395 xor decomposition is very inefficient #425
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Dimosts
reviewed
Oct 16, 2023
| # there are multiple decompositions possible, Recursively using sum allows it to be efficient for all solvers. | ||
| decomp = [sum(self.args[:2]) == 1] | ||
| if len(self.args) > 2: | ||
| decomp = Xor([decomp,self.args[2:]]).decompose()[0] |
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Ah, you made it recursive, nice one. It's indeed clear that it will create far less constraints
Dimosts
approved these changes
Oct 16, 2023
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Came up with this one, it avoids nested conjunctions/disjunctions, in favour of nested sums.
Seems to lead to a smaller amount of constraints after flattening.