1 Introduction
Supermodular games, also known as the games of strategic complements, are formalized by Topkis in 1979 [23] and have been extensively studied in the literature, such as Bernstein and Federgruen [2][1], Cachon [4], Cachon and Lariviere [5], Fudenberg and Tirole [11], Lippman and McCardle [16], Milgrom and Roberts [17][18], Milgrom and Shannon [19], Topkis [24], and Vives [25][26]. In supermodular games, the utility function of every player has increasing differences. Then the best response of a player is a nondecreasing function of other players’ strategies. For example, if firm A’s competing firm B starts spending more money on research it becomes more advisable for firm A to do the same.
Supermodular games arise in many applied models. They cover most static market models. For example, the investment games, Bertrand oligopoly, Cournot oligopoly all can be modeled as supermodular games. Many models in operations research have also been analyzed as supermodular games. For example, supply chain analysis, revenue management games, price and service competition, inventory competition etc. Recently, the problem of power control in cellular CDMA wireless network is also modeled as a supermodular game.
The existence of a pure Nash equilibrium in any supermodular game is proved by Tarski’s fixed point theorem [21]. The wellknown Tarski’s fixed point theorem (Tarski) asserts that, if is a complete lattice and is orderpreserving from into itself, then there exists some such that .
This theorem plays a crucial role in the study of supermodular games for economic analysis and has other important applications. To compute a Nash equilibrium of a supermodular game, a generic approach is to convert it into the computation of a fixed point of an order preserving mapping. Recently, an algorithm has been proposed in Echenique [10] to find all pure strategy Nash equilibria of a supermodular game, which motivated to the study in this paper.
An efficient computational algorithm for finding a Nash equilibrium has been a recognized important technical advantage in applications. Further, it is sometimes desirable to know if an alreadyfound equilibrium for such applications is unique or not, for the decision whether additional resource should be spent to improve the already found solution. There were some interesting complexity results in algorithmic game theory research along this line, on determining whether or not a game has a unique equilibrium point. For the bimatrix game, Gilboa and Zemel
[12] showed that it is NPhard to determine whether or not there is a second Nash equilibrium. For this problem, computing even one equilibrium (which is know to exist), is already difficult and no polynomial time algorithms are known: Nash equilibrium for the bimatrix game is known to be PPADcomplete [9]. Similar cases are known for other problems such as the market equilibrium computation (Codenotti et al.)[3].In this work, we first consider the fixed point computation of order preserving functions over a complete lattice, both for finding a solution and for determining the uniqueness of an alreadyfound solution. Then we study the computational problems for finding one pure Nash equilibrium and determining the uniqueness of the equilibrium in supermodular games. We are interested in both the oracle function model and the polynomial function model. For both the fixed point problem and supermodular games, the domain space can be huge. Most interesting discussions consider a succinct representation (see Section 2.2) of the lattice such that the input size is related to . It is enough for the representation of a variable in a lattice of size . Both the oracle function model and the polynomial time function model return the function value on a lattice node where is of size . They differ in the ways the functions are computed. The polynomial time function model computes by an explicitly given algorithm, in time polynomial of . The oracle model, on the other hand, always returns the value in one oracle step. More details comparing those two models can be found in Section 2.3.
1.1 Main Results and Related Work
A partially order set is defined with as a binary relation on the set such that is reflexive, transitive, and antisymmetric. A lattice is a partially ordered set , in which any two elements and have a least upper bound (supremum), , and a greatest lower bound (infimum), , in the set. A lattice is complete if every nonempty subset of has a supremum and an infimum in . Let be a mapping from to itself. is orderpreserving if for any and of with .
We focus on the componentwise ordering and lexicographic ordering finite lattices. Let , where and
are two finite vectors of
with . We denote the componentwise ordering and the lexicographic ordering as and respectively. Clearly, is a finite lattice with componentwise ordering and is a finite lattice with lexicographic ordering.Let and be an order preserving mapping from into itself under the componentwise ordering and the lexicographic ordering respectively.
1.1.1 Tarski’s Fixed Points: Oracle Function Model
When and are given as oracle functions, we develop a complete understand for finding a Tarski’s fixed point as well as determining uniqueness of the Tarski’s fixed point in both the lexicographic ordering and the componentwise ordering lattices.
We develop an algorithm of time complexity to find a Tarski’s fixed point on the componentwise ordering lattice , for any constant dimension . This algorithm is based on the binary search method. We first present the algorithm when . Follows the similar principle, this algorithm can be generalized to any constant dimension. This is the first known polynomial time algorithm for finding the Tarski’s fixed point in terms of the componentwise ordering. In literature, we only have a polynomial time algorithm for the total order lattices (Chang et al.) [6].
Recently, Mihalis, Kusha and Papadimitriou stated in a private communication that they proved a lower bound of in the oracle function model for finding a Tarski s fixed point in the two dimensional case, and conjectured a lower bound of for general (Christos H. Papadimitriou, private communication, March, 2019). Together with our upper bound results, they establish a matching bound of for finding a Tarski s fixed point in the two dimensional case.
On the other hand, given a general lattice with one already known fixed point, to find out whether it is unique will take time for any algorithm. For componentwise ordering lattice, we derive a matching bound for determining the uniqueness of the fixed point, where and . In addition, we prove this matching bound for both deterministic algorithm and randomized algorithm.
For a lexicographic ordering lattice, it can be viewed as a componentwise ordering lattice with dimension one by an appropriate polynomial time transformation to change the oracle function for the dimension space to an oracle function on the dimension space. All the above results can be transplanted onto the lexicographic ordering lattice with a set of related parameters.
In literature, a polynomial time algorithm is known only for the total order lattices. When the lattice has a total order, i.e., all the point in the lattice is comparable, there is a matching bound of , where an lower bound is known for general lattices (when the lattice is given as an oracle) in Chang et al.[6].
1.1.2 Tarski’s Fixed Points: Polynomial Function Model
Under the polynomial time function model, our polynomial time algorithm applies when the dimension is any finite constant. When the dimension is used as a part of the input size in unary, we first present a polynomialtime reduction of a 3SAT problem to an order preserving mapping from a componentwise ordering lattice into itself. As a result of this reduction, we obtain that, given as a polynomial time function, determining whether has a unique fixed point in is a CoNP hard problem. Furthermore, even when the dimension is one, we also show that determining the uniqueness of Tarski’s fixed point in a lexicographic lattice is CoNP hard though there exists a polynomialtime algorithm for computing a Tarski’s fixed point in a lexicographic lattice in any dimension.
1.1.3 Supermodular Games
For supermodular games, we develop an algorithm to find a pure Nash equilbirum in polynomial time in the oracle function model, where is the total number of players, is the number of strategies of player and . It is the first polynomial time algorithm when is a constant. Thus a pure Nash equilibirum can be found in time in the polynomial function model, where . In the polynomial function model, we prove determining the uniqueness is CoNPhard.
In literature, Robinson(1951) [20] introduce the iterative method to solve a game and Topkis(1979) [23] use this method to find a pure Nash equilibrium in supermodular game which takes time . The first nontrivial algorithm for finding pure Nash equilibria is proposed by Echenique in 2007[10]. However, the algorithm takes expenontial time to find the first pure equilibrium in the worst case.
Polynomial Function  Oracle Function  

Componentwise  
Lexicographic 
Polynomial Function  Oracle Function  

Componentwise  CoNPComplete  
Lexicographic  CoNPComplete 
1.2 Organization
The rest of the paper is organized as follows. First, in Section 2, we present definitions as well as the difference of the polynomial function model and the oracle function model. We develop polynomial time algorithms in oracle function model for componentwise ordering and lexicographic ordering in Section 3. In Section 4, we derive the matching bound for determining the uniqueness of Tarski’s fixed point under the oracle function model. We prove coNP hardness for determining the uniqueness of Tarski’s fixed point under the polynomial function model in Section 5. In Section 6, we develop the computational results for finding one pure Nash equilibrium and determining the uniqueness of the equilibrium in supermdular games. We conclude with discussion and remarks on our results and open problems in Section 7.
2 Preliminaries
In this section, we first introduce the formal definitions of the related concepts as well as the Tarski’s fixed point theorem. We next compare the difference between the oracle function model and the polynomial function model.
2.1 The Lattice and Tarski’s Fixed Point Theorem
Definition 1.
(Partial Order vs. Total Order) A relationship on a set is a partial order if it satisfies reflexivity (); antisymmetry ( and implies ); transitivity ( and implies ). It is a total order if : either or .
Definition 2.
(Lattice) is a lattice if

is a partial ordered set;

There are two operations: meet and join on any pair of elements of such that and
The lattice is complete lattice if for any subset , there is a unique meet and a unique join: and .
For simplicity, we use for a lattice when no ambiguity exists on . We should specify whenever it is necessary.
Definition 3.
(Order Preserving Function) A function on a lattice is order preserving if implies .
Theorem 1.
(Tarski’s Fixed Point Theorem)[21]. If is a complete lattice and an increasing from to itself, there exists some such that , which is a fixed point of .
This theorem guarantees the existence of fixed points of any orderpreserving function on any nonempty complete lattice.
Definition 4.
(Lexicographic Ordering Function). Given a set of points on a dimensional space , the lexicographic ordering function is defined as:
, if either or , for , and for some .
Definition 5.
(Componentwise Ordering Function). Given a set of points on a dimensional space, the componentwise ordering function is defined as:
, if .
2.2 Big Input Data and Succinct Representation
For the problems we consider in this work, there are usually nodes where is a constant and is an input parameter. Therefore, the input size is exponential in the input parameter . We need to represent such input data succinctly. As an example, for the set , the input can be described as all the integers : . Each such integer can be written by up to bits.
When a computational problem involved a function such as . There is always a question how this function is given as an input? As an exmaple, let represent the parity of the integer . Then as an input to the computational problem, can be an circuit that takes the last bit of the input . Therefore, the size of is a polynomial in the number of bits, , of the input data. In general, however, the input functions are not that simple. We should define two models of functions for succinct representation with input involved with functions on big dataset in our computational problems.
2.3 The Oracle Function Model Versus the Polynomial Time Function Model
The two succinctly represented function models are the oracle functions and the polynomial functions.
For the oracle model, we treat the function as a black box that outputs the function value for every domain variable once a request is sent in to the oracle. The output of the oracle is arbitrary on the first query but it cannot change a function value after a query is already made to the oracle on the same variable. For exmaple, let . Let be an oracle function. When we ask for , the oracle could answer anything, either or . Suppose the oracle answers in the first query in one run of our algorithm. Later, if we need to use again in the same run of the algorithm, it must be the same . Equivalently, we may assume that the function values are stored in the harddisk. After a query, it is saved in the memory cache. Later uses of the same query will be the value in the memory cache and there is no need to check with the harddisk again. It is important to note that, the oracle funciton model contains all the functions where is its domain and is its range. This is very different from the polynomial function we are going to introduce next.
For the polynomial function model, the input function is an algorithm that gives the answer for the function value on the input data. The algorithm returns the answer in time polynomial in the input parameter . Alternatively, the polynomial time algorithm can be replaced by a polynomial size logical circuits consisting of gates of Boolean variables.
Clearly oracle function admits much more functions than those computable in polynomial time. Therefore a problem is usually much harder under the oracle function model than under the polynomial time function model.
3 Polynomial Time Algorithm under Oracle Function Model
In this section, we consider the complexity of finding a Tarski’s fixed point in any constant dimension with the function value given by an oracle. Chang et al. [6] proved that a fixed point can be found in time polynomial when the given lattice is total order.
Define , where and are two finite vectors of with .
Theorem 2.
(Chang et al.)[6] When is given as an input and the order preserving function is given as an oracle, a Tarski’s fixed point can be found in time on a finite lattice when is a total order on .
Since any two vectors in the lexicographic ordering is comparable, the lexicographic ordering is a total order. We have
Corollary 1.
When is given as an input and the order preserving function is given as an oracle, a Tarski’s fixed point can be found in time on a finite lattice when is a lexicographic ordering in .
The proof is rather standard utilizing the total order property of the lexicographic ordering. As the componentwise ordering lattice cannot be modelled as a total order, it leaves open the oracle complexity of finding a fixed point in componentwise ordering lattice. Here we show that this problem is also polynomial time solvable, by designing a polynomial algorithm to find a fixed point of in time given componentwise ordering lattice .
The algorithm exploits the order properties of the componentwise lattice and applying the binary search method with a dimension reduction technique. To illustrate the main ideas, we first consider the 2D case before moving on to the general case.
WLOG, we assume is a square centred at point . The componentwise ordering is denoted as .
Algorithm 3.1.
Point_check() (A polynomial algorithm for 2D lattice)

Input:

2dimensional lattice , (Input size to the oracle is since the input size for both dimensions to the oracle is . )

Oracle function . is a order preserving function. and if


Point_check()

Let be the center point in . Let be the left most point in such that . Let be the right most point in such that .

If ,return();end;

If ,. Point_check();

If , . Point_check();

If and , Binary_Search();

If and , Binary_Search();



Binary_Search()

Let

If ,return();end;

If , . Point_check();

If , . Point_check();

If and , Binary_Search();

If and , Binary_Search();


Theorem 3.
When the order preserving function is given as an oracle, a Tarski’s fixed point can be found in time on a finite 2D lattice formed by integer points of a box with side length by using Algorithm 3.1 Point_check.
Proof.
Start from a lattice of size , we first prove that in at most steps the above algorithm either finds the fixed point or reduces the input lattice to size .
[H]A polynomial algorithm for 2D Lattice [50]Fig1.pdf[Point_check()] [50]Fig2.pdf[Binary_Search()] Consider the algorithm Point_check().

Case I: If , is the fixed point which is found in 1 step.

Case II: If , since is a order preserving function, , we have . Let . Define . Then is a order preserving function on the complete lattice . By Tarski’s fixed point theorem, there must exist a fixed point in . Next we only need to check which is only size of .

Case III: If , similar to the analysis in Case II, we only need to consider which is only size of in the next step.

Case IV: If and , we prove that Binary_Search() finds a fixed point or reduce the size of the lattice by half in steps. Since is a order preserving function, adjacent points , it is impossible that and . Thus, on a line segment where , if and , there must exist a point such that . On the other hand, we have and by the boundary condition , therefore, there must exist a point such that . This point can be found in time by using binary search. If , similar to the analysis in Case II, we only need to consider which is at most size of in the next step. If , we only need to consider which is at most size of in the next step. If , then is the fixed point.

Case V: If and , similarly, we can prove that Binary_Search()finds a fixed point or reduce the size of the lattice by half in steps.
The size of the lattice is reduced by half in every steps. Therefore, the algorithm finds a fixed point in at most steps.∎
∎
The above algorithm can be generalized to any constant dimensional lattice with , where and are two finite vectors of with . We reduce a dimension problem to a dimension one. Assume we have an algorithm for a dimensional problem with time complexity . Let the algorithm be .
Consider a dimensional lattice . Choose the central point in , and denote it by . Take the section of
by a hyperplane parallel to
passing through . Denote it as . Clearly, it is a dimensional lattice. We define a new oracle function on , based on the oracle function on . Define , if . We apply the algorithm to obtain a Tarski’s fixed point in time . Let the fixed point be denoted by . Therefore, or , where is some constant, is a dimensional unit vector with 1 on its th position.In either case, we obtain a new box with size no more than half of the original box defined by , such that maps all points in into and is order preserving. We can apply the algorithm recursively on . The base case can be handle easily. Therefore the total time is
It follows that .
Formally, the polynomial time algorithm for finding a Tarski’s fixed point in a ddimensional componentwise ordering lattice is described as follows.
Algorithm 3.2.
Fixed_point() (A polynomial algorithm for any constant dimensional lattice)

Input:

A dimensional lattice , WLOG, (Input size to the oracle is since the input size for both dimensions to the oracle is .).

An oracle function . is a order preserving function. and if .


Fixed_point()

If

Let be the center point in .

Let .

Let .

Fixed_point().

If , ; Fixed_point();

If , ; Fixed_point();

If , return ; end;


If , let be the left end point and be the right end point. binary_search().


binary_search()

If , output ;

else if , output ;

else

If , binary_search();

If , binary_search();

else output .


Theorem 4.
When the order preserving function is given as an oracle, a Tarski’s fixed point can be found in time on a componentwise ordering lattice .
4 Determining Uniqueness under Oracle Function Model
It has been a natural question to check whether there is another fixed point after finding the first one, such as in the applications for finding all Nash equilibria (Echenique)[10]. In this section we develop a lower bound that, given a general lattice with one already known fixed point, finding whether it is unique will take time for any algorithm. Even for the componentwise ordering lattice, we also derive a matching bounds for determining the uniqueness of the fixed point even for randomized algorithms. The technique builds on and further reveals crucial properties of mathematical structures for fixed points.
Theorem 5.
Given a lattice , an order preserving function and a fixed point , it takes time for any deterministic algorithm to decide whether there is a unique fixed point.
Proof.
Consider the lattice on a real line: . Let , define and for all except a possible fixed point . or which is not known until we query . Given a deterministic algorithm , define be the th item queried in its effort to find . Our adversary will answer whenever asks for until the last item when the adversary answers . Clearly this derives a lower bound of .
∎∎
For a randomized algorithm , let
be the probability
queries on its th query. Let be the total number of queries makes. We have:Therefore, there exist such that
The adversary will place , which is queried with probability when we choose . Therefore, we have
Theorem 6.
Given a lattice , an order preserving function and a fixed point , it takes time for any randomized algorithm to decide whether there is a unique fixed point with probability at least .
As we noted before, for a lexicographic ordering lattice, it can be viewed as a total ordering lattice or componentwise ordering lattice with dimension one by an appropriate polynomial time transformation to change the oracle function for the dimension space to an oracle function on the dimension space. Therefore,
Corollary 2.
Given a lattice , an order preserving function and given a fixed point , it takes time both for any deterministic algorithm and for any randomized algorithm to decide whether there is a unique fixed point with probability at least .
Next we consider a componentwise lattice.
Theorem 7.
Given the componentwise lattice of dimensions, an order preserving function and a fixed point , the deterministic oracle complexity is to decide whether there is a unique fixed point.
Proof.
For dimension , let . For , let for any nonzero vector . Define where is a unit vector in th coordinate. Therefore, is well defined on nonzero vectors in the lattice . One example of two dimension case is demonstrated in Fig.1. The fixed point is denoted by the red color. The direction of all the other points are defined by the function .
The adversary will set to be except at certain points (to be decided according to the algorithm) where it may hide a zero point.

Proof of the Lower Bound:
First consider such that . It constitute a solution of dimension. By inductive hypothesis, it requires time to decide whether or not there is one zero point at .
Second, when there is no such zero point, we need to decide if there is a zero point at with . Fixing any , we will set, for all with , whenever none of such is queried, and set otherwise. This will take queries.
One may note that the adversary always answers a nonzero value. In fact, for any pair and not query, the adversary can make without violating the order preserving property.

Proof of the Upper Bound:
We design an algorithm which always queries the componentwise maximum point of the lattice . We should have . We are done if it is zero. Otherwise, there must exist some , such that . The problem is reduced to a smaller lattice which has a total sum of side lengths at most . The claim follows.
∎∎
For the randomized lower bound, it follows in the same way as in the onedimensional case for general lattice. We can always set for all with and if none of such is queried.
Corollary 3.
Given the componentwise lattice of dimensions, an order preserving function and a fixed point , it takes time for any randomized algorithm to decide whether there is a unique fixed point with probability at least .
5 Determining Uniqueness under Polynomial Function Model
In this section, we consider the dimension as a part of the input size in unary and develop a hardness proof for the polynomial function model for determining the uniqueness of a given fixed point. We start with a polynomialtime reduction from the 3SAT problem which is NPcomplete to one of finding a second Tarski’s fixed point, by deriving an order preserving mapping from a componentwise ordering lattice into itself, with a given fixed point. Therefore, given as a polynomial time function with a known fixed point, determining whether has another fixed point in is an NPhard problem. In other words, determining the uniqueness of a Tarski’s fixed point is coNPhard.
Furthermore, even for the case when the dimension is one, the uniqueness problem is still coNPhard. This can be done by designing a polynomialtime reduction from the 3SAT problem to the uniqueness of Tarski’s fixed point in a lexicographic lattice. As the lexicographic order defines a total order, it can be reduced to a one dimensional problem by finding a polynomial time algorithm for the order function calculation. It then follows that determining the uniqueness of Tarski’s fixed point in a lexicographic lattice is CoNP hard though there exists a polynomialtime algorithm for finding one Tarski’s fixed point in a lexicographic lattice in any dimension.
We start with one of the NPcomplete problems, 3SAT, defined as follows.
Definition 6.
(3CNFformula) A literal is a boolean variable. A clause is several literals connected with ’s. A boolean formula is in conjuctive normal form (CNF) if it is made of clauses connected with ’s. If every clause has exactly 3 literals, the CNFformula is called 3CNFformula.
Definition 7.
(3SAT Problem)
Input: boolean variables
clauses , each consisting of three literals from
the set .
Output: An assignment of to the boolean variables ,
such that the 3CNFformula , i.e.,
there is at least one true literal for every clause.
Theorem 8.
[13] 3SAT is NPcomplete
For both lexicographic ordering and componentwise ordering, the CoNPhardness results can be derived from a reduction from 3SAT problem.
5.1 Proof of CoNPhard in Lexicographic Ordering
Corollary 4.
Given lattice and an order preserving mapping as a polynomial function, determining that has a unique fixed point in is a CoNP hard problem.
Proof.
Consider a 3SAT problem with a 3CNFformula . We define the function as follows: and , we rewrite in binary form . Let if , and otherwise.
Then is an order preserving function on the lexicographic ordering lattice . If we find a fixed point and on lattice , we find an assignment such that for the 3SAT problem. Since 3SAT problem is NPhard, find a second Tarski’s fixed point in lexicographic ordering lattice is NPhard. Therefore, determining the uniqueness is CoNPhard. ∎∎
5.2 Proof of CoNPhard in Componentwise Ordering
Corollary 5.
Given lattice and an order preserving mapping as a polynomial function, determining that has a unique fixed point in is a CoNP hard problem.
Proof.
Again we consider a 3SAT problem with a 3CNFformula . For any node in a ddimensional componnentwise ordering lattice, we define the function as follows:

, where .

.

, we rewrite in binary form . if , and otherwise.
Then is an order preserving function on the componnentwise ordering lattice . If we find a fixed point and on lattice , we find an assignment such that for the 3SAT problem. Since 3SAT problem is NPhard, find a second Tarski’s fixed point in componentwise ordering lattice is NPhard. Therefore, determining the uniqueness is CoNPhard. ∎∎
6 Finding Equilibria in Supermodular Games
In previous sections, we solve the computational problems of the Tarski’s fixed point. We are still interested in how to find a pure Nash equilibrium and how to determine the uniqueness of the pure Nash in a supermodular game. As the strong connection between the equilibrium of supermodular games and the Tarski’s fixed point, the question is whether the previous results hold for supermodular games.
In this section, we develop a polynomial time algorithm to find a pure Nash equilibirum which is more efficient than the algorithm we design for Tarski’s fixed point before. But the CoNPhardness result still holds for supermodular games.
6.1 Supermodular Games and Tarski’s Fixed Points
We will start with the formal definition of supermodular games.
Definition 8.
(Supermodular Games) Let be a finite supermodular game with players:

is a finite subset of ;

has increasing differences in , where is the strategy set of all other players except player . I.e.,
In the following discussion, W.O.L.G, we assume The model can be viewed as a discretized version of a game with continuous strategy spaces, where each is an interval. Then is a componentwise lattice (see Example 1).
Example 1.
(Supermodular game and componentwise lattice) Consider a supermodular game of two players: . Then is a componentwise ordering lattice as shown in Fig. 2.
Let denote ’s bestresponse function in .
Denote as the greatest element and as the least element in .
In supermodular games, by Topkis’ theorem[22], we have,
Let be the least bestresponse function of the game, then is order preserving.
Tarski’s fixed point theorem guarantees the existence of fixed points of any order preserving function on any nonempty complete lattice. A supermodular game is a complete lattice and the least bestresponse function is orderpreserving from to itself. Therefore, there exists an equilibrium point such that .
6.2 Equilibrium Computation in Supermodular Games
Recall , where and is the bestresponse function of the supermodular game. We assume the strategy set for each player is . In Tarski’s fixed point theorem, the only requirement for function is orderpreserving. In supermodular game, is not only orderpreserving but also need to be consistency, since for the same , the value of should be the same. Therefore. if there exists an algorithm that finds a Tarski’s fixed point for any componentwise lattice with orderpreserving function in time T, can finds an equilibrium in supermodular game in time . However, not vice versa.
W.L.O.G., assume .
Theorem 9.
When the best response function is given as an oracle, a pure Nash equilibrium can be found in time in a supermodular game .
Proof.
The algorithm is similar to the proof of Theorem 4 for finding one Tarski’s fixed point. The only difference here is for the 2D case. On a box, we start with the node , where can be any integer. We query the value of . Assume the value is . Next we query the value of . Because in the previous query we have already known that , we must have .

If , is a pure Nash equilibrium.

If
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