Dinesman's multiple-dwelling problem: Difference between revisions

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@@ Line 1,856: / Line 1,856: @@
   print dinesman -- print all solutions (only one): [(3,2,4,5,1)]</lang>
-Or as a list comprehension:
+Or as a list comprehension (syntactic sugar for a list monad):
 <lang haskell>import Data.List (permutations)
@@ Line 1,862: / Line 1,862: @@
 main =
   print
-    [ ( "Baker lives on " ++ show b
+    [ ( "Baker lives on " <> show b,
-      , "Cooper lives on " ++ show c
+        "Cooper lives on " <> show c,
-      , "Fletcher lives on " ++ show f
+        "Fletcher lives on " <> show f,
-      , "Miller lives on " ++ show m
+        "Miller lives on " <> show m,
-      , "Smith lives on " ++ show s)
+        "Smith lives on " <> show s
+      )
-    | [b, c, f, m, s] <- permutations [1 .. 5]
+      | [b, c, f, m, s] <- permutations [1 .. 5],
-    , b /= 5
-    , c /= 1
+        b /= 5,
-    , f /= 1
+        c /= 1,
-    , f /= 5
+        f /= 1,
-    , m > c
+        f /= 5,
-    , abs (s - f) > 1
+        m > c,
-    , abs (c - f) > 1 ]</lang>
+        abs (s - f) > 1,
+        abs (c - f) > 1
+    ]</lang>
 {{out}}
 <pre>[("Baker lives on 3","Cooper lives on 2","Fletcher lives on 4","Miller lives on 5","Smith lives on 1")]</pre>