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0 pointscomex5y ago0 comments

puts on pedant’s hat

That wouldn’t actually make P = NP, because P is defined mathematically from first principles rather than being based on the rules of the universe. However, it would shift the class of practically computable problems from being ‘like P’ (but with limits on the runtime and memory usage achievable), to being ‘like P^NP’, P with an NP oracle, which is surprisingly a different complexity class from NP. At least, that’s what I gather based on Wikipedia and Stack Overflow [1]; I’m not an expert on the subject so I could be mistaken.

[1] https://cs.stackexchange.com/questions/2712/does-npnp-np

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qsort5y ago

puts on hat from the second level of the pedant hierarchy

Assuming P!=NP!=co-NP, that computational model is equivalent to either NP or co-NP depending on the multiverse semantics you are using, not P^NP. If you can send the computation through multiple universes and get back a 'YES' or 'NO' answer, depending on whether the 'YES' or 'NO' is quantified existentially or universally on the leaves of the computation tree, you get NP or co-NP. Intuitively speaking, imagine you have a 3-SAT instance; you can send one assignment of truth values to the variables to each parallel universe for computation. If the multiverse answers you 'YES' if there exists one universe that says 'YES', that solves NP problems in P-time. If the multiverse answers you 'YES' if all universes say 'YES', that solves co-NP problems in P time.

jbay8085y ago

Very interesting! Thank you for putting on your pedant hat!

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