KenKen puzzle addends: REDUX A (corrected) non-recursive algorithm

Question

This question relates to those parts of the KenKen Latin Square puzzles which ask you to find all possible combinations of ncells numbers with values x such that 1 <= x &

Answer 1

And here is another recursive, generator-based solution, but this time using some simple math to calculate ranges at each step, avoiding needless recursion:

def latinSquares(max_val, target_sum, n_cells):
  if n_cells == 1:
    assert(max_val >= target_sum >= 1)
    return ((target_sum,),)
  else:
    lower_bound = max(-(-target_sum / n_cells), 1)
    upper_bound = min(max_val, target_sum - n_cells + 1)
    assert(lower_bound <= upper_bound)
    return ((v,) + w for v in xrange(upper_bound, lower_bound - 1, -1)
                     for w in latinSquares(v, target_sum - v, n_cells - 1))

This code will fail with an AssertionError if you supply parameters that are impossible to satisfy; this is a side-effect of my "correctness criterion" that we never do an unnecessary recursion. If you don't want that side-effect, remove the assertions.

Note the use of -(-x/y) to round up after division. There may be a more pythonic way to write that. Note also I'm using generator expressions instead of yield.

for m in latinSquares(6,12,4):
  print m