Timeline for Is Toom's rule robust under local but non-on-site noise?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 28, 2021 at 10:20 | vote | accept | Andi Bauer | ||
| Sep 28, 2021 at 6:29 | comment | added | Ilkka Törmä | @AndiBauer Yes, you're correct. You can think of $\mu(y)$ as a noisy or perturbed version of $f$ that produces a random itinerary (aka spacetime diagram) with initial state $y$. In the first equation, $\cdots$ is indeed the set of itineraries where the local rule of $f$ is not followed on any coordinate of $V$. The condition is that its probability decays exponentially with the size of $V$, but otherwise $\mu$ can have local and non-local correlations, and the positions of errors on each time step can even depend on the contents of the configuration. | |
| Sep 27, 2021 at 21:55 | comment | added | Andi Bauer | Ok, I think the latter now. And then in your first equation $\mu(\ldots)<\epsilon^{|V|}$, the $\ldots$ is the set of space-time configurations which differ from those of $f$ on $V$? Am I getting that right? | |
| Sep 27, 2021 at 21:48 | comment | added | Andi Bauer | Thanks for your answer. I'm coming from a different background and have trouble parsing it, hence a few basic questions: $\mu\in M_\epsilon$ is the probability distribution over different noise configurations, consisting of the space-time points where bit-flips occur, right? Or should I think of $\mu$ as a probability distribution over the space-time configurations of the noisy cellular automaton already? | |
| Sep 27, 2021 at 21:13 | history | edited | Ilkka Törmä | CC BY-SA 4.0 |
Typo in result
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| Sep 27, 2021 at 21:08 | history | answered | Ilkka Törmä | CC BY-SA 4.0 |