Tuesday, December 11, 2012

"number" game

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In the last blog, we know that there are "number" games n+1=(n| ) and -(n+1)=( |n). It is easy to see that in all positive integers, L will be the winner. Indeed, for any games of the form ((X| )| ), L will be the winner, as R can never be able to make a move.

How about negative integers ( |n)? If L is the first player, he loses as he cannot make a move. If R is the first player, the next configuration is the game n, and R is the loser. Therefore the second player will always be the winner.

Lets consider ( |-n). Who wins?
If L is the first player, he loses. If R is the first player, the next configuration is the game -n with L makes the first move -- but for the game -n, the one who makes the first move loses--and so L loses again. It is a game that R wins.

Lets consider (n|-m). Who wins?
If L is the first player, the next configuration is n, and so L wins. If R is the first player, the next configuration is -m with L makes the first move, and so R wins. It is a game that the first player wins.

Lets consider a little bit more complicated ( |n,-m). Who wins?
If L is the first player, he loses. If R is the first player, he can choose the next configuration to be n or -m (and L makes the first move): If R chooses n, he loses; if R chooses -m, he wins.
It is a game that there is no definite winner, but R has a winning strategy - choosing -m if he is the second player.

Lets consider (n|n, -m). Who wins?
If L is the first player, he wins. If R is the first player, he wins only if he chooses -m.
It is a game of which the first player has a winning strategy.

Lets consider (( |n,m),( |-n)|n, -m). Who wins?
Again, it is a game that the first player has a winner strategy: If L is the first, he should choose ( |n,m); If R is the first, he should choose -m.

It is not that complicated, right.....?