Timeline for Shifting an irrational binary sequence
Current License: CC BY-SA 4.0
10 events
when toggle format | what | by | license | comment | |
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Dec 10 at 4:05 | answer | added | Ville Salo | timeline score: 5 | |
Dec 9 at 22:33 | history | became hot network question | |||
Dec 9 at 16:27 | vote | accept | Dominic van der Zypen | ||
Dec 9 at 15:47 | answer | added | Martin M. W. | timeline score: 12 | |
Dec 9 at 15:26 | comment | added | Dominic van der Zypen | Thanks @VilleSalo, an answer would be very much appreciated! | |
Dec 9 at 15:07 | comment | added | Ville Salo | AND is not surjective, so not 2-to-1. I can write an answer at some point. A key point for doing one-sided is that xor is left-permutive, I guess that's what I meant when I said it's one-sided. | |
Dec 9 at 14:51 | comment | added | Dominic van der Zypen | Is ${\tt AND}$ , $\land$, a two-to-one cellular automaton? If yes, the argument would not work. If we take $s = {\tt 01\, 001\, 0001 \, 00001}\ldots$ then $s \land \ell_1(s)$ is the constant 0 sequence. A similar argument can be made for ${\tt OR}$, $\vee$. Maybe you can elaborate your argument in an answer - many thanks. | |
Dec 9 at 14:47 | comment | added | Ville Salo | (This argument is clearer for two-sided sequences, but the CA is one-sided so it works out) | |
Dec 9 at 14:40 | comment | added | Ville Salo | No, xor is a two-to-one cellular automaton. A periodic point generates a finite subshift under the shift, so the preimage subshift is finite, so has only periodic points. | |
Dec 9 at 14:31 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |