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From: jbs@mit-eddie.UUCP (Jeff Siegal)
Newsgroups: net.puzzle
Subject: Re: pennies puzzle (SPOILER)
Message-ID: <250@mit-eddie.UUCP>
Date: Fri, 21-Feb-86 01:29:24 EST
Article-I.D.: mit-eddi.250
Posted: Fri Feb 21 01:29:24 1986
Date-Received: Mon, 24-Feb-86 07:15:08 EST
References: <953@houxa.UUCP> <904@whuxlm.UUCP>
Reply-To: jbs@eddie.UUCP (Jeff Siegal)
Organization: MIT, Cambridge, MA
Lines: 27

In article <904@whuxlm.UUCP> dim@whuxlm.UUCP (McCooey David I) writes:
>> Imagine a two-player game, in which each of the players begins
>> with an infinite number of pennies.  There exists a round table,
>> and each player in his turn places a penny on the table.  (Turns
>> are alternated).  The game ends when there is no more room on the
>> table for any pennies.  The person who last put a penny on the table
>> is declared the winner.
>> 
>> Question:  Given that one of these players has a winning strategy,
>>    	which player (the first, or the second) can always win?  
>> 	Prove your answer by giving the strategy.
>> 
>The second player can always win if he uses the following strategy:
>
>	Considering the center of the table as the "origin", always
>	place his penny at a spot reflected through the origin from
>	where his opponent just placed his last penny.
>
>Using this strategy, the second player will always have a spot to place his
>penny because he is simply mirroring the actions of the first player.

Almost.  The first player can place his penny _AT_ the center of the
table.  Then he can always mirror the actions of his opponent.

Therefore, the first player can always win.

Jeff Siegal - MIT EECS