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Recall some terminology: Let $\mathsf P$ be a finite set of propositional atoms, and let $\Phi$ be a proposition over $P$ that is generated from $\top$, $\bot$, $\neg$, $\wedge$, and $\vee$. Then:

  • A satisfying assignment for $\Phi$ is a function $\varsigma:\mathsf P\to\{\bot,\top\}$ such that $\varsigma\vDash\Phi$ holds with the usual rules for propositional validity. Intuitively: we replace every $p\in \mathsf P$ with $\varsigma(p)$ in $\Phi$, and reduce using truth-tables.
  • Write AllSAT for the computational problem of, given $\Phi$, finding all of its satisfying assignments.

Question: What is known about AllSAT in the special case that $\Phi$ encodes a set of clauses, one for each $p\in \mathsf P$, each having the form

"$p$ is true if at most $n_p\geq 0$ of the elements from $W_p\subseteq\mathsf P$ are false"?

I am particularly interested in references for:

  1. Mathematical characterisations of the solutions.
  2. Algorithms for computing these solutions.
  3. The complexity of these algorithms.

Note: We usually further constrain $\Phi$ such that $n_p < \#W_p /2$ for every $p$.

Example: Take $\mathsf P=\{a, b, c\}$ and take $$ \Phi = ((b \vee c) \rightarrow a) \wedge ((c \wedge a \wedge b) \rightarrow b) \wedge ((a\vee b\vee c) \rightarrow c) . $$ This $\Phi$ encodes three clauses:

  1. $a$ is true if at most one element from $\{b,c\}$ is false; so $n_a=1$ and $W_a=\{b,c\}$.
  2. $b$ is true if at most zero elements from $\{a,b,c\}$ are false; so $n_b=0$ and $W_b=\{a,b,c\}$.
  3. $c$ is true if at most two elements from $\{a,b,c\}$ are false; so $n_c=2$ and $W_c=\{a,b,c\}$.

Thank you.

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  • $\begingroup$ Circuits of the form $(a\land b)\rightarrow c$ are encoding 3 SAT clauses $(\overline a\lor\overline b\lor c)$. $\endgroup$
    – rus9384
    Commented Dec 11, 2023 at 20:11
  • $\begingroup$ You can check NP-hardness (of satisfiability) using Schaefer's dichotomy theorem. $\endgroup$ Commented Dec 11, 2023 at 20:53

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