4 ms·
I think you misunderstood something. The problem itself is sometimes referred to as the "P=NP" problem, but that's just a name, it is also called the "P vs. NP"
by bonoboTP 7y ago
I think you misunderstood something. The problem itself is sometimes referred to as the "P=NP" problem, but that's just a name, it is also called the "P vs. NP" problem. If you can prove either P!=NP or P==NP, you get the prize.
Can you give a reference to back up your interpretation?
- majos 7y agoYour interpretation is correct. The official problem statement [1] is "does P = NP?", so either "P = NP" or "P != NP" is a valid answer. [1] https://www.claymath.org/sites/default/files/pvsnp.pdf https://www.claymath.org/sites/default/files/pvsnp.pdf
- klyrs 7y ago> Can you give a reference to back up your interpretation? Alas, no. I do recall being astonished at my claim, and then being convinced by a colleague (which was backed up by plain language on the CMI page, in my memory...) but now that I'm re-reading (current and archive.org'd) I cannot find such a thing. Disturbing. Yet relieving. Fuck my memory.