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- What is the $3$-SAT problem? - Mathematics Stack Exchange
3SAT is the case where each clause has exactly 3 terms EDIT (to include some information on the point of studying 3SAT): If someone gives you an assignment of values to the variables, it is very easy to check to see whether that assignment makes all the clauses true; in other words, you can efficiently check any alleged solution
- np complete - How do I reduce 3-SAT to a 3-SAT NAE problem . . .
For the pedantic's sake, we first have a polynomial reduction $3SAT \leq_p s3SAT$, where the later has strictly 3 terms per clause, not less (as accepted by the former) This is achieved by taking an instance of $3SAT$ and mapping the clauses respectively:
- algorithms - How exactly does a Max 2 Sat reduce to a 3 Sat . . .
Note: I've also asked this question on StackOverflow here I've been reading this article which tries and explains how the max 2 sat problem is essentially a 3-sat problem and is NP-hard
- computer science - Can a literal be repeated in SAT or 3SAT . . .
$\begingroup$ The convention may differ In the foundational paper of Karp, 3SAT is at most 3 literal per clause, but as Wikipedia note it, "Some authors restrict k-SAT to CNF formulas with exactly k literals ", and it shows how the two problems are polynomial time equivalent
- discrete mathematics - How can I formulate the 3-SAT problem as a 0-1 . . .
Using this translation strategy, you can add a new linear constraint to the ILP for every clause in the 3SAT problem The two problems are now equivalent: there's an integer solution to this ILP if and only if there's a boolean solution to the original 3SAT problem An example instance of a 3SAT decision problem:
- Reduction from 3SAT to NAE3SAT - Mathematics Stack Exchange
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- how do you prove that 3-SAT is NP-complete?
3-SAT is NP-Complete because SAT is - any SAT formula can be rewritten as a conjunctive statement of literal clauses with 3 literals, and the satisifiability of the new statement will be identical to that of the original formula
- Why doesnt implication graph work for 3SAT as it does for 2SAT?
I tried google and this site, but I couldn't find anything which explains why this method to solve $2SAT$ in polynomial time won't work for $3SAT$ I know it can't, otherwise $3SAT$ wouldn't be NP-Complete, but I'd like to understand why
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