Day 9: Disk Fragmenter

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3 points
*

PYTHON

Execution Time: Part1 = 0.02 seconds. Part2 = ~2.1 seconds. total = ~2.1 seconds

Aiming for simplicity over speed. This is pretty fast for not employing simple tricks like trees and all that.

code

because of text limit and this code being slow, I put it in a topaz paste: [ link ]

Edit:

New version that is using a dictionary to keep track of the next empty slot that fits the current index.

Execution Time: Part1 = 0.02 seconds. Part2 = ~0.08 seconds. total = ~0.08 seconds 80 ms

code

you can also find this code in the Topaz link: [ link ]

Edit: final revision. I just realized that the calculating for “last_consecutive_full_partition” was not necessary and very slow. if I know all the next available slots, and can end early once my current index dips below all next available slots then the last_consecutive_full_partition will never be reached. This drops the time now to less than ~0.1 seconds

Probably Final Edit: I found someone’s O(n) code for OCaml. I tried to convert it to be faith fully in pure python. seems to work really really fast. 30-50 ms time for most inputs. seems to scale linearly too

FastCode
def int_of_char(x):
    return ord(x) - ord('0')

# Represent content as tuples:
# ('Empty', size) or ('File', id, size)
def parse(line):
    arr = []
    for i in range(len(line)):
        c = int_of_char(line[i])
        if i % 2 == 0:
            arr.append(('File', i // 2, c))
        else:
            arr.append(('Empty', c))
    return arr

def int_sum(low, high):
    return (high - low + 1) * (high + low) // 2

def size(elem):
    t = elem[0]
    if t == 'Empty':
        return elem[1]
    else:
        return elem[2]

def part1(array):
    total = 0
    left = 0
    pos = 0
    right = len(array) - 1

    while left < right:
        if array[left][0] == 'File':
            # File
            _, fid, fsize = array[left]
            total += fid * int_sum(pos, pos + fsize - 1)
            pos += fsize
            left += 1
        else:
            # Empty
            _, esize = array[left]
            if array[right][0] == 'Empty':
                right -= 1
            else:
                # Right is File
                _, fid, fsize = array[right]
                if esize >= fsize:
                    array[left] = ('Empty', esize - fsize)
                    total += fid * int_sum(pos, pos + fsize - 1)
                    pos += fsize
                    right -= 1
                else:
                    array[right] = ('File', fid, fsize - esize)
                    total += fid * int_sum(pos, pos + esize - 1)
                    pos += esize
                    left += 1

    # If one element remains (left == right)
    if left == right and left < len(array):
        if array[left][0] == 'File':
            _, fid, fsize = array[left]
            total += fid * int_sum(pos, pos + fsize - 1)

    return total

def positions(arr):
    total = 0
    res = []
    for e in arr:
        res.append(total)
        total += size(e)
    return res

def array_fold_right_i(f, arr, acc):
    pos = len(arr) - 1
    for elt in reversed(arr):
        acc = f(elt, pos, acc)
        pos -= 1
    return acc

def part2(array):
    def find_empty(size_needed, max_pos, pos):
        while pos <= max_pos:
            if array[pos][0] == 'File':
                raise Exception("Unexpected: only empty at odd positions")
            # Empty
            _, esize = array[pos]
            if esize >= size_needed:
                array[pos] = ('Empty', esize - size_needed)
                return pos
            pos += 2
        return None

    emptys = [1 if i < 10 else None for i in range(10)]
    pos_arr = positions(array)

    def fold_fun(elt, i, total):
        if elt[0] == 'Empty':
            return total
        # File
        _, fid, fsize = elt
        init_pos = emptys[fsize]
        if init_pos is None:
            new_pos = pos_arr[i]
        else:
            opt = find_empty(fsize, i, init_pos)
            if opt is None:
                new_pos = pos_arr[i]
            else:
                new_pos = pos_arr[opt]
                pos_arr[opt] += fsize
                emptys[fsize] = opt
        return total + fid * int_sum(new_pos, new_pos + fsize - 1)

    return array_fold_right_i(fold_fun, array, 0)

def main():
    with open('largest_test', 'r') as f:
        line = f.read().replace('\r', '').replace('\n', '')
    arr = parse(line)
    arr_copy = arr[:]
    p1 = part1(arr_copy)
    print("Part 1 :", p1)
    p2 = part2(arr)
    print("Part 2 :", p2)

if __name__ == "__main__":
    main()

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2 points

So cool, I was very hyped when I managed to squeeze out the last bit of performance, hope you are too. Especially surprised you managed it with python, even without the simple tricks like trees ;)

I wanted to try it myself, can confirm it runs in under 0.1s in performance mode on my laptop, I am amazed though I must admin I don’t understand your newest revision. 🙈

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3 points
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Just to let you know, I posted the fastest python version I could come up with. Which took heavy inspiration from [ link to github ]

supposedly O(n) linear time, and does seem to work really fast.

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2 points
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Thanks! your Haskell solution is extremely fast and I don’t understand your solution, too. 🙈 lol

My latest revision just keeps a dict with lists of known empty slots with the length being the dict key, including partially filled slots. I iteratively find the slot that has the lowest index number and make sure the lists are properly ordered from lowest to highest index number.

looking at the challenge example/description, it shows a first pass only type of “fragmenting”. we can be confident that if something did not fit, it can just stay in the same spot even if another slot frees up enough space for it to fit. so just checking if current index is lower than the lowest index number of any of the slot lengths would just be enough to stop early. That is why I got rid of last_consecutive_full_partition because it was slowing it down by up to 2 seconds.

in example, even if 5555, 6666, or 8888 can fit in the new spot created by moving 44, they are staying put. Thus a first pass only sort from back to front.

00...111...2...333.44.5555.6666.777.888899
0099.111...2...333.44.5555.6666.777.8888..
0099.1117772...333.44.5555.6666.....8888..
0099.111777244.333....5555.6666.....8888..
00992111777.44.333....5555.6666.....8888..
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2 points

Thank you for the detailed explanation!, it made me realize that our solutions are very similar. Instead of keeping a Dict[Int, List[Int]] where the value list is ordered I have a Dict[Int, Tree[Int]] which allows for easy (and fast!) lookup due to the nature of trees. (Also lists in haskell are horrible to mutate)

I also apply the your technique of only processing each file once, instead of calculating the checksum afterwards on the entire list of file blocks I calculate it all the time whenever I process a file. Using some maths I managed to reduce the sum to a constant expression.

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2 points
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I only now found your edit after I had finished my previous comment. I think splitting into two lists may be good: one List of Files and one of Empty Blocks, I think this may not work with your checksumming so maybe not.

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