Data update

Task/N-queens-problem/C/n-queens-problem-5.c

@@ -0,0 +1,69 @@
+#include <stdio.h>
+#define MAXN 31
+
+int nqueens(int n)
+{
+  int q0,q1;
+  int cols[MAXN], diagl[MAXN], diagr[MAXN], posibs[MAXN]; // Our backtracking 'stack'
+  int num=0;
+  //
+  // The top level is two fors, to save one bit of symmetry in the enumeration by forcing second queen to
+  // be AFTER the first queen.
+  //
+  for (q0=0; q0<n-2; q0++) {
+    for (q1=q0+2; q1<n; q1++){
+      int bit0 = 1<<q0;
+      int bit1 = 1<<q1;
+      int d=0; // d is our depth in the backtrack stack
+      cols[0] = bit0 | bit1 | (-1<<n); // The -1 here is used to fill all 'coloumn' bits after n ...
+      diagl[0]= (bit0<<1 | bit1)<<1;
+      diagr[0]= (bit0>>1 | bit1)>>1;
+
+      //  The variable posib contains the bitmask of possibilities we still have to try in a given row ...
+      int posib = ~(cols[0] | diagl[0] | diagr[0]);
+
+      while (d >= 0) {
+        while(posib) {
+          int bit = posib & -posib; // The standard trick for getting the rightmost bit in the mask
+          int ncols= cols[d] | bit;
+          int ndiagl = (diagl[d] | bit) << 1;
+          int ndiagr = (diagr[d] | bit) >> 1;
+          int nposib = ~(ncols | ndiagl | ndiagr);
+          posib^=bit; // Eliminate the tried possibility.
+
+          // The following is the main additional trick here, as recognizing solution can not be done using stack level (d),
+          // since we save the depth+backtrack time at the end of the enumeration loop. However by noticing all coloumns are
+          // filled (comparison to -1) we know a solution was reached ...
+          // Notice also that avoiding an if on the ncols==-1 comparison is more efficient!
+          num += ncols==-1;
+
+          if (nposib) {
+            if (posib) { // This if saves stack depth + backtrack operations when we passed the last possibility in a row.
+              posibs[d++] = posib; // Go lower in stack ..
+            }
+            cols[d] = ncols;
+            diagl[d] = ndiagl;
+            diagr[d] = ndiagr;
+            posib = nposib;
+          }
+        }
+        posib = posibs[--d]; // backtrack ...
+      }
+    }
+  }
+  return num*2;
+}
+
+
+main(int ac , char **av)
+{
+  if(ac != 2) {
+    printf("usage: nq n\n");
+    return 1;
+  }
+  int n = atoi(av[1]);
+  if(n<1 || n > MAXN) {
+    printf("n must be between 2 and 31!\n");
+  }
+  printf("Number of solution for %d is %d\n",n,nqueens(n));
+}