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 <= maxval and x(1) + ... + x(ncells) = targetsum. Having tested several of the more promising answers, I'm going to award the answer-prize to Lennart Regebro, because:

1. his routine is as fast as mine (+-5%), and

2. he pointed out that my original routine had a bug somewhere, which led me to see what it was really trying to do. Thanks, Lennart.

chrispy contributed an algorithm that seems equivalent to Lennart's, but 5 hrs later, sooo, first to the wire gets it.

A remark: Alex Martelli's bare-bones recursive algorithm is an example of making every possible combination and throwing them all at a sieve and seeing which go through the holes. This approach takes 20+ times longer than Lennart's or mine. (Jack up the input to max_val = 100, n_cells = 5, target_sum = 250 and on my box it's 18 secs vs. 8+ mins.) Moral: Not generating every possible combination is good.

Another remark: Lennart's and my routines generate the same answers in the same order. Are they in fact the same algorithm seen from different angles? I don't know.

Something occurs to me. If you sort the answers, starting, say, with (8,8,2,1,1) and ending with (4,4,4,4,4) (what you get with max_val=8, n_cells=5, target_sum=20), the series forms kind of a "slowest descent", with the first ones being "hot" and the last one being "cold" and the greatest possible number of stages in between. Is this related to "informational entropy"? What's the proper metric for looking at it? Is there an algorithm that producs the combinations in descending (or ascending) order of heat? (This one doesn't, as far as I can see, although it's close over short stretches, looking at normalized std. dev.)

Here's the Python routine:

#!/usr/bin/env python
#filename: makeAddCombos.07.py -- stripped for StackOverflow

def initialize_combo( max_val, n_cells, target_sum):
    """returns combo
    Starting from left, fills combo to max_val or an intermediate value from 1 up.  
    E.g.:  Given max_val = 5, n_cells=4, target_sum = 11, creates [5,4,1,1].
    """
    combo = []
    #Put 1 in each cell.
    combo += [1] * n_cells
    need = target_sum - sum(combo)
    #Fill as many cells as possible to max_val.
    n_full_cells = need //(max_val - 1)
    top_up = max_val - 1
    for i in range( n_full_cells): combo[i] += top_up
    need = target_sum - sum(combo)
    # Then add the rest to next item.
    if need > 0:
        combo[n_full_cells] += need
    return combo
#def initialize_combo()

def scrunch_left( combo):
    """returns (new_combo,done)
    done   Boolean; if True, ignore new_combo, all done;
            if Falso, new_combo is valid.

    Starts a new combo list.  Scanning from right to left, looks for first
    element at least 2 greater than right-end element.  
    If one is found, decrements it, then scrunches all available counts on its
    right up against its right-hand side.  Returns the modified combo.
    If none found, (that is, either no step or single step of 1), process
    done.
    """
    new_combo = []
    right_end = combo[-1]
    length = len(combo)
    c_range = range(length-1, -1, -1)
    found_step_gt_1 = False
    for index in c_range:
        value = combo[index]
        if (value - right_end) > 1:
            found_step_gt_1 = True
            break
    if not found_step_gt_1:
        return ( new_combo,True)

    if index > 0:
        new_combo += combo[:index]
    ceil = combo[index] - 1
    new_combo += [ceil]
    new_combo += [1] * ((length - 1) - index)
    need = sum(combo[index:]) - sum(new_combo[index:])
    fill_height = ceil - 1
    ndivf = need // fill_height
    nmodf = need % fill_height
    if ndivf > 0:
        for j in range(index + 1, index + ndivf + 1):
            new_combo[j] += fill_height
    if nmodf > 0:
        new_combo[index + ndivf + 1] += nmodf
    return (new_combo, False)
#def scrunch_left()

def make_combos_n_cells_ge_two( combos, max_val, n_cells, target_sum):
    """
    Build combos, list of tuples of 2 or more addends.
    """
    combo = initialize_combo( max_val, n_cells, target_sum)
    combos.append( tuple( combo))
    while True:
        (combo, done) = scrunch_left( combo)
        if done:
            break
        else:
            combos.append( tuple( combo))
    return combos
#def make_combos_n_cells_ge_two()

if __name__ == '__main__':

    combos = []
    max_val     = 8
    n_cells     = 5
    target_sum  = 20
    if n_cells == 1: combos.append( (target_sum,))
    else:
        combos = make_combos_n_cells_ge_two( combos, max_val, n_cells, target_sum)
    import pprint
    pprint.pprint( combos)

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 = <list> + <list> 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))