NetNews Usenet Archive 1992 #20/NN_1992_20.iso/spool/sci/math/11319

Internet Message Format | 1992-09-12 | 1.7 KB

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From: [e-mail] (David Barton)
Newsgroups: sci.math
Subject: Re: Chess Problem
Message-ID: <[e-mail]>
Date: 12 Sep 92 18:54:13 GMT
References: <[e-mail]> <[e-mail]>
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In-Reply-To: [e-mail]'s message of 12 Sep 92 05:36:28 GMT
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In article <[e-mail]> [e-mail]
(Carter Bennett) writes:

      So there are 4426165368 possible ways to place the 8 rooks on the
   board.  My understanding is that no two rooks can attack each other
   only when they are arranged such that they line up on one of the two
   diagonals of the board.  Meaning there are 2 ways out of 4426165368
   possible setups that meet the criteria.  Putting that in normalized odds:

No.  Each rook has to be in a different row AND a different column
from every other rook.  There are far more than two ways to do this.

Let's see.  The first rook can go anywhere; it eliminates fifteen
squares from possible next placement, leaving 49.  The second rook has
49 out of 64 squares to choose from and eliminates thirteen squares
from the next possible placement set, leaving 36.  Continuing:

  p = 49/64 * 36/64 * 25/64 * 16/64 * 9/64 * 4/64 * 1/64

or, whipping out my handy dandy emacs calculator:

  p = 5.77565515416e-6

or

  p = .00000577565515416

If I can do this, with my distance from my prob/stat course, it can't
be THAT hard.

                    Dave Barton
                    [e-mail]