Comment by Aunche
10 hours ago
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.