Newsgroups: sci.math
From: hoey@pooh.tec.army.mil (Dan Hoey)
Date: 1995/10/08
Subject: Re: Game Strategy

: whi...@mail.powerup.com.au wrote:

: > "Two players, in turn, take balls out of two boxes.  At any one move, each
: > of the players can take an arbitrary non-zero number of balls that is less
: > than 11 from any of the boxes, but only from one of the boxes.  The winner
: > is the player who takes the last ball or balls.  Show that the starting
: > player can always win if one of the boxes contains 100 balls and the other
: > 83."

Procedure for playing a normal impartial game:

1. Learn to play Nim.
2. Learn to convert any normal impartial game to Nim.

It's harder for misere play in some cases, but not this one.

: > I have developed a strategy for one box:  firstly take enough balls from a
: > box to leave it with a multiple of 11, and then for after each turn, make
: > sure the box contains a multiple of 11 (a 'safe' move).

Jonathan E. Hardis (jhar...@tcs.wap.org) wrote:

: 11?  Here's a hint.  At the end of the game, one of the boxes must be
: empty, and the other must contain 12 balls.  The rules are that your
: opponent must take out between 1 and 11 balls, not zero.

Here's another hint: if you read the post you're responding to, you
might not look so ridiculous.  Players remove between 1 and 10 balls,
not 11.  Whitby's one-box strategy is correct.

Dan Hoey                [posted and mailed]
Hoey@AIC.NRL.Navy.Mil