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Questions related to the (computational) complexity of solving problems

7
votes
Your problem is NP-hard, as can be seen by a (known, I believe) reduction from CNF-SAT: Reduce from (for example) $3$-SAT as follows. Given a formula $F$, create a formula for your special NAE-SAT by …
answered Jun 26 '17 by user53923
2
votes
If such a constant-factor approximation algorithm would exist, and it would run in polynomial-time, you could use it to solve CNF-SAT in polynomial time. Since CNF-SAT is NP-hard, this may give you th …
answered Feb 8 '17 by user53923
0
votes
As you remark, inputs that halt in a rejecting state should also take polynomial time for $M$ to be a polynomial-time DTM. Note that the running time $T_M$ as defined in your question is indeed defi …
answered Dec 7 '17 by user53923
2
votes
Suppose every clause contains all but c variables. I think the problem is in $P$, which I believe can be seem from your example with $c=0$. In your case, every clause "forbids" exactly one satisfying …
answered Mar 16 '18 by user53923
1
vote
This variant of 1-in-3 SAT is polynomial time solvable. I will try to show this using the fact that 2-CNF-SAT is polynomial time solvable: wikipedia 2-SAT. (There may be better ways or more efficient …
answered Nov 29 '17 by user53923
3
votes
To start with possible NP-hardness (where for each problem, we want a matching/independent set of size at least $k$): Independent set is NP-hard on "normal" graphs (and also on hypergraphs) Maximum …
answered Feb 6 '17 by user53923
4
votes
I believe the claim is more commonly written as $coNP \subseteq NP/poly$, but that of course does not immediately answer your question. The fact that this implies that the polynomial hierarchy coll …
answered Jun 19 by user53923