this post was submitted on 22 Jun 2024
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Not everything can be done in constant time, that's O(k) (sh.itjust.works)
submitted 6 months ago by [email protected] to c/[email protected]
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[–] [email protected] 4 points 6 months ago (1 children)

By “certain distance function”, I mean a specific function that forces the problem to be O(n!^2).

But fear not! I have an idea of such function.

So the idea of such function is the hamming distance of a hash (like sha256). The hash is computed iterably by h[n] = hash(concat(h[n - 1], l[n])).

This ensures:

  • We can save partial results of the hashing, so we don’t need to iterate through the entire lists for each permutation. Otherwise we would get another factor n in time complexity.
  • The cost of computing the hamming distance is O(1).
  • Order matters. We can’t cheat and come up with a way to exclude some permutations.

No idea of the practical use of such algorithm. Probably completely useless.

[–] [email protected] 2 points 6 months ago (3 children)

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

[–] [email protected] 2 points 6 months ago* (last edited 6 months ago) (1 children)

So in your code you do the following for each permutation:

for (int i = 0; i<n;i++) {

You’re iterating through the entire list for each permutation, which yields an O(n x n!) time complexity. My idea was an attempt to avoid that extra factor n.

I’m not sure how std implements permutations, but the way I want them is:

1 2 3 4 5

1 2 3 5 4

1 2 4 3 5

1 2 4 5 3

1 2 5 3 4

1 2 5 4 3

1 3 2 4 5

etc.

Note that the last 2 numbers change every iteration, third last number change every 2 iterations, fourth last iteration change every 2 x 3 iterations. The first number in this example change every 2 x 3 x 4 iterations.

This gives us an idea how often we need to calculate how often each hash need to be updated. We don’t need to calculate the hash for 1 2 3 between the first and second iteration for example.

The first hash will be updated 5 times. Second hash 5 x 4 times. Third 5 x 4 x 3 times. Fourth 5 x 4 x 3 x 2 times. Fifth 5 x 4 x 3 x 2 x 1 times.

So the time complexity should be the number of times we need to calculate the hash function, which is O(n + n (n - 1) + n (n - 1) (n - 2) + … + n!) = O(n!) times.

EDIT: on a second afterthought, I’m not sure this is a legal simplification. It might be the case that it’s actually O(n x n!), as there are n growing number of terms. But in that case shouldn’t all permutation algorithms be O(n x n!)?

EDIT 2: found this link https://stackoverflow.com/a/39126141

The time complexity can be simplified as O(2.71828 x n!), which makes it O(n!), so it’s a legal simplification! (Although I thought wrong, but I arrived to the correct conclusion)

END EDIT.

We do the same for the second list (for each permission), which makes it O(n!^2).

Finally we do the hamming distance, but this is done between constant length hashes, so it’s going to be constant time O(1) in this context.

Maybe I can try my own implementation once I have access to a proper computer.

[–] [email protected] 1 points 6 months ago (1 children)

you forgot about updating the hashes of items after items which were modified , so while it could be slightly faster than O((n!×n)²) , not by much as my data shows .

in other words , every time you update the first hash you also need to update all the hashes after it , etcetera

so the complexity is O(n×n + n×(n-1)×(n-1)+...+n!×1) , though I dont know how to simplify that

[–] [email protected] 2 points 6 months ago* (last edited 6 months ago) (1 children)

My implementation: https://pastebin.com/3PskMZqz

Results at bottom of file.

I’m taking into account that when I update a hash, all the hashes to the right of it should also be updated.

Number of hashes is about 2.71828 x n! as predicted. The time seems to be proportional to n! as well (n = 12 is about 12 times slower than n = 11, which in turn is about 11 times slower than n = 10).

Interestingly this program turned out to be a fun and inefficient way of calculating the digits of e.

[–] [email protected] 2 points 6 months ago (1 children)

Agh I made a mistake in my code:

if (recalc || numbers[i] != (hashstate[i] & 0xffffffff)) {
	hashstate[i] = hasher.hash(((uint64_t)p << 32) | numbers[i]);
}

Since I decided to pack the hashes and previous number values into a single array and then forgot to actually properly format the values, the hash counts generated by my code were nonsense. Not sure why I did that honestly.

Also, my data analysis was trash, since even with the correct data, which as you noted is in a lineal correlation with n!, my reasoning suggests that its growing faster than it is.

Here is a plot of the incorrect ratios compared to the correct ones, which is the proper analysis and also clearly shows something is wrong.

Desmos graph showing two data sets, one growing linearly labeled incorrect and one converging to e labeled #hashes

Anyway, and this is totally unrelated to me losing an internet argument and not coping well with that, I optimized my solution a lot and turns out its actually faster to only preform the check you are doing once or twice and narrow it down from there. The checks I'm doing are for the last two elements and the midpoint (though I tried moving that about with seemingly no effect ???) with the end check going to a branch without a loop. I'm not exactly sure why, despite the hour or two I spent profiling, though my guess is that it has something to do with caching?

Also FYI I compared performance with -O3 and after modifying your implementation to use sdbm and to actually use the previous hash instead of the previous value (plus misc changes, see patch).

[–] [email protected] 2 points 6 months ago

It has been a pleasure having this internet argument with you. I learned a bit, and you learned a bit. It’s a win win :)

[–] [email protected] 2 points 6 months ago

Good effort of actually implementing it. I was pretty confident my solution is correct, but I’m not as confident anymore. I will think about it for a bit more.

[–] [email protected] 1 points 6 months ago* (last edited 6 months ago)

actually all of my effort was wasted since calculating the hamming distance between two lists of n hashes has a complexity of O(n) not O(1) agh

I realized this right after walking away from my devices from this to eat something :(

edit : you can calculate the hamming distance one element at a time just after rehashing that element so nevermind