e5f4633c39
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102 lines
2.7 KiB
Plaintext
102 lines
2.7 KiB
Plaintext
% 4-queens problem
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queens4([[S11, S12, S13, S14],
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[S21, S22, S23, S24],
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[S31, S32, S33, S34],
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[S41, S42, S43, S44]
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]) :-
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%% rows
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card(1,1,[S11, S12, S13, S14]),
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card(1,1,[S21, S22, S23, S24]),
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card(1,1,[S31, S32, S33, S34]),
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card(1,1,[S41, S42, S43, S44]),
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%% columns
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card(1,1,[S11, S21, S31, S41]),
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card(1,1,[S12, S22, S32, S42]),
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card(1,1,[S13, S23, S33, S43]),
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card(1,1,[S14, S24, S34, S44]),
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%% diag left-right
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card(0,1,[S14]),
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card(0,1,[S13, S24]),
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card(0,1,[S12, S23, S34]),
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card(0,1,[S11, S22, S33, S44]),
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card(0,1,[S21, S32, S43]),
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card(0,1,[S31, S42]),
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card(0,1,[S41]),
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%% diag right-left
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card(0,1,[S11]),
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card(0,1,[S12, S21]),
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card(0,1,[S13, S22, S31]),
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card(0,1,[S14, S23, S32, S41]),
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card(0,1,[S24, S33, S42]),
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card(0,1,[S34, S43]),
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card(0,1,[S44]).
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/*
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Article 4689 of comp.lang.prolog:
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From: leonardo@dcs.qmw.ac.uk (Mike Hopkins)
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Subject: Re: Solving 4 queens using boolean constraint
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Message-ID: <1992Apr6.140627.10533@dcs.qmw.ac.uk>
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Date: 6 Apr 92 14:06:27 GMT
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References: <1992Apr6.105730.13467@corax.udac.uu.se>
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The problem insists that each row and column contains exactly one
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queens: therefore the program should be:
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fourQueens(q(r(S11, S12, S13, S14),
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r(S21, S22, S23, S24),
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r(S31, S32, S33, S34),
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r(S41, S42, S43, S44))) :-
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%% rows
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bool:sat(card([1],[S11, S12, S13, S14])),
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bool:sat(card([1],[S21, S22, S23, S24])),
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bool:sat(card([1],[S31, S32, S33, S34])),
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bool:sat(card([1],[S41, S42, S43, S44])),
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%% columns
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bool:sat(card([1],[S11, S21, S31, S41])),
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bool:sat(card([1],[S12, S22, S32, S42])),
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bool:sat(card([1],[S13, S23, S33, S43])),
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bool:sat(card([1],[S14, S24, S34, S44])),
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%% diag left-right
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bool:sat(card([0-1],[S14])),
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bool:sat(card([0-1],[S13, S24])),
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bool:sat(card([0-1],[S12, S23, S34])),
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bool:sat(card([0-1],[S11, S22, S33, S44])),
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bool:sat(card([0-1],[S21, S32, S43])),
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bool:sat(card([0-1],[S31, S42])),
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bool:sat(card([0-1],[S41])),
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%% diag right-left
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bool:sat(card([0-1],[S11])),
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bool:sat(card([0-1],[S12, S21])),
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bool:sat(card([0-1],[S13, S22, S31])),
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bool:sat(card([0-1],[S14, S23, S32, S41])),
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bool:sat(card([0-1],[S24, S33, S42])),
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bool:sat(card([0-1],[S34, S43])),
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bool:sat(card([0-1],[S44])).
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This then gives the following result:
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| ?- fourQueens(A).
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A = q(r(0,_C,_B,0),r(_B,0,0,_A),r(_A,0,0,_B),r(0,_B,_A,0)),
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bool:sat(_C=\=_B),
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bool:sat(_A=\=_B) ? ;
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no
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| ?-
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This therefore represents the desired two solutions!
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===================================================
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Mike Hopkins
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Dept. of Computer Science, Queen Mary and Westfield College,
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Mile End Road, London E1 4NS, UK
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Tel: 071-975-5241
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ARPA: leonardo%cs.qmw.ac.uk@nsfnet-relay.ac.uk
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BITNET: leonardo%uk.ac.qmw.cs@UKACRL.BITNET
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===================================================
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*/
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