Dinesman's multiple-dwelling problem: Difference between revisions

=={{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.