Day 9: Disk Fragmenter

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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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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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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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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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yeah, I was a bit exhausted thinking in a high level abstract way. I do think that if I do the checksum at the same time I could shave off a few more milliseconds. though it is at like the limits of speed, especially for python with limited data types(no trees lol). Decently fast enough for me :)

edit: I also just tested it and splitting into two lists gave no decent speed up and made it slower. really iterating backwards is fast with that list slice. I can’t think of another way to speed it up past it can do rn

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