Parameterized Two-Player Nash Equilibrium
Danny Hermelin, Chien-Chung Huang, Stefan Kratsch, Magnus Wahlstrom
aa r X i v : . [ c s . CC ] J un Parameterized Two-Player Nash Equilibrium
Danny Hermelin, Chien-Chung Huang, Stefan Kratsch, and Magnus Wahlstr¨om
Max-Planck-Institute for Informatics, Saarbr¨ucken, Germany { hermelin,villars,skratsch,wahl } @mpi-inf.mpg.de Abstract.
We study the computation of Nash equilibria in a two-playernormal form game from the perspective of parameterized complexity.Recent results proved hardness for a number of variants, when parame-terized by the support size. We complement those results, by identifyingthree cases in which the problem becomes fixed-parameter tractable.These cases occur in the previously studied settings of sparse gamesand unbalanced games as well as in the newly considered case of locallybounded treewidth games that generalizes both these two cases.
Algorithmic game theory is a quite recent yet rapidly developing discipline thatlies at the intersection of computer science and game theory. The emergence ofthe internet has given rise to numerous applications in this area such as onlineauctions, online advertising, and search engine page ranking, where humans andcomputers interact with each other as selfish agents negotiating to maximizetheir own payoff utilities. The amount of research spent in attempting to devisecomputational models and algorithms for studying these types of interactions hasbeen overwhelming in recent years; unsurprisingly perhaps, when one considersthe economical rewards available in this venture.The central problem in algorithmic game theory is that of computing a
Nashequilibrium , a set of strategies for each player in a given game, where no playercan gain by changing his strategy when all other players strategies remain fixed.This problem is so important because Nash equilibria provide a good way to pre-dict the outcomes of many of the scenarios described above, and other scenariosas well. Furthermore, Nash’s Theorem states that for any finite game a mixedNash equilibrium always exists. However, for this concept to be meaningful forpredicting behaviors of rational agents which are in many cases computers, anatural prerequisite is for it to be computable. This led researchers such as Pa-padimitriou to dub the problem of computing Nash equilibria as one of the mostimportant complexity problems of our time [25].The initial breakthrough in determining the complexity of computing Nashequilibria was made by Daskalakis, Goldberg, and Papadimitriou [11, 20]. Thesetwo papers introduced a reduction technique which was used by the authors forshowing that computing a Nash equilibrium in a four player game is PPAD-complete. Shortly afterwards, this hardness result was simultaneously extendedto three player games by Daskalakis and Papadimitriou [15], and by Chen and
Deng [5]. The case of two player (bimatrix) games was finally cracked a yearlater by Chen and Deng [6], who proved it to be PPAD-complete. This impliedthe existence of a polynomial-time algorithm for the core case of bimatrix gamesto be unlikely.Since the result of Chen and Deng [6], the focus on computing Nash equilibriain bimatrix games was directed either towards finding approximate Nash equilib-ria [3, 7–9, 12, 13, 23], or towards finding special cases where exact equilibria canbe computed in polynomial time [2, 8, 10, 21, 23]. Nevertheless, for general bima-trix games the best known algorithm for computing either approximate or exactequilibria essentially tries all possibilities for the support of both players (the setof strategies played with non-zero probability), which can be assumed to be atmost logarithmic in the approximate case [23]. Once the support of both playersis known, one can compute a Nash equilibrium by solving a linear-program.
Theorem 1 ([24]).
A Nash equilibrium in a bimatrix game, where the supportsizes are bounded by k , can be computed in n O ( k ) time. Due to the central role that the algorithm of Theorem 1 plays in comput-ing exact and approximate Nash equilibria, it is natural to ask whether onecan improve on its running-time substantially. In particular, can we remove thedependency on the support size from the exponent? The standard frameworkfor answering such questions is that of parameterized complexity theory [16, 19].Estivill-Castro and Parsa initiated the study of computing Nash equilibria in thiscontext [18] . They showed that when the support size is taken as a parameter,the problem is W[2]-hard even in certain restricted settings. The implication oftheir result is a negative answer to the above question. In particular, combin-ing their reduction with the results of Chen et al. [4] gives a sharp contrast toTheorem 1 above.
Theorem 2 ([18]).
Unless
FPT=W[1] , there is no n o ( k ) time algorithm forcomputing a Nash equilibrium with support size at most k in a bimatrix game. The consequence of Theorem 2 above is devastating in the sense that forlarge enough games that have equilibriums with reasonably small supports, thetask of computing equilibria already becomes infeasible. The main motivationof this paper is to find scenarios where one can circumvent this. Our goal isthus to identify natural parameters which govern the complexity of computingNash equilibria, and which can help in devising feasible algorithms. We believethat this direction can prove to be fruitful in the quest for understanding thecomputational limitations of this fundamental problem. Indeed, prior to ourwork, Kalyanaraman and Umans [21] provided a fixed-parameter algorithm forfinding equilibrium in bimatrix games whose matrices have small rank (and someadditional constraints).Our techniques are based on considering a natural graph-theoretic represen-tation of bimatrix games. This is done by taking the union of the underlyingboolean matrix of the two given payoff matrices, and considering this matrixas the biadjacency matrix of a bipartite graph. A similar approach was taken by [10], and in particular by [2] who considered games that have an underly-ing planar graph structure. Our work complements both these results as will beexplained further on.A natural class of games that has a convenient interpretation in the graph-theoretic context is the class of ℓ -sparse games [8, 10, 14]. Here each column androw in both payoff matrices of the game have at most ℓ non-zero entries. Aninitial tempting approach in these types of games would be to try to devise aparameterized algorithm with ℓ taken as a single parameter. However, Chen,Deng, and Teng [8] showed that unless PPAD = P, there is no algorithm forcomputing an ε -approximate equilibrium for a 10-sparse game in time polynomialboth in ε and n . Thus, such an FPT algorithm cannot exist unless PPAD is in P.We complement this result by showing that if ℓ is taken as a parameter, and thesize of the supports is taken as an additional parameter, then computing Nashequilibrium is fixed-parameter tractable. Theorem 3.
A Nash equilibrium in a ℓ -sparse bimatrix game, where the supportsizes is bounded by k , can be computed in ℓ O ( kℓ ) · n O (1) time. Note that the above result also complements the polynomial-time algorithmsgiven in [8, 10] for 2-sparse games. While in these algorithms there was no as-sumption made on the size of support of the equilibrium to be found, bothalgorithms could handle only win-lose games [1, 9], games with boolean payoffmatrices.Our second result is concerned with k -unbalanced games , games where therow player has a small set of k strategies [21, 23]. Lipton, Markakis, and Mehta [23]observed that in such games there is always an equilibrium where the row playerplays a strategy with support size at most k + 1. Thus, by applying Theorem 1one can find a Nash equilibrium in n O ( k ) time for these types of games. Canthis result be improved to an algorithm running in f ( k ) · n O (1) time? We givea partial answer to this question, by showing that if the number ℓ of differentpayoffs of the row player is taken as an additional parameter, the problem indeedbecomes fixed-parameter tractable. Theorem 4.
A Nash equilibrium in a k -unbalanced bimatrix game, where therow player has ℓ different payoff values, can be computed in ℓ O ( k ) · n O (1) time. Our last result considers bimatrix games whose corresponding graph has aconvenient structural property, namely the property of having locally boundedtreewidth. Note that both ℓ -sparse games and k -unbalanced games have cor-responding graphs with this property, as well as the games with an underly-ing planar graph structure considered by [2]. We show that in games of locallybounded treewidth, where the payoff matrices have at most ℓ different values,one can compute a Nash equilibrium in f ( k, ℓ ) · n O (1) . Although this might seemas a generalization of both of our results mentioned above, the reader shouldnote that here we have a stricter requirement on the number of different valuesin the payoff matrices, and the running-time dependency on both parametersincreases much faster. Theorem 5.
A Nash equilibrium in a locally bounded treewidth game, where thesupport sizes are bounded by k , and the payoff matrices have at most ℓ differentvalues, can be computed computed in f ( k, ℓ ) · n O (1) time for some computablefunction f () . The paper is organized as follows: We begin with some preliminaries in Sec-tion 2. In Section 3 we consider ℓ -sparse games and prove Theorem 3. Section 4addresses locally bounded treewidth games and proves Theorem 5. Finally, inSection 5 we prove Theorem 4 regarding k -unbalanced games. Let G := ( A, B ) be a bimatrix game, where
A, B ∈ Q n × n are the payoff matrices of the row and the column players respectively. The row (column) player hasa strategy space consisting of the rows (columns) [ n ] := { · · · n } . (For ease ofnotation, except in unbalanced games, we assume that both players have thesame number of strategies; different numbers of strategies do not affect any ofour results.) The row (column) player chooses a strategy profile x ( resp. y ), whichis a probability distribution over his strategy space, that is, x i , y j ≥ ∀ i, j , andfurthermore P ni =1 x i = 1 and P nj =1 y j = 1. The expected outcomes of the gamefor the row and the column players are x T Ay and x T By respectively.The players are rational, always aiming for maximizing their expected payoffs.They have reached a Nash equilibrium if the current strategies x and y are suchthat neither player has a deviating strategy x ′ resp. y ′ such that x ′ T Ay > x T Ay resp. x T By ′ > x T By , i.e., if neither of them can improve his payoff independentlyof the other. The following proposition gives an equivalent condition for a pairof strategies to be an equilibrium. Lemma 1. ([24, Chapter 3])
The pair of strategy vectors ( x, y ) is a Nash equi-librium for the bimatrix game ( A, B ) if and only if(i) x s > ⇒ ( Ay ) s ≥ ( Ay ) j for all j = s ;(ii) y s > ⇒ ( x T B ) s ≥ ( x T B ) j for all j = s . The support of a strategy vector x is defined as the set S ( x ) = { i | x i > } .Note that the above proposition implies that if ( x, y ) is a Nash equilibrium,in the column vector Ay , the entries in S ( x ) are equivalent and no less thanall other entries not in S ( x ); symmetrically, in the row vector x T B , the entriesin S ( y ) are equivalent and no less than other entries not in S ( y ). It is knownthat, given possible supports I, J ⊆ [ n ] it can be efficiently decided whetherthere is a matching Nash equilibrium, and the corresponding strategy vectorscan be computed via linear programming (Theorem 1).The following graph associated with a bimatrix game is useful for presentingour algorithms in Sections 3 and 4. Definition 1.
Let G = ( A, B ) be a bimatrix game with A, B ∈ Q n × n . Theundirected (and bipartite) graph G = G ( G ) associated with G is defined to be the bipartite graph with vertex classes V r , V c := [ n ] , referred to as row resp. column vertices. There is an edge between i ∈ V r and j ∈ V c iff A i,j = 0 or B i,j = 0 . As a last bit of notation: For
I, J ⊆ [ n ], and any n × n matrix A , we use A I,J to denote the submatrix composed of rows in I and columns in J . We alsouse A I, ∗ as a shorthand for A I, [ n ] . Thus, A i, ∗ means the i ’th row of A . In this section we present the proof for Theorem 3. Throughout the section we let G := ( A, B ) denote our given bimatrix game, where A and B are rational valuematrices with at most ℓ non-zero entries per row or column. We will presentan algorithm for finding an Nash equilibrium where the support sizes of bothplayers are at most k (and k is taken as a parameter). The high-level strategy isto show that it suffices to search for equilibria that induce one or two connectedcomponents in the associated graph G = G ( G ). This permits us to find candidatesupport sets by enumerating subgraphs of G (on one or two components).We begin by introducing minimal equilibria: Definition 2.
A Nash equilibrium ( x, y ) is minimal if for any Nash equilib-rium ( x ′ , y ′ ) with S ( x ′ ) ⊆ S ( x ) and S ( y ′ ) ⊆ S ( y ) , we have S ( x ′ ) = S ( x ) and S ( y ′ ) = S ( y ) . Our algorithm iterates through all possible support sizes k , k ≤ k in increas-ing order to determine whether there exists an equilibrium ( x, y ) with | S ( x ) | = k and S ( y ) = k . To avoid cumbersome notation, we will assume that k = k = k (extending this to general case will be immediate). Thus at a given iteration, thealgorithm can assume that no equilibrium exists with smaller supports, i.e. itcan restrict its search to minimal equilibriums. This fact will prove crucial lateron. Furthermore, since our game is ℓ -sparse, our algorithm only needs to searchfor equilibriums where both player receive non-negative payoffs. Lemma 2. If G = ( A, B ) is an ℓ -sparse game, where A, B ∈ Q n × n and n > ℓk ,then in any Nash equilibrium with support at most k × k , both players receivenon-negative payoffs. For an equilibrium ( x, y ), let the extended support of x be the rows S ( x ) ∪ N ( S ( y )), and similarly for y , where the neighborhood N ( I ) is taken over thegraph G := G ( G ) of the game. Note that any row not in the extended supportof x would have payoff constantly zero given the current strategy of y , and thusis not important for the existence of an equilibrium. We will show that for aminimal equilibrium ( x, y ), the extended supports of x and y induce a subgraphof G which has at most two connected components. This will be done in twosteps: The first is the special case where A S ( x ) ,S ( y ) = B S ( x ) ,S ( y ) = , while thesecond corresponds to the remaining cases. Lemma 3. If ( x, y ) is a minimal Nash equilibrium for a game ( A, B ) with A S ( x ) ,S ( y ) = B S ( x ) ,S ( y ) = , then the subgraphs induced by N [ S ( x )] and N [ S ( y )] in the graph associated with the game are both connected.Proof. Let G x := G [ N [ S ( x )]] be the subgraph of G induced by N [ S ( x )], andaiming towards a contradiction, suppose that G x is disconnected. Let C be aconnected component in G x , and write p := P i ∈ V r ( C ) x i to denote the prob-ability that a row strategy in C is played according to x . Now define a newrow strategy by setting ˆ x i = x i /p if i ∈ V x , and ˆ x i = 0 otherwise. We arguethat (ˆ x, y ) is a Nash equilibrium, contradicting the fact that ( x, y ) is minimal.Obviously, the expected payoff in (ˆ x, y ) is zero for both players, as A S (ˆ x ) ,S ( y ) = B S (ˆ x ) ,S ( y ) = . Furthermore, there is no row i such that ( Ay ) i >
0, since thestrategy y is unchanged and the original strategy pair ( x, y ) is an equilibrium.Now assume that there is a column j such that (ˆ x T B ) j >
0. Then B i,j = 0for some i ∈ S (ˆ x ), and by the connectivity assumption we must have B i,j = 0for all i ∈ S ( x ) \ S (ˆ x ), but then ( x T B ) j = p (ˆ x T B ) j >
0, which contradicts theassumption that ( x, y ) is a Nash equilibrium. By Lemma 1, it follows that (ˆ x, y )is a Nash equilibrium. ⊓⊔ Lemma 4. If ( x, y ) is a minimal Nash equilibrium for a game ( A, B ) with either A S ( x ) ,S ( y ) = or B S ( x ) ,S ( y ) = , and with a non-negative payoff for both players,then the subgraph induced by N [ S ( x ) ∪ S ( y )] is connected.Proof. Let H be the subgraph induced by N [ S ( x ) ∪ S ( y )] and suppose that H is disconnected. We will derive a contradiction by showing that ( x, y ) is notminimal.Let C be a connected component in H intersecting both S ( x ) and S ( y ), andlet V ˆ x := V r ( C ) ∩ S ( x ) and V ˆ y := V c ( C ) ∩ S ( y ) denote the set of row and columnstrategies in C , respectively. We define a new pair of strategy profiles (ˆ x, ˆ y ) where S (ˆ x ) = V ˆ x and S (ˆ y ) = V ˆ y , by normalizing ( x, y ) onto V ˆ x and V ˆ y . That is, we let p := P i ∈ V ˆ x x i , and set ˆ x i = x i /p if i ∈ V ˆ x , and ˆ x i = 0 otherwise. Similarly, welet q := P j ∈ V ˆ y y i , and set ˆ y accordingly. As either S (ˆ x ) ⊂ S ( x ) or S (ˆ y ) ⊂ S ( y ),to prove the lemma it suffices to argue that (ˆ x, ˆ y ) is an equilibrium.Consider a row strategy i ∈ [ n ]. We claim( A ˆ y ) i = ( ( Ay ) i /q if A i,V ˆ y = ,0 otherwise. (1)The second case is clear. For the first case, assume that A i,V ˆ y = . Now,if A i,j = 0 for j ∈ S ( y ) \ V ˆ y , then there would be an edge in H from thevertex corresponding to row i , which is in C , to the vertex corresponding tocolumn j , which is not in C , contradicting that C is a connected component.Thus ( Ay ) i = X j ∈ S ( y ) A i,j y j = X j ∈ V ˆ y A i,j ( q ˆ y j ) + X j ∈ S ( y ) \ V ˆ y A i,j y j = q ( A ˆ y ) i + 0and the claim follows. Now let s ∈ S (ˆ x ), and consider some arbitrary row strategy i ∈ [ n ]. As-sume by way of contradiction that ( A ˆ y ) s < ( A ˆ y ) i . It is clear that A i,V ˆ y = ,thus ( A ˆ y ) i = ( Ay ) i /q ; we consider the cases for row s . If A s,V ˆ y = , thenby (1), we have ( A ˆ y ) s = ( Ay ) s /q , implying ( Ay ) s < ( Ay ) i . On the other hand,if A s,V ˆ y = but A s,S ( y ) = , then C would not be a connected component of H (since a neighbor of s would be missing).Thus A s,S ( y ) = and ( Ay ) s = ( A ˆ y ) s = 0, and ( Ay ) i = q ( A ˆ y ) i >
0. In bothcases we contradict that ( x, y ) is an equilibrium. Thus we fulfill condition (i)of Lemma 1, and by symmetry we also fulfill condition (ii). We have shownthat (ˆ x, ˆ y ) is an equilibrium, contradicting the minimality of ( x, y ). ⊓⊔ As an immediate corollary of Lemma 3 and 4, we get that the subgraph in G ( G ) induced by the extended support of a minimal equilibrium has at most twoconnected components. In the following lemma we show that in graphs of smallmaximum degree, we can find all such subgraphs quite efficiently. This will allowus to find a small, minimal equilibrium by checking all sets of rows and columnsthat would be candidates for being the extended supports of one. Lemma 5.
Let G be a graph on n vertices and with maximum degree ∆ = ∆ ( G ) .In time ( ∆ + 1) t · n c + O (1) one can enumerate all subgraphs on t vertices thatconsist of c connected components.Proof. We first show how to enumerate all connected subgraphs on t vertices intime ( ∆ + 1) t · n O (1) by a branching algorithm. At any point selected verticeswill be active or passive. When a vertex is selected it will first be active and laterbe set to passive. Selecting a vertex and making it active respectively setting avertex to passive is called an event .First, we branch on the choice of one out of n starting vertices in G and setit active. Then until we have selected t vertices we do the following: We considerthe least recently added active vertex and branch on one of at most ∆ + 1events, namely selecting one of its at most ∆ neighbors (and making it active) orsetting the vertex itself to passive. We terminate when we have selected t vertices(and output the corresponding subgraph) or when there are no more activevertices. Clearly, on each branch of this algorithm there are at most 2 t events.Thus the branching tree has at most ( ∆ + 1) t leaves, implying a total runtimeof ( ∆ +1) t · n O (1) . Observe that for every connected subgraph on t vertices thereis a sequence of events such that the graph occurs in the enumeration.The generalization to c components is straightforward: When there are nomore active vertices but we have not yet selected c components, then we selectone of the less than n remaining vertices of G as a new active vertex, startinga new component. Selecting new starting vertices adds a factor of n c to ourruntime. ⊓⊔ We are now in position to describe our entire algorithm. It first iteratesthrough all possible sizes of extended support in increasing order. In each itera-tion, it enumerates all subgraphs that might correspond to the extended supportof a minimal equilibrium. It then checks all ways of selecting a support from the given subgraph, and for each such selection it uses the algorithm behind Theo-rem 1 to check whether there is an equilibrium on the support. If no equilibriumis found throughout the whole process, the algorithm reports that there existsno equilibrium with support size at most k in G . The running time is boundedby ℓ O ( kℓ ) n O (1) from Lemma 5, times (cid:0) kℓ + kk (cid:1) = 2 O ( kℓ ) ways of selecting the sup-port, times n O (1) for checking for an equilibrium. In total, we get a running timeof ℓ O ( kℓ ) n O (1) .Finally, completeness comes from the exhaustiveness of Lemma 5 and thestructure given by Lemmas 3 and 4. In the case that the payoffs of our games are non-negative, i.e.,
A, B ∈ Q n × n ≥ ,we can reduce our running time to be polynomial in ℓ , for ℓ -sparse games. Webegin with a strengthening of Lemmas 3 and 4. Lemma 6.
Let G = ( A, B ) be a bimatrix game with A, B ∈ Q n × n ≥ , and G bethe graph associated with G . If ( x, y ) is a minimal Nash equilibrium for G , theneither | S ( x ) | = | S ( y ) | = 1 , or G [ S ( x ) ∪ S ( y )] is connected.Proof. Let G S := G [ S ( x ) ∪ S ( y )]; assume that G S is not connected. If the ex-pected outcome is zero, then (since the entries are non-negative) every entryin A S ( x ) , ∗ and B ∗ ,S ( y ) is zero, and we get an equilibrium by selecting any sin-gle row i ∈ S ( x ) and column j ∈ S ( y ). Otherwise, every row of A S ( x ) ,S ( y ) andevery column of B S ( x ) ,S ( y ) contains some positive entry. Let C be a connectedcomponent of G S on row vertices V ˆ x and column vertices V ˆ y , and define a newpair of strategy profiles (ˆ x, ˆ y ) where S (ˆ x ) = V ˆ x and S (ˆ y ) = V ˆ y , by normalizing( x, y ) onto V ˆ x and V ˆ y as in Lemma 4.We will argue that (ˆ x, ˆ y ) is an equilibrium.Let s ∈ V ˆ x , and assume by way of contradiction that for some row i ∈ [ n ], wehave ( A ˆ y ) i > ( A ˆ y ) s . Let ( A ˆ y ) s = c ; by non-connectivity of G S , ( Ay ) s = qc .Further let ( A ˆ y ) i = c and ( Ay ) i = qc + (1 − q ) c . Now ( Ay ) s = qc < qc ≤ qc + (1 − q ) c = ( Ay ) i , contradicting our assumptions; the last inequality isbecause the entries are non-negative. Repeating the argument symmetrically, wefind that (ˆ x, ˆ y ) is a Nash equilibrium. ⊓⊔ Thus, to find an equilibrium in G = ( A, B ), it suffices to search for occurrencesof the support, rather than the extended support. Invoking Lemma 5 directlywith a bound of 2 k vertices gives a running time of ℓ O ( k ) n O (1) . Let G = ( A, B ) be a given game with
A, B ∈ P n × n , with P ⊂ Q , | P | ≤ ℓ , andlet G = G ( G ) the graph associated with G . In this section we will present analgorithm that finds an equilibrium with support sizes at most k when G comesfrom a graph class with locally bounded treewidth. Note that this is a partial extension of the results of the previous section, as graphs of bounded degree havelocally bounded treewidth, while on the other hand we assume that there is abounded set P of only ℓ different payoff values which can occur in the games.(The case P = { , } would correspond to win-lose games.) Definition 3 ([17]).
A graph class has locally bounded treewidth if there isa function f : N → N such that for every graph G := ( V, E ) of the class, anyvertex v ∈ V , and any d ∈ N , the subgraph of G induced by all vertices withindistance at most d from v has treewidth at most f ( d ) . We refer readers to [19] for more details on the notion of treewidth and locallybounded treewidth. The crucial property of locally bounded treewidth graphsin our context is that first-order queries can be answered in FPT time on suchgraphs when the parameter is the size of first-order formula [19, Chapter 12.2].For ease of presentation we show how to find an equilibrium where bothplayers have support size k (the algorithm can be easily adapted to supportsizes k , k ≤ k ). Let I and J be two subsets of k elements in [ n ]. We say that twomatrices A ∗ , B ∗ ∈ Q k × k occur in G at ( I, J ) if A ∗ = A I,J and B ∗ = B I,J . Thepair ( A ∗ , B ∗ ) forms an equilibrium pattern if there exists an equilibrium ( x, y )where ( A ∗ , B ∗ ) occur in G at ( S ( x ) , S ( y )). Our algorithm will try all possible ℓ k pairs of matrices ( A ∗ , B ∗ ), and for each such pair it will determine whether it isan equilibrium pattern.When does a pair of matrices ( A ∗ , B ∗ ) form an equilibrium pattern? The firstobvious condition is that it occurs in G at some pair of position sets ( I, J ). Fur-thermore, by definition of a Nash equilibrium, there is a pair of strategies ( x, y )with S ( x ) = I and S ( y ) = J , such that neither player has a better alternative.The difficulty here lies in the fact that, even given the support S ( y ) of the col-umn player, there may be too many possible strategies for the row player thathave supports different from I . To circumvent this, we define equivalence of rowswith respect to supports S ( y ), and of columns with respect to supports S ( x ). Definition 4.
Let
I, J ⊆ [ n ] . Two rows i , i ∈ [ n ] are J -equivalent if A i ,J = A i ,J . Similarly, two columns j , j ∈ [ n ] are I -equivalent if B I,j = B I,j . Lemma 7.
Let J be the support of the column player. For any row strategy x there is a row strategy ˆ x such that:(i) the support S (ˆ x ) contains at most one row from each J -equivalence class(ii) and for any column strategy y with support J we have ˆ x T Ay = x T Ay .The same is true for column strategies, given a support I of the row player. For each possible equilibrium pattern ( A ∗ , B ∗ ) we do the following. Foreach choice of rows A † ⊆ P × k that do not occur in A ∗ and each choice ofcolumns B † ⊆ P k × that do not occur in B ∗ , we create two matrices C = (cid:18) A ∗ A † (cid:19) and D = (cid:18) B ∗ B † (cid:19) . We use Theorem 1 to see if there is an equilibrium ( x, y ) in the game (
C, D )with S ( x ) = S ( y ) = [ k ]. If there is such an equilibrium, then we proceed asfollows to find an occurrence of ( A ∗ , B ∗ ) that avoids the rows and columnswhich were not chosen. For this let F be the rows which occur neither in A ∗ nor in A † and let F be the columns which occur neither in B ∗ nor in B † . Wesay that F and F are forbidden for ( A ∗ , B ∗ ). We note that given ( A ∗ , B ∗ ), aset of rows F ⊆ P × k , and a set of columns F ⊆ P k × , one can write a first-order formula of size bounded by some function in k and | P | = ℓ to determinewhether ( A ∗ , B ∗ ) has an occurrence which avoids F and F . Example 1.
Consider a win-lose game (
A, B ) encoded into relations A and B such that A ( r, c ) is true iff A r,c = 1, and likewise for B . Then a 2 × ∃ r , r , c , c A ( r , c ) ∧ ¬ A ( r , c ) ∧ ¬ A ( r , c ) ∧ A ( r , c ) ∧ B ( r , c ) ∧ ¬ B ( r , c ) ∧ ¬ B ( r , c ) ∧ B ( r , c ) ∧∀ r ′ ( ¬ A ( r ′ , c ) ∨ ¬ A ( r ′ , c )) ∧ ∀ c ′ ( ¬ B ( r , c ′ ) ∨ ¬ B ( r , c ′ )) . In general, with ℓ different values, there would be ℓ − A i and B i encoding the game, where A i ( r, c ) is true if A r,c = z i , for every z i ∈ P exceptthe zero value.Since the number of possible choices for F and F is bounded by some functionin k and ℓ , and for each such choice we can determine whether F and F is aforbidden pair for ( A ∗ , B ∗ ) in polynomial-time, the total time for determiningwhether ( A ∗ , B ∗ ) is an equilibrium pattern is FPT in k and ℓ . Since the numberof pairs ( A ∗ , B ∗ ) is also bounded by a function in k , the total running time ofour entire algorithm is also FPT in k and ℓ .To complete the proof of Theorem 5, let us briefly argue completeness. As-sume that there is any equilibrium with support sizes equal to k , let I and J bethe supports, and let A ∗ and B ∗ be corresponding sub-matrices. Observe thatwe may set all entries in columns outside J of A to zero without harm, ditto forrows outside I in B . According to Lemma 7 it suffices to keep one copy of eachrow outside A ∗ in A (also discard the corresponding zero-row in B to keep thesize the same). The same is of course true for columns outside B ∗ in B . Exceptfor a permutation this is equal to one of the games ( C, D ) that we considered.Therefore our algorithm will find such an equilibrium if one exists.
In this section we briefly consider k -unbalanced bimatrix games. A bimatrixgame ( A, B ) is k -unbalanced if A, B ∈ Q k × n ≥ for some k << n [21, 23] (i.e.,the row player has a significantly smaller number of strategies than the columnplayer). We will show that a Nash equilibrium can be computed in FPT-timewith respect to k and ℓ , where ℓ denotes the number of different payoffs that therow player has, i.e., ℓ := |{ A i,j : 1 ≤ i ≤ k, ≤ j ≤ n }| . Similar to Definition 4 we define two column strategies i, j ∈ [ n ] to be equiv-alent if A ∗ ,i = A ∗ ,j . (However, notice that unlike Definition 4, here equivalenceof column strategies is defined with respect to the row player payoff.) Lemma 8.
For each equilibrium there is an equilibrium where the column playerplays at most one column from each equivalence class.
Using Lemma 8 we can easily devise an FPT algorithm for computing aNash equilibrium in our setting. The algorithm simply guesses the support ofthe row player and column player, and then uses the method of Theorem 1 todetermine whether there exists a Nash equilibrium corresponding to these sets ofsupports. Observe that there are at most ℓ k column-strategy equivalence classes.Furthermore, according to Lipton, Markakis, and Mehta [23], in a k -unbalancedgame there always exists an equilibrium where the column player has supportsize at most k + 1. Thus the number of guesses the algorithm makes is boundedby 2 k · (cid:0) ℓ k k +1 (cid:1) = ℓ O ( k ) , and for each such guess, the amount of time required ispolynomial. This completes the proof of Theorem 4. References
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A Omitted proofs
A.1 Section 4
Proof (Lemma 7).
We prove the lemma row strategies given a support J ofthe column player. The proof for column strategies is similar. Let x be a rowstrategy which includes two strategies i and i that are J -equivalent. It sufficesto show that there is a row strategy ˆ x with S (ˆ x ) = S ( x ) \ i and ˆ x T Ay = x T Ay for any column strategy y with support J . For this, take ˆ x to be the strategydefined by ˆ x i := x i + x i , ˆ x i := 0, and ˆ x i = x i for all i ∈ { , . . . , n } \ { i , i } .Let y be any column strategy with S ( y ) = J . By definition of J -equivalence, wehave ( Ay ) i = ( Ay ) i . Thus, we get that(ˆ x T Ay ) i + (ˆ x T Ay ) i = ( x T Ay ) i + ( x T Ay ) i . Furthermore, since ˆ x and x equal on all entries i ∈ [ n ] \ { i , i } , we know that X i = i ,i (ˆ x T Ay ) i = X i = i ,i ( x T Ay ) i , and thus(ˆ x T Ay ) = X i = i ,i (ˆ x T Ay ) i + (ˆ x T Ay ) i + (ˆ x T Ay ) i = X i = i ,i ( x T Ay ) i + ( x T Ay ) i + ( x T Ay ) i = ( x T Ay ) . Therefore ˆ x T Ay = x T Ay for all column strategies y with S ( y ) = J . ⊓⊔ A.2 Section 5
Proof (Lemma 8).
Let ( x, y ) be an equilibrium, and suppose that S ( y ) includestwo equivalent strategies i and j . To prove the lemma it suffices to show thatthe there exists an equilibrium ( x, y ∗ ) with S ( y ∗ ) = S ( y ) \ { j } . For this, let ustake y ∗ to be the strategy vector defined by y ∗ i := y i + y j , y ∗ j := 0, and y ∗ x = y x for all x ∈ { , . . . , n }\{ i, j } . Clearly S ( y ∗ ) = S ( y ) \{ j } . Thus, for each s ∈ S ( y ∗ )we have ( x T B ) s ≥ ( x T B ) j , ∀ j = s, since this condition holds for all s ∈ S ( y ) ⊇ S ( y ∗ ) by Lemma 1. Furthermore, dueto the definition of equivalence, a simple calculation shows that ( Ay ) s = ( Ay ∗ ) s for all s ∈ { , . . . , k } . Therefore, for each s ∈ S ( x ), we get again using Lemma 1that ( Ay ∗ ) s = ( Ay ) s ≥ ( Ay ) j , ∀ j = s. It follows that ( x, y ∗ ) satisfies both conditions of Lemma 1, and so it is indeedan equilibrium.) satisfies both conditions of Lemma 1, and so it is indeedan equilibrium.