 Welcome back to the NPTEL course on game theory. In the previous session we introduced the labels for the vertices of the best response polyhedron and corresponding normalized polytope. So and we stated also one lemma without a proof, we will get back to that lemma and we continue discussing the Lemm Cahousen algorithm. So the lemma is the following thing. In a non-degenerate game for a pair of extreme points x, y we have the size of the labels of x is m and size of the labels of y is n. So let us look at the proof. So we assume the game to be non-degenerate, game is non-degenerate. Therefore support of x bar, let me recall x is basically an extreme point of the polytope and x bar is the corresponding scaled vector which is a probability vector which corresponds to an equilibrium mixed strategy. So this x bar is basically x bar is nothing but the normalized vector of x. So support of x bar, this many pure strategies can be best response to x bar. So x bar can have at the most support of x bar pure strategies. And similarly at most support of y bar pure strategies can be best response y bar. So now if we use this fact here then what we can get is the size of Lx has to be less than or equals to the size of S1 minus support of x this one plus size of the support of x. So this is nothing but m. Similarly the size of Ly has to be less than or equals to the size of S2 minus support of y plus the size of the support of y this is nothing but n. So therefore the labels of x can be at most m, labels of y can be at most n. But we have these polytopes are full dimensions and xy is vertex or extreme points. Therefore mod Lx has to be greater than or equals to m and mod Ly greater than or equals to n which immediately implies mod Lx is m, mod Ly is n. So the sizes of this labels has are exactly m and n. So this proves the lemma. Now let us look at the next thing its theorem which says that a pair of extreme points xy of this polytope p1 cross p2 minus 00 is fully labeled if and only if the corresponding normalized vector x bar y bar is a Nash equilibrium. So let us look at the proof of this fact. So xy is a extreme point let us say let x, y in p1 cross p2 minus single term 00 is fully labeled. Let us assume this is fully labeled let t1 is equals to the support of x, t2 is equals to support of y. So what we need to show is that this xy is going to be an equilibrium. So let us prove this fact. So let us go to the for every k in t1. So x does not have label k as xk greater than 0. Therefore if x does not have this label k this immediately implies y must have label k this implies a yk must be 1. Therefore a y bar k has to be 1 over y transpose for this k a y bar the kth entry of that must be 1 by y transpose 1. This is basically the we have introduced this is the dot product between y and this vector of 1's. For others for other case a y bar k has to be less than equals to let me put k prime here for all k prime in s1. So therefore at that k if x does not have the label k then xk greater than 0. So therefore the one of the inequalities has to be tight therefore a yk has to be 1. For other these things a y bar k prime must be less than equals to 1 by y transpose 1 this is all coming from this thing. So this implies k is best response y bar. So if k is in t1 then let us look at the next thing further for k is not in t1 x does have label k then by the previous lemma what we have is that a label cannot appear twice in fully labeled pair which implies that y does not have label k implies k is not best response like y bar. So this proves that for all k in t1 you have this best response condition and if k is not in t1 then k cannot be best supplied to y bar. So this is there therefore x bar y bar is going to be Nash Equilibrium. So here I have removed I have not done certain details but which can be finished by yourself. Let us look at the converse. Suppose x bar y bar the corresponding normalized one is Nash. So this immediately implies s1 minus support of x union support of y is contained in label of x this is all coming from the definition and s2 minus support of y union support of x is contained in ly. So therefore lx union ly when you take it this is going to be simply s1 union s2 that is saying that they are fully labeled. So this proves the theorem. So the essence of this theorem is that the extreme points of this polytope p1 cross p2 the extreme points which are different from origin they are exactly the Nash Equilibrium. So this is exactly what one uses in proving the providing an algorithm for computing Nash Equilibrium. So let me go to this thing. So now we come to the Lemke-Hausen algorithm. So the idea is start from origin so this is fully labeled and from origin we go to what we do is that pivot alternately in p1 and p2 until a completely labeled pair is found. Starting from origin you go to the next extreme point which is connected to 0 0 and then you go on and look for the completely labeled pair. So this is basically the idea here. So let us introduce few notations for this. So let me call v1 is nothing but the set of extreme points of p1 and similarly v2 is set of extreme points of p2 and then ei is basically set of edges between adjacent extreme points. So what it means is that e1 is nothing but x, x prime in v1 cross v1 such that the Lx the intersection of these labels of x and x prime must be exactly m minus 1 the size of this. So that means the labels of Lx and labels of Lx prime there is only one difference. So that e2 is going to be y, y prime in v2 cross v2 such that the labels of y intersection labels of y prime they must be n minus 1. So this is the sum notation that we require and in fact some more notation needed v is equals to v1 cross v2 then e is nothing but the following thing x, y, x prime y belongs to v cross v such that x, x prime is in e1 union x, y, x, y prime in v cross v such that y, y prime is in e2. So what we are saying that one in this first thing the y in the second polytope is fixed and x to x prime you are changing it. So you are only changing in the polytope one and in the second thing you are going you are essentially pivoting between the second polytope x is fixed here y is changed here y and y prime comes. So the whole idea here is that if we restrict our attention to extreme points that are almost fully labeled with the possible exception of particular label i then there is always a unique way in which we can proceed. So for this let me introduce again another notation let us take a label i from S1 union S2 let vi so is nothing but x, y in v such that Lx union Ly is contained in it contains S1 union S2 minus i except i all other labels will be there in Lx union Ly and ei is going to be the edges here e intersection vi cross vi. So once you set up this notation so the final theorem is basically the following thing let i is a label then this vi contains 00 so origin is always there as well as every x, y in v such that corresponding normalized this thing is Nash equilibrium. So that means this vi whatever we have defined here this contains origin as well as the set of Nash equilibrium that is the first thing. Then assuming non-degeneracy that actually we made it as a blanket assumption so we are assuming this thing this is a non-degeneracy is automatically made to us. Then 00 and the elements of vi corresponding to an equilibrium have degree 1 in the graph vi, ei and all other nodes in vi have degree 2. So what it is saying is that 00 has only one neighbor so from 00 I can go to one only one other neighbor and similarly the equilibrium if the point x, y is corresponding to an equilibrium Nash equilibrium then that also has a degree 1 that means from there you can only go to one other node and if it is not a Nash equilibrium then for all other nodes have degree 2 that means they have at least 2 parts. So let us try to prove this. So the first part is obvious so 00 and all pairs corresponding to Nash equilibrium are fully labeled therefore first part is obvious so we only need to prove the second part. So for the second part consider x, y in vi and let x bar, y bar is the corresponding normalized strategies. So because x, y these are all extreme points that we said we already proved in the previous one of the previous lemma we have this size of Lx is m similarly the size of Ly is n. So this we already have proved it so if x, y is equals to 00 or x bar, y bar is Nash equilibrium then x, y is fully labeled and Lx intersection Ly is going to be empty set. So this is if this happens. So let us look for that the neighbors of x, y are those elements in vi that replace i with some other label correct because they are the neighbors so therefore one of the label is going to be replaced so that the elements in the vi that replace i with some other label. So basically those where other constraints holds with equality instead of one corresponding to now since only x or y has label i only one of them either x will have i label or y will have label but not both we may only replace yes it from one of them. So dropping the label i we obtain a new label and by non-degeneracy this label is going to be unique so you should see that why this non-degeneracy is a very, very crucial here. So because of the non-degeneracy only when you are dropping this label i you will be able to get only one another label only okay there cannot be multiple this thing okay. So what happens otherwise what happens is the following thing if Lx intersection Ly is j for some duplicate label j in S1 union S2 then the neighbors of x, y are obtained by replacing j with another label this replacement can be done either in P1 or in P2 and for each of them there is exactly one neighbor by the same reasoning as before so this proves this theorem. So the essence of this theorem is that this points here in this vi let us go here. So if you take first thing is vi contains a region and all the points corresponding to Nash equilibrium and then under the degeneracy 00 and the elements of vi correspond to an equilibrium have degree 1 and other nodes have degree 2. So this is basically the next point in fact this is the crucial to the Lemke-Housen algorithm. So let us so now what we have shown is that this graph vi for each i in S1 union S2 consists of paths and cycles that are pairwise designed and the ends of paths correspond to the pair of one is origin and to the equilibrium of the underlying game. So because of this degree 2 and this degree 1 so either they are cycling or the ends of the paths will be either 00 or they must correspond they must end at an equilibrium of the underlying game. So this is essentially the Lemke-Housen algorithm. So in the Lemke-Housen algorithm essentially starts with 00 and it looks for a path. So from 00 you can go to another along this graph which we have introduced and then at some point of time it will stop because it cannot move any further because that is the degree 1 once you reach a degree 1 point then you end with that and that is giving you a Nash equilibrium. In fact this also provides you a proof of existence of a Nash equilibrium. So this is the very important algorithm for solving by matrix games and in fact in the I plan to actually uploads some video where we will work out an example to see how this algorithm works and all. So that will be like a supplementary video to this video where I will tell you how to put everything in a concrete example. And there is another thing is that this algorithm belongs to the class of path following algorithms or homotopy based algorithms. So this is a very important algorithm for computing the Nash equilibrium of by matrix games. In the worst case this algorithm can also take exponentially many steps. So with this we will conclude this session and in the next session we will start looking at evolutionary stability. Thank you all.