There are 2012 lamps arranged on a table. Two persons Diana and Ethan play the following game.
In each move the player flips the switch of one lamp, but he or she must never get back an arrangement of the lit lamps that has already been on the table. A player who cannot move loses.

Diana makes the first move, followed by Ethan. Who has a winning strategy?

Assuming best play, Diana (d) can always win, and Ethan (e) always lose.

Assume that there are just two (n=2) lights. We can use the decimal equivalent of the binary value of the lamps (1=lit, 0=off) to detail the 'moves', starting from position S= start.

S=0, d=2,e=3, d=1, e loses.

S=3, d=2,e=0,d=1, e loses.

S=2, d has a choice:(i) d=0, e=1,d=3, e loses, or (ii) d=3,e=1,d=0, e loses.

Obviously, S = 1 yields the same result; and these are all the alternatives.

Now assume that there are n=3 lights. In the same way, e always loses after the 2^n=8 possible combinations have been exhausted, e.g. S=0, d=4,e=6,d=2,e=3,d=1,e=5,d=7, e loses.

Now assume that there are n=2012 lights. There are 2^2012 possible combinations, and after a long game, d reaches the last possible combination, and e loses.

I don't imagine it's a game either would want to play again.

Edited on April 18, 2015, 12:06 am

Posted by broll on 2015-04-17 23:53:38