 Welcome back to NPTEL course on game theory. In the previous session I introduced the support of mixed strategies and then using the support enumeration how we can try to compute a Nash equilibrium. Now, of course if you look at it we the algorithm actually asks to verify the some set of inequalities for every k sized subsets. So it is a lot more complex in terms of computing because it requires exponential you need to verify all subsets of size k and k is running full. So is there a better way of doing it? So that is the question which Lemkehausen algorithm answers. So we will now go back to start looking at this Lemkehausen algorithm then we start understanding how the algorithm works. So in the previous class I also have introduced this polyhedron polytopes and other things and I have not said why they are required. So in this session we will see them. So just to recall that non-degenerate condition is something that we have made. So I will recall once again the definition of this thing. So you take a bi-matrix km a, b and then I call this as a non-degenerate if for every xy in delta 1 cross delta 2 support of x the size of it is less is greater than equals to support of y. If y is best response to x, if y is the best response to x then the support of x must be bigger than support of y. Similarly support of y should be bigger than or equals to support of x if x is best response to y. So in fact this is essentially the idea. So for an equilibrium the support of x and support of y are going to be the same. So that is a reason that support enumeration algorithm works. So now we will introduce certain polytopes. So before that so another assumption that we will make is that a and b 0 column. So this is actually not a really a serious assumption. The reason is that suppose if a and b have some 0 column what I can always do is that I can add all the entries by a fixed quantity take a prime to be aij plus some alpha. Similarly b prime you take it to be bij plus beta let us take it because I have added constant to all the entries same constant to all the entries. If you really look at the equilibrium structure will not change. So whatever is equilibrium for a and b original game is equilibrium to a prime and b prime. Similarly any equilibrium for a prime b prime is also equilibrium for a and b. So which is a trivial exercise just if you write down the best response condition that becomes a simple fact. So in that sense this matrices a and b having no 0 column is a kind of a not really a serious assumption. So but we assume these conditions throughout this session. So therefore now we made two assumptions one is the non-degeneracy and other is no 0 column. Now let me define the certain sets. So p1 bar is the set of all x, v in rm cross r such that x greater than equals to 0 b transpose x less than equals to v1, x transpose 1 is 1. So I will introduce this is basically a vector of 1s this is also vector of 1s appropriate dimensions. Similarly p2 bar I will define it as yu in rn cross r such that y is greater than equals to 0, ay is less than or equals to u1 and y transpose is 1. So again this 1 is vector of 1s all of them. So what is really we are saying here. So this is x greater than equals to 0 b transpose x less than or equals to v. So what is this meaning? So let us try to understand this one. So b transpose x less than or equals to v sequence of 1s. Now if I take the x comma by this is nothing but b transpose x y so because these are all positive these things so this is going to be b transpose x is less than equals to v1 so therefore this is less than or equals to v. So what it says is that so if you are looking at x the player 1s this thing player 1s mixed strategy if he is playing x the player 2 cannot get more than v. So that is exactly this condition b transpose x less than or equals to v says. So the player 1s best possibility is not more than v. He will get at the most v we do not know what exactly he will get it. So in a similar fashion this says is that if the player 2 is using y mixed strategies and player 1 will not get more than u that is exactly what this condition says. So these are known as the best response polyhedrons. So the p1 bar p2 bar are the best response polyhedrons. Now the next question that I would like to ask is that is it a polytope? So if you really look at it this v because this is only a v is an upper bound by anything bigger than 1v will also work. So therefore this is only a polyhedral not really polytope. So in a sense this best response polyhedron p1 bar contains all the mixed strategies of player 1 and it also contains a scalar v which is going to be the maximum pay that player 2 will get. And similarly in the p2 bar this is the all the mixed strategies of player 2 together with a value u and the player 1 cannot get more than u. So this is basically the best response polyhedron. So in a sense this v is upper bound for player 1's v is an upper bound for player 2 payoff u is an upper bound for player 1's payoff here. So what will happen here is that. Now let us look at p1 bar. Let us say one of the constraints we have is xk greater than equals to 0. Suppose this let us say assume xk is equals to 0. Suppose if we assume that this polyhedron you take some x and let us assume that xk is equals to 0. That means what this k is in S1 is not played in x. So that is the meaning of it. Then similarly let us say B transpose x let us say some l is less than equals to u and suppose this happens with equality that is B transpose x l is u. So that means the lth coordinate of B transpose x is exactly u. What it means is that the l in S2 is best response of player 2 to x because if player 1 is playing x player 2's best response is nothing but is pure best response is l. So the polyhedron p1 bar lives in Rm plus 1 this is inside Rm plus 1 and it has one equality constraints. In fact you can verify that it is m dimensional polyhedron. So I will not go into proving this dimensionality thing which I will ask you to verify but for the future we will not require the proof of this fact so I will omit these details. So this immediately tells me that any extreme point if you take it if you take any extreme point of p1 bar if you take it. So the inequality constraints because the dimension I said it is m. So therefore m of the inequality constraints have to be satisfied with equality. So therefore an extreme point x v corresponds to a situation where support of x is equals to k first some k less than equals to m and at least k pure strategies of player 2 are best responses to x. So when we take x v let us say it is an extreme point and if it support is k what we are saying that first is k has to be less than equals to m because the dimension is cannot be more than m so k has to be less than equals to m and because at least k pure best responses of player 2 because k of this inequality k of this inequalities have to be satisfied with equality. So because the dimension is m so any extreme point in any extreme point if the support of x is k k has to be less than equals to m and because of that one because the inequalities have to be strict at those k point this thing. So wherever that inequality is equality that becomes a best response. So that is exactly what I am trying I am saying here that so at least k pure strategies of player 2 are best responses to x so this is a this comes essentially from that by the definition of p 1 bar. So this particular thing becomes in equality wherever this is equal that becomes a best response to x so that is basically of course we are assuming non degeneracy. So the non degeneracy we are assuming because of this so the exactly k will be there will be k pure best responses of player 2. So this is the non degeneracy is going to give it of course analogous statement we can make it for the other p 2 bar so let us start for you. So what I will do is that so if we really as I said this polyhedron is not bounded so what now I will do is that make something simpler. Let me recall what is p 1 bar is we have this x v such that the most important thing is B transpose x less than or equals to v and other conditions. So I would like to avoid this v so and then what I am now define is that I do not make the probability distribution I take any x in Rm such that B transpose x is less than or equals to 1 that is the vector of 1s comma and x is a non zero vector. So what we have now done is that so I have made the unit pay in B transpose x is cannot be more than 1 no entry of B transpose x is less than or equals to more than 1 and all non negative vectors so it is a when I divide this B transpose x this v if I brought bring this side x by v now x by v need not be a probability vector so that is a simply a non zero vector that is exactly what I have done while in defining this p. So this p allows me to reduce this p I put p 1 and p 2 is for the player 2 such that of course y greater than equals to 0 and then Ay less than equals to 1. So in a way these are these vectors are normalized so instead of now mixed strategies we have now looked at this normalized. So these are they are not mixed strategies so they are not mixed strategies and most important thing is that R is in 0 0 0 belongs to p 1 as well as of course the 0 vector in R n belongs to p 1 0 vector in R n belongs to p 2 so therefore the vectors in x will also have a 0 vector but non-zero vector when you take it I can always make a probability vector so this is the idea that we use it. So what we have said is that so if I take x comma v in p 1 bar then what I have is that this x by v is going to be in p so this is the main point here. So let us take x belongs to p 1 bar is an extreme point of course assume this is not equals to 0 then x by x transpose 1 1 by x transpose 1 is extreme point okay. So here is a small error here this is not x is in p 1 if it is an extreme point non-zero extreme point then x by x transpose 1 comma 1 by x transpose 1 is extreme point of p 1 bar a similar result you can say about the p 2 bar and p 2 okay. So what I am trying to say here is that the extreme points of p 1 corresponds to extreme points of p 1 bar and similarly extreme points of p 2 corresponds to extreme points of p 2 bar so this is going to play an important role in the future okay. So now our next goal is to understand the how to characterize these extreme points so let us go through that. So for any point x and y for each x in p 1 y in p 2 I want to define what is called Lx this is basically the indices I will define in the following way k in S1 such that xk is 0 that means the in this particular thing corresponds is essentially the support of x and j in S2 such that B transpose x j is 1. So wherever the B transpose x is that inequality corresponds to B transpose x is equality. So we are looking at the wherever that inequalities are equalities so this is the label of x similarly the label corresponding to y is nothing but L in S2 such that y L is equals to 0 this is the support of y union i in S1 such that Ay is 1 okay. So this is what we are saying here is that first we started with the support of y and then the wherever that inequality corresponding to Ay less than equals to 1 that is the inequality wherever that inequality is strict look at those things and that we start calling it as Ly these are known as a labels of x and y okay. So we say that x comma y belongs to p 1 cross p 2 is fully labeled if Lx union Ly is S1 union S2. So if the pair of strategies in p 1 and p 2 p 1 cross p 2 if they are called fully labeled if Lx union Ly is nothing but S1 union S2. In fact now we have a following interesting lemma in a non-degenerate game for a pair of extreme points so we have Lx size of Lx is m size of Ly is n we will prove this in the next session and continue to go for the Lemke-Hausen-Algard we will stop here at this moment we will continue in the next session thank you.