Comment by demibabs
12 hours ago
No I get that, but I don't get how proving 3SAT can be proved in subexponential time does not imply all NP-complete problems can be as well, since they can all be reduced to 3SAT in polynomial time.
12 hours ago
No I get that, but I don't get how proving 3SAT can be proved in subexponential time does not imply all NP-complete problems can be as well, since they can all be reduced to 3SAT in polynomial time.
Suppose 3SAT can be solved in 2^(n^0.9) time (this is not the actual bound in paper). And another NP problem can be reduced to 3SAT in polynomial time, but the reduction increases the problem size from n to n^2. Then after the reduction the other NP problem can solved in 2^(n^1.8) time, so it does not mean all np problems are sub exponential.
Hey, this is a common error. (2^(n^0.9))^2 is not 2^(n^1.8) but 2^(2 * n^0.9).
Suppose 3sat is solved in 2^sqrt(n). That's subexponential. An n^2 reduction from sat to 3sat is polynomial. However, an algorithm using this reduction with the discovered solution, would run in (2^sqrt(n))^2 which is just 2^n.
(2^sqrt(n))^2 = 2^sqrt(n) * 2^sqrt(n) = 2^(sqrt(n) + sqrt(n)) = 2^(2 sqrt(n))
but
2^(sqrt(n)^2) = 2^n
That last statement is not correct. (2^sqrt(n))^2 equals 2^(2 * sqrt(n)). If you just take n=1 in your expression you can see the error, your (2^sqrt(n))^2 yields 4, while 2^n is just 2.
Yeah but you're fixing the wrong thing. The problem after reduction to 3sat is (2^sqrt(n^2)). The altered n becomes the input to 3sat.
This also means your other comment here is wrong to suggest any error: https://news.ycombinator.com/item?id=50254142
...that's...not how exponents work
The result is right, they wrote the parentheses in the wrong spot, don't be so shocked.
Basically 3SAT algo isn't sub-exponential enough to absorb the arbitrary poly cost for the reduction
So 3SAT+poly reduction ends up being exponential
Gotcha, thanks
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.
Misunderstood your question sorry, I have no idea!