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4

Your problem is known as Weighted-2-satisfiability, and is known to be NP-complete. The easiest way to see that is by reduction from Vertex Cover (exercise). Note that the link above is about whether there is a satisfying assignment having at most $k$ true variables. This is equivalent to your problem (exercise).


3

Given a CNF with $n$ variables and $m$ clauses, we denote by $A_i$ the set of truth assignments that do not satisfy clause $i$. Then the number of satisfiable assignments is $$2^n-\left|\bigcup_{i=1}^m A_i\right|.$$ Since each assignment makes at most one clause false, $A_i$ and $A_j$ are disjoint for $i\neq j$. Also note $|A_i|=2^{n-n_i}$ where $n_i$ is ...


3

It is NP-hard. Given an instance of your problem, the sum of the integers in the optimal subset $N'$ is at least $B$ (which implies that it must actually be exactly $B$) if and only if the corresponding subset sum instance has answer "yes".


2

As j_random_hacker suggested, we can reduce MAX-2-SAT to this problem. Given an instance of MAX-2-SAT with $n$ variables $x_1,\ldots,x_n$ and $m$ clauses, we can encode it as the following constraints: For each variable $x_i$, add $m$ identical constraints: $x_i\le 0$ (if identical constraints are not allowed, we can use $x_i\le 0,x_i\le 0.1,x_i\le 0.01,\...


2

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 ...


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