 Welcome to this session, in the previous session we discussed Germlos theorem which says one of the player has a winning strategy or the game ends in a draw. Now in this session we look at some examples and determine whether which player has a winning strategy. In particular we look at the games where either win or loss happens, there is no draw possibility. So we start with the game of chump. So recall the chump, in a chump there is a rectangular array of cells, each cell is considered as a chocolate and this corner is a poison. Now the player stick turns alternatively, remember that if a player selects a cell, say if a player selects a cell, all the cells right side and above are taken by that player. So now if you continue this game, in fact it is quite easy for us to realise that this game will never end in a draw, it will always end in a either a win or loss. So which player wins this game? So let us see, the in fact chump is a game where first player has a winning strategy. So let us look at it, how do we play this game? So we can start with the following thing, let us consider instead of rectangle let us consider a square. In the first move player 1 will mark this square, therefore all this will disappear from the board. In the next player 2, he can mark any square in this row or in this row. So whatever player 2 can mark here, corresponding a symmetric square in the other side, what it means is that suppose if player 2 marks here, so it is symmetric thing there is a move for player 1 here or if player 2 marks here, player 1 has a place to mark here. If player 2 marks here, player 1 now can mark here. So in conclusion what I am saying is that wherever player 2 is making a move player 1 has a corresponding move therefore whenever player 2 has a move player 1 always has a move. Therefore in the end we conclude that player 1 is going to win this game, thus the game of chump has a winning strategy for player 1 even when the board is given by a square. Now what about a rectangular, how do we play this game? Then the same idea but let us look at it, player 1 let us assume the blue player let us takes this then player 2 let us assume he will take something let us say whatever let us take he takes this that means this entire thing is now under recall realize that if player 2 is going to be going to be winner in this game by taking this position what player 1 can do here is that instead of taking this one player 1 could start with this position. So this is the idea that is there, so in a sense let us write down formally. Suppose the formal argument is that suppose player 2 has a winning strategy if player 2 has a winning strategy what it means is that we consider 2 games there is let us assume that let us try writing 2 games here. So remember as we assume player 2 has winning strategy, so let us this is a dummy game let us think about this player 1 in this dummy game player 1 first takes this one player 1 takes this one, so this position then player 2 responds his winning move. So let us say this is his winning move in this dummy game now what player 2 is doing in this game the player 1 will do it here. So in this game this is the real game player 1 exactly follows the player 2 strategy here he will take here this one and then this game goes on. Now we can see that if player 2 has a winning strategy here in this game then the player 1 has to win here, so there is a contradiction. So this immediately tells you that player 2 cannot win therefore player 1 has a winning strategy. So this is the idea of this proof for this theorem in a chump game we are using basically a strategy stealing argument here whatever player 2 is doing in this dummy game we are following this. This is something like suppose if you are playing with both Kasparov and Karpov so let us say you play with one of them first and then replicate that strategy in the game with Karpov then you are sure to win one of those 2 games or at least you will end a draw you will so that is you are using this argument here. So the next game that we will consider is hex. So let us look at it. So what is the game of hex? We have seen it already in this thing it is a regular hexagons arranged in a diamond ship in such a way that there is a same number of hexagons along each side of the board. So this is an example of the board in fact in the earlier session we have seen how this is played. So there is a blue player this side this is a blue player his goal is basically to mark to color these hexagons in such a way that he gets a path from this side to this side and this is a red player he will try to color these hexagons in such a way that there is a path from this side to this side. So who has a been extra strategy in this game? So let us check this one. So let us look at the idea of this. So as I said this two players alternatively play like in tic-tac-toe each player in his turn places his symbol blue or red colored according to the player on one of the hex curve. Winner is the player who first obtains a connected path of adjacent hexagons stretching between the sides of that player's label. This is invented by Piet Heijn a Danish scientist mathematician writer and poet and this is actually rediscovered by John Nash and it became a very popular thing in this thing. So let us check how do we solve this game? Does this game has a winning strategy for either player or can there be a draw? This is the question. So let us try to prove this. So first let us see this game cannot end in a draw the idea is very simple. So the let us go back to the slides this game what you do is that suppose let us say one of the player is going to color it other player let us say he makes the hole. If you make holes what is really going to happen is that if you keep let us say blue player is coloring with blue this coloring these hexagons with blue and let us assume that red player is cutting making a hole set the rectangle. Now at the end of the day either the when all the hexagons are taken by one of the players the sheet either will remain connected or disconnected. If it is disconnected that means the person who made the holes has a path from one side to the other otherwise the other player has the win. So that is all. So what we do is that we imagine that this playing board of hex is made of paper when ever red moves he colors the hexagon of his choice and whenever blue moves he cuts his hexagon of his choice. Now either the board will remain a single piece or at least two pieces. This completes the proof. This proof is quite intuitive and in fact the one of the underlying idea in this proof is known as is Jordan-Kau theorem which is actually quite an important result in mathematics and in fact we can also use this hex game to prove the Jordan-Kau theorem. So let us go to the next result. So the first thing is that this theorem tells you that this hex game can never end in a draw. So let us look at the next part. Who has a winning strategy? Now what the idea is that the first player has a winning strategy. How does this go? Let us look at this. Let us go back to the picture once again. So how we do is very simple thing. Let us say blue player, let us say blue is the first player initially he picks any blue player picks any hexagon. So he picks any hexagon and then he keeps this as if he has not played. What we are really doing is that we are using the same argument like what we have seen earlier the strategy stealing argument. So we know that, okay so this is like a contradiction. Let us look at the proof. So let us say first player actually can adapt. Let us look at this here. What I am saying here is that the first player to adapt the second player's winning strategy as follows. We are going to use this thing. So the first player on his first move just colors in an arbitrary chosen hexagon. Subsequently for each move by the other player the first player responds with the appropriate move dictated by second player's winning strategy. This is called the stealing the strategy argument. So what exactly we are doing it? So player 1 initially makes a random move, okay random I mean is that he picks any of the hexagon and he colors it and then he will wait for the second player's strategy. Whatever the second player is playing assuming that the second player has a winning strategy. So he thinks that like in the previous case you consider another boat separately kept it. He makes some move and then the second player is making his winning move, winning he is following his winning strategy that second player's winning strategy the first player is going to follow in the first boat. So it is what happens here is that at some time it is possible that the player 1 may need to mark hexagon which is already marked initially. Then what happens is that if that if he has to mark that that means he is already in a good shape it is not a problem. So it does in this thing and then he arbitrarily chooses something else and then this procedure is repeated and that eventually gives you this the proof that if the second player has a winning strategy the player 1 can also win the game in a sense it is a contradiction and it proves this proves that the player 2 cannot have a winning strategy and hence player 1 wins this game. So this is a again once again we are using this strategy stealing argument and which proves this game of hex. So now this game of hex is a very very interesting game in for several reasons. So what we are going to do here is give you some arguments how this hex game can be used to prove Brouwer's fixed point theorem. The Brouwer's fixed point theorem is a very very important result which we are going to use in this course. So let us go with that. So before that I need to explain what is a Brouwer fixed point theorem. So let us what is this? So let us take f is a function from an interval 01 to another interval 0 comma 1. Then what it says is that there exists a point of course I assume always continuous function there exists a point x in 01 such that f of x is equals to x such a point x is called fixed point not this is a one dimensional thing but you can also do it in a two dimensional thing if f is a function from 01 cross 01 to 01 cross 01. Then and of course continuous function then there exists x that is of course x1 comma x2 in 01 cross 01 such that f of x is equals to x. So this is a two dimensional version and like that I can go for n dimensions. So we will not prove this n dimension here we will give a two dimensional version and instead of 01 cross 01 I can actually take any k convex and compact set and f is from k to k then of course continuous then there exists x in k such that f x is x. So this is basically the Brouwer's fixed point theorem. So let us prove this fact. So we use this hex game and then let us check how we can prove this. So what we do is that we consider this hex board and we consider a dual version of it. So this is the hex board we have and each of these hexagons we take the central points the central points and connect them and if you connect the central points you actually get a this kind of a diagram. Now coloring this hexagon is nothing but coloring this vertex in that sense coloring these hexagons is nothing but the coloring the vertices. Now if I have a path here means I will have a path from one side to other side along these vertices and edges. So this is basically this is known as the dual lattice of the hex board of the hex board. Now as I said coloring these hexagons is equivalent to coloring the nodes. So these nodes can actually be described by the following thing this one can check it if you take u is equals to 01 and v is equals to this two nodes basically we are identifying everything you can see that these are hexagons and if I initial give some points you can say that u if you start u is equals to 01 and v is equals to root 3 by 2 half then the two nodes and you make this lattice you consider this a u plus b v a comma b is in z then the two nodes x y are neighbor if and only if norm x minus y norm x minus y here is x 1 minus y 1 square plus x 2 minus y 2 square and square root of it. So standard Euclidean norm we are looking at it. So this becomes this. So now let us check the next part. Now what we do is that this board this dual lattice I will convert this into this kind of a picture. So now it is a square it is this is not like something like 01 cross 01. So I consider this transformation g which is given by this. So we this point remember what the points u and v are given by given in the previous slide so look at this. So u is given by 01, u is 01 and v is root 3 by 2 half that is exactly what we are using it and that is where we have used this. This u is that point v is this point and I am considering a transformation g where the point u is transformed to this and the point v is mapped to 01 and extend this linearly then if you draw the complete map this picture actually transforms to this picture. Now the points are all fixed here. So now the game of x can be thought as a game on the square of this lattice. So how we can go further now? So let us check that one. So we start with a continuous function f which is as I said from a unit square to itself keep this in mind and we need to show that f has a fixed point what we do is that we start with some epsilon and then choose a point x such that this happens. So we need to show that for any epsilon if I can find a point x such that fx minus x is there sufficiently small this gives you an approximation of a fixed point. This approximation as epsilon goes to 0 will converge to a fixed point that is actually an argument from of compactness which I will leave it as an exercise but let us try to produce an approximate fixed point. So what we do is that for a given epsilon we choose a delta such that this happens. So why is this happening? This is happening because f is continuous on closed and bounded set which is 0, 1 cross 0, 1 that means this is a compact set. So that means f is a continuous function on a compact set implies f is uniformly continuous and hence this happens. Now we consider the regions h plus to be this set of all x comma y such that f1 xy minus x greater than equals to epsilon. So what is f1? Remember f is a continuous from square to square that means the first component is the f1, second component is f2. So h plus is basically all points x comma y such that f1 xy minus x is bigger than or equals to epsilon, h minus is xy such that x minus f1 xy is less than equals to greater than equals to epsilon, v plus is with the second coordinate and y. So now once this is done we have defined this h plus, h minus, v plus, v minus. So let us look at the next part. Now observe that no vertex of h plus is adjacent to a vertex of h minus. So this follows from the definitions of h plus and h minus. In fact no vertex of h plus will have a first coordinate equal to 1 and no vertex of h minus will have a first coordinate equal to 0. These all come from the definitions of this h plus and h minus. Therefore if a player, one of the player is marking his cells only from h plus and h minus it will not form a winning set because they are not vertices therefore we cannot form a connected path. So thus in particular it cannot connect two sides joining x1 is equals to 0 on one side and x1 is equals to 1 on the other side. In a similar fashion we can see that no vertex in v plus is adjacent to v minus and in a same fashion v plus union v minus cannot be a winning set for any player who wish to join the sides x2 is equals to 0 to the side x2 is equals to 1. Thus the Hex theorem because in the Hex theorem one of the player has a winning strategy this implies that the union of all these sets h plus, h minus, v plus, v minus they cannot cover all the vertices which are defined in our construction. Therefore there must be a point which is not there in any of these h plus, h minus, v plus, v minus and that vertex by the very definition of these sets will satisfy this property and this property immediately says that that point x is going to be a approximate fixed point. Now as I already said the compactness arguments now provide a fixed point for the function F. Thus this game of Hex provides a proof for Brouwer fixed point theorem. Of course we have done only in the two dimensions this argument can be extended to higher dimensions which we will not do in this course. So we end this session here and we will continue in the next session.