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k-SAT is a special case of SAT. Since SAT is NP-complete, I don't understand why we don't have k-SAT is NP-complete for whatever values of k. In the class, my professor used polynomial reduction from SAT to prove that 3-SAT is NP-complete. I don't understand why we need to prove for that, shouldn't a special case must follow the rule for general case?

old man
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Well, your claim is clearly not true for 1- or 2-SAT.

1-SAT problem (i.e., where you have at most one literal!) is obviously linear in the number of clauses and variables, as the polarity in each clause would tell you how to pick the variables.

2-SAT is trickier, but you can solve it in polynomial time: https://www.geeksforgeeks.org/2-satisfiability-2-sat-problem/

For any k > 3, k-SAT is obviously NP-complete; as any algorithm for those can be used to solve 3-SAT, as your professor seems to have discussed.

So, the fundamental confusion here is that while k-SAT is an instance of the general SAT problem, it does not mean all k-SAT problems are equally difficult. In a sense, 3-SAT is the "simplest" instance of SAT that is NP-complete, and thus forms a good basis to reduce other problems to. Hope that helps!

alias
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  • thanks for pointing out the 1-SAT and 2-SAT cases. That makes sense. But I am still not quite understand the 3-SAT. In our class, we are given that CNF-SAT is NP-complete. To prove that 3-SAT is also NP-complete, my professor reduced CNF-SAT to 3-SAT. That implies that 3-SAT is at least as hard as CNF-SAT. This constrast with my intution that 3-SAT is the simplest case of general SAT. – old man Apr 21 '20 at 06:28
  • Think of "simplest" in that sentence as the "lowest" *k* such that it's NP-complete. So, 4-SAT, 5-SAT etc. are just as hard as 3-SAT, it just happens that 3-SAT is the simplest because it's the most restrictive in that family. Otherwise they are all NP-complete, literally meaning solving one in polynomial time would solve all the others. So in that sense they are all equally simple and/or hard, so long as k>=3. – alias Apr 21 '20 at 17:20