Dinesman's multiple-dwelling problem: Difference between revisions

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 =={{header|Picat}}==
+===Constraint modelling===
-First a constraint modelling version (using Picat's cp module), then a version using permutations.
 <lang picat>import util.
 import cp.
-go =>
-   dinesman_cp,
-   dinesman2,
-   nl.
-% CP approach
 dinesman_cp =>
    println(dinesman_cp),
@@ Line 2,769: / Line 2,761: @@
    solve(X),
-   println([baker=Baker, cooper=Cooper, fletcher=Fletcher, miller=Miller, smith=Smith]).
+   println([baker=Baker, cooper=Cooper, fletcher=Fletcher, miller=Miller, smith=Smith]).</lang>
+{{out}}
+<pre>[baker = 3,cooper = 2,fletcher = 4,miller = 5,smith = 1]</pre>
+===Using permutations===
-%
+<lang Picat>%
-% The constraints (non CP approach)
-%
 % floors: 1: bottom .. 5: top floor
 %
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 adjacent(A,B) => abs(A-B) == 1.
-% Non-CP approach, using permutations
 dinesman2 =>
    println(dinesman2),
@@ Line 2,795: / Line 2,787: @@
 </lang>
+{{out}}
-Output:
+<pre>[baker = 3,cooper = 2,fletcher = 4,miller = 5,smith = 1]</pre>
-<pre>
-dinesman_cp
-[baker = 3,cooper = 2,fletcher = 4,miller = 5,smith = 1]
-dinesman1
-[baker = 3,cooper = 2,fletcher = 4,miller = 5,smith = 1]
-</pre>
 =={{header|PicoLisp}}==