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.^{1}^{1}1Nash 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 PPADcomplete [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 zerosum games^{2}^{2}2A game is called a zerosum 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 zerosum 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 .^{3}^{3}3See 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 constantsum game is strategically equivalent to a zerosum game. However, the zerosum game has rank zero, while the constantsum 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 zerosum games [5, p. 245] and constantsum games [5, p. 346].
Closely related to our work is the class of strategically zerosum games defined by Moulin and Vial in [6]. For the bimatrix case, they provide a complete characterization of strategically zerosum 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 zerosum, and, if so, calculate an equivalent zerosum 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 zerosum 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 .
1.2 Our Contribution
Given a nonzerosum 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 bitlength 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 nonzerosum 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 zerosum 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 j^{th} column of and to denote the i^{th} 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 twoplayer 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 .^{4}^{4}4Some 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:
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 correspondence^{5}^{5}5A 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 2player 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.^{6}^{6}6See 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:
(1)  
(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.
Defining , , and , we rewrite (3) as:
(4) 
In addition, let us define and the subspace , which we define as
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 .^{7}^{7}7By 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

.

is a rank1 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.
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 rank1 game that is strategically equivalent to the rankk 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 rank1 matrix. By repeated applications of the formula, one can obtain a matrix decomposition as the sum of multiple rank1 matrices. ^{8}^{8}8For 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 rank1 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
(5) 
has rank exactly one less than the rank of .
Proof.
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:
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
Firstoff, 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 matrixvector 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:
Then, we also have:
Then we have that . Then, letting and , we have and
Therefore, we have
(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 Rank1 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 rank1 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:

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

Choose any element of such that and form , . Then, .
Theorem 8.
Given the matrix pencil, , with , assume that and . Construct vectors and as follows:

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

Let .

Let .
Then there exists such that if and only if either:

and ; or

.
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 welldefined. 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 welldefined. 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:
(7)  
(8) 
Since is a scalar variable, it is clear from (8) that (7) only has a solution (or possibly multiple solutions), , if simultaneously satisfies singlevariable polynomials, where each polynomial is of degree at most . Thus, one could solve all singlevariable 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
(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.
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.
Lemma 11.
If , then and .
4.4 Solving for
Let us now present our main result for solving for and calculating the equivalent rank game.
Theorem 12.
Consider the game and formulate
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