In a read-twice opposite CNF formula each variable appears twice, once positive and once negative.
I'm interested in the $\oplus\text{Rtw-Opp-CNF}$ problem, which consists in computing the parity of the number of satisfying assignments of a read-twice opposite CNF formula.
I was unable to find any reference about the complexity of such problem. The closest I was able to find is that the counting version $\#\text{Rtw-Opp-CNF}$ is $\#\text{P}$-complete (see section 6.3 in this paper).
Thanks in advance for your help.
Update 10th April 2016
- In this paper, the $\oplus\text{Rtw-Opp-SAT}$ problem is shown to be $\oplus\text{P}$-complete, however the formula produced by reduction from $3\text{SAT}$ is not in CNF, and as soon as you try to convert it back into CNF you get a read-thrice formula.
- The monotone version $\oplus\text{Rtw-Mon-CNF}$ is shown to be $\oplus\text{P}$-complete in this paper. In such paper, $\oplus\text{Rtw-Opp-CNF}$ is quickly mentioned at the end of section 4: Valiant says it is degenerate. It is not clear to me what being degenerate exactly means, nor what does it imply in terms of hardness.
Update 12th April 2016
It would be also very interesting to know if anyone has ever studied the complexity of the $\Delta\text{Rtw-Opp-CNF}$ problem. Given a read-twice opposite CNF formula, such problem asks to compute the difference between the number of satisfying assignments having an odd number of variables set to true and the number of satisfying assignments having an even number of variables set to true. I've not found any literature about it.
Update 29th May 2016
As pointed out by Emil Jeřábek in his comment, it is not true that Valiant said that the problem $\oplus\text{Rtw-Opp-CNF}$ is degenerate. He only said that a more restricted version of such problem, $\oplus\text{Pl-Rtw-Opp-3CNF}$, is degenerate. In the meanwhile, I continue to not know what degenerate exactly means, but at least now it seems clear that it is a synonym of lack of expressive power.