Timeline for Can knot diagrams be monotonically simplified using under moves?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| S Jun 10, 2019 at 10:58 | history | suggested | Glorfindel | CC BY-SA 4.0 |
broken image fixed (click 'rendered output' or 'side-by-side' to see the difference; image retrieved via Wayback Machine); for more info, see https://meta.mathoverflow.net/a/4058/70594
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| Jun 10, 2019 at 8:10 | review | Suggested edits | |||
| S Jun 10, 2019 at 10:58 | |||||
| Nov 5, 2014 at 15:53 | comment | added | Dylan Thurston | That's pretty nice, thanks for the pointer. The program looks pretty effective, and I'm curious what it does on that satellite trefoil. | |
| Nov 4, 2014 at 21:23 | comment | added | Michael | Things that are obvious to humans are not necessarily obvious to computers, so there is a possibility for a connected sum to get tangled. However, you are right, that would be a too easy a test. How about this one: consider the usual embedding of a torus in $R^3$ and let the trefoil (or any torus knot) run over that torus as usual. Then take that torus and tangle it into the thick version of the Gordian unknot. That would lead to a seriously tangled trefoil. | |
| Nov 4, 2014 at 19:43 | comment | added | Bruno Martelli | I don't know (I still haven't seen the program running), but if you perform the diagram connected sum I suppose it simplifies the knot exactly as before. One should try some hard version of the trefoil knot... | |
| Nov 4, 2014 at 17:08 | comment | added | Michael | What happens when you run this program on a heavily twisted version of non-trivial knot, something like (Gordian unknot)#Trefoil? Does the program monotonically decrease the crossing number until you get to a more or less canonical representation of the knot? | |
| Nov 4, 2014 at 14:22 | history | edited | Bruno Martelli | CC BY-SA 3.0 |
picture
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| Nov 4, 2014 at 14:15 | history | answered | Bruno Martelli | CC BY-SA 3.0 |