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3 votes
0 answers
255 views

Is there a version of 3-SAT that is NP-complete but grows like $2^n$ instead of $2^{n \choose 3}$?

If I have $n$ variables and I want to write down all 3-SAT problems, the number of problems is $2^{8{n \choose 3}}$, since each clause has 3 variables and each variable can be negated or not. But ...
3 votes
0 answers
125 views

Positive boolean satisfiability problem : finding minimal solutions

Consider, over a finite set of boolean variables $X$, a Boolean system in CNF (conjunctive normal form) whose clauses only contain non-negated literals. For every assignment of the variables which ...
-1 votes
1 answer
445 views

What is wrong with the argument that zero permanent is polynomial?

This Lecture summarizes some well known facts about $\#P$ completeness of permanent. Given a CNF formula $\phi$ on $n$ variables, they construct matrix $A$ such that: $$perm(A)=4^{3m} \#SAT(\phi)$$ ...
4 votes
1 answer
362 views

Lower bound on the number of solutions of 2SAT

To compute the number of solutions of a 2SAT is a hard problem. Is there some nontrivial lower or upper bound on this number in terms of a “coarse-grained” description of the Boolean formula, for ...
16 votes
4 answers
1k views

Representing mathematical statements as SAT instances

The following problem (call it THEOREMS) belongs to class NP. Input: Mathematical statement $S$ (written in some formal system such as ZFC) and positive integer $n$ written in unary. Output: "Yes" if ...
18 votes
7 answers
3k views

SAT and Arithmetic Geometry

This is an agglomeration of several questions, linked by a single observation: SAT is equivalent to determining the existence of roots for a system of polynomial equations over $\mathbb{F}_2$ (note ...
8 votes
0 answers
237 views

Size of 3-SAT assignments

Let $F(N,M)$ be the set of 3-SAT formula with $N$ variables and $M$ clauses. For a given formula $f\in F(N,M)$, we can ask for the set $s_f$ of truth assignments that satisfy $f$. (If $f$ is ...
3 votes
1 answer
708 views

Is strictly harder than NP-hard cryptography possible?

Looks like there is cryptography based on NP-hard problem, e.g. McEliece cryptosystem. The algorithm is an asymmetric encryption algorithm and is based on the hardness of decoding a general linear ...
1 vote
0 answers
36 views

Is there an algorithm for this constrained Hypergraph optimization problem?

I'm currently developing an algorithm for computing knot coloring invariants and got to the following question: Given a set $S$ and a certain hyper-graph $H \subseteq S^3 $, find a decomposition $S = ...
2 votes
0 answers
66 views

$3$SAT generation with prescribed number of solutions

Given $n,k\in\Bbb N$ with $0\leq k \leq 2^n$ can we generate an uniformly random instance among all possible solutions of an $n$ variable $3$-SAT instance and exactly $k$ solutions in $poly(n\log k)$ ...
2 votes
0 answers
192 views

relationship of max-sat and min-cut in theory and practice [closed]

I have been using MAX-SAT solver to obtain the exact ground state of ising spin glass model: For 1D periodic model, for systems with 50 binary variables and interaction range of 15th nearest ...
6 votes
0 answers
154 views

Complexity of finding three perfect matchings with no edge in common in a bridgeless cubic graph

According to a conjecture: Conjecture (Fan & Raspaud, 1994) Every bridgeless cubic graph contains three perfect matchings with no edge in common. Equivalent statement here Main question: ...
10 votes
1 answer
675 views

How hard is Heyting satisfiability, i.e. the constructive version of SAT? In particular, is 2-HSAT NL-complete or is it harder?

First of all, is it clear what I mean by $k$-HSAT? I'm assuming that for $k>2$, $k$-HSAT is NP-complete, but the details of the reductions between $k$-HSAT and $k$-SAT aren't obvious to me. I'm ...
1 vote
0 answers
89 views

Deciding / Approximating Parity of Small Depth Decision Trees

Let C be a circuit such that: C: $\{0,1\}^n$ to $\{0,1\}$ the top most gate is a parity gate all the inputs to the parity gate are small depth decision trees there is a total of $2^{ log^k n}$ ...