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Achava Nakhash, the Loving Snake  
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 More options Nov 7 2009, 2:04 am
Newsgroups: sci.math
From: "Achava Nakhash, the Loving Snake" <ach...@hotmail.com>
Date: Fri, 6 Nov 2009 18:04:48 -0800 (PST)
Local: Sat, Nov 7 2009 2:04 am
Subject: Re: Game with coins
On Nov 6, 2:31 pm, jane <jane1...@rambler.ru> wrote:

> There are 10 piles of coins (possibly with different amount of coins). Each of the two players can take arbitrary number of coins either from the leftmost or from the rightmost pile of coins. The one who cannot make a move loses.
> Is there a winning strategy for some of the player in this game?

> Thanks in advance.
> jane.

Yes.

All right, that is probabaly not the answer you were looking for.  I
am not going to give you the answer, but I will give you a technique
for working out the winning strategy for many such games.  I learned
this from a class with Elwyn Berlekamp around 1974, and I do not know
any references, but the key phrase is Sprague-Grundy number.

Every position is assigned a number which is a positive integer.  The
empty position, in which the player to move loses, is assigned 0.  To
find the number for any position, you look at every position that can
be generated from the given position.  Since this is an inductive,
recursive sort of procedure, all of these child positions already
have numbers.  The current position is assigned the least integer that
hasn't already been assigned to one of the child positions.  Thus if
the numbers of the child positions do not include 0, the given
position is assigned 0.  If they are, say, 0, 1, 3, 6, 7, then the
given position is assigned 2.

It is easy to see that the second player to move has a winning stategy
if and only if the Sprague-Grundy number of the position is 0.  Figure
out which positions have number 0.  The strategy is then to always
move so that the resulting position has SG number 0.

Good luck!

Regards,
Achava


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