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Causal invariance and rule reversal

I am wondering, if one reverses a rule in a causally invariant Wolfram model (i.e. changes it from A->B to B->A), whether the model remains causally invariant. Does anyone know an answer to the question?

POSTED BY: Pavlo Bulanchuk
2 Replies

Ok, yes, very good. If we have at least two rules, then the reverse rules do not have to be causally invariant. I also came up with a single rule example: Consider the rule AB->AA. It is causally invariant for all starting states (can prove). If we start from ABA, we end with AAA in a single step. If we reverse the rule, then AAA can go into ABA or AAB.

I suspect, if we start from a graph (as opposed to a string), we can construct something very similar, although it seems more difficult to prove

POSTED BY: Pavlo Bulanchuk

I recal lthe definition

Causal Invariance: A property of multiway graphs whereby all possible paths yield the isomorphic causal graphs. When causal invariance exists, every branch in the multiway system must eventually merge. Causal invariance is a core property associated with relativistic invariance, quantum objectivity, etc. In the theory of term rewriting, a closely related property is confluence. In a terminating system, causal invariance implies that whatever path is taken, the "answer" will always be the same.

Consider the rules B-> A, C -> A acting on a word composed exclusively of letters from the alphabet {A, B, C}. This system is evidently causally invariant, because all the brances will merge to the same word after the substitution of all B and all C by A. On the other hand, the system obtained by reversing the rules A-> B, A -> C will be non-causally invariant for any word containing at least one "A". For example, applying these rules to A, we obtain two brances B and C, which cannot merge, since there is not rule to transform neither B nor C.

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