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I am studying for NP problems.
To prove k-CNF-SAT is NP-hard, there must exists something that can be reduced to k-CNF-SAT. So what I thought is to reduce 3-CNF-SAT to k-CNF-SAT and reduce k-CNF-SAT to 3-CNF-SAT both proves that it is NP-hard.

I know that 3-CNF-SAT is NP-Complete, because of its number of literals, but this property seems dedicate no effect to proof.
Thanks for any suggestion.

Note: k-CNF-SAT is a conjunctive normal form which k > 3.
And my question is same as the title, I have no idea about this question, since I just only know 3-SAT is a special case of k-SAT.

Raphael
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proX
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2 Answers2

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To reduce $k$-CNF to CNF: every instance of $k$-CNF is an instance of CNF, so the reduction is trivial.

To reduce CNF to $k$-CNF, where $k \ge 3$: convert the CNF formula to a circuit that computes it; then convert the circuit to a 3-CNF formula using the Tseitin transformation. (If your definition of $k$-CNF requires that there be exactly $k$ literals in every clause, then you can pad: introduce $k$ new variables $y_1,\dots,y_k$; for every clause that has less than $k$ literals, add $\lor y_1 \lor y_2 \lor \dots$ to pad it out to exactly $k$ literals; and then add clauses that force all of $y_1,\dots,y_k$ to be false.)

If this is too terse, see our reference question and a good textbook to learn the fundamentals of reductions.

Community
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D.W.
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A clause in 3-CNF can be converted to k-CNF by adding extra padding:

(l1 V l2 V l3) can be converted to (l1 V l2 V l3 V y) ∧ (l1 V l2 V l3 V y')

Keep adding this extra padding until each clause contains k literals.

Same way, a k-CNF clause can be broken until each clause contains 3 literals.

(l1 V l2 V l3 V l4) can be broken into (l1 V l2 V y) ∧ (l3 V l4 V y') by resolution.

Ridhi Jain
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