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Dinesman's multiple-dwelling problem: Difference between revisions
Dinesman's multiple-dwelling problem (view source)
Revision as of 12:06, 30 June 2011
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-> NIL</pre>
=={{header|Prolog}}==
Prolog is THE language for this kind of puzzle !<br>
Works with SWI-Prolog and library(clpfd) written by '''Markus Triska'''.
<lang Prolog>:- use_module(library(clpfd)).
:- dynamic top/1, bottom/1.
% Baker does not live on the top floor
rule1(L) :-
member((baker, F), L),
top(Top),
F #\= Top.
% Cooper does not live on the bottom floor.
rule2(L) :-
member((cooper, F), L),
bottom(Bottom),
F #\= Bottom.
% Fletcher does not live on either the top or the bottom floor.
rule3(L) :-
member((fletcher, F), L),
top(Top),
bottom(Bottom),
F #\= Top,
F #\= Bottom.
% Miller lives on a higher floor than does Cooper.
rule4(L) :-
member((miller, Fm), L),
member((cooper, Fc), L),
Fm #> Fc.
% Smith does not live on a floor adjacent to Fletcher's.
rule5(L) :-
member((smith, Fs), L),
member((fletcher, Ff), L),
abs(Fs-Ff) #> 1.
% Fletcher does not live on a floor adjacent to Cooper's.
rule6(L) :-
member((cooper, Fc), L),
member((fletcher, Ff), L),
abs(Fc-Ff) #> 1.
init(L) :-
% we need to define top and bottom
assert(bottom(1)),
length(L, Top),
assert(top(Top)),
% we say that they are all in differents floors
bagof(F, X^member((X, F), L), LF),
LF ins 1..Top,
all_different(LF),
% Baker does not live on the top floor
rule1(L),
% Cooper does not live on the bottom floor.
rule2(L),
% Fletcher does not live on either the top or the bottom floor.
rule3(L),
% Miller lives on a higher floor than does Cooper.
rule4(L),
% Smith does not live on a floor adjacent to Fletcher's.
rule5(L),
% Fletcher does not live on a floor adjacent to Cooper's.
rule6(L).
solve(L) :-
bagof(F, X^member((X, F), L), LF),
label(LF).
dinners :-
retractall(top(_)), retractall(bottom(_)),
L = [(baker, _Fb), (cooper, _Fc), (fletcher, _Ff), (miller, _Fm), (smith, _Fs)],
init(L),
solve(L),
maplist(writeln, L).
</lang>
Output :
<pre>?- dinners.
baker,3
cooper,2
fletcher,4
miller,5
smith,1
true ;
false.
</pre>
true ==> predicate succeeded. <br>
false ==> no other solution.
=={{header|PureBasic}}==
<lang PureBasic>Prototype cond(Array t(1))
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