Parameterized Two-Player Nash Equilibrium Danny Hermelin, Chien-Chung Huang, Stefan Kratsch, and Magnus Wahlstrom..

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Parameterized Two-Player Nash Equilibrium Danny Hermelin, Chien-Chung Huang, Stefan Kratsch, and Magnus Wahlstrom..

 Played by two players: Row and Column –Two payoff matrices. A,B  Q n  n. Bimatrix Game Row chooses i Column chooses j Row payoff A[i,j] = -2 Column payoff B[i,j] = 0

 Example: –Rock, paper, scissors: Bimatrix Game  This example is a zero-sum game: –Row and column payoffs sum up to zero.  General bimatrix games are not necessarily such. –In fact, the interesting cases (to us) are not zero-sum.

 Players can play mixed strategies. –Distribution over rows and columns. Bimatrix Game Row chooses distribution x Column chooses distribution y Row expected payoff x T Ay = 0 Column expected payoff x T By = 1 1/2 0 x 10 0 y

 Neither player can improve their payoff, assuming the other player plays the same. Nash Equilibrium Row can improve by switching to row 2. Not Nash !

 Neither player can improve their payoff, assuming the other player plays the same. Nash Equilibrium Theorem (Nash) : Any bimatrix rational game has a mixed equilibrium. Nash !

 The Nash Equilibrium (NE) problem: Given a bimatrix rational game, find an equilibrium.  NP-completeness theory does not apply because solution always exists.  PPAD-complete by a series of papers: –Daskalakis, Goldberg, and Papadimitriou [STOC’06,STOC’06]. –Daskalakis and Papadimitriou [ECCC’05] –Chen and Deng [ECCC’05] –Chen and Deng [FOCS’06]  The 3-SAT of algorithmic game theory ! Computing Nash Equilibrium

 Support: Set of strategies played with non-zero probability.  When support of both players is known, NE is easy. Computing Nash Equilibrium  Solve LP with the following constraints: ‒ x s > 0  (Ay) s  (Ay) j for all j  s. ‒ y s > 0  (x T B) s  (x T B) j for all j  s

Computing Nash Equilibrium Theorem : NE can be solved in n O(k) time, when the supports of each player are bounded by k. –Can this be improved substantially? –Can we remove k out of the exponent? Theorem (Estivill-Castro, Parsa) : NE cannot be solved in n o(k) time unless FPT=W[1]. GOAL: find interesting special cases that circumvent this

Graph Representation of Bimatrix Games  Bipartite graph on rows and columns  (i,j) is an edge  A[i,j]  0 or B[i,j]  0

1. l -sparse games: –Degrees  l. 2. k-unbalanced games: –One side has  k vertices. 3. Locally bounded treewidth: –Every d-neighborhood has treewidth  f(d). –Generalizes both previous cases. Interesting Special Cases  l l (1)  k k (2)(3) previously studied games

Our Results Theorem : NE in l -sparse games, where the support is bounded by k, can be solved in l O(k l ) n O(1) time. –Without the restriction on the support size the problem is PPAD- complete [Chen, Deng, and Teng ‘06]. Theorem : NE in locally bounded treewidth games, where the support is bounded by k, and both payoff matrices have l different values, can be solved in f( l, k) n O(1) time for some computable f(). –General k-sparse games is not known to be FPT. –But how do we show its not ? Theorem : NE in k-unbalanced games, where the row player’s payoff matrix has l different values, can be solved in l O(k ) n O(1) time. 2

l -Sparse Games  Recall l := max-degree and k:= support size.  Two easy observations: 1.Enough to search for minimal equilibriums. 2.If n > k l, then both players receive non-negative payoffs on any k  k equilibrium. Definition: An equilibrium (x,y) is minimal if for any equilibrium (x’,y’) with S(x’)  S(x) and S(y’)  S(y), we have S(x’) = S(x) and S(y’) = S(y). If a player get negative payoff and n > k l, there will always be a zero-payoff strategy to switch to.

l -Sparse Games Definition: The extended support of (x,y) is S(x)  N(S(y)) for the row player, and S(y)  N(S(x)) for the column player S(x) S(y) N(S(y)) extended support of row The size of the extended support of each player  k + k l.

l -Sparse Games  Main technical lemma: Lemma: If (x,y) is minimal equilibrium, then the subgraph H  G induced by the extended supports has at most 2 connected components. Proof sketch: 1. Prove separately for the case where A s(x),s(y) = 0 and B s(x),s(y) = 0, and for the case when one of these matrices is not all-zero. 2.In the latter case, normalize probabilities on some connected component of H. 3. In the former case, argue the same on G[N(S(x))] and G[N(S(y))].

l -Sparse Games  Folklore FPT lemma: Lemma: Let G be a graph on n vertices of maximum degree . Then one can enumerate all induced subgraphs H on h vertices and c connected components in H  G in  O(h) n O(c) time. Proof sketch: 1. Guess c vertices S in G to be the targets of vertices in different connected components of H. 2.Branch on the h-neighborhood of S to enumerate all H  G. 3.The size of each branch-tree is  O(h).

l -Sparse Games  The algorithm: 1.Guess the number h of strategies in both extended support. 2.Guess the number of connected components c  {1,2} in the corresponding induced subgraph. 3.Enumerate all induced subgraphs on h vertices and c connected components. 4.For each such subgraph, the supports of both players are known. Thus, one can use LP to determine if it corresponds to an equilbrium.

l -Sparse Games  Extensions: 1.We can improve running-time to l O(k l ) n O(1) in case both payoff matrices are non-negative. 2.Another route to a well-known PTAS. 3.Connectivity lemma can be used to show that the problem has no “polynomial kernel”.

Open questions 1.k-unbalanced games with an arbitrary number of payoffs. 2.Bounded treewidth games with an arbitrary number of payoffs. 3.Parameterized analog of the PPAD class. Conjecture: NE parameterized by k in k-unbalanced games is Para-PPAD-Complete.