Data commit

Task/Hamming-numbers/PL-I/hamming-numbers.pli

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+(subscriptrange):
+Hamming: procedure options (main); /* 14 November 2013 with fixes 2021 */
+   declare (H(2000), p2, p3, p5, twoTo31, Hm, tenP(11)) decimal(12)fixed;
+   declare (i, j, k, m, d, w) fixed binary;
+
+   /* Quicksorts in-place the array of integers H, from lb to ub */
+   quicksortH: procedure( lb, ub ) recursive;
+      declare ( lb,   ub    )binary(15)fixed;
+      declare ( left, right )binary(15)fixed;
+      declare ( pivot, swap )decimal(12)fixed;
+      declare sorting        bit(1);
+      if ub > lb then do
+         /* more than one element, so must sort */
+         left    = lb;
+         right   = ub;
+         /* choosing the middle element of the array as the pivot */
+         pivot   = H( left + ( ( right + 1 ) - left ) / 2 );
+         sorting = '1'b;
+         do while( sorting );
+            do while( left  <= ub & H( left  ) < pivot ); left  = left  + 1; end;
+            do while( right >= lb & H( right ) > pivot ); right = right - 1; end;
+            sorting = ( left <= right );
+            if sorting then do;
+               swap       = H( left  );
+               H( left  ) = H( right );
+               H( right ) = swap;
+               left       = left  + 1;
+               right      = right - 1;
+            end;
+         end;
+         call quicksortH( lb,   right );
+         call quicksortH( left, ub    );
+      end;
+   end quicksortH ;
+
+   /* find 2^31 - the limit for Hamming numbers we need to find */
+   twoTo31 = 2;
+   do i = 2 to 31;
+      twoTo31 = twoTo31 * 2;
+   end;
+   /* calculate powers of 10 so we can check the number of digits */
+   /* the numbers will have */
+   tenP( 1 ) = 10;
+   do i = 2 to 11;
+      tenP( i ) = 10 * tenP( i - 1 );
+   end;
+
+   /* find the numbers */
+   m = 0;
+   p5 = 1;
+   do k = 0 to 13;
+      p3 = 1;
+      do j = 0 to 19;
+         Hm = 0;
+         p2 = 1;
+         do i = 0 to 31 while( Hm < twoTo31 );
+            /* count the number of digits p2 * p3 * p5 will have */
+            d = 0;
+            do w = 1 to 11 while( tenP(w) < p2 ); d = d + 1; end;
+            do w = 1 to 11 while( tenP(w) < p3 ); d = d + 1; end;
+            do w = 1 to 11 while( tenP(w) < p5 ); d = d + 1; end;
+            if d < 11 then do;
+               /* the product will be small enough */
+               Hm = p2 * p3 * p5;
+               if Hm < twoTo31 then do;
+                  m = m + 1;
+                  H(m) = Hm;
+               end;
+            end;
+            p2 = p2 * 2;
+         end;
+         p3 = p3 * 3;
+      end;
+      p5 = p5 * 5;
+   end;
+
+   /* sort the numbers */
+   call quicksortH( 1, m );
+
+   put skip list( 'The first 20 Hamming numbers:' );
+   do i = 1 to 20;
+      put skip list (H(i));
+   end;
+   put skip list( 'Hamming number 1691:' );
+   put skip list (H(1691));
+
+end Hamming;