Mugging and craving for dota has left me little time to post any new entries recently, but here is the latest post.

I haven recently looking into several problems in Maths, one pertaining to game theory and the other is number theory. Ok, playing 'Chopsticks' with some friends and my brother reignited a mathematical interest to the game. That is, is there a optimal strategy by any player, and what is the outcome when both players play optimally. First of all, lets have a recap on the rules of the game. The rules have been restated in symbolic form, to avoid all those material confusion.

1) Both players start with 2 piles, containing 1 object each.
2) When its the player turn, there are 2 legal operations.
-Lets label this player's piles as C and D.
-1st operation: Where x < (C or D); the player can move so that (C-x, D+x), or (C+x, D-x)
-2nd operation: The player can move so that: (D+C, C) or (C+D, D)
3) After each turn, update your hand, by taking modulus 5.
4) If both piles having nothing left, the player loses.

A bit of experimentation can reveal that there can only be 3 results from this game; either player A wins, or player B wins, or both players are trapped in infinite loop such that the game is a draw. I think when both players play optimally, the result is a draw. The reasoning is as follows. A starts first. A has a winning strategy that prevents him from losing. So the outcome for him is either win or draw. When its B turn, B knows for him, the outcome is either draw or lose. So B will choose his strategy such that he can minimise his losses. In this case, B will opt for the draw. So ultimately, both players will be locked in a infinite loop and be forced to concede to a draw.

Of course this says nothing about the optimal strategy itself. I am currently still searching for it. The best way to do this is to do an exhautive tree search, but this is proving tedious due to the great amount of possibilities. When I have the strategy, I will post it up here.

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I am also occupied somewhat with the problem of the Collatz Conjecture. This conjecture goes like this: Take any number, if its even, divide by 2. If its odd, multiply by 3 and add 1. Repeat the above steps with this new number. The conjecture states that all numbers eventually lead to the steps, 4,2,1,4,2,1, ad infinitum. Try it yourself.

First of all, we need to see that this involves 2 rules, a lengthening rule, one that makes numbers bigger, and a shortening rule, one that makes numbers smaller. It is impossible to predict on the outset whether a combination of lengthening and shortening rules lead to a specific number or lead to inifinity.

Next observe that all number of the form 2^x lead automatically to the loop just by dividing by 2. In other words, if you somehow hit on to this branch of numbers, there is no escape for you.

A bit of experimentation shows that starting with any number, you use more shortening rules than lengthening rules. I cant proof this, but I think it has to do with the property of the numbers itself. Still investigating.


In the meantime, we will all have to cross our fingers and hope for the best for Promos. Be sure to remember to study smart. No us spending 12 straight hours memorising everything, and not comprehending anything at all.