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

Your algorithm seems pretty good at first blush, and I don't think OO or another language would improve the code. I can't say if recursion would have helped but I admire the non-recursive approach. I bet it was harder to get working and it's harder to read but it likely is more efficient and it's definitely quite clever. To be honest I didn't analyze the algorithm in detail but it certainly looks like something that took a long while to get working correctly. I bet there were lots of off-by-1 errors and weird edge cases you had to think through, eh?

Given all that, basically all I tried to do was pretty up your code as best I could by replacing the numerous C-isms with more idiomatic Python-isms. Often times what requires a loop in C can be done in one line in Python. Also I tried to rename things to follow Python naming conventions better and cleaned up the comments a bit. Hope I don't offend you with any of my changes. You can take what you want and leave the rest. :-)

Here are the notes I took as I worked:

• Changed the code that initializes tmp to a bunch of 1's to the more idiomatic tmp = [1] * n_cells.
• Changed for loop that sums up tmp_sum to idiomatic sum(tmp).
• Then replaced all the loops with a tmp = + one-liner.
• Moved raise doneException to init_tmp_new_ceiling and got rid of the succeeded flag.
• The check in init_tmp_new_ceiling actually seems unnecessary. Removing it, the only raises left were in make_combos_n_cells, so I just changed those to regular returns and dropped doneException entirely.
• Normalized mix of 4 spaces and 8 spaces for indentation.
• Removed unnecessary parentheses around your if conditions.
• tmp[p2] - tmp[p1] == 0 is the same thing as tmp[p2] == tmp[p1].
• Changed while True: if new_ceiling_flag: break to while not new_ceiling_flag.
• You don't need to initialize variables to 0 at the top of your functions.
• Removed combos list and changed function to yield its tuples as they are generated.
• Renamed tmp to combo.
• Renamed new_ceiling_flag to ceiling_changed.

And here's the code for your perusal:

def initial_combo(ceiling=5, target_sum=13, num_cells=4):
    """
    Returns a list of possible addends, probably to be modified further.
    Starts a new combo list, then, starting from left, fills items to ceiling
    or intermediate between 1 and ceiling or just 1.  E.g.:
    Given ceiling = 5, target_sum = 13, num_cells = 4: creates [5,5,2,1].
    """
    num_full_cells = (target_sum - num_cells) // (ceiling - 1)

    combo = [ceiling] * num_full_cells \
          + [1]       * (num_cells - num_full_cells)

    if num_cells > num_full_cells:
        combo[num_full_cells] += target_sum - sum(combo)

    return combo

def all_combos(ceiling, target_sum, num_cells):
    # p0   points at the rightmost item and moves left under some conditions
    # p1   starts out at rightmost items and steps left
    # p2   starts out immediately to the left of p1 and steps left as p1 does
    #      So, combo[p2] and combo[p1] always point at a pair of adjacent items.
    # d    combo[p2] - combo[p1]; immediate difference
    # cd   combo[p2] - combo[p0]; cumulative difference

    # The ceiling decreases by 1 each iteration.
    while True:
        combo = initial_combo(ceiling, target_sum, num_cells)
        yield tuple(combo)

        ceiling_changed = False

        # Generate all of the remaining combos with this ceiling.
        while not ceiling_changed:
            p2, p1, p0 = -2, -1, -1

            while combo[p2] == combo[p1] and abs(p2) <= num_cells:
                # 3,3,3,3
                if abs(p2) == num_cells:
                    return

                p2 -= 1
                p1 -= 1
                p0 -= 1

            cd = 0

            # slide_ptrs_left loop
            while abs(p2) <= num_cells:
                d   = combo[p2] - combo[p1]
                cd += d

                # 5,5,3,3 or 5,5,4,3
                if cd > 1:
                    if abs(p2) < num_cells:
                        # 5,5,3,3 --> 5,4,4,3
                        if d > 1:
                            combo[p2] -= 1
                            combo[p1] += 1
                        # d == 1; 5,5,4,3 --> 5,4,4,4
                        else:
                            combo[p2] -= 1
                            combo[p0] += 1

                        yield tuple(combo)

                    # abs(p2) == num_cells; 5,4,4,3
                    else:
                        ceiling -= 1
                        ceiling_changed = True

                    # Resume at make_combo_same_ceiling while
                    # and follow branch.
                    break

                # 4,3,3,3 or 4,4,3,3
                elif cd == 1:
                    if abs(p2) == num_cells:
                        return

                    p1 -= 1
                    p2 -= 1

if __name__ == '__main__':
    print list(all_combos(ceiling=6, target_sum=12, num_cells=4))