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.

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stbede 5 hours ago | parent | next [-]

(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

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raegis 5 hours ago | parent | prev | next [-]

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.

	▲	Dylan16807 an hour ago \| parent [-]
		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

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jballanc 6 hours ago | parent | prev [-]

...that's...not how exponents work

	▲	Dylan16807 an hour ago \| parent [-]
		The result is right, they wrote the parentheses in the wrong spot, don't be so shocked.