N Queens Problem in ParaSail

Here is a (parallel) solution to the "N Queens" problem in ParaSail, where we try to place N queens on an NxN chess board such that none of them can take each other. This takes the idea of using the "continue" statement as a kind of implicit recursion to its natural conclusion.  This presumes you can turn a "normal" data structure like Vector<> into a concurrent data structure by using the keyword "concurrent," which presumably means that locking is used on all operations to support concurrency.  It is debatable whether this use of a "continue" statement to effectively start the next iteration of a loop almost like a recursive call is easier or harder to understand than true recursion. 

interface N_Queens <N : Univ_Integer := 8> is
    // Place N queens on a checkerboard so that none of them can
    // "take" each other.
    type Row is new Integer<1, N>;
     
    function Place_Queens() -> Vector<Vector<Row>> 
      {for all I in Place_Queens#range : Length(Place_Queens[I]) == N};
end interface N_Queens;

class N_Queens <N : Univ_Integer := 8> is
    type Column is new Integer<1, N>;
    type Sum is Vector<Boolean, Index => Integer<2, 2*N>>;
    type Diff is Vector<Boolean, Index => Integer<1-N, N-1>>;
    
exports
    
    function Place_Queens() -> Vector<Vector<Row>> 
      {for all I in Place_Queens#range : Length(Place_Queens[I]) == N} is
      // Place N queens on checkerboard so that none of them can "take" each other
      var Solutions : concurrent Vector<Vector<Row>> := [];
      
      *Outer_Loop*
      for (C : Column := 1; Trial : Vector<Row> := [];
        Diag1 : SSum := [.. => #false]; Diag2 : Diff := [.. => #false]) loop
          // Iterate over the columns
        
          for R in Row concurrent loop
              // Iterate over the rows
              if not Diag1[R + C] and then not Diag2[R - C] then
                  // Found a Row/Column combination that is not on any diagonal
                  // already occupied.
                  if C < N then
                      // Keep going since haven't reached Nth column.
                      continue loop Outer_Loop with (C => C+1, Trial => Trial | R,
                        Diag1 => Diag1 | [(R+C) => #true],
                        Diag2 => Diag2 | [(R-C) => #true]);
                  else
                      // All done, remember trial result.
                      Solutions |= Trial;
                  end if;
              end if;
          end loop;
          
          
      end loop Outer_Loop;
      return Solutions;
      
    end function Place_Queens;
end class N_Queens;