this post was submitted on 12 May 2024
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[–] SorteKanin@feddit.dk 82 points 6 months ago* (last edited 6 months ago) (5 children)

If the solution to a problem is easy to check for correctness, must the problem be easy to solve?

For instance, it is easy to check if a filled sudoku grid is a valid solution. Must it therefore be easy to solve sudokus?

Most people would probably intuitively answer "no", and most computer scientists agree, but this has still not been proven, so we actually don't know.

[–] kromem@lemmy.world 36 points 6 months ago* (last edited 6 months ago) (1 children)

Well, there's counterfactual examples of this, so it must not be true.

In pretty much every single relationship worldwide, one person can very easily determine if the recommendation from the other for where to eat or what to watch is correct or not.

And yet successfully figuring out where to eat or what to watch is nigh impossible.

[–] arthur@lemmy.zip 3 points 6 months ago* (last edited 6 months ago)

I think there's a ~~fighter~~ further* problem, it may be true and we just don't know the easy way to do it.

[–] sunbeam60@lemmy.one 10 points 6 months ago* (last edited 6 months ago)

That’s actually the simplest and clearest description of the P/NP problem I’ve ever read.

[–] azulavoir@sh.itjust.works 8 points 6 months ago

I think my favorite troll statement to a mathematician/comp scientist is:

"Actually, P > NP - there exist problems where it's harder to verify a solution than to arrive at one"

[–] Risus_Nex@lemmy.world 7 points 6 months ago (2 children)

Isn't it proof enough? Using the Sudoku example: there are certainly different levels of difficulties, depending on how many numbers are set in the beginning and other parameters. Checking if the solved answer is correct, is always the same "difficulty" - thus there is no correlation between the difficulty of the puzzle at the beginning and checking the Correctness. Some people might not be able to solve it, but they certainly can check if the solution is right

[–] SorteKanin@feddit.dk 32 points 6 months ago (1 children)

Isn’t it proof enough?

Unfortunately no. The question is a simplification of the P versus NP problem.

The problem lies in having to prove that no method exists that is easy. How do you prove that no matter what method you use to solve the sudoku, it can never be done easily? You'll need to somehow prove that no such method exists, but that is rather hard. In principle, it could be that there is some undiscovered easy way to solve sudokus that we don't know about yet.

I'm using sudokus as an example here, but it could be a generic problem. There's also a certain formalism about what "easy" means but I won't get into it further, it is a rather complicated area.

Interestingly, it involves formal languages a lot, which is funny as you wouldn't think computer science and linguistics have a lot in common, but they do in a lot of ways actually.

[–] Phen@lemmy.eco.br 2 points 6 months ago (2 children)

You can solve any sudoku easily by trying every possible combination and seeing if they are correct. It'll take a long time, but it's fairly easy.

[–] SorteKanin@feddit.dk 20 points 6 months ago

Well it just so happens that the definition of "easy" in the actual problem is essentially "fast". So under that definition, checking every single possible solution is not an "easy" method.

[–] Nomecks@lemmy.ca 10 points 6 months ago* (last edited 6 months ago)

What if the sudoku is 1 milllion lines by 1 million lines? How about a trillion by a trillion? The answer is still easy to check, but it takes exponentially longer to solve the board as the board gets larger. That's the jist of the problem: Is there a universal solution to a problem like this that can solve any size sudoku before the heat death of the universe?

[–] quilan@lemmy.world 15 points 6 months ago* (last edited 6 months ago)

For the purposes of OPs problem (P v NP), it considers not particular solutions, but general algorithmic approaches. Thus, we consider things as either Hard (exponential time, by size of input), or Easy (only polynomial time, by size of input).

A number of important problems fall into this general class of Hard problems: Sudoku, Traveling Salesman, Bin Packing, etc. These all have initial setups where solving them takes exponential time.

On the other hand, as an example of an easy problem, consider sorting a list of numbers. It's really easy to determine if a lost is sorted, and it's always relatively fast/easy to sort the list, no matter what setup it had initially.

[–] DrJenkem@lemmy.blugatch.tube 2 points 6 months ago* (last edited 6 months ago) (1 children)

Most people would probably intuitively answer "no", and most computer scientists agree, but this has still not been proven, so we actually don't know.

I disagree, I think most computer scientists believe that P != NP, at least when it comes to classical computers. If we believed that P = NP, then why would we bother with encryption?

EDIT: nvm, I misread it.

[–] SorteKanin@feddit.dk 5 points 6 months ago* (last edited 6 months ago) (1 children)

I think you've misunderstood πŸ˜…. Answering "no" to that question corresponds to P != NP (there are problems that are easy to verify but not easy to solve), while "yes" means P = NP (if a solution is easy to check, the problem must be easy to solve). So I am saying most people and most scientists believe P != NP exactly as you say.

[–] DrJenkem@lemmy.blugatch.tube 4 points 6 months ago* (last edited 6 months ago)

Reading comprehension is hard my bad.

Edit: wait no, it's "easy" I'm just dumb.