This repository has been archived on 2023-08-20. You can view files and clone it, but cannot push or open issues or pull requests.
yap-6.3/CHR/chr/examples/examples-queens.bool
vsc e5f4633c39 This commit was generated by cvs2svn to compensate for changes in r4,
which included commits to RCS files with non-trunk default branches.


git-svn-id: https://yap.svn.sf.net/svnroot/yap/trunk@5 b08c6af1-5177-4d33-ba66-4b1c6b8b522a
2001-04-09 19:54:03 +00:00

102 lines
2.7 KiB
Plaintext

% 4-queens problem
queens4([[S11, S12, S13, S14],
[S21, S22, S23, S24],
[S31, S32, S33, S34],
[S41, S42, S43, S44]
]) :-
%% rows
card(1,1,[S11, S12, S13, S14]),
card(1,1,[S21, S22, S23, S24]),
card(1,1,[S31, S32, S33, S34]),
card(1,1,[S41, S42, S43, S44]),
%% columns
card(1,1,[S11, S21, S31, S41]),
card(1,1,[S12, S22, S32, S42]),
card(1,1,[S13, S23, S33, S43]),
card(1,1,[S14, S24, S34, S44]),
%% diag left-right
card(0,1,[S14]),
card(0,1,[S13, S24]),
card(0,1,[S12, S23, S34]),
card(0,1,[S11, S22, S33, S44]),
card(0,1,[S21, S32, S43]),
card(0,1,[S31, S42]),
card(0,1,[S41]),
%% diag right-left
card(0,1,[S11]),
card(0,1,[S12, S21]),
card(0,1,[S13, S22, S31]),
card(0,1,[S14, S23, S32, S41]),
card(0,1,[S24, S33, S42]),
card(0,1,[S34, S43]),
card(0,1,[S44]).
/*
Article 4689 of comp.lang.prolog:
From: leonardo@dcs.qmw.ac.uk (Mike Hopkins)
Subject: Re: Solving 4 queens using boolean constraint
Message-ID: <1992Apr6.140627.10533@dcs.qmw.ac.uk>
Date: 6 Apr 92 14:06:27 GMT
References: <1992Apr6.105730.13467@corax.udac.uu.se>
The problem insists that each row and column contains exactly one
queens: therefore the program should be:
fourQueens(q(r(S11, S12, S13, S14),
r(S21, S22, S23, S24),
r(S31, S32, S33, S34),
r(S41, S42, S43, S44))) :-
%% rows
bool:sat(card([1],[S11, S12, S13, S14])),
bool:sat(card([1],[S21, S22, S23, S24])),
bool:sat(card([1],[S31, S32, S33, S34])),
bool:sat(card([1],[S41, S42, S43, S44])),
%% columns
bool:sat(card([1],[S11, S21, S31, S41])),
bool:sat(card([1],[S12, S22, S32, S42])),
bool:sat(card([1],[S13, S23, S33, S43])),
bool:sat(card([1],[S14, S24, S34, S44])),
%% diag left-right
bool:sat(card([0-1],[S14])),
bool:sat(card([0-1],[S13, S24])),
bool:sat(card([0-1],[S12, S23, S34])),
bool:sat(card([0-1],[S11, S22, S33, S44])),
bool:sat(card([0-1],[S21, S32, S43])),
bool:sat(card([0-1],[S31, S42])),
bool:sat(card([0-1],[S41])),
%% diag right-left
bool:sat(card([0-1],[S11])),
bool:sat(card([0-1],[S12, S21])),
bool:sat(card([0-1],[S13, S22, S31])),
bool:sat(card([0-1],[S14, S23, S32, S41])),
bool:sat(card([0-1],[S24, S33, S42])),
bool:sat(card([0-1],[S34, S43])),
bool:sat(card([0-1],[S44])).
This then gives the following result:
| ?- fourQueens(A).
A = q(r(0,_C,_B,0),r(_B,0,0,_A),r(_A,0,0,_B),r(0,_B,_A,0)),
bool:sat(_C=\=_B),
bool:sat(_A=\=_B) ? ;
no
| ?-
This therefore represents the desired two solutions!
===================================================
Mike Hopkins
Dept. of Computer Science, Queen Mary and Westfield College,
Mile End Road, London E1 4NS, UK
Tel: 071-975-5241
ARPA: leonardo%cs.qmw.ac.uk@nsfnet-relay.ac.uk
BITNET: leonardo%uk.ac.qmw.cs@UKACRL.BITNET
===================================================
*/