points by wbl 3 days ago | 0 comments
No. NP is always poly time reduceable to 3SAT
[–] mswphd 3 days ago
practically, a poly-time reduction can increase teh size of hte problem instance. So maybe you transform a 3-color instance (graph on n vertices) into a n^2 sized SAT instance. You can then solve this in time 2^{n^2 / sqrt{log{n^2}}}, but this is not sub-exponential.
[–] jibalt 3 days ago
Your "No" is wrong. As he said, "Conversions to other problems can still end up exponential", so (NP is poly time reduceable to 3SAT + 3SAT is sub-exponential) does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
[–] LPisGood 3 days ago
Subexponential(poly(n)) may be exponential in n
[–] dataflow 3 days ago
Did you reduce in the wrong direction or am I misunderstanding what you're trying to say?
[–] cochne 3 days ago
[–] adgjlsfhk1 3 days ago
[–] jibalt 3 days ago
[–] adgjlsfhk1 3 days ago
Also for NP problems that can't linear reduce to kSAT just having the knowledge that SAT is sub-exponential (and the tools), makes it a lot more likely that we see a whole lot more problems become exponential.
[–] calf 3 days ago