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Dec 19, 2010 at 21:37 vote accept Mohammad Al-Turkistany
Dec 13, 2010 at 3:02 comment added Boaz Barak Indeed, for these kinds of reasons Impagliazzo, Paturi and Zane argued that when asking about $2^{\Omega(n)}$ vs $2^{o(n)}$ complexity you need to set $n$ to be the size of the witness (which you need to define as part of the problem). In the $k$-clique case the witness is of size $\log \binom{|V|}{k} \sim k\log |V|$ for small $k$, while, as you say, you can assume w.l.o.g there are at least $k|V|$ edges and the input size is much larger than the witness size.
Dec 8, 2010 at 16:24 history answered Luca Trevisan CC BY-SA 2.5