. In fact, I sha estabish that the agorithm wi be inear in the worst case. When wi best-responses advance quicky? Consider an n-payer game, and suppose that payer i has strategy-space [0, 1]. We can appy Agorithm 2 by first discretizing i s strategy-space to {k/k : k = 0,... K}. Figure 4 iustrates the situation. On the top part of the figure is payer i s payoff function s i u i (s i, s i ), hoding opponents strategies fixed at s i. In the midde of the figure is i s best-response to s i, when i is restricted to choosing s i or a arger strategy her strategy-space in Γ r (s i, s i ). We seect the smaest best response when there is more than one. For strategies s i [0, s 1 i ], s 1 i is the best-response. For strategies s i (s 1 i, s 2 i ), it is a bestresponse to choose s i ; the soution is at a corner. For s i [s 2 i, s 3 i ], on the other hand, i s best response is to choose s 3 i. Finay, for s i (s 3 i, 1], it is again optima for i to be at a corner, and s i is the unique best-response in the restricted game. The bottom of Figure 4 shows β i,γ r (s i,s i )(s i, s i ) s i. I now show that best-responses advance quicky over sets of s i such that β i,γ r (s i,s i )(s i, s i ) s i > 0 (indicated as Fast regions in Figure 4), and sowy over sets of s i such that β i,γ r (s i,s i )(s i, s i ) s i = 0 (indicated as Sow regions in Figure 4). Let k (K)/K be i s smaest best-response to s i in the discretized version of the game i.e. the game where i has strategy-space {k/k : k = 0,... K}. Assume that i s payoff function is continuous, then the maximum theorem ensures that inf β i,γ r (s i,s i )(s i, s i ) im inf K K k (K)/K, as the set of maximizers parameterized by K is upper hemicontinuous. First, et s i be such that inf β i,γ r (s i,s i )(s i, s i ) s i > 0. Then k (K) s i grows at rate at east K. So the number of strategies that we eiminate from future best-response cacuations, k (K), grows at rate at east K, and thus best-responses advance quicky. In fact (see 8.2) it wi in the worst case advance ineary. Second, et s i be such that inf β i,γ r (s i,s i )(s i, s i ) s i = 0. As K grows, k (K)/K wi approach s i, so the number of strategies we eiminate from future cacuations wi be, in the imit, zero. In practice, the agorithm wi

20 20 F. ECHENIQUE u i (s i, s i ) s 1 i s 2 i s 3 i s i β Γ r (s)(s) s i β Γ r (s)(s) s i Fast Sow Fast Sow s i Figure 4. Sow and fast regions.. advance one-step-at-a-time, unti an increase in some other payer s strategy pus the agorithm away from the region.

21 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 21 What do we earn about Agorithm 2? We, it wi advance quicky over fast regions, and require more time in sow regions. The resut depends on the game: it depends on the size of the regions, on how much time it spends in each region, and on whether the agorithm wi actuay end up in these regions. For exampe, the argest equiibrium often impies that the agorithm does not need to search over the sow regions on the upper side of payers strategy spaces. Normay, fast regions are important enough, and the agorithm is fast enough over these regions (faster than inear), that the agorithm finds a equiibria very quicky Bounding how much best-responses change with K. The argument in Section 8.1 says that sowness depends on how sowy best responses advance. But it is too strong to require that best responses advance quicky gobay normay there are some strategy-profies at which best responses advance sowy. One possibe soution is to contro the size of the regions in strategy space where best responses must advance sowy. I deveop a different soution beow. I consider a famiy of games, and assume that, at each s in strategy space, the average best-response over the famiy of games does not advance sowy. It foows that Agorithm 2 is on average inear. Let (Ω, F, P ) be a probabiity space. Let (Γ ω, ω Ω) be a famiy of GSC such that Γ ω has n payers, for a ω Ω, and each payer i has strategy-space {1, 2,... K}. 4 Assume that there is γ (0, 1) such that, for any strategy-profie s, γ(k s i ) E inf β i,γ r ω (s)(s) s i. Note that S i is finite, so E inf β i,γ r ω (s)(s) is we-defined. Let T (ω) be the time required by Agorithm 2 to find a candidate equiibria. I use the accounting procedure in (Aho, Hopcroft, and Uman, 1974, p ) to cacuate T (ω), which here amounts to counting the number of payoff-function evauations that the agorithm performs. Strategy spaces are fixed, so T (ω) has finite range, and thus E(T ) is we-defined. Proposition 9. E(T ) is O(nK/γ 2 ). Proof. Let L be the number of iterations in Agorithm 2. Each iteration, = 1, 2,... L, has I iterations in the RT-agorithm. Each iteration h, h = 1,... I, impies a best-response cacuation for each payer i. Fix one payer. Let x h be the strategy that is a best response in iteration of Agorithm 2 and iteration h of the corresponding RT agorithm. Let K be the number of strategies we need to consider for that payer in iteration ; note that 4 I restrict the payers strategies to save on notation. One can easiy extend the argument to more genera strategy spaces.

22 22 F. ECHENIQUE K = K x I 1 1. If the cacuated best response in iteration h 1 was xh 1, the best-response cacuation is done in K x h 1 time. We thus need to bound L I (K x h ). =1 h=0 Consider the -th iteration of the RT agorithm, where each payer has K strategies. I sha prove that by induction. First, I 1 E (K x h ) EK /γ h=0 E { I 1 h=0 (K x h ) = E I 2 h=0 (K x h ) + E [ (K x I 1 ) x ]} I 2 { I } 2 E h=0 (K x h ) + (1 γ)(k x I 2 ), as E [ (K x I 1 ) x ] I 2 (1 γ)(k x I 2 ). Second, et I 1 (I 1) (m+1) m E (K x h ) E (K x h ) + (1 γ) h (K x I m ). h=0 h=0 Then { (I 1) (m+1) E h=0 (K x h ) + } m h=0 (1 γ)h (K x I m ) { = (I 1) (m+1) E h=0 (K x h ) + [ m h=0 (1 γ)h E (K x I m h=0 ) x I (m+1) { (I 1) (m+1) E h=0 (K x h ) + } m h=0 (1 γ)h (K x I m ). Induction on m now proves I 1 I 1 [ E (K x h ) EK (1 γ) h = EK ] 1 (1 γ) I 1 /g K /γ h=0 h=0 Now, by a simiar cacuation E L K /γ E =1 L =1 (1 γ) 1 j=0 I j K/γ E L (1 γ) I K/γ, =1 ]}

23 FINDING ALL EQUILIBRIA IN GAMES OF STRATEGIC COMPLEMENTS 23 where I = max {I : = 1,... L}. Then L 1 (1 γ)i(l+1) E K /γ E K/γ K/γ 2. 1 (1 γ) =1 But there are n payers, so the expected time used by the agorithm is bounded by nk/γ Two-payer games with strict preferences Let Γ be a two-payer game where payers have strict preferences and d 1 = d 2 = 1. I present a simpe version of Agorithm 2 that finds a the equiibria of Γ. I can bound the compexity of this simpe version of Agorithm 2. Agorithm 3. Find s = inf E using T (inf S), and s = sup E using T (sup S). Let Ê = {s, s}. The set of possibe states of the agorithm is S, the agorithm starts at state s. Let the state of the agorithm be m S. Whie m s, repeat the foowing sub-routine to obtain a new state m. Subroutine If m + (1, 1) s, then do steps 1-4: (1) Let s be a maxima eement in } { s Ê : s m + (1, 1). (2) Run T r (m + (1, 1)); et ŝ be the strategy profie at which it stops. (3) Check that no payer j wants to deviate from ŝ j to a strategy in the interva [ s j, (m + (1, 1)) j ]. If no payer wants to deviate, add ŝ to Ê. (4) Let m = ŝ. Say that Agorithm 3 makes an iteration each time it does steps 1-4. Say that Agorithm 3 makes a payoff-function evauation each time it cacuates u 1 or u 2. Let r be the time required to make a payoff-function evauation. Note that r is independent of K 1 and K 2. Let K = min {K 1, K 2 } and K = max {K 1, K 2 }. Theorem 10. Agorithm 3 finds a Nash equiibria in O(rK 2 ) time, and does at most K iterations. Proof. First I prove that Agorithm 3 is we-behaved and stops when it says that it stops. Let M S be the set of states visited by Agorithm 3. Note that, for a m M, m + (1, 1) m, for a m obtained at a ater iteration of the subroutine. Further, at each iteration of the subroutine there is a unique m found. So, if m, m M then either m is found after m and

24 24 F. ECHENIQUE m + (1, 1) m, or m is found after m and m + (1, 1) m. Then, M is totay ordered, and for any m, m M, if m m then either m+(1, 1) m or m +(1, 1) m. I sha prove beow that m s impies that m+(1, 1) s; so if m s then the agorithm does not stop at m. Since M is finite, the agorithm stops in a finite number of steps, and it stops when the state is s. I need to prove the foowing Caim. If s E(Γ r ( s)), then either s = inf E(Γ r ( s)), or inf E(Γ r ( s)) + (1, 1) s. Proof of the caim. Suppose that s E(Γ r ( s)), and that s ŝ = inf E(Γ r ( s)). By Lemma 4, ŝ s. Suppose without oss of generaity that s 1 ŝ 1. Now, ŝ 1 β Γ r ( s)(ŝ 2 ) so s 2 = ŝ 2 woud impy that β Γ r ( s)(ŝ 2 ) has at east two different eements, s 1 and ŝ 1. Impossibe since payers in Γ have strict preferences. It must be then that s 2 ŝ 2. But then ŝ s impies ŝ 1 < s 1 and ŝ 2 < s 2, so ŝ + (1, 1) s. This proves the caim. I now prove that M m s impies that m + (1, 1) s: et m M {inf S (1, 1)} be the state from which m was obtained by T r (m +(1, 1)). There must be such an m by definition of m: either m is found in step 2 of the agorithm, or m = s, and thus m was found by T (inf S) = T r (inf S). Now, m = inf E(Γ r (m + (1, 1))) and s E(Γ r (m + (1, 1))), so the caim and m s impies that m + (1, 1) s. Second, I prove that Ê = E. The proof that Ê E is very simiar to the proof that Ê E in Theorem 7, so I omit it. I sha prove that E Ê. Let s E and suppose, by way of contradiction, that s / Ê. Let m be some state of the agorithm such that m s, we must have m s or s woud be added to Ê, since s E impies that s passes the test in step 3. The caim impies that m + (1, 1) s, as m s for the same reason that m s. Now induct on M: M s s, and if, at some state m, m s, then m + (1, 1) s for the state m that the state transits to. By induction, we must have s + (1, 1) s. A contradiction, as s E impies that s s. Now I sha prove that the agorithm needs ess than K iterations. First, each iteration of Agorithm 3 produces one and ony one eement of M, so there are no more iterations than there are eements in M. Second, M {1,... K 1 } {1,... K 2 }, and for each m, m M, m m then either m + (1, 1) m or m + (1, 1) m. Thus M cannot have more eements than either {1,... K 1 } or {1,... K 2 }. Thus, M has not more than min {K 1, K 2 } = K eements. Now I sha prove that Agorithm 3 needs no more than 2K 2 payoff-function evauations. If K 1 K 2, et us change the game: add stricty dominated strategies to the payer with the smaest K i unti that payer has as many strategies as the other payer. The set of equiibria do not change, and besides