Dinesman's multiple-dwelling problem: Difference between revisions

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 =={{header|D}}==
+<lang d>import std.stdio, std.range, std.math, std.traits;
-{{works with|D|2}}
-<lang d>import std.stdio, std.range, std.math;
+// From http://rosettacode.org/wiki/Permutations#D
-auto permutations(size_t n) {
+T[][] permutations(T)(T[] items) {
-    auto seq = array(iota(n));
-    size_t[][] res;
+    T[][] result;
-    void perms(size_t[] s, size_t[] pre = []) {
+    void perms(T[] s, T[] prefix=[]) {
-        if (s.length) {
+        if (s.length)
-            foreach (i, c; s) {
+            foreach (i, c; s)
-               perms(s[0 .. i] ~ s[i + 1 .. $], pre ~ c);
+               perms(s[0 .. i] ~ s[i+1 .. $], prefix ~ c);
-            }
+        else
-        } else res ~= pre;
+            result ~= prefix;
     }
-    perms(seq);
+    perms(items);
-    return res;
+    return result;
 }
-size_t[] solve(T...)(size_t len, T fun) {
+T[] solve(T, F...)(T[] items, F predicates) {
-    auto perms = permutations(len);
+  foreach (perm; permutations(items)) {
+    foreach (pred; predicates)
-    outer:
+      if (!pred(perm))
-    foreach (p; perms) {
-        foreach (fn; fun)
+        break;
+    return perm;
-            if (!fn(cast(int[])p))
+  }
-                continue outer;
-        return p;
+  return [];
-    }
-    return null;
 }
 void main() {
+  enum P { Baker, Cooper, Fletcher, Miller, Smith }
-    enum Floors = 5u;
-    enum {Baker, Cooper, Fletcher, Miller, Smith};
+  auto c1= (P[] s){ return s[P.Baker] != s.length-1; };
-    auto c1 = (int[] s){ return s[Baker] != s.length-1; };
+  auto c2= (P[] s){ return s[P.Cooper] != 0; };
-    auto c2 = (int[] s){ return s[Cooper] != 0; };
+  auto c3= (P[] s){ return s[P.Fletcher] && s[P.Fletcher] != s.length-1; };
-    auto c3 = (int[] s){ return s[Fletcher] != 0 && s[Fletcher] != s.length-1; };
+  auto c4= (P[] s){ return s[P.Miller] > s[P.Cooper]; };
-    auto c4 = (int[] s){ return s[Miller] > s[Cooper]; };
+  auto c5= (P[] s){ return abs(s[P.Smith] - s[P.Fletcher]) != 1; };
-    auto c5 = (int[] s){ return abs(s[Smith] - s[Fletcher]) != 1; };
+  auto c6= (P[] s){ return abs(s[P.Cooper] - s[P.Fletcher]) != 1; };
-    auto c6 = (int[] s){ return abs(s[Cooper] - s[Fletcher]) != 1; };
+  // c1, c2, ... can't be an array
-    if (auto sol = solve(Floors, c1, c2, c3, c4, c5, c6))
+  auto solution = solve([EnumMembers!P], c1, c2, c3, c4, c5, c6);
+  if (solution.length)
-        writeln(sol);
+    writeln(solution);
 }</lang>
+Output:
-<pre>[2, 1, 3, 4, 0]</pre>
+<pre>[Baker, Cooper, Fletcher, Miller, Smith]</pre>
 =={{header|Haskell}}==