home *** CD-ROM | disk | FTP | other *** search
- Path: sparky!uunet!mcsun!uknet!pavo.csi.cam.ac.uk!camcus!cet1
- From: cet1@cus.cam.ac.uk (C.E. Thompson)
- Newsgroups: sci.math
- Subject: Re: Chess Problem (8 bishops)
- Message-ID: <1992Sep14.160743.24477@infodev.cam.ac.uk>
- Date: 14 Sep 92 16:07:43 GMT
- References: <1992Sep12.222402.14408@nmt.edu>
- Sender: news@infodev.cam.ac.uk (USENET news)
- Organization: U of Cambridge, England
- Lines: 43
- Nntp-Posting-Host: grus.cus.cam.ac.uk
-
- In article <1992Sep12.222402.14408@nmt.edu>, jmarlan@titan.nmt.edu
- (Jon Marlan) writes:
- |>
- |> In article <BuGA8t.DL8@cmptrc.lonestar.org> carter@cmptrc.lonestar.org
- |> (Carter Bennett) writes:
-
- It was, indeed, the saving grace in an otherwise utterly useless posting.
-
- |> > FOLLOW-UP QUESTION FOR EXTRA CREDIT:
- |> >
- |> > Determine the odds for doing the same with eight bishops! ;-)
- |>
- |> I'll give this one a try.
-
- [Details omitted: as others have already pointed out your errors.]
-
- There are 22522960 ways of arranging 8 bishops on a chessboard so that
- none attacks any of the others. The probability in question is therefore
- 22522960/4426165368 = 1/196.5179...
-
- From the fact that there are (1,32,356,1704,3532,2816,632,16) ways
- of arranging 0..7 non-attacking bishops on the 32 black squares
- (or, of course, the white ones) it is easy to deduce that there are
- (1,64,1736,26192,242856,1444928,5599888,14082528,22522960,22057472,
- 12448832,3672448,489536,20224,256) ways of arranging 0..14 non-attacking
- bishops on the full board.
-
- I admit to using a program to get the first set of numbers, but it is
- perfectly possible, if a little tedious, to do by hand. I find it
- helpful to think of arranging non-attacking rooks on a board shaped
-
- X
- X X X
- X X X X X
- X X X X X X X
- X X X X X X X
- X X X X X
- X X X
- X
-
- Chris Thompson
- JANET: cet1@uk.ac.cam.phx
- Internet: cet1@phx.cam.ac.uk
-