One possibility would be for you to get a professor to contact me and tell me that it’s for research..

]]>(x | y | c) & (z | ~c | x_f) & (x_f | x_t | x_t)

Alright, we added the third clause there, even if it really lives way on end of F’.

Well, if (x | y | z) couldn’t be made true in F, the only way for its correspondent in F’ to be made true starts with binding c to true for the first conjunct.

Saying “there is no way to make a clause true” is different from saying “the variables in this clause are set to false”. So, for example, it’s very possible for the variable z to be set to true, and to use that setting to make the formula satisfiable.

]]>(x | y | c) & (z | ~c | x_f) & (x_f | x_t | x_t)

Alright, we added the third clause there, even if it really lives way on end of F’.

Well, if (x | y | z) couldn’t be made true in F, the only way for its correspondent in F’ to be made true starts with binding c to true for the first conjunct. That means x_f must be true in the second conjunct. Is that possible? Yes, it is, if x_t is bound to false. So this conjunction does have an assignment which imparts it with the “not all equal” property, and so it does not force F’ to be out of NAE3SAT.

Well, if this conjunction doesn’t do the job of putting F’ out of NAE3AT, what other part of F’ might? If there is not necessarily any such segment in F’, then the mapping is not valid, correct? This problem would seem to generalize to any all-false clause you could find in any assignment to an unsatisfiable F, which I think is something Ian was driving at.

I thought for a moment that this issue might not be fatal. (And I’m still hopeful you’ll show me it isn’t!) I had missed your instruction, “We replace all instances of the constant value true with x_T”. But I do not think this is any help here (though I am not sure if you’re claiming it is). The mapping can only send 3CNF formulas to 3CNF formulas; it doesn’t know anything about assignments in itself, and so there’s no mechanism to make the substitution there. And while maybe it helps to know about such a substitution if our goal is to show F’ is in NAE3SAT, it does not seem to help us when our goal is to show F’ is not in NAE3SAT, since our “adversary” can make any assignment it wants there to foil us.

]]>I just recently discovered your website and it has been a huge help in my research. I’m a masters student at Unicamp – Brazil and I work with the complexity of several graph labelling problems.

Keep up the good work!

]]>It would be right if we have two edges (d,g) and (c,h).

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