Comment by imadr
11 hours ago
P=NP -> 3SAT has a polynomial time algorithm
P=/=NP -> 3SAT has no polynomial-time algorithm
ETH is true -> 3SAT has no subexponential time algorithm
ETH is false -> 3SAT has a subexponential time algorithm
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ETH true -> P=/=NP
ETH false (!!!) -> P ?= NP remain unsolved
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).
2 replies →
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.
1 reply →
...that's...not how exponents work
1 reply →
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!