Comment by csheehan10
9 hours ago
To achieve this you would need to show that 3SAT is solvable in 2^{n^c} time for every c>0, not just a single c<1.
In this case they have c = 1-0.5(\log \log n)/(\log n) which is approaching 1 for large n.
9 hours ago
To achieve this you would need to show that 3SAT is solvable in 2^{n^c} time for every c>0, not just a single c<1.
In this case they have c = 1-0.5(\log \log n)/(\log n) which is approaching 1 for large n.
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