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Did you read the remark at the end of my comment? In the practical cases I was exploring, that combinatorial explosion does not happen. It's relaxed in the sens
by sbazerque 2y ago
Did you read the remark at the end of my comment? In the practical cases I was exploring, that combinatorial explosion does not happen. It's relaxed in the sense that it is coordination-free.
- j-pb 2y agoNot sure what you mean. I'm talking about the "relaxed" P' being defined via the power set of S. 2^S= {s | s ⊆ S} Now if all your P is only a mapping then P'(S) = {<s, P(s)> | s ∈ S} but then your "coordination free" P was monotonic anyways.