Complexity of Iterated Dominance and a New Definition of Eliminability

Transcription

1 Complexity of Iterated Dominance and a New Definition of Eliminability Vincent Conitzer and Tuomas Sandholm Carnegie Mellon University 5000 Forbes Avenue Pittsburgh, PA {conitzer, sandholm}@cs.cmu.edu April 8, 2004 Abstract While the Nash equilibrium solution concept is studied more and more intensely in our community, the perhaps more elementary concept of iterated dominance has received much less attention. This paper makes two main contributions. First, we prove some basic complexity results about iterated dominance. We show that determining whether there is some path that eliminates a given strategy is NP-complete with weak dominance. This allows us to also show that determining whether there is a path that leads to a unique solution is NP-complete. Both of these results hold regardless of whether dominance by mixed strategies is allowed. (The second result was already known for the case of dominance by pure strategies only, in a slightly less restricted setting [4].) As our second contribution, we present a new definition of elimination of strategies, and show that it spans a spectrum of solution concepts from strict dominance to Nash equilibrium. We point out that checking whether a strategy is eliminable relative to the sets of all strategies is conp-complete under this definition. 1 Introduction In settings with self-interested agents, the optimal action for one agent may depend on the actions taken by other agents. In such settings, the agents require tools from game theory to rationally decide on an action. Game theory offers various formal models of strategic settings (the best-known of which is a game in normal (or matrix) form, specifying a payoff (utility) for each agent for each combination of strategies that the agents choose), as well as solution concepts, which, given a game, specify which outcomes are reasonable (under various assumptions of rationality and common knowledge). Probably the best-known (and, in our community, certainly the most-studied) solution concept is that of Nash equilibrium. A Nash equilibrium specifies a strategy for each player, in such a way that no player has an incentive to (unilaterally) deviate from 1

2 the prescribed strategy. The complexity of constructing a Nash equilibrium in matrix games has been labeled one of the two most important open problems on the boundary of P today [8]. The best-known algorithm for computing a Nash equilibrium, the Lemke-Howson algorithm [5], has a worst-case exponential running time [9]. Perhaps surprisingly, the more elementary solution concepts of dominance and iterated dominance have received much less attention in the computer science community. (The main computational study of these concepts has taken place in a paper in the game theory community [4].) A strategy strictly dominates another strategy if it performs strictly better against all opponent strategies, and weakly dominates it if it performs at least as well against all opponent strategies, and strictly better against at least one. The idea is that dominated strategies can be eliminated from consideration. In iterated dominance, the elimination proceeds in rounds, and becomes easier as more strategies are eliminated: in any given round, the dominating strategy no longer needs to perform better than or as well as the dominated strategy against opponent strategies that were eliminated in earlier rounds. One of the main reasons that these solution concepts have received less attention is the fact that they are often too strong, in the following sense. Oftentimes it is not the case that all but one strategy can be eliminated this way for each player, and thus there is no solution (consisting of a single strategy for each player) that is consistent with the concept. 1 In contrast, a Nash equilibrium always exists in finite games when mixed strategies (probability distributions over pure strategies) are allowed [7]. Unfortunately, the Nash equilibrium concept is often too weak: typically multiple Nash equilibria exist. The problem of equilibrium selection is a major one: in general, if the players are choosing their strategy according to different Nash equilibria, the result is not a Nash equilibrium. 2 Another interpretation of this is that the two solution concepts ((iterated) dominance and Nash equilibrium) actually suffer from the same problem: multiple strategies for a player may be consistent with it (because none of them can be dominated, or all of them occur in some Nash equilibrium). Ideally, we would have a sensible solution concept that prescribes a single solution for every game, but such a concept has eluded game theorists to this day. In this paper, we study some fundamental complexity questions concerning iterated dominance. We also propose a new elimination concept that, in terms of how powerful it is in eliminating strategies, spans a spectrum from strict dominance to Nash equilibrium. 2 Definitions In this section, we briefly review dominance and iterated dominance. Player i s strategy σ i is said to be strictly dominated by player i s strategy σ i if for any vector of strategies σ i for the other players, u i (σ i,σ i ) >u i (σ i,σ i). The definition is the same for weak dominance, except the inequality is not required to be strict but it should be strict for at least one vector σ i (so that two strategies cannot dominate each other). In this definition, it is sometimes allowed for σ i to be a mixed strategy (probability distribution over pure strategies), in which case the associated payoffs are the expected payoffs. For the purposes of this definition, the other players ( i) can be thought of as a single player. In iterated dominance, dominated strategies are eliminated, and no longer have any 1 Perhaps the one place where dominance is often studied is in the mechanism design literature, where the objective is to design the game itself. The freedom to design the game typically makes it possible to ensure that the game can be solved by dominance. 2 An exception is the case of two-player zero-sum games, where one player s loss is always the other s gain. Here, any combination of Nash equilibrium strategies (in this context also known as maximin strategies) is a Nash equilibrium. 2

3 effect on future dominance relations. For example, σ 1 may originally not dominate σ 1 because the latter performs better against σ 2; but then, once σ 2 is eliminated because it is dominated by σ 2, σ 1 dominates σ 1, and the latter can be eliminated. Either strict or weak dominance can be used in this definition. More detailed discussions and examples can be found in standard texts on game theory or microeconomics [2, 6]. 3 Complexity of iterated dominance We now move on to iterated dominance. It is well-known that iterated strict dominance is path-independent (in the end the remaining strategies will be the same regardless of the order in which strategies are eliminated), and the elimination procedure can be executed in polynomial time (even when dominance by mixed strategies is allowed, because a simple linear program can be set up to determine whether a given strategy is dominated). In contrast, iterated weak dominance is path-dependent, and here we will investigate two questions: whether a given strategy is eliminated in some path, and whether there is a path to a unique solution. Definition 1 Given a game in matrix form and a distinguished strategy σ, IWD- STRATEGY-ELIMINATION asks whether there is some path of iterated weak dominance that eliminates σ. Given a game in matrix form, IWD-UNIQUE-SOLUTION asks whether there is some path of iterated weak dominance that leads to a unique solution (one strategy left per player). The following lemma shows a special case in which allowing for weak dominance by mixed strategies does not help. We will prove our hardness results in this setting, so that they will hold whether or not dominance by mixed strategies is allowed. Lemma 1 Suppose the only payoffs in a game are {0, 1}. Then every pure strategy that is weakly dominated by a mixed strategy is also weakly dominated by a pure strategy. Proof: Suppose pure strategy σ is weakly dominated by mixed strategy σ.ifσgets a payoff of 1 against some opponent strategy (or vector of opponent strategies if there are more than 2 players), then all the pure strategies that σ places positive probability on must also get a payoff of 1 against that opponent strategy (or else the expected payoff would be smaller than 1). Moreover, at least one of the pure strategies that σ places positive probability on must get a payoff of 1 against an opponent strategy that σ gets 0 against (or else the inequality would never be strict). It follows that this pure strategy weakly dominates σ. We are now ready to prove the main results of this section. Theorem 1 IWD-STRATEGY-ELIMINATION is NP-complete, even with 2 players, and with 0 and 1 being the only utilities occurring in the matrix whether or not dominance by mixed strategies is allowed. Proof: The problem is in NP because given a sequence of strategies to be eliminated, we can easily check whether this is a valid sequence of eliminations (even when dominance by mixed strategies is allowed, because a simple linear program can be set up to determine whether a given strategy is dominated). To show that the problem is NP-hard, we reduce an arbitrary SAT instance (given by a nonempty set of clauses C over a nonempty set of variables V, with corresponding literals L = {+v : v V } { v :v V}) to the following IWD-STRATEGY-ELIMINATION instance. (In this instance, we will specify that certain strategies are uneliminable. A strategy σ r 3

4 can be made uneliminable, even when 0 and 1 are the only allowed payoffs, by adding another strategy σ r and another opponent strategy σ c, so that: 1. σ r and σ r are the only strategies that give the row player a utility of 1 against σ c.2.σ r and σ r always give the row player the same utility. 3. σ c is the only strategy that gives the column player a utility of 1 against σ r, but otherwise σ c always gives the column player utility 0. This makes it impossible to eliminate any of these three strategies. We will not explicitly specify the additional strategies to make the proof more legible.) In this proof, we will denote row player strategies by s, and column player strategies by t, to improve legibility. Let the row player s pure strategy set be given as follows. For every variable v V, the row player has corresponding strategies s 1 +v,s 2 +v,s 1 v,s 2 v. Additionally, the row player has the following 2 strategies: s 1 0 and s 2 0, where s 2 0 = σ r (that is, it is the strategy we seek to eliminate). Finally, for every clause c C, the row player has corresponding strategies s 1 c (uneliminable) and s 2 c. Let the column player s pure strategy set be given as follows. For every variable v V, the column player has a corresponding strategy t v. For every clause c C, the column player has a corresponding strategy t c, and additionally, for every literal l L that occurs in c, a strategy t c,l. The column player has a single additional strategy t 1. For every variable v V, the column player has corresponding strategies t +v,t v (both uneliminable). Finally, the column player has two strategies t 1 0 (uneliminable) and t 2 0. The utility function for the row player is given as follows: u r (s 2 +v,t v )=1for all v V ; u r (s 1 v,t v )=1for all v V ; u r (s 1 +v,t 1 )=1for all v V ; u r (s 2 v,t 1 )=1for all v V ; u r (s b +v,t +v )=1for all v V and b {1,2}; u r (s b v,t v )=1for all v V and b {1,2}; u r (s 2 0,t c )=1for all c C; u r (s b 0,t 1 0)=1for all b {1,2}; u r (s 1 0,t 2 0)=1; and the row player s utility is 0 in every other case. The utility function for the column player is given as follows: u c (s 2 l,t c)=1for all c C and l L where l c (literal l occurs in clause c); u c (s 2 l 2,t c,l1 )=1for all c C and l 1,l 2 L, l 1 l 2 where l 2 c; u c (s 1 c,t c )=1for all c C; u c (s b c,t c,l )=1for all c C, l L, and b {1,2}; and the column player s utility is 0 in every other case. We now show the two instances are equivalent. First, suppose there is a solution to the SAT instance: that is, a truth-value assignment to the variables in V such that all clauses are satisfied. Then, consider the following sequence of eliminations in our game: 1. For every variable v that is set to true in the assignment, eliminate t v (which gives the column player utility 0 everywhere). 2. Then, for every variable v that is set to true in the assignment, eliminate s 2 +v using s 1 +v (which is possible because t v has been eliminated, and because t 1 has not been eliminated (yet)). 3. Now eliminate t 1 (which gives the column player utility 0 everywhere). 4. Next, for every variable v that is set to false in the assignment, eliminate s 2 v using s 1 v (which is possible because t 1 has been eliminated, and because t v has not been eliminated (yet)). 5. For every clause c which has the variable corresponding to one of its positive literals l =+vset to true in the assignment, eliminate t c using t c,l (which is possible because s 2 l has been eliminated, and s2 c has not been eliminated (yet)). 6. For every clause c which has the variable corresponding to one of its negative literals l = v set to false in the assignment, eliminate t c using t c,l (which is possible 4

5 because s 2 l has been eliminated, and s2 c has not been eliminated (yet)). 7. Because the assignment satisfied the formula, all the t c have now been eliminated. Thus, we can eliminate s 2 0 = σr using s 1 0. It follows that there is a solution to the IWD-STRATEGY-ELIMINATION instance. Now suppose there is a solution to the IWD-STRATEGY-ELIMINATION instance. By Lemma 1, we can assume that all the dominances are by pure strategies. We first observe that only s 1 0 can eliminate s 2 0 = σr, because it is the only other strategy that gets the row player a utility of 1 against t 1 0, and t 1 0 is uneliminable. However, because s 2 0 performs better than s 1 0 against the t c strategies, it follows that all of the t c strategies must be eliminated. For each c C, the strategy t c can only be eliminated by one of the strategies t c,l (with the same c), because these are the only other strategies that get the column player a utility of 1 against s 1 c, and s 1 c is uneliminable. But, in order for some t c,l to eliminate t c, s 2 l must be eliminated first. Only s 1 l can eliminate s 2 l, because it is the only other strategy that gets the row player a utility of 1 against t l, and t l is uneliminable. We next show that for every v V only one of s 2 +v,s 2 v can be eliminated. This is because in order for s 1 +v to eliminate s 2 +v, t v needs to have been eliminated and t 1, not (so t v must be eliminated before t 1 ); but in order for s 1 v to eliminate s 2 v, t 1 needs to have been eliminated and t v, not (so t 1 must be eliminated before t v ). So, set v to true if s 2 +v is eliminated, and to false otherwise Because by the above, for every clause c, one of the s 2 l with l c must be eliminated, it follows that this is a satisfying assignment to the SAT instance. Using Theorem 1, it is now (relatively) easy to show that IWD-UNIQUE-SOLUTION is also NP-complete. Theorem 2 IWD-UNIQUE-SOLUTION is NP-complete, even with 2 players, and with 0 and 1 being the only utilities occurring in the matrix whether or not dominance by mixed strategies is allowed. Proof: Again, the problem is in NP because we can nondeterministically choose the sequence of eliminations and verify whether it is correct. To show NP-hardness, we reduce an arbitrary IWD-STRATEGY-ELIMINATION instance to the following IWD- UNIQUE-SOLUTION instance. Let all the strategies for each player from the original instance remain part of the new instance, and let the payoffs resulting from the players playing a pair of these strategies be the same. We add three additional strategies σ 1 r,σ 2 r,σ 3 r for the row player, and three additional strategies σ 1 c,σ 2 c,σ 3 c for the column player. Let the additional payoffs be as follows: u r (σ r,σ j c)=1for all σ r / {σ 1 r,σ 2 r,σ 3 r}and j {2,3}; u r (σ i r,σ c )=1for all i {1,2,3}and σ c / {σ 2 c,σ 3 c}; u r (σ i r,σ 2 c)=1for all i {2,3}; u r (σ 1 r,σ 3 c)=1; and the row player s utility is 0 in all other cases involving a new strategy. u c (σ 3 r,σ c )=1for all σ c / {σ 1 c,σ 2 c,σ 3 c}; u c (σ r,σ j c)=1for all j {2,3}(σ r is the strategy to be eliminated in the original instance); u c (σ i r,σ 1 c)=1for all i {1,2}; u r (σ 1 r,σ 2 c)=1; u r (σ 2 r,σ 3 c)=1; and the column player s utility is 0 in all other cases involving a new strategy. We proceed to show the two instances are equivalent. First suppose there exists a solution to the original IWD-STRATEGY-ELIMINATION instance. Then, perform the same sequence of eliminations to eliminate σ r in the new IWD-UNIQUE-SOLUTION instance. (This is possible because at any stage, any weak 5

6 dominance for the row player in the original instance is still a weak dominance in the new instance, because the two strategies payoffs for the row player are the same when the column player plays one of the new strategies; and the same is true for the column player.) Once σ r is eliminated, let σ 1 c eliminate σ 2 c. (It performs better against σ 2 r.) Then, let σ 1 r eliminate all the other remaining strategies for the row player. (It always performs better against either σ 1 c or σ 3 c.) Finally, σ 1 c is the unique best response against σ 1 r among the column player s remaining strategies, so let it eliminate all the other remaining strategies for the column player. Thus, there exists a solution to the IWD-UNIQUE-SOLUTION instance. Now suppose there exists a solution to the IWD-UNIQUE-SOLUTION instance. By Lemma 1, we can assume that all the dominances are by pure strategies. We will show that none of the new strategies (σ 1 r,σ 2 r,σ 3 r,σ 1 c,σ 2 c,σ 3 c) can either eliminate another strategy, or be eliminated before σ r is eliminated. Thus, there must be a sequence of eliminations ending in the elimination of σ r, which does not involve any of the new strategies, and is therefore a valid sequence of eliminations in the original game (because all original strategies perform the same against each new strategy). We now show that this is true by exhausting all possibilities for the first elimination before σ r is eliminated that involves a new strategy. None of the σ i r can be eliminated by a σ r / {σ 1 r,σ 2 r,σ 3 r}, because the σ i r perform better against σ 1 c. σ 1 r cannot eliminate any other strategy, because it always performs poorer against σ 2 c. σ 2 r and σ 3 r are equivalent from the row player s perspective (and thus cannot eliminate each other), and cannot eliminate any other strategy because they always perform poorer against σ 3 c. None of the σ j c can be eliminated by a σ c / {σ 1 c,σ 2 c,σ 3 c}, because the σ j c always perform better against either σ 1 r or σ 2 r. σ 1 c cannot eliminate any other strategy, because it always performs poorer against either σ r or σ 3 r. σ 2 c cannot eliminate any other strategy, because it always performs poorer against σ 2 r or σ 3 r. σ 3 c cannot eliminate any other strategy, because it always performs poorer against σ 1 r or σ 3 r. Thus, there exists a solution to the IWD-STRATEGY-ELIMINATION instance. A slightly weaker version of the part of Theorem 2 concerning dominance by pure strategies only is the main result by Gilboa, Kalai, and Zemel [4]. 3 4 A new concept of eliminability The concept of (iterated) dominance is often too strong (it cannot eliminate enough strategies). But, if possible, we would like a stronger argument for choosing a strategy (or eliminating another strategy) than Nash equilibrium. Hence, it would be desirable to have a solution concept that is inbetween the two in strength. In this section, we will provide such a concept. In fact, our concept spans an entire spectrum of strength between Nash equilibrium and strict dominance, and in the extremes can be made to coincide with either of these two solution concepts. We will think of a concept as stronger if it can eliminate fewer strategies. We would expect this to mean (presuming that the concept is sensible) that if a strategy can be eliminated with a stronger concept, this is a stronger argument that the strategy should not be played. Also, with a stronger concept it is more difficult to eliminate everything but a single solution; so if a solution can still be obtained this way, it presumably satisfies the criteria of weaker concepts as well. 4 3 Besides not proving the result for dominance by mixed strategies, the original result was weaker because it required payoffs {0, 1, 2, 3, 4, 5, 6, 7, 8} rather than just {0, 1} (so our Lemma 1 canot be applied to it to get the result for mixed strategies). 4 Admittedly, this concept of strength remains a bit counterintuitive in some cases; for instance, it may be tempting to say that Nash equilibrium is a more powerful concept than (iterated) dominance because it is able to eliminate more strategies from consideration. The reader is advised to be careful. 6

7 4.1 Example Because the definition of our concept is complex, we will first illustrate it with an example. Consider the following (partially specified) game. σc 1 σc 2 σc 3 σc 4 σr 1?,??, 2?, 0?, 0 σr 2 2,? 2, 2 2, 0 2, 0 σr 3 0,? 0, 2 3, 0 0, 3 σr 4 0,? 0, 2 0, 3 3, 0 A quick look at this game reveals that strategies σr 3 and σr 4 are both almost dominated by σr but 2 they perform better than σr 2 on σc 3 and σc 4, respectively. Similarly, strategies σc 3 and σc 4 are both almost dominated by σc 2 but they perform better than σc 2 on σr 4 and σc 3, respectively. So we are unable to eliminate any strategies using (even weak) dominance. Now consider the following reasoning. In order for it to be worth it for the row player to ever play σr 3 rather than σr, 2 the column player should play σc 3 at least 2 3 of the time. (If it is exactly 2 3, then switching from σ2 r to σr 3 will cost the row player 2 exactly 1 3 of the time, but the row player will gain 1 exactly 2 3 of the time, so the expected benefit is 0.) But, similarly, in order for it to be worth it for the column player to ever play σc 3, the row player should play σr 4 at least 2 3 of the time. But again, in order for it to be worth it for the row player to ever play σr, 4 the column player should play σc 4 at least 2 3 of the time. Thus, if both the row and column player accurately assess the probability that the other places on these strategies, and their strategies are rational with respect to these assessments (as would be the case in a Nash equilibrium); then, if the row player puts positive probability on σr, 3 by the previous reasoning, the column player should be playing σc 3 at least 2 3 of the time, and σ4 c at least 2 3 of the time. Of course, this is impossible; so, in a sense, the row player should not be playing σr. 3 It may appear that all we have shown is that σr 3 is not played in any Nash equilibrium. But, to some extent, our argument for not playing σr 3 did not make use of the full elimination power of the Nash equilibrium concept. Most notably, we only reasoned about a small part of the game: we never mentioned strategies σr 1 and σc 1, and we did not even specify most of the payoffs for these strategies. (It is easy to extend this example so that our argument only uses an arbitrarily small fraction of the strategies and of the payoffs in the matrix, for instance by adding many copies of σr 1 and σc 1.) The locality of the reasoning is more akin to the notion of dominance, which is perhaps the extreme case of local reasoning only two strategies are mentioned in it. So, in this sense, our argument is somewhere inbetween dominance and Nash equilibrium in strength. 4.2 Definition We are now ready to give the formal definition. To make the definition a bit simpler, we define its negation when a strategy is not eliminable relative to certain sets of strategies. Definition 2 Given a two-player game in matrix form, subsets D r,e r of the row player s pure strategies Σ r, and subsets D c,e c of the column player s pure strategies Σ c, we say that strategy e r E r is not eliminable relative to D r,e r,d c,e c,if there exist functions (partial mixed strategies) p r : E r [0, 1] and p c : E c [0, 1] with p r (e r) > 0, p r (e r ) 1, and p c (e c ) 1, such that e r E r e c E c 7

8 For any e r E r with p r (e r ) > 0, for any mixed strategy d r placing positive probability only on strategies in D r, there is some pure strategy σ c Σ c E c for the column player such that (letting p σ denote the mixed strategy that results from placing the remaining (unused) probability from the partial mixed strategy p on σ) u r (e r,p c σ c ) u r (d r,p c σ c ). (Or, if p c already uses up all the probability, we simply have u r (e r,p c ) u r (d r,p c ) no σ c needs to be chosen.) The same for the column player: for any e c E c with p c (e c ) > 0, for any mixed strategy d c placing positive probability only on strategies in D c, there is some pure strategy σ r Σ r E r for the row player such that u c (p r σ r,e c ) u c (p r σ r,d c ). (Or, if p r already uses up all the probability, u c (p r,e c ) u c (p r,d c ) no σ r needs to be chosen.) In the example from the previous subsection, we may set D r = {σ 2 r}, D c = {σ 2 c }, E r = {σ 3 r,σ 4 r},e c ={σ 3 c,σ 4 c}, and e r = σ 3 r. Then, by the reasoning that we did, it is impossible to set p r and p c so that the condition is satisfied, and hence σ 3 r is eliminable relative to these sets. 4.3 The spectrum of strength In this subsection we show that our concept spans a spectrum of strength all the way from Nash equilibrium (when the sets D r,e r,d c,e c are chosen as large as possible), to strict dominance (when the sets are chosen as small as possible). We first show that the Nash equilibrium concept is weaker than the elimination concept introduced here in the sense that the elimination concept can never eliminate a strategy that is in some Nash equilibrium. So, if a strategy can be eliminated by the elimination concept, it can be eliminated by the Nash equilibrium concept. Theorem 3 If there is some Nash equilibrium that places positive probability on pure strategy σ r, then σ r is not eliminable relative to any D r,e r,d c,e c. Proof: Let σ r be the row player s strategy in the Nash equilibrium, and let σ c be the column player s strategy in the Nash equilibrium. For any D r,e r,d c,e c with σr E r, simply let p r coincide with σ r on E r that is, let p r be the probabilities that the row player places on the strategies in E r in the equilibrium. (Thus, p r (σr ) > 0). Similarly, let p c coincide with σ c on E c. We will prove that the condition for noneliminability is satisfied for the row player; the case of the column player follows by symmetry. For any e r E r with p r (e r ) > 0, for any mixed strategy d r,we have u r (e r,σ c) u r (d r,σ c) 0, by the Nash equilibrium condition. Now, let σ c arg max σ Σc E c (u r (e r,p c σ) u r (d r,p c σ)). Then we must have u r (e r,p c σ c ) u r (d r,p c σ c ) u r (e r,σ c) u r (d r,σ c) 0(because p c σ c and σ c coincide on E c, and for the former, the remainder of the distribution is chosen to maximize this expression). It follows that σr is not eliminable relative to any D r,e r,d c,e c. We next show that by choosing the sets D r,e r,d c,e c as large as possible, we can make the elimination concept coincide with the Nash equilibrium concept. Theorem 4 Let D r = E r =Σ r and D c = E c =Σ c. Then e r is eliminable relative to these sets if and only if there is no Nash equilibrium that places positive probability on e r. 8

9 Proof: The only if direction follows from Theorem 3. For the if direction, suppose e r is not eliminable relative to D r = E r =Σ r and D c = E c =Σ c. The partial distributions p r and p c with p r (e r) > 0 that show that e r is not eliminable must use up all the probability (the probabilities must sum to one), because there are no strategies outside E c =Σ c and E r =Σ r to place any remaining probability on. Hence, we must have, for any strategy e r E r =Σ r with p r (e r ) > 0, that for any mixed strategy d r, u r (e r,p c ) u r (d r,p c )(and the same for the column player). But these are precisely the conditions for p r and p c to constitute a Nash equilibrium. It follows that there is a Nash equilibrium with positive probability on e r. Incidentally, this leads to the following corollary showing that applying our concept can be computationally hard: Corollary 1 Deciding whether a given strategy is eliminable relative to D r = E r = Σ r and D c = E c =Σ c is conp-complete, even in symmetric 2-player games. Proof: By Theorem 4, this is simply the converse of asking whether there exists a Nash equilibrium with positive probability on the strategy. This is known to be NPcomplete [3], even in this restricted setting [1]. We now show that the concept of strict dominance is stronger than the elimination concept introduced here in the sense that the elimination concept can always eliminate a strictly dominated strategy (as long as the dominating strategy is in D r.) Theorem 5 If pure strategy σ r is strictly dominated by some mixed strategy d r, then σ r is eliminable relative to any D r,e r,d c,e c as long as σ r E r, and all the pure strategies that d r places positive probability on are in D r. Proof: To show that σ r is not eliminable relative to these sets, we must have p r (σ r) > 0, and thus we must demonstrate an inequality of the form u r (σ r,p c σ c ) u r (d r,p c σ c )or u r (σ r,p c ) u r (d r,p c )(because d r only places positive probability on strategies in D r ). But this is impossible, because by strict dominance, u r (σ r,σ c )<u r (d r,σ c ) for any mixed strategy σ c. We next show that by choosing the sets E r,e c as small as possible, we can make the elimination concept coincide with the strict dominance concept. Theorem 6 Let E c = {} and E r = {e r }. Then e r is eliminable relative to D r,e r,d c,e c if and only if it is strictly dominated by some mixed strategy placing positive probability only on elements of D r. Proof: The if direction follows from Theorem 5. For the only if direction, suppose that e r is eliminable relative to these sets. That means that there exists a mixed strategy d r placing positive probability only on strategies in D r such that for any σ c, u(e r,σ c )<u(d r,σ c )(because E c = {} and E r = {e r }, this is the only way in which an attempt to prove that e r is not eliminable could fail). But this is exactly the condition for d r to strictly dominate e r. 4.4 Advantages of the new concept In our opinion, the main benefit of the new concept is that, when a strategy cannot be eliminated by dominance (but it can be eliminated by the Nash equilibrium concept), 9

10 the new concept may present a stronger argument than Nash equilibrium for eliminating the strategy, using sets D r,e r,d c,e c smaller than the entire strategy set. To get the strongest possible argument for eliminating a strategy, the sets D r,e r,d c,e c should be chosen as small as possible while still having the strategy be eliminable relative to these sets. Our definition also allows for iterated elimination of strategies. 5 Conclusions While the Nash equilibrium solution concept is studied more and more intensely in our community, the perhaps more elementary concept of iterated dominance has received much less attention. In this paper, we have made two main contributions. First, we proved some basic complexity results about iterated dominance. We showed that determining whether there is some path that eliminates a given strategy is NP-complete with weak dominance. This allowed us to also show that determining whether there is a path that leads to a unique solution is NP-complete. Both of these results hold regardless of whether dominance by mixed strategies is allowed. (The second result was already known for the case of dominance by pure strategies only, in a slightly less restricted setting [4].) As our second contribution, we presented a new definition of elimination of strategies, and showed that it spans a spectrum of solution concepts from strict dominance to Nash equilibrium. We pointed out that checking whether a strategy is eliminable relative to the sets of all strategies is conp-complete under this definition. References [1] Vincent Conitzer and Tuomas Sandholm. Complexity results about Nash equilibria. In Proceedings of the Eighteenth International Joint Conference on Artificial Intelligence (IJCAI), [2] Drew Fudenberg and Jean Tirole. Game Theory. MIT Press, [3] I. Gilboa and E. Zemel. Nash and correlated equilibria: Some complexity considerations. Games and Economic Behavior, [4] Itzhak Gilboa, Ehud Kalai, and Eitan Zemel. The complexity of eliminating dominated strategies. Mathematics of Operation Research, 18: , [5] C. E. Lemke and J. T. Howson. Equilibrium points of bimatrix games. Journal of the Society of Industrial and Applied Mathematics, 12: , [6] Andreu Mas-Colell, Michael Whinston, and Jerry R. Green. Microeconomic Theory. Oxford University Press, [7] John Nash. Equilibrium points in n-person games. Proc. of the National Academy of Sciences, 36:48 49, [8] Christos Papadimitriou. Algorithms, games and the Internet. In STOC, pages , [9] R. Savani and B. von Stengel. Long Lemke-Howson paths. Technical Report LSE-CDAM , London School of Economics,