4 ms·
If this is true (and the proof has no flaw), it would be huge and pleasant (many expect it, although there are some who think it's a problem that might be unpro
by zeptomu 9y ago
If this is true (and the proof has no flaw), it would be huge and pleasant (many expect it, although there are some who think it's a problem that might be unprovable in our current logic framework). So let's see how it works out.
However the author missed that there is a special case for N == 1, where actually P == NP (sorry, I could not resist).
- naturalgradient 9y agoDon't forget the case where P == 0
- dom0 9y agoYou're both wrong. P is the union of all TIME complexity classes defined by n^k for every natural k. NP is the union of all NTIME complexity classes defined by n^k for every natural k.