# A Manifold of Polynomial Time Solvable Bimatrix Games

This paper identifies a manifold in the space of bimatrix games which contains games that are strategically equivalent to rank-1 games through a positive affine transformation. It also presents an algorithm that can compute, in polynomial time, one such rank-1 game which is strategically equivalent to the original game. Through this approach, we substantially expand the class of games that are solvable in polynomial time. It is hoped that this approach can be further developed in conjunction with other notions of strategic equivalence to compute exact or approximate Nash equilibria in a wide variety of bimatrix games.

## Authors

• 3 publications
• ### Fast Algorithms for Rank-1 Bimatrix Games

The rank of a bimatrix game is the matrix rank of the sum of the two pay...
12/11/2018 ∙ by Bharat Adsul, et al. ∙ 0

• ### Simple Games versus Weighted Voting Games: Bounding the Critical Threshold Value

A simple game (N,v) is given by a set N of n players and a partition of ...
10/20/2018 ∙ by Frits Hof, et al. ∙ 0

• ### Manipulating Tournaments in Cup and Round Robin Competitions

In sports competitions, teams can manipulate the result by, for instance...
11/09/2009 ∙ by Tyrel Russell, et al. ∙ 0

• ### On the Computation of Strategically Equivalent Rank-0 Games

It has been well established that in a bimatrix game, the rank of the ma...
03/31/2019 ∙ by Joseph L. Heyman, et al. ∙ 0

• ### (weak) Calibration is Computationally Hard

We show that the existence of a computationally efficient calibration al...
02/20/2012 ∙ by Elad Hazan, et al. ∙ 0

• ### Approximate Solutions to a Class of Reachability Games

In this paper, we present a method for finding approximate Nash equilibr...
11/01/2020 ∙ by David Fridovich-Keil, et al. ∙ 0

• ### Winning the War by (Strategically) Losing Battles: Settling the Complexity of Grundy-Values in Undirected Geography

We settle two long-standing complexity-theoretical questions-open since ...
06/03/2021 ∙ by Kyle Burke, et al. ∙ 0

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Introduction

The study of game theory – the model of strategic interaction between rational agents – has wide ranging applicability within the fields of economics, engineering, and computer science. Computing a Nash equilibrium (NE) in a

-player finite game is one of the fundamental problems in game theory.111Nash equilibrium is an acceptable and a widely used solution concept for games; we define it later in the paper. Due to the well known theorem by Nash in 1951, we know that every finite game has a solution, possibly in mixed strategies [1]. However, computation of a NE has been shown to be hard. For , [2, 3] proved that computing a NE is Polynomial Parity Argument, Directed Version (PPAD) complete. Indeed, even the -player case has been shown to be PPAD-complete [4].

In this work we focus on finite, -player bimatrix games in which the payoffs to the players can be represented as two matrices, and . With the hardness of computing a NE in the bimatrix case well established, identifying classes of games whose solutions can be computed efficiently is an active area of research. As an example in bimatrix games, one can compute the solution of zero-sum games222A game is called a zero-sum game if the sum of payoffs of the players is zero for all actions of the players.

in polynomial time using the minimax theorem and a simple linear program

[5, p. 152]. Another subclass of games that admits a polynomial time solution is the class of strategically zero-sum games introduced by Moulin and Vial in 1978 [6].

More recently, in a series of works [7, 8] the authors have developed polynomial time algorithms for solving another subclass of bimatrix games called rank- games. Rank- games are defined as games where the sum of the two payoff matrices is a rank- matrix. Here we note that this subclass lies in a space of .

We say that two games, and , are strategically equivalent if the optimal strategies of every player in corresponds to the optimal strategies of every player in .333See Section 2 for a formal definition and the specific form of strategic equivalence that we study in this work. Unfortunately, many operations that preserve the strategic equivalence of bimatrix games modify the rank of the game. For example, the well studied constant-sum game is strategically equivalent to a zero-sum game. However, the zero-sum game has rank zero, while the constant-sum game is a rank- game. Since the rank of a game influences the most suitable solution technique one should be particularly interested in determining if a given game is strategically equivalent to a game , where the rank of is less than the rank of .

In this contribution, we do just that. We show that, in polynomial time, it is possible to determine if the game lies in a manifold of size . If so, we show that the game is strategically equivalent to some rank- game . Furthermore, we show that one can also calculate this rank- game in polynomial time. As we will show, our algorithm is surprisingly straightforward and significantly expands the space of games which can be solved in polynomial time.

### 1.1 Prior Work

Many papers in the literature have explored the concept of equivalent classes of games. One such concept is strategic equivalence, those games that share exactly the same set of NE. Indeed, for a classical example, von Neumann and Morgenstern studied strategically equivalent -person zero-sum games [5, p. 245] and constant-sum games [5, p. 346].

Closely related to our work is the class of strategically zero-sum games defined by Moulin and Vial in [6]. For the bimatrix case, they provide a complete characterization of strategically zero-sum games [6, Theorem 2]. While the authors do not analyze the algorithmic implications of their characterization, Kontogiannis and Spirakis do analyze the approach in their work on mutually concave games. They find that the characterization in [6, Theorem 2] can determine whether a bimatrix game is strategically zero-sum, and, if so, calculate an equivalent zero-sum game in time .

In [9], Kontogiannis and Spirakis formulate a quadratic program where the optimal solutions of the quadratic program constitute a subset of the correlated equilibria of a bimatrix game. Furthermore, they then show that this subset of correlated equilibria are exactly the NE of the game. In order to show polynomial tractability of the quadratic program, they define a class of mutually concave bimatrix games. Surprisingly, they find that their characterization of mutually concave bimatrix games is equivalent to Moulin and Vial’s characterization of strategically zero-sum bimatrix games [9, Corollary 2]. They then propose a parameterized version of the Mangasarian and Stone quadratic program that is guaranteed to be convex for a mutually concave game and has time complexity for [9, Theorem 2]. With the problem of solving a mutually concave game shown to be tractable, the authors then proceed to show that recognizing a mutually concave game can be done in time .

As far as we are aware, no other authors have specifically studied games that are strategically equivalent to rank- games. With the recently developed fast algorithms [7, 8] for solving rank- games, now seems like an appropriate time to do just that.

### 1.2 Our Contribution

Given a nonzero-sum bimatrix game, , we develop the SER1 algorithm that determines whether or not the given game lies in a manifold of size . If so, then the game is strategically equivalent to some rank- game . Our approach relies on the classical linear algebra result known as the Wedderburn Rank Reduction Theorem and the theory of matrix pencils. However, we show that it is possible to avoid many of the traditional solution concepts applied to matrix pencils and determine if there exists a solution such that the pencil is a rank- pencil via solving a single quadratic expression. Thus, our approach is surprisingly simple and leads to an amazingly fast algorithm. In Section 5 we show that both the determination of strategic equivalence and the computation of the strategically equivalent rank- game can be done in time , where is the bit-length of the largest absolute value of entries in , and is the complexity of multiplication. This equivalent rank- game can then be solved in polynomial time [7, 8]. Since the two games are strategically equivalent, the NE strategies of the equivalent rank- game are exactly the NE strategies of the original nonzero-sum bimatrix game .

We thus show a significant expansion to the class of bimatrix games that can be solved in polynomial time.

An astute observer whom is familiar with the field would likely recognize that our problem can easily subsume the strategically zero-sum case. However, with multiple existing polynomial time algorithms for that case [6, 9, 10], we choose to specifically focus on strategically equivalent rank- games.

### 1.3 Notation

We use and

to denote, respectively, the all ones and all zeros vectors of length

. All vectors are annotated by bold font, e.g , and all vectors are treated as column vectors.

is the set of probability distributions over

, where . Let , , denote the vector with in the th position and ’s elsewhere.

Consider a matrix . We use to indicate the rank of the matrix . indicates the subspace spanned by the columns of the matrix , also known as the column space of the matrix . We indicate the nullspace of the matrix , the space containing all solutions to , as . In addition, we use to denote the jth column of and to denote the ith row of .

## 2 Preliminaries

In this section, we recall some basic definitions in bimatrix games and the definition of strategic equivalence in bimatrix games.

We consider here a two player game, in which player 1 (the row player) has actions and player 2 (the column player) has actions. Player 1’s set of pure strategies is denoted by and player 2’s set of pure strategies is . If the players play pure strategies , then player 1 receives a payoff of and player 2 receives .

We let represent the payoff matrix of player 1 and represent the payoff matrix of player 2. As the two-player finite game can be represented by two matrices, this game is commonly referred to as a bimatrix game. The bimatrix game is then defined by the tuple . Define the matrix as the sum of the two payoff matrices, . We define the rank of a game as .444Some authors define the rank of the game to be the maximum of the rank of the two matrices and , but this is not the case here.

Players may also play mixed strategies, which correspond to a probability distribution over the available set of pure strategies. Player 1 has mixed strategies and player 2 has mixed strategies , where and . Using the notation introduced above, player 1 has expected payoff and player 2 has expected payoff .

A Nash Equilibrium is defined as a tuple of strategies such that each player’s strategy is an optimal response to the other player’s strategy. In other words, neither player can benefit, in expectation, by unilaterally deviating from the Nash Equilibrium. This is made precise in the following definition.

###### Definition 1 (Nash Equilibrium [1]).

We refer to the pair of strategies as a Nash Equilibrium (NE) if and only if:

 p∗TAq∗≥pTAq∗∀p∈Δm;p∗TBq∗≥p∗TBq ∀q∈Δn.

It is a well known fact due to Nash [1] that every bimatrix game with finite set of pure strategies has at least one NE in mixed strategies. However, one can define games in which multiple NE exist in mixed strategies. Let be the Nash equilibrium correspondence555A correspondence is a set valued map [11, p. 555].: Given the matrices , denotes the set of all Nash equilibria of the game . Note that due to the result in [1], is nonempty for every .

We say that two games are strategically equivalent if both games have the same set of players, the same set of strategies per player, and the same set of Nash equilibria. The following definition formalizes this concept.

###### Definition 2.

The 2-player finite games and are strategically equivalent iff .

We now have a well known Lemma on strategic equivalence in bimatrix games that is typically stated without proof.666See e.g [6, 9]. For completeness, the proof is in Appendix A.

###### Lemma 1.

Let be two matrices. Let and where , , , and . Then the game is strategically equivalent to .

## 3 Problem Formulation and Main Results

From Lemma 1, we conclude that if there exists , , , and such that:

 ~A =α1A+β11muT, (1) ~B =−α2A+α2rcT+β2v1Tn (2)

then is strategically equivalent to the rank- game via a positive affine transformation (PAT). Here we note that there may be other transforms, even nonlinear transforms, which maintain the set of NE for a specific game. We don’t consider those transforms and focus on the class of PATs. Throughout this work, any mention of strategic equivalence refers to strategic equivalence via a PAT.

Combining (1) and (2), we have:

 ~A =−α1α2~B+α1rcT+β11muT+α1α2β2v1Tn. (3)

Defining , , and , we rewrite (3) as:

 ~A+γ~B=β11muT+γβ2v1Tn+α1rcT=D+^r^cT. (4)

In addition, let us define and the subspace , which we define as

 Mm×n(R)={M∈Rm×n∣∣ there exists u∈Rn, and v∈Rm such that M=1muT+v1Tn}.

From (4), it is clear that if is strategically equivalent to the rank- game via a PAT, then there exists a such that . However, while is a necessary condition, it is not sufficient. For sufficiency, we also require the existence of , , and such that .777By the definition of , we have . This implies that .

Thus, what we have shown above is the following result:

###### Proposition 2.

is strategically equivalent to through a PAT if and only if , where .

###### Proof.

The proof follows from the preceding discussions. ∎

In what follows, we show the converse holds. Furthermore, in Section 4 we devise methods for determining , , and . Finally, in Section 5 we construct an algorithm that efficiently implements the results from Section 4.

For a matrix , decompose it in a manner such that , where and . Let denote the matrix . This function can be computed efficiently.

###### Assumption 3.

The game satisfies

1. .

2. is a rank-1 matrix.

###### Theorem 4.

If Assumption 3 holds, then there exists a matrix and vectors , such that the bimatrix game is strategically equivalent to the rank- game .

###### Proof.

This result is a direct consequence of Theorem 6 presented in Subsection 4.1 and Theorem 19 presented in Appendix C. ∎

We briefly note here that, even if one were to be given , and a game that is strategically equivalent to it would be impossible to exactly recover without also knowing , and . Thus, our technique calculates another game, which is rank- and strategically equivalent to both and .

Let us now turn our attention to showing the necessary conditions under which Assumption 3 holds true and devising methods to test those conditions.

## 4 Proof of the Main Result

In this section, we prove the main theoretical results presented in Section 3. Let us now turn our attention to decomposing the game via the Wedderburn Rank Reduction Formula.

### 4.1 Reducing the Rank of a Game via the Wedderburn Rank Reduction Formula

In this subsection, we show that if there exists a such that where , then it may be possible, under certain conditions, to calculate a rank-1 game that is strategically equivalent to the rank-k game . In what follows, we first present the Wedderburn Rank Reduction formula upon which our technique is based, and then we proceed to state our result, leaving the proof to Appendix C.1.

The Wedderburn Rank Reduction formula is a classical technique in linear algebra that allows one to reduce the rank of a matrix by subtracting a specifically formulated rank-1 matrix. By repeated applications of the formula, one can obtain a matrix decomposition as the sum of multiple rank-1 matrices. 888For further reading on the Wedderburn rank reduction formula, we refer the reader to Wedderburn’s original book [12, p. 69], or to the excellent treatment of the topic by Chu et al. [13].

We now proceed to state Wedderburn’s original theorem. Following that, we show how one can exploit the flexibility of the decomposition to extract specifically formulated rank-1 matrices that allow us, when certain conditions hold true, to reduce the rank of a bimatrix game.

###### Theorem 5 ([12, p. 69][13]).

Let be an arbitrary matrix, not identically zero. Then, there exists vectors and such that . Then, setting for convenience, the matrix

 C2\coloneqqC1−w−11C1x1yT1C1 (5)

has rank exactly one less than the rank of .

###### Proof.

The original proof of (5) is due to Wedderburn [12, p. 69]. See Appendix B for our restated version. ∎

Let us now proceed to show how one can apply Theorem 5 in order to reduce the rank of a bimatrix game. For ease of exposition, we break the result into two theorems–Theorem 6 and Theorem 19 (presented in Appendix C).

###### Theorem 6.

Consider the game . If there exists such that where with , , , , then there exists and such that:

1. and . Let and compute using (5) as follows:

 ~C2=~C−w−11~Cx1yT1~C=~C−1m^uT.
2. and . Let and compute using (5) as follows:

 ~C3=~C2−w−12~C2x2yT2~C2=~C−^v1Tn=~C−1m^uT−^v1Tn.

Define and . Then, and the bimatrix game is strategically equivalent to the rank- game .

###### Proof.

See Appendix C. ∎

In Theorem 6 we assumed the existence of such that is in our desired form. In the next subsection, we show an efficient decomposition of that we can conduct without knowing . Following that, we then show a technique for solving for such a (if it exists).

### 4.2 An Efficient Decomposition of ~C

First-off, let us assume that there exists a such that , where with . See Appendix D for the cases of . Therefore, we know that there exists a vector such that and a such that . Examining Theorem 6, we see that the term always appears as a term in the matrix-vector product . Therefore, we can replace all appearances of the term with and there is no need to explicitly calculate .

An algorithmic efficient manner for selecting is to let for some . Then we have that:

 w−11=1yT1~Cx1=1eTi1m=1.

Then, we also have:

 =w−11~Cx1yT1~C=1meTi~C=1meTi(~A+γ~B) =(1m~A(i)+γ1m~B(i)).

Then we have that . Then, letting and , we have and

 =~C2ej1Tn=[~A+γ~B−(1m~A(i)+γ1m~B(i))]ej1Tn =(~A(j)−1m~ai,j)1Tn+γ(~B(j)−1m~bi,j)1Tn.

Therefore, we have

 ~C3=~A+γ~B−(1m~A(i)+γ1m~B(i))−((~A(j)−1m~ai,j)1Tn+γ(~B(j)−1m~bi,j)1Tn). (6)

For notational simplicity, let and . Clearly, we have for all .

###### Theorem 7.

Consider the game , select any , and formulate as in (6). Let and . If there exists such that , then the game is strategically equivalent to the rank- game .

###### Proof.

The proof follows from the preceding discussion and Theorem 6. ∎

Thus, what we’ve shown is that our problem is equivalent to forming a rather simple matrix decomposition to calculate , and then determining whether there exists a scalar parameter such that . Let us now turn our attention to the theory of matrix pencils and develop an efficient procedure for calculating .

### 4.3 Rank-1 Matrix Pencils

We would be remiss if we did not mention that fact that problems in the form , with matrices of known values and an unknown scalar parameter, are typically referred to as linear matrix pencils (or just pencils)[14, p. 24],[15]. Therefore, our problem of finding a such that is intimately connected to the theory of matrix pencils and the

generalized eigenvalues

of matrix pencils. However, we eschew the traditional techniques for solving matrix pencils in order to avoid the higher than necessary computational complexity and possible numerical difficulties. Instead, in this subsection, we develop a series of results that allow us to determine whether or not there exists a such that by solving for the roots of a single polynomial equation (at worse a quadratic) and then conducting at most two matrix comparisons.

Let us begin by stating some facts about rank-1 matrices that will allow us to easily ascertain when a given matrix is rank- and to solve for values of , when they exist, such that the matrix pencil, , is a rank- pencil.

###### Fact 1.

The matrix in is rank- if and only if and the following hold true:

1. Every row (column) of is a scalar multiple of every other row (column) of .

2. Choose any element of such that and form , . Then, .

###### Theorem 8.

Given the matrix pencil, , with , assume that and . Construct -vectors and as follows:

1. Choose any such that . Such an is guaranteed to exist since .

2. Let .

3. Let .

Then there exists such that if and only if either:

1. and ; or

2. .

###### Proof.

We first prove the forward direction. Suppose there exists such that . We split the proof of the forward direction into two cases:

Case : Suppose that . Since , this implies . Furthermore, and implies . In addition, note that implies that is undefined; therefore, the expression is undefined and cannot hold true.

Case : Now, suppose . Then . Also, , so is well-defined. Then follows from Fact 1.

We next prove the reverse direction. Suppose and . Then and . Of course, as in above, since is undefined at by definition, we conclude that .

Now, suppose , which implies that is well-defined. This implies that and/or . Furthermore, since is the solution to a system of linear or quadratic equations with real coefficients, we have . Then, follows from Fact 1. ∎

Since it is rather trivial to determine whether or not and , we’ll assume throughout the rest of this subsection that . Let us now examine the matrix equality , introduce some additional notation, and state some lemmas that allow us to determine whether or not there exists a finite such that .

With and as defined in Theorem 8, let us write the following system of equations:

 A+λB =rj(λ)ci(λ)T (7) ⎡⎢ ⎢ ⎢⎣a1,1+λb1,1…a1,n+λb1,n⋮⋱⋮am,1+λbm,1…am,n+λbm,n⎤⎥ ⎥ ⎥⎦ =⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣(a1,j+λb1,j)(ai,1+λbi,1)ai,j+λbi,j…(a1,j+λb1,j)(ai,n+λbi,n)ai,j+λbi,j⋮⋱⋮(am,j+λbm,j)(ai,1+λbi,1)ai,j+λbi,j…(am,j+λbm,j)(ai,n+λbi,n)ai,j+λbi,j⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦ (8)

Since is a scalar variable, it is clear from (8) that (7) only has a solution (or possibly multiple solutions), , if simultaneously satisfies single-variable polynomials, where each polynomial is of degree at most . Thus, one could solve all single-variable polynomials and then check whether or not every solution has a common value. While this procedure is somewhat efficient, we’ll show below that at most of the polynomials have finite solutions and therefore contribute any meaningful information. In addition, we’ll show that it is sufficient to identify one polynomial that is not the zero polynomial and then conduct a matrix checking problem for the solution(s) of that polynomial.

Let us now introduce notation for the polynomial represented by row and column in (8). For , let

 fs,t(i,j;λ)=as,t+λbs,t−(as,j+λbs,j)(ai,t+λbi,t)ai,j+λbi,j (9)

From (9), it is clear that when or , then and are the zero polynomial. In other words, trivially holds true for all for one entire column and one entire row of (8). Since these expressions hold true for all , they lend no information for determining whether or not there exists such that . Thus, we can disregard these polynomials and only consider the remaining polynomials. Let us now show that at least one of the remaining polynomials is not the zero polynomial. The proof of the Lemma is deferred to Appendix A.

###### Lemma 9.

Let , and assume that . Given the matrix pencil, , construct -vectors and as in Theorem 8 and as in (9). If , then there exists at least one pair such that is not the zero polynomial.

###### Theorem 10.

Let , and assume that with . Given the matrix pencil, , construct -vectors and as in Theorem 8. Pick any such that is not the zero polynomial. Let be solutions to . Then, there exists such that if and only if or/and .

###### Proof.

Let us first note that, by Lemma 9, there exists at least one pair such that is not the zero polynomial.

Now, suppose there exists such that . Then, by Theorem 8, . Thus, for all . In particular, . Therefore, either or/and . It then follows that or/and .

In the other direction, let be solutions to . Note that may be a linear equation. In that case, for simplicity, let . Now, suppose and let . Then and by Theorem 8. The case of is similar and therefore omitted. ∎

Let us use to represent the solution set obtained from Theorem 10. Note, has a maximum cardinality of and may, of course, be empty.

Throughout the presentation in this subsection, we’ve assumed . For completeness, we present the next trivial lemma without proof.

If , then and .

### 4.4 Solving for γ∗

Let us now present our main result for solving for and calculating the equivalent rank- game.

###### Theorem 12.

Consider the game and formulate