It's also fast enough for low n (up to 8). Why write more code when you can get your result with less? Maybe the time he's saved with this approach can now be used writing a deduplicator (to cull out mirrored and rotated solutions).

Maybe the point of the article was not only to show how fast PyPy is, but how convenient itertools.permutation() is. You can't show the convenience of itertools by not using it.

However... You know what? When I was asked to write a n-queens solver recently, I also wrote a backtracker. It just doesn't mean I did it better than this guy.

Yes, it's twice as long and harder to read. Improvements, anyone?

  def hit_last(p):
      x = p[-1]
      i = len(p)-1
      for j in range(i):
          y = p[j]
          if x == y or i - j == abs(x - y):
              return True
      return False

  def n_queen(n):
      p = [0]
      while True:  # Begin iter w/ board good, except maybe last Q
          full = len(p) == n
          hit = hit_last(p)
          if full or hit:  # Will we backtrack?
              if not hit:  # Found solution?
                  yield p
              while len(p) > 0 and p[-1] == n-1:
                  p.pop()
              if len(p) == 0:
                  return
              p[-1] += 1
          else:
              p.append(0)

    alex@alex-gaynor-laptop:/tmp$ time python queens.py 

    real    0m16.837s
    user    0m16.820s
    sys 0m0.000s
    alex@alex-gaynor-laptop:/tmp$ time pypy queens.py 

    real    0m2.548s
    user    0m2.530s
    sys 0m0.010s