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I am working on exact algorithms for an NP-hard problem $P$. I was able to get a $(1.75^n$) time algorithm for split graphs. When it comes to bipartite graphs, the problem becomes hard to tackle. Now, I want to prove that there is no sub-exponential algorithm for the problem on bipartite graphs. So to prove this, I have to provide a reduction from 3-SAT, which doesn't admit a $2^{o(n)}$ algorithm. I have reduced 3-SAT ($m$ clauses, $n$ varibales) to $P$ ($G'$ with $n'$ vertices) such that
(1) the reduction is done polynomial time.
(2) the size of the reduced instance $n'$ is $m+n^3$.
(3) 3-SAT is satisfiable if and only if $G'$ satisfy the necessary condition for the problem $P$
(4) $G'$ is bipartite
Now, can i say that the problem $P$ does not admit a $2^{o(n)}$ time algorithm for bipartite graphs? My main question is: does the reduction has anything to do with the size of the reduced instance, $n'$.

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Yes, obviously the lower bounds you get depend on the size of the reduction. Second thing is, you have no lower bounds as we don't know if P = NP. You have a conditional lower bound.

Under ETH, your problem cannot be solved in time $2^{o(\sqrt[3]{n})}$.

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