MinekPo1 [She/Her]

nya !!! :3333 gay uwu

I’m in a bad place rn so if I’m getting into an argument please tell me to disconnect for a bit as I dont deal with shit like that well :3

  • 8 Posts
  • 298 Comments
Joined 1 year ago
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Cake day: June 14th, 2023

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  • honestly I was very suspicious that you could get away with only calling the hash function once per permutation , but I couldn’t think how to prove one way or another.

    so I implemented it, first in python for prototyping then in c++ for longer runs… well only half of it, ie iterating over permutations and computing the hash, but not doing anything with it. unfortunately my implementation is O(n²) anyway, unsure if there is a way to optimize it, whatever. code

    as of writing I have results for lists of n ∈ 1 … 13 (13 took 18 minutes, 12 took about 1 minute, cant be bothered to run it for longer) and the number of hashes does not follow n! as your reasoning suggests, but closer to n! ⋅ n.

    desmos graph showing three graphs, labeled #hashes, n factorial and n factorial times n

    link for the desmos page

    anyway with your proposed function it doesn’t seem to be possible to achieve O(n!²) complexity

    also dont be so negative about your own creation. you could write an entire paper about this problem imho and have a problem with your name on it. though I would rather not have to title a paper “complexity of the magic lobster party problem” so yeah


  • unless the problem space includes all possible functions f , function f must itself have a complexity of at least n to use every number from both lists , else we can ignore some elements of either of the lists , therby lowering the complexity below O(n!²)

    if the problem space does include all possible functions f , I feel like it will still be faster complexity wise to find what elements the function is dependant on than to assume it depends on every element , therefore either the problem cannot be solved in O(n!²) or it can be solved quicker










  • to be fair , neither the free software movement nor the open source movement (which are distinct ideologically) are explicitly socialist . in a way , especially the free software movement , they embody an extention of liberalism .

    both of these movements focus on the individuals freedom and take issue not with developers/companies being systemically incentivized to develop closed source / nonfree software , but with individual developers/companies doing so . thus the solution taken is limited to the individual not to systemic change .






  • autistic complaining about units

    ok so like I don’t know if I’ve ever seen a more confusing use of units . at least you haven’t used the p infix instead of the / in bandwith units .

    like you used both upper case and lowercase in units but like I can’t say if it was intentional or not ? especially as the letter that is uppercased should be uppercased ?

    anyway

    1Mb

    is theoretically correct but you likely ment either one megabyte (1 MB) or one megibyte (MiB) rather than one megabit (1 Mb)

    ~325mb/s

    95mb/s

    and

    9mb/s

    I will presume you did not intend to write ~325 milibits per second , but ~325 megabits per seconds , though if you have used the 333 333 request count as in the segment you quoted , though to be fair op also made a mistake I think , the number they gave should be 3 exabits per second (3 Eb/s) or 380 terabytes per seconds (TB/s) , but that’s because they calculated the number of requests you can make from a 1 gigabit (which is what I assume they ment by gbit) wrong , forgetting to account that a byte is 8 bits , you can only make 416 666 of 4 kB (sorry I’m not checking what would happen if they ment kibibytes sorry I underestimated how demanding this would be but I’m to deep in it now so I’m gonna take that cop-out) requests a second , giving 380 terabits per second (380 Tb/s) or 3.04 terabytes per second (3.04 TB/s) , assuming the entire packet is exactly 114 megabytes (114 MB) which is about 108.7 megibytes (108.7 MiB) . so anyway

    packet size theoretical bandwidth
    1 Mb 416.7 Gb/s 52.1 GB/s
    1 MB 3.3 Tb/s 416.7 GB/s
    1 MiB 3.3 Tb/s 416.7 GB/s
    300 kb 125.0 Gb/s 15.6 GB/s
    300 kB 1000.0 Gb/s 125.0 GB/s
    300 kiB 1000.0 Gb/s 125.0 GB/s
    30 kb 12.5 Gb/s 1.6 GB/s
    30 kB 100.0 Gb/s 12.5 GB/s
    30 kiB 100.0 Gb/s 12.5 GB/s

    hope that table is ok and all cause im in a rush yeah bye