Timeline for One OEIS after another
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2017 at 13:42 | comment | added | user202729 | FYI, I have a Python program (modify from yours) that can calculate exactly terms up to 14. (that is, 6 first terms 1, 1, 1, 2, 5, 14, others take too much time for me) --- If you use Py(th)on for this answer instead, probably others can use Proton in next answers? --- However, because this is your answer, you would probably prefer to work it out yourself. | |
| Dec 4, 2017 at 14:25 | comment | added | hyperneutrino♦ | @NieDzejkob I'll actually probably see if I can fix Proton first before retrying this. I'll probably switch to using the builtin if possible. | |
| Dec 4, 2017 at 14:03 | comment | added | Maya | @HyperNeutrino using a builtin function that's more than 20 times faster must help. The only problem is the generator -> list conversion bug in Proton. | |
| Dec 4, 2017 at 13:19 | comment | added | hyperneutrino♦ | @NieDzejkob Hm strange. I may need to just brute-force this then because I don't think optimizations will help me much anymore (or be correct for that matter :P). I'll try to fix it before most people notice :P | |
| Dec 4, 2017 at 8:27 | comment | added | Maya | I tried rewriting this in Python, I cross-checked it like 10 times now. I couldn't find a bug, but the result for input 5 (n = 8) is 44, instead of the 14 expected. I left the Proton program running overnight, but it didn't finish. Using itertools' permutations and set union and intersection should speed it up a lot, but there seems to be a bug in the Proton interpreter. TL;DR: This answer may need corrections... | |
| Oct 10, 2017 at 15:27 | comment | added | Christian Sievers | So you intended the graphs to be planar, and the same generation seems to work. Then everything seems fine! | |
| Oct 10, 2017 at 15:06 | comment | added | hyperneutrino♦ | @ChristianSievers math.stackexchange.com/a/2463430/457091 | |
| Oct 10, 2017 at 15:03 | comment | added | Christian Sievers | Maybe it doesn't matter, maybe your code does something I didn't see. Are your graphs supposed to be triangulations of the sphere? Where did you get your ideas? As for correct results, the paper also implies that (at least if you have a sphere with all the faces) you don't need the splitting of pentagons before n=12. | |
| Oct 10, 2017 at 14:51 | comment | added | hyperneutrino♦ | @ChristianSievers I don't think so? I mean the result for 3, 4, 5, 6 are all correct lol | |
| Oct 10, 2017 at 14:37 | comment | added | Christian Sievers | The generation looks like what is suggested by Bowen and Fisk in "Generation of triangulations of the sphere", but since you start with a single triangle you seem to miss one face? | |
| Oct 10, 2017 at 12:08 | history | edited | hyperneutrino♦ | CC BY-SA 3.0 | deleted 4 characters in body |
| Oct 10, 2017 at 12:01 | comment | added | hyperneutrino♦ | @PeterTaylor Assuming I understand the approach you're describing, yes, it looks for triangles and places a vertex adjacent to all 3 vertices, or two adjacent cycles and deletes the common edge and places a vertex adjacent to all 4, same for 3 triangles on a pentagon. I think that's one you're describing. | |
| Oct 10, 2017 at 6:50 | comment | added | Peter Taylor | Is this the approach of splitting triangles, squares, and pentagons as per plantri? Looks like it might be, but some of the syntax is unfamiliar. | |
| Oct 10, 2017 at 3:21 | history | edited | hyperneutrino♦ | CC BY-SA 3.0 | added 9 characters in body |
| Oct 10, 2017 at 2:58 | comment | added | hyperneutrino♦ | @Stephen Currently bugfixing lol | |
| Oct 10, 2017 at 2:54 | comment | added | Stephen | Wait, you actually did it? If you don't write a paper with these freaking programs and go talk to some professor, you're passing up on something cool :P | |
| Oct 10, 2017 at 2:33 | history | answered | hyperneutrino♦ | CC BY-SA 3.0 |