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 2012 Lamps (Posted on 2015-04-17)
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?

 No Solution Yet Submitted by K Sengupta No Rating

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 re(4): Possible solution Comment 5 of 5 |
(In reply to re(3): Possible solution by Jer)

My first post may have been a bit condensed. Elaborating:

S=0, d=2,e=3, d=1, e loses. In binary (0=off, 1 = on)

S 00 = 0
d 10 = 2
e 11 = 3
d 01 = 1
e loses, etc. It's just a shorthand way of representing the 'play', to demonstrate some rather obvious conclusions.

Whether a cutoff can exist is an interesting question. I haven't found an example, but that doesn't mean one doesn't exist.

A problem for another day, perhaps.

Edited on April 18, 2015, 11:36 pm
 Posted by broll on 2015-04-18 23:34:00

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