Consider the following 3-SAT variant defined over the variables $x_1,\ldots,x_n$. In the $k$P$k$N-3SAT problem each variable $x_j$, $j \in [n]$, occurs exactly $k$ times as a positive literal in $\phi$, and exactly $k$ times as a negative literal in $\phi$, where $\phi$ is a 3-CNF formula. The problem is then to decide if such a formula has a satisfying assignment.

Is the $k$P$k$N-3SAT problem NP-complete?

In the $m$P$n$N-SAT problem each positive literal occurs exactly $m$ times in $\phi$, and each negative literal occurs exactly $m$ times in $\phi$, where $\phi$ is a CNF formula. It was shown in [1] that $2$P$1$N-SAT is NP-complete. This hints that the $k$P$k$N-3SAT problem is hard as well.

The $1$P$1$N-SAT is apparently easy, see a related question and answer here. Is $k$P$k$N-3SAT perhaps hard already for $k \geq 2$?

[1] Yoshinaka, Ryo. "Higher-order matching in the linear lambda calculus in the absence of constants is NP-complete." Term Rewriting and Applications. Springer Berlin Heidelberg, 2005. 235-249.


2 Answers 2


1) 2P2N-3SAT is NP-complete

(Edit to take into account all comments below)

Take a 2P1N-SAT instance, check the all-1 vector is not a model, then add the clause with all variables negated to the formula to transform it into a 2P2N-SAT instance.

For each 1-clause $(a)$ in the new 2P2N formula, replace it by the three following 3-clauses : $(a \lor x \lor x)(\lnot x \lor y \lor y)(\lnot x \lor \lnot y \lor \lnot y)$ where $x,y$ are fresh 2P2N variables (i.e. appearing twice as positive literals and twice as negative literals).
For each 2-clause $(a \lor b)$ in the formula, replace it by the six following 3-clauses : $(a \lor b \lor x)(\lnot x \lor y \lor y)(\lnot x \lor \lnot y \lor \lnot y)(x \lor z \lor \lnot z)(z \lor t \lor \lnot t)(\lnot z \lor t \lor \lnot t)$ where $x,y,z,t$ are fresh 2P2N variables.
Keep all 3-clauses in the formula.
For each $k$-clause ($k>3$) in the formula, transform it into a set of 3-clauses by the usual way, which does not change the number of occurences of positive (negative) literals of the taken $k$-clause, but add only extra variables which appear once as a positive literal and once as a negative literal. For each such new variable $x$, add the two following clauses to the 3-CNF formula : $(x\lor y \lor \lnot y)$ and $(\lnot x \lor \ y\lor \lnot y)$ where $y$ is a fresh 2P2N variable.

All these transformations can obviously be done in polynomial time and transform the 2P1N-SAT instance into a 2P2N-3SAT instance. The NP-complete 2P1N-SAT problem is then reducible to the 2P2N-3SAT problem.

2) The extension to $k$P$k$N-3SAT can be done in the same spirit (a bit longer) to prove it is NP-complete as well.


As confirmed independently in another answer, 2P2N-3SAT is already NP-complete. Another name for the fragment you are interested in is (3, B$k$)-SAT where "B" stands for Balanced and indicates that there are the same number of positive and negative occurences of the variables, $k$ is the number of positive and the number of negative occurences, and the 3 indicates the size of clauses: we are indeed in a fragment of 3SAT.

In particular, 2P2N-3SAT is the same as (3, B2)-SAT and this fragment was proven NP-complete as Theorem 1 in [1].

A related natural question is: given that 2P2N-3SAT is NP-complete, and given that 2P1N-SAT is NP-complete, what about about 2P1N-3SAT? This one turns out to be trivial: all instances are always satisfiable, Theorem 2.4 in [2]. If you allow some clauses to have fewer than 3 variables, then the satisfiability problem is NP-complete as observed in Theorem 2.1 in [2].

[1] Piotr Berman, Marek Karpinski, Alexander D. Scott. Approximation hardness of short symmetric instances of MAX-3SAT, Electronic Colloquium on Computational Complexity, Technical Report TR03-049 (2003)
[2] Tovey, Craig A. A simplified NP-complete satisfiability problem, Discrete applied mathematics 8.1 (1984): 85-89.


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