 Welcome back to this NPTEL course on game theory. In the previous session we have proved the Spray-Grundy theorem and we mentioned that the Spray-Grundy theorem identifies any impartial game with a NIM game. We will see how this happens. So let us look at the certain things. Remember in any NIM game let us say with single heap. So this we denote this by S star of N. This is known as the number. So what we would like to say now is that any game, any impartial game if you take G and we have its Grundy value G of G and look at the corresponding star of GG. So we would like to say somehow these two are equivalent. So how do we do it? So we will discuss about this aspect now in this lecture. So this requires for us to introduce certain facts but before that let me mention here the following thing. So let us look at the NIM game with k-piles of sizes N1, N2, Nk let us say. So this is P position if and only if the NIM sum the individual file sizes is 0. So this is a fact that we have proved even while proving the strategy for NIM games. So if you can either go back and see that video and see this or take it as an exercise and do it. So this is something very important. Basically we would like to look at it. So the most important fact I would like to say is that the NIM game of multiple piles is same as like the sum of the individual games. So that is the spray Grundy theorem. So in fact here now I would like to define if G is one game and let us say H is another game. What do we mean by G is same as H? This is something that we need to define. So what exactly this means? So let us take any other impartial game. Let me denoted by k if G plus k H plus k have same outcome then we say that G is equal to H. So two impartial games G and H we are saying them that they are equivalent or they are equal equivalent if sum with any another impartial game they should have a same outcomes. That means if the G plus k has N position H plus k also will be N position if G plus k is P then H plus k should also be P and they should happen with any other impartial game k and if that happens we say that G and H are equivalent. Now we would like to look at some important points. So take a G is an impartial game then look at G plus G. So what can we say about G plus C? Who is going to win? So this is in fact this idea we have explained earlier. So you are going to play two games who is going to win? Suppose if the game G is going to be 1 by player 1 what about G plus G? Can we say that G plus G is N R P? So interesting fact here is that G plus G is always. So why is this? Basically the idea is that if a player has a optimal strategy has a winning strategy what you are going to do is that you imitate it repeat the same strategy whatever the other player is playing you just simply follow in your game and then you will know that you are going to get to this P position. So this is a very nice symmetry argument and this proves that G plus G will always be P. This is one fact then another fact is that if K is P position then G is going to be equivalent to G plus K. So why is this true? This is true mainly because if you take some of two P positions this will always be a P position. So sum of two P positions is always P position and sum of N and P is. So if G is P K is P so therefore G plus K is again P if J is N and N plus P is again going to be N. So therefore this thing is there but why is that sum of two P positions is always P? So this is an interesting exercise I will say. So one should certainly look this take this as exercise and try doing this one. So this proves that G is always equivalent to G plus K as long as K is P. So what is you take a G and take its gringy value and take the corresponding number what can we say about G plus? So very interesting thing is that its gringy value therefore this is a P position. So G plus the number of GG is always a P position. Therefore G is equivalent to G plus star GG from the previous fact which is same as G plus G plus and now this is equivalent to so we have used multiple facts here. So G plus G is always a P position that we have proved it here and therefore this is P. So therefore this is going to be simply. So therefore any game is equivalent to a NIM game. So this is basically the consequence of the spray-grindy theorem. So therefore every impartial game is equivalent to. So this immediately implies that if we know how to play a NIM game we can always play that any impartial game by using this spray-grindy theorem. So this introduces this tells us why this spray-grindy theorem is a very very important theorem and this is a in fact characterizes every impartial game. Now let us look at one more example in the previous session we have discussed one example. Now let us look at another example. So we keep we consider one pile game with the rule and we can remove any even number of chips it is not the whole pile. So I can remove any number of chips even number of chips but I cannot remove whole that is important or we can remove provided it has odd number of chips. So now we have changed the rules a little we instead of removing any number of chips we are saying that we can remove any number of even number of even chips even number of chips but we cannot remove entirely if the entire pile size is even but at the same time we can remove the entire pile provided it is odd. So now here what are the terminal positions? Obviously if there are only 0 that is a terminal position not only this there is another terminal position which is 2 when there are 2 chips are there I cannot remove anything. So therefore 2 is also a terminal this thing. So therefore the Grundy value let us say x and the corresponding Grundy value so if there is a 0 the Grundy value is going to be 0 if it is 1 the Grundy value is going to be 1 and if it is 2 the Grundy value is going to be 0 and if it is 3 the Grundy value is 2 with 4 it will be 1 5 it will be 3 and you can go on. In fact we can calculate that g of 2k is going to be k minus 1 and g of 2k minus 1 is going to be k for k grad equals to 1 inductively one can calculate that whenever this is an even number given by 2k then this is simply going to be k minus 1 and when it is an odd number it is given by this k where odd number is given by 2k minus 1. So this is going to be the this thing. Suppose now let us look at this game is played 3 piles of sizes 10, 13, 20. Therefore g of 10 is going to be 4, g of 13 is going to be 7, g of 20 is going to be 9. Therefore if I take the Nimsum 4 plus 7 plus 9 this is going to be 1, 0 this is not equals to 0. So recall the definitions of this Nimsum and make sure that this is not 0. Therefore this is a N position. Therefore what is going to be the if you are in an N position what is the winning move you need to make it to 0. Once again recall the NIM game strategy go back there and see it the winning move is always to make it to 0. So therefore you want to make it to 0 therefore sg value the spring degree value this 9 we want to make it to 3. If I make this into 3 then this thing that means we need to remove for this we may remove 12 chips from the pile of 20 leaving 8 and g of 8 is 3 keep that in mind. So why do we want this to be 3 if I put 3 here that Nimsum is going to be 0 and therefore g of 8 I want to make it spring degree value to be 3 to make it 3 what we need to look at it. So therefore g of 8 is going to be 3 so that is exactly this one. So this is a another example of how you can use this spring degree theorem. So far we have seen impartial games with normal play. Now we would like to see a major play game. So recall what is a normal play? In a normal play the player who made the last move is basically the winner here in a major play he is going to be losing. So let us look at a very interesting example which is known as a silver coinage. The silver coinage the name silver is basically the very famous mathematician J.J. Silvester on his this thing we call this as a silver coinage game this game is basically the following thing. Here people the players alternatively of course we only take natural numbers what are the rules? A player cannot pick a number which is sum of previously used numbers. So let me illustrate with an example. So let us say the player 1, player 2 let us say in the first round player 1 has picked 3. Now player 2 he cannot take he cannot pick 3, he cannot pick 3 plus 3 cannot please pick 3 plus 3 plus 3 like this. So he cannot pick from this cannot pick. So let us say he will he can pick 7, 7 is certainly not in this. Now once 3 and 7 are picked what player 1 cannot pick from of course 3, 3 plus 3, 3 plus 3 plus 3 and like this and also not only that 7, 7 plus 7, 7 plus 7 plus 7 this thing only that 3 plus 7, 3 plus 7 plus 7, 3 plus 7 plus 7 plus 7 like this and not only that plus 7, 3 plus 3 plus 7 like this and go on like this. In a nutshell he cannot pick m into 3 plus n into 7 if for m, n belongs to m. So he cannot pick any of the number which can be written as a combination of 3 and 7 positive integer combination of 3 and 7. So for example he can pick 5, once he picks 5 the same thing happens 3, 7, 5 and things like that. So there is no way he can pick the other numbers. Now let us say for example the player 2 now can pick 4, once he picks 4 what are all the possibilities that are available to player 1. So lot of numbers let me not mention what are all the things which cannot be done but let us assume in the next round let us say player 1 picks 2. Once he picks 2 the most interesting thing that happens here is that all the numbers now can be written as a combination of 2, 3, 5, 7, 4. So the only possibility that will be left for player 2 will be only 1 no other and once he picks a 1 there are no more moves. Therefore whoever picks 1 because any number is a multiple of 1 so therefore after that no chance. So now what would you look like to consider here? Should we consider a normal play or measure play? So let us look at if 1 is a choice for a player so he naturally will pick at the very first move therefore the game makes no sense. So therefore the normal play here makes no sense. So therefore picking 1 is not legal move here this is the main rule. If a player picks 1 that means he is ruling out every other possibility. So therefore picking 1 means he is going to be a loser so you do not want the players to pick. So the game is essentially not whoever picks is loser. So there is one is not a legal move in that sense and this. So now this game is very very different from the normal play. In fact in the previous games like champ and other things you can also say them as a measure but the silver coin is actually ensures that the measure play is a natural option. That is main interesting thing here. Now let us look at few interesting points with this game because what is the length of this game? How long can this game go on? For example people can any large number you take it and then if you pick that large number and players can alternate and can reduce one by one and then go on picking the numbers. Therefore the play can take any amount of time any long it can be any long the length can be anything. Therefore this game is not like a previous games but there is an interesting thing which is an interesting result which says that once there are if I pick 3 player 1 picks 3 player 2 picks 7 then what happens is that any number bigger than or equal to 6 that is 12 is not possible. 12 is not possible when already player 1 and player 2 picks 3 and 7. So therefore what is happening is that this is automatically imposing some kind of a finiteness. So this result is a very interesting result in fact it requires a proof which we will discuss in the next session but before concluding this session I would like to ask everyone to try thinking about proving this fact why 13 is not possible. For example that is easy 13 is 2 into 3 6 plus 7 13 14 is 7 into 7 plus 7 so 2 into 7 14 we can actually go on doing this. So how do we prove this and all that becomes the next sessions topic. So the main interesting thing here is that the normal play and major play they are not exactly the opposite and this is one example where the major play is a natural and not under normal play the game stops immediately. So we will continue this in the next session. Thank you. Any number bigger than or equal to 2 into 6 that is 12 is not possible.