this post was submitted on 09 Dec 2024
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Advent Of Code

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Day 9: Disk Fragmenter

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  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

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[โ€“] Acters 3 points 4 weeks ago* (last edited 4 weeks ago) (12 children)

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.

codebecause 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

codeyou 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

:::spoiler 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()

[โ€“] VegOwOtenks 2 points 4 weeks ago (11 children)

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. ๐Ÿ™ˆ

[โ€“] Acters 2 points 4 weeks ago* (last edited 4 weeks ago) (9 children)

Thanks! your Haskell solution is extremely fast and I don't understand your solution, too. ๐Ÿ™ˆ 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..
[โ€“] VegOwOtenks 2 points 4 weeks ago* (last edited 4 weeks ago)

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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