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tgflynn | 2 years ago
> Doing this enables us to represent each Di as a variable sequence, but with all the negative literals being removed. See Table I for illustration.
What justification does he have for throwing out negated variables ? If you do that the problem likely becomes trivial, but has nothing to do with the original problem.
pfedak|2 years ago
It does create issues with determining which expressions are satisfied, since two different paths can meet at a merged trie vertex higher up, which requires the extra bookkeeping that kills it.
tgflynn|2 years ago