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What happens to cryptography if P NP?
It’s not an encryption algorithm, but indistinguishability obfuscation exists if P=NP. In general, modern cryptography does not exist if P=NP. Russell Impagliazzo wrote a paper which meditated on the implications of P=NP and other fundamental questions in complexity by positing some possible “worlds” we might live in.
What would happen if we solved P NP?
If P=NP, then all of the NP problems can be solved deterministically in Polynomial time. This is because the NP problems are all essentially the same problem, just stated in different terms.
Does P NP break cryptography?
These information-theory-secure cryptosystems cannot be broken regardless of how much computer power is applied, so a proof of P = NP would not affect their security. It doesn’t.
What happens to NP-hard if P NP?
No. A problem is Np-Hard if all NP problems are reducible to an instance of that problem in polynomial time. Some NP-Hard problems cannot be solved in nondeterministic polynomial time, and are not in NP. Then these problems will not be polynomial time solvable regardless of whether or not P=NP.
Is Bitcoin NP hard?
As it turns out, this requires solving two optimization problems, both of which are NP-hard! …
What happens if P != NP?
If P equals NP, every NP problem would contain a hidden shortcut, allowing computers to quickly find perfect solutions to them. But if P does not equal NP, then no such shortcuts exist, and computers’ problem-solving powers will remain fundamentally and permanently limited.
Is Bitcoin NP-hard?
Is there proof that P = NP in cryptography?
Short version, as I understand it: public key cryptography assumes that, for all practical purposes, P≠NP. Proof that P=NP would lead to the field’s inevitable demise, since it would eventually make computing someone’s private key, given their public one, doable in a reasonable amount of time.
Is the NP problem insolvable in P time?
Most asymmetric keys rely on NP problems being insolvable in P time, it’s just never been proven that this set of problems don’t have P time solutions. So basically if the proof holds it removes the sneaking suspicion that some bright sparck can find a way of factoring primes quickly and breaking all our odds. 66• August 9, 2010 3:38 PM
Is the proof that factoring is NP-complete true?
Actually, that is not true. The proof is based on the idea that SAT (hence NP-complete problems) is hard, and therefore NP != P. However, factoring was never proved to be NP-complete. The proof would only imply the impossibility of easy factoring IF factoring is NP-Complete, which no one knows.