🚀 go-pugleaf

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
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