April 16, 2025 The Halting Problem is a terrible example of NP-Harder It's a justifiable copout, but it's still a copout. Short one this time because I have a lot going on this week. In computation complexity, NP is the class of all decision problems (yes/no) where a potential proof (or "witness") for "yes" can be verified in polynomial time. For example, "does this set of numbers have a subset that sums to zero" is in NP. If the answer is "yes", you can prove it by presenting a set of numbers. We would then verify the witness by 1) checking that all the numbers are present in the set (~linear time) and 2) adding up all the numbers (also linear). NP-complete is the class of "hardest possible" NP problems. Subset sum is NP-complete. NP-hard is the set all problems at least as hard as NP-complete. Notably, NP-hard is not a subset of NP, as it contains problems that are harder than NP-complete. A natural question to ask is "like what?" And the canonical example of "NP-harder" is the halting problem (HALT): does program P halt on input C? As the argument goes, it's undecidable, so obviously not in NP. I think this is a bad example for two reasons: All NP requires is that witnesses for "yes" can be verified in polynomial time. It does not require anything for the "no" case! And even though HP is undecidable, there is a decidable way to verify a "yes": let the witness be "it halts in N steps", then run the program for that many steps and see if it halted by then. To prove HALT is not in NP, you have to show that this verification process grows faster than polynomially. It does (as busy beaver is uncomputable), but this all makes the example needlessly confusing. "What's bigger than a dog? THE MOON" Really (2) bothers me a lot more than (1) because it's just so inelegant. It suggests that NP-complete is the upper bound of "solvable" problems, and after that you're in full-on undecidability. I'd rather show intuitive problems that are harder than NP but not that much harder...
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