Timeline for Iteration cycles of Z_n weights in path graphs: Why cycles of length 182 for a 6-node path?
Current License: CC BY-SA 3.0
34 events
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| Aug 29, 2017 at 10:41 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Struck sentence no longer relevant. (This should be my last edit.)
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| Aug 28, 2017 at 11:38 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Clarified.
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| Aug 27, 2017 at 15:01 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Removed TeX markup in title.
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| Aug 27, 2017 at 13:56 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Toned it down a notch. :-)
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| Aug 27, 2017 at 7:36 | answer | added | Peter Heinig | timeline score: 1 | |
| Aug 27, 2017 at 6:56 | comment | added | Peter Heinig | While the explanatory value of, or even explainability of cellular automata is notoriously controversial (though they are certainly wonderful things), I think it is a relevant comment that your discovery is strongly reminiscent of what is ill-advisedly called an elementary cellular automaton. Why 'ill-advised'? Because the modifier 'elementary', added to 'cellular automaton', does not correspond to how the modifier 'elementary' is used in modern mathematics (namely, to indicate first-order logic, not dimensionality). | |
| Aug 27, 2017 at 5:30 | comment | added | Gerhard Paseman | Similarly for powers pp of 2 I am seeing (pp-1)pp/2 in the sequence, and maybe (pp-1)phi(pp) for prime powers in general. Can this be proven or at least numerically confirmed? Gerhard "This Who Can't Prove, Conjecture" Paseman, 2017.08.26. | |
| Aug 27, 2017 at 5:25 | comment | added | Gerhard Paseman | Let p be an odd prime, and let A be (the matrix representing) left shift drop (introducing a zero) of a vector of length p, and B be right shift. So A^p = B^p = O and ABA=A and BAB=B The data by Moritz Firsching suggest (A+B)^(2p-1) = A+B mod p. Can this be proved? Gerhard "Calling All Combinatorialists And Matricians" Paseman, 2017.08.26. | |
| Aug 27, 2017 at 1:43 | comment | added | Gerry Myerson | $182=2\cdot7\cdot13$, $7\equiv13\equiv1\bmod6$; $48422=2\cdot11\cdot31\cdot71$, $11\equiv31\equiv71\equiv1\bmod{10}$. | |
| Aug 26, 2017 at 22:03 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Aug 26, 2017 at 20:37 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
added 241 characters in body
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| Aug 26, 2017 at 19:40 | answer | added | Moritz Firsching | timeline score: 3 | |
| Aug 26, 2017 at 19:20 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Aug 26, 2017 at 19:06 | comment | added | Joseph O'Rourke | @MoritzFirsching: Your computational abilities (& processors) $\gg$ mine. :-) | |
| Aug 26, 2017 at 19:03 | comment | added | Moritz Firsching | @JosephO'Rourke, the sequence continues like $30172506, 36, 7812, 4067052$ for $n=18, 19, 20, 21$. It seems in general numbers with small factors give larger values. | |
| Aug 26, 2017 at 18:46 | comment | added | Joseph O'Rourke | @MoritzFirsching: $48422$ is impressive! Yes, I've tried to find various derived sequences in the OEIS without luck. | |
| Aug 26, 2017 at 18:42 | comment | added | Moritz Firsching | for $n=10$, and the identity permutation, the process takes 48422 step until it cycles. In fact, taking powers of the adjacency matrix mod $n$ until it becomes cyclic, you need to take this many steps: $1, 1, 2, 4, 6, 8, 182, 12, 28, 48, 48422, 20, 1638, 24, 1200, 6240, 120, 32,...$. (not (yet) in OEIS). For the even cases, like 6, you seem to get back to the identity. | |
| Aug 26, 2017 at 17:14 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Aug 26, 2017 at 17:10 | comment | added | Joseph O'Rourke | @GerhardPaseman: Also verified $n=8$, but I will stop there. Re-edited to reflect this. Thanks for your help. | |
| Aug 26, 2017 at 17:08 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
Update re exhaustive.
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| Aug 26, 2017 at 17:05 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
deleted 1 character in body
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| Aug 26, 2017 at 16:59 | comment | added | Peter Heinig | I have been agonizing over whether to mention this at all, since it is so tangential, yet let me say that this reminds me of an interesting niche-subject within flow-theory of graphs: zero-sum flows, about which you can find much by searching. Roughly, these are flow-like-assignments to the edges, the flow-condition being defined without an orientation on the edges. This is quite different from classic flows on graphs, and I think that this will not help you much, yet it is a subject where weights are being added at each vertex (and modulo a fixed modulus). | |
| Aug 26, 2017 at 16:56 | comment | added | Joseph O'Rourke | @GerhardPaseman: Exhaustive search verified the cycles displayed up to $n=7$. In particular, for $n=6$, there is just the one cycle of length $182$. | |
| Aug 26, 2017 at 16:56 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
Emphasize non exhaustive search
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| Aug 26, 2017 at 16:54 | comment | added | Gerhard Paseman | OK. I will edit your post to indicate that your results are not yet exhaustive. Gerhard "To Make A Good Impression" Paseman, 2017.08.26. | |
| Aug 26, 2017 at 16:41 | comment | added | Joseph O'Rourke | @GerhardPaseman: Will do an exhaustive search later... | |
| Aug 26, 2017 at 16:26 | comment | added | Gerhard Paseman | Have you done an exhaustive search? Or used symmetry to do a comprehensive search? I suspect this anomaly may be related to the automorphism group of Alt(6). Gerhard "Number Six Is Exceedingly Symmetric" Paseman, 2017.08.26. | |
| Aug 26, 2017 at 16:21 | comment | added | Gerhard Paseman | This reminds me (for odd n) of additive permutations. I will look for connections and report back. Gerhard "Going On A Combinatorial Path" Paseman, 2017.08.26. | |
| Aug 26, 2017 at 16:13 | comment | added | Joseph O'Rourke | @MoritzFirsching: Right, I just start with a random permutation, and then await a cycle. | |
| Aug 26, 2017 at 16:12 | comment | added | Moritz Firsching | I see, so the original permutation does not need to be part of the cycle. | |
| Aug 26, 2017 at 16:00 | history | edited | Joseph O'Rourke | CC BY-SA 3.0 |
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| Aug 26, 2017 at 15:44 | comment | added | Moritz Firsching | What is the permutation that gives a cycle length of $6$ for $n=7$? | |
| Aug 26, 2017 at 14:20 | comment | added | Brendan McKay | Note that the operation is that of multiplying by the adjacency matrix mod $n$, so multiple steps are multiplication by powers of the adjacency mod $n$. | |
| Aug 26, 2017 at 14:03 | history | asked | Joseph O'Rourke | CC BY-SA 3.0 |