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Dec 29, 2012 at 9:15 vote accept Numerator
Jul 21, 2012 at 19:05 comment added Xodarap @Numerator: you are doing $2^{\log n}=n$ checks, where $n$ is the length of the input.
Jul 12, 2012 at 11:37 comment added Luke Mathieson Remember that to be in P you want an algorithm that runs in polynomial time in the size of the input. In this case, if we denote the size of the input as N we know that $\text{#clauses } \leq N$. Hence we also have $n=O(log N)$ so the $2^n$ assignments only amount to a polynomial in the overall input size $N$. Don't let texts trick you when they use the variable $n$, it's just a variable, not a special magic number that is always the best measure for the size of the input. Sorry about the formatting, I'm typing this on my phone.
Jul 12, 2012 at 10:09 comment added Numerator But a $2^n$ checks is still defined as plynoimal time?
Jul 12, 2012 at 9:04 history answered Luke Mathieson CC BY-SA 3.0