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I like GPT-4's explanation of P vs NP. Prompt: "Give me good computer science arguments for why P = NP and why P != NP" Response: "The P vs. NP problem is a
by jdlyga 4y ago
I like GPT-4's explanation of P vs NP.
Prompt:
"Give me good computer science arguments for why P = NP and why P != NP"
Response:
"The P vs. NP problem is a major open question in computer science, which asks whether problems whose solutions can be quickly verified by a computer can also be quickly solved by a computer. Both sides of the argument have been extensively debated, but the problem remains unsolved. Here are some arguments for both possibilities:
Arguments for P = NP:
Lack of counterexamples: Despite extensive research, no concrete counterexample has been found to definitively prove that P != NP. This lack of evidence may suggest that P = NP could still be true.
Potential algorithmic breakthroughs: If P = NP, this implies that a polynomial-time algorithm exists for every problem in NP, even though such algorithms may not be known yet. There is always the possibility that new algorithmic techniques or breakthroughs could be discovered, leading to polynomial-time solutions for NP problems.
Simplicity and symmetry: From a theoretical perspective, some argue that P = NP is a more elegant and symmetric outcome, as it would mean all problems that can be verified in polynomial time can also be solved in polynomial time.
Arguments for P != NP:
Exponential lower bounds: Some computational problems have proven exponential lower bounds, indicating that these problems are fundamentally difficult to solve. While these problems may not be in NP, they provide evidence that not all problems can be solved efficiently.
The lack of efficient algorithms: Despite extensive research and many attempts, no efficient (i.e., polynomial-time) algorithms have been found for any of the NP-complete problems, which are among the hardest problems in NP. This suggests that P and NP might indeed be separate complexity classes.
The existence of NP-complete problems: NP-complete problems are those that, if any one of them can be solved in polynomial time, then all NP problems can be solved in polynomial time. They are considered the hardest problems in NP. The existence of such problems, and the lack of polynomial-time algorithms for them, supports the belief that P != NP.
Cryptography: Modern cryptography relies on the assumption that certain problems are hard to solve, such as factoring large numbers. If P = NP, then these cryptographic schemes would become insecure, as the difficult problems on which they are based could be solved efficiently. The widespread use and success of cryptography provide some evidence that P != NP.
It is important to note that these arguments are speculative, and no definitive proof has been found for either side of the debate. The P vs. NP problem remains a significant open question in computer science."
- CaptainNegative 4y agoIt's kind of disappointing that GPT-4 defines NP-completeness in terms of Turing/Cook reductions while Karp reductions have been the preferred definition for the last half century or so. For example, under GPT-4's definition there is no distinction between NP-Completeness and Co-NP-Completeness. I guess that speaks to the volumes of laymen explanations out there.
- danbruc 4y agoWhat in the response indicates Turing or Cook reductions? I would say the answer does not go into nearly enough technical details to conclude anything like that, it does not even mention [polynomial time] reductions. Even among the people seeking some understanding of the P vs NP problem this is a level of technical detail essentially relevant to almost none of them. I studied computer science and if we ever discussed the distinction between different kinds of polynomial time reductions, then this got erased from my memory long ago. And if you really need this level of technical detail, then you can probably ask for it.
- CaptainNegative 4y ago> NP-complete problems are those that, if any one of them can be solved in polynomial time, then all NP problems can be solved in polynomial time. That is (nearly) correct under Cook reductions but not Karp reductions. I say "nearly" because "if language L can be solved (decided) in polynomial time, then so can any problem in NP" is a vacuously true statement for any L not in P, so by this definition nearly all languages (a measure 1 fraction of them) are also NP-complete despite most of them not being in any particularly interesting complexity class. If we escape this degeneracy by replacing the antecedent with "given access to an efficient way to decide L", then we've recreated NP hardness under Cook reductions.
- danbruc 4y agoIf I understand this correctly as I mostly read this just today, then Cook reductions - which is the name for polynomial Turing reductions - are the more general class whereas Karp reductions - which are polynomial many-one reductions - are a more restricted class which - to a first approximation - only allows one oracle query and must return the result unmodified. So if I am concerned with what can be done in polynomial time, then why would I limit myself to Karp reductions? If I want to finely separate classes of problems, then Karp reductions are probably better suited as the reduction builds a much more direct link between the two problems. That is (nearly) correct under Cook reductions but not Karp reductions. It is however still not clear to me why this is not true under Karp reductions, because there are not necessarily Karp reductions for some problems that have Cook reductions? If so, are there two different NP-complete classes, one per reduction?