 In the last session we discussed this silver coinage game. We will now again introduce this and study this game and what are the best moves and all some results that are there we will present them. So let us introduce the game once again. So the game is between 2 players, each player in their turn has to announce a number of course we are only looking at the natural numbers. They announce a number when the in the next player's turn he cannot, he or she cannot announce a number which can be expressed in terms of previously announced numbers. So let us look at an example let us say a player in the round 1 let us say player 1 has chosen number 2. Then what happens is that the player 2 whatever he chooses it cannot be 2, it cannot be 4, cannot be 6, cannot be 8 or any of them this thing because 4 is nothing but 2 plus 2, 6 is nothing but 2 plus 2 plus 2 like wise. So player 2 let us say he can announce maybe 7. Now once player 2 announced 7 in the round 2 player 1 he can announce a number which is not there in 2, 4, 6, 8 like this and it should not be inside this multiples of 7, 7, 14, 21 like this. Not only that it cannot be inside 2 plus 7, 4 plus 7, 6 plus 7 like this like. So in other words the number it cannot belongs to any number x into 2 plus y into 7 where x comma y are greater than equals to 0. So a player can only announce a number which is not a combination of the previously announced numbers. Now what exactly is the interesting thing about it? For example let us say if a player announces 1 every number is multiple of 1 and hence there is the other player the next player cannot make any move. That means if a player announces 1 that means the next player has no more moves. So the game will not be interesting if we consider a normal play. So this game is considered as a major play. That means player to choose 1 is loser. That means if a player to choose 1 is going to be a loser. In fact if the players if he has no more moves to make other than 1 he is going to become a loser. So we are considering a major play here. Now the interesting thing is that how long does this continue? How long does this game continue? The very interesting fact here is that as long as we want why? Let us look at, suppose the players can choose let us say 10000 the 1 player can choose 10000 and the next player can choose 91900 and then the next player can choose 9998 and it goes on 321. So here the length is 10000. This is one possibility of this game tree and the length here is 10000. Can this be bigger? The answer is in fact yes. So let us look at this another possibility. So for example it can go 2 power 10000, 2 power 9999 and like this it can go to 2 cube 2 square and then 2. Then after that we can actually look at the odd numbers. So this is also one possible way of this thing. So therefore the length can be made anything. So these are the games which we called as unboundedly. So whatever one player chooses if we want to make the length of this game any number we can make it. So therefore this thing and at the start itself the game can have any number of moves. So therefore it is a kind of unboundedly unbounded. So now but the question that comes is if this game is going to be like this why is this interesting? So the interest is mainly because of the following result due to Sylvester. What this Sylvester theorem says is that suppose take 2 numbers AB as I said we are in this thing and they are co-prime, they are relatively prime for the GCD of AB is 1. Then AB minus A minus B is the largest integer not representable by XA plus Y by of course X, Y goes to 0. So once you choose 2 numbers which are relatively prime whose GCD is 1 then AB minus A minus B is the largest integer which cannot be representable by XA plus YB. That means the length of this is going to be finite once you choose like this. So this is basically a result due to Sylvester. In fact this problem is a even is known as a Frobenius problem I will come to that later but let us look at the proof of this result it is not very hard. So let us see the proof of this fact. So now the GCD of AB is 1 so in our we go back to our school mathematics we have proved the following result which says that there exist integers X0, Y0 they are the integers they could be non-negative as well as negative such that A X0 plus B Y0 is equals to 1. So this is something that what this immediately implies for any n there exist X1, Y1 in Z such that A X1 plus B Y1 is equals to n. So whatever number n we choose there will be some corresponding X1 and Y1 such that A X1 plus B Y1 is equals to n. It is a simple it is X1 is nothing but n into X0 Y0 is Y1 is nothing but n into Y0 then this will work. Now what we want to show is that AB minus A minus B is the largest integer not represented. So let us try this one. So let n is bigger than AB minus A minus B that means to write it another way this is nothing but A minus 1 into B minus 1 so this is a this thing take this one. So now look at the solutions of A X1 plus B Y1 is equals to n we already know that there are solution but we can also write it take X to be okay let me not put here this X to be X1 minus TB Y to be Y plus TA if we do this one then what will happen is that A X plus B Y is nothing but A X1 minus TAB plus BY1 plus TAB which is nothing but A X1 plus BY1 which is n. So therefore the solutions of A X plus BY is equals to n or X is equals to X1 minus TB Y is equals to Y1 plus TA okay. Now let us look at the T let TB smallest such that Y1 plus TA is greater than equals to 0 because Y1 is positive A is positive so therefore you look at the first T for which this is going to be greater than equals to 0. So of course we do not know Y1 whether Y1 is a non-negative or not by definition if you look back Y1 is just a solution here we do not know it is just simply an integer therefore we cannot say whether Y1 is negative or positive but we can because we are adding a positive number so eventually this becomes a non-negative and then look at the T which is smallest such that this happens okay such a T now exists. Now look at this one A into X1 minus TB plus B into X1 plus TA this is equals to n which is of course by definition bigger than equals to A minus 1 into B minus 1 because that is this thing. So T is smallest therefore Y1 plus TA has to be less than equals to A minus 1. If it is Y1 plus TA is not less than equals to A minus 1 that means it is bigger than equals to A then T can be made smaller but T is the smallest so therefore Y1 plus TA has to be less than equals to A minus 1. So now we use this fact therefore what we now get is that A into X1 minus TB is bigger than or equals to okay there is a small typo here this is Y1 so Y1 plus TA is less than equals to A minus 1 so therefore I use that here so this becomes A minus 1 into B minus 1 minus B into A minus 1 because B into Y1 plus TA is less than equals to A minus 1 therefore B into Y1 plus TA is less than equals to B into A minus 1 if I bring this B Y plus TA to this set then minus of B Y1 plus TA that is using this inequality what we get is exactly this. Now if you simplify this fact what we get here is minus A plus 1 so let me rewrite here what we have got A into X1 minus TB is greater than equals to minus A plus 1 now this immediately tells me that X1 minus TB is bigger than or equals to minus 1 plus 1 over A therefore X1 minus TB is greater than equals to 0 because X1 minus TB is an integer all of them are integers and minus 1 plus 1 over A 1 over A is something bigger than minus 1 this is strictly greater than minus 1 therefore X1 minus TB has to be greater than equals to 0 this immediately tells me that n is nothing but A into X1 minus TB plus B into Y1 plus TB and this is X1 minus TB is greater than equals to 0 and Y1 plus TB is also greater than equals to 0 therefore n is expressible as a combination of A and B that means any number bigger than or equals to A minus 1 into B minus 1 can be written as a combination of A and B. Now the next part that requires here is to show that A minus 1 into B minus 1 that is nothing but AB minus A minus B plus 1 so look at AB minus A minus B this we want to show that this is not representable by A and B if we prove that then what we can we say that AB minus A minus B is the largest integer not representable by A and B so let us prove this fact. Suppose we go for a contradiction suppose AB minus A minus B is nothing but AX plus BY for some XY greater than equals to 0 let us say this happens then what will happen here is that this is a multiple of A this is a multiple of A therefore when I divide by A what I will get is that minus B is congruent to BY mod A because when I divide A will divide these 2 numbers are same therefore this minus this is a multiple of A so AB minus A minus B minus AX plus BY which is 0 is a multiple of A but AB minus A is a multiple AX is a multiple therefore minus B minus BY is multiple of A therefore minus B is congruent to BY mod A. Now A and B have are relatively co-prime so therefore the GCD of AB is 1 this immediately implies Y is congruent to minus 1 mod A so minus B is congruent to BY mod A and A and B are relatively prime and B is common here therefore if GCD of AB is 1 then this is same as minus 1 congruent to Y mod A that is exactly what is written here. In a similar fashion we can say that X is congruent to minus 1 mod B both are there so because Y is congruent to minus 1 mod A what we can really say here is that Y is bigger than or equals to A minus 1 because Y is a non-negative number so Y plus 1 is a divisible by A if Y plus 1 is divisible by A then Y has to be at least A minus 1 it cannot be less than A minus 1 in a similar fashion this particular thing gives me that X is bigger than or equals to B minus 1. Now look at AX plus BY so AB are all positive numbers so therefore this is greater than equals to A into B minus 1 plus B into A minus 1. If I rewrite this one what I have got here 2AB minus A minus B so what we have got is that AX plus BY is bigger than or equals to 2AB minus A minus B but this is strictly greater than AB minus A minus B this is a contradiction because AX plus BY is nothing but AB minus A minus B and that is the assumption that we have made therefore what we got is it is strictly greater than therefore it is a contradiction and hence AB minus A minus B cannot be represented by combinations of A and B. So this proves the Sylvester theorem as I said this is known as a Frobenius problem in fact the Frobenius problem asked for not just 2 numbers A, B but in general it asked for multiple numbers when you have A1, A2, Ak some k numbers and then you would like to answer what is the largest number not representable by this A1, A2, Ak and in fact this problem is very closely related with the theory of numerical semi-groups and of course we will not go into those details we will only look at we have only looked at this two case and this two case is proved by is proven by Sylvester and because of this Sylvester result this game is known as a silver coinage. Now in fact for the multiple case this problem is still open what is the largest number it is still there is no closed form available so in that sense this is an interesting problem to pursue for them. Now this theorem the Sylvester's theorem tells us that this game once at some point of time if people choose two numbers which are relatively prime then the game will not go on long it will end in a finite time but the question is once we know that this eventually this is going to stop in a finite time we need to understand who is going to win but this is again a problem which is largely open. So what we will see is that some examples where we can decisively conclude but in general this problem is a open problem. So let us say if a player 1 let us say announces 2 if a player 1 announces 2 immediately the player 2 can announce 3 after that no more move available except 1 therefore choosing 2 is not good similarly choosing 2 choosing 3 is also not good that means if I choose 3 you can choose 2 and the game ends so choosing 3 is not a good for me this thing so choosing 3 is not a good option in a sense 2 and 3 if someone plays a 2 the other will play 3 and if someone plays 3 other will play 2 so this is not a very good thing in that sense. So now let us look at another thing so let us say suppose player 1 plays 4 so if player 1 plays 4 what if player 2 plays 5 what will happen to this one so let us write down some this thing I will write down this way because player 1 has played 4 and the other player has played 5 so 4 and 5 are already ruled out so in fact so what happens is that the 0 5 10 this thing these are these numbers are not possible similarly 4 9 14 these are not possible similarly 8 13 these are not possible like this all these numbers we can see that they are not possible therefore only thing that remains is 2 3 6 7 11. So in 2 3 6 7 what are the good choices to play of course obviously I said choosing a 2 is not good for me so therefore I would not like to choose 2 what if if I choose 6 if I choose 6 let us look at it then player other player can choose 7 if I choose 6 immediately the 11 is not 6 plus 5 11 11 will also become impossible to play therefore 7 is there and after that I am forced to choose 2. Similarly if I choose 7 once again if I choose 7 7 plus 4 is 11 so 11 is not possible and in fact all the other numbers will also be impossible so therefore the only possibility is choosing 6 may be so in other words so 6 7 are also going to be a the choices if I choose 7 the other player he will choose 6 because instead of 6 if I choose 3 for example then instead of 7 instead of 6 each 3 a good choice if the player chooses 3 then he will immediately lose because the other player chooses 2 so therefore 3 is not possible there so 6 is there and then again goes back to so therefore 7 is also not good so the only move then will be 11 choosing 11 is the best option so like this this game one can proceed and look at what are the best moves possible so this game apart from showing this small small results this game is in general largely open in fact it is not very clear what exactly are the best moves in this game this is an open problem in fact the best place to look at about this game is the book by Burlcamp Conway and Richard Guy so they have a book called Winning Ways in fact this is 4 volumes so this book has a detailed description about that and in that game this game is a very interesting game which is completely different from a novel play okay now once we have done with this one the next thing that we would like to see here is that where we can go further so if we really look back in this last 9 sessions including this we have been discussing mostly about impartial games what about the partisan games in fact we have seen one partisan game which is a domineering so that is an example of a part partisan games the partisan games play a different kind of difficulties and in this course we will not discuss about this partisan game and instead we refer to this books the 4 volume book Winning Ways for this thing. Here one exercise would like to point out here is that in the domineering we have player 1 is playing horizontally and player 2 is playing vertically now the interesting question here that we can ask here is that suppose if the players if we make it impartial in the sense that the players can either place either a horizontal or vertical then how does the game change so this is an interesting exercise and with this exercise we will stop this communitarian game theory then we will switch to the zero sum games the classical game theory which I have introduced in the course introduction session so we will discuss about the zero sum games and then go on. Thank you we will meet in the next session.