Apr 21, 2022 · Given that 3SAT is $\textsf{NP}$-complete, I think showing that 2SAT is in $\textsf{P}$ on its own does little to help build intuition as to how far apart these problems are. The $\textsf{NL}$-completeness result is nice, since we now have some intuitive contrast between $\textsf{NP}$ and $\textsf{NL}$: namely, 3SAT vs. 2SAT.

I have read this paper and the wikipedia summary here and even this homework problem, but I do not understand how 2SAT is in P. Here is my concern: the idea is that we start by setting a random variable. Then we set any implied variables. Repeat, and if we eventually reach a contradiction, then the problem has no solution.

Jan 17, 2022 · This procedure is actually just a Fourier-Motzkin elimination, in the context of boolean algebra. You can also apply it for more general SAT formulas (consider this as an exercise). But the 2SAT formulas behave well for it, because each step transforms a 2SAT formula into another 2SAT formula with one less variable.

I would also like to point out that NL=(!=)P doesn't concern polynomial time reductions since we wouldn't gain anything by that. 2SAT is linear time solvable, so we need an at most logspace reduction for it to mean anything useful. $\endgroup$

But this doesn't really matter.). In any case, a cheap way to reduce HORNSAT to 2SAT is to solve your instance of HORNSAT in polynomial time, i.e., to check satisfiability, and then return a trivially satisfiable instance of 2SAT or a trivially not satisfiable instance of 2SAT.

Oct 13, 2015 · Finding all truth assignment to 2SAT. Ask Question Asked 9 years, 4 months ago. Modified 2 years, 6 months ...

Mar 10, 2018 · In $2SAT$ this works for each clause to create implications that solve the clause. In $3SAT$ a similar implication is not possible. At first it appears that we can create an implication for $(a \lor b \lor c)$ like this $(\lnot a,\lnot b) \to (c)$.

Feb 12, 2019 · For some time I've been trying to understand reduction of 3-SAT to MAX 2-SAT. I reviewed most of most popular books about computational complexity (Thomas Cormen, Papadimitriou) but I can't find an

Jul 9, 2015 · If a single 2-SAT clause contains the same variable twice, then either both literals are equal (in which there is only one way to set the variable, which reduces the 2-SAT formula), or one literal is the negation of the other, in which case any way you set the variable will satisfy that clause and thus that clause can be deleted.

A bit informally stated, but I hope the point comes across. Your proof shows that 3SAT formulas of one clause cannot be represented as 2SAT formulas, and I didn't assume anything about the other ones when writing that comment. Do you, by "most", mean that there are many more 3SAT formulas that cannot be represented as 2SAT ones? $\endgroup$ –