Timeline for Counting shortest paths on a triangular grid
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 14, 2020 at 13:58 | answer | added | user92069 | timeline score: 0 | |
| Apr 14, 2020 at 9:03 | answer | added | Kevin Cruijssen | timeline score: 0 | |
| Apr 14, 2020 at 4:31 | answer | added | Mitchell Spector | timeline score: 1 | |
| Apr 12, 2020 at 19:19 | answer | added | Arnauld | timeline score: 0 | |
| Apr 12, 2020 at 16:26 | answer | added | Jonathan Allan | timeline score: 3 | |
| Apr 12, 2020 at 15:21 | history | became hot network question | |||
| Apr 12, 2020 at 15:05 | comment | added | Mitchell Spector | @jonrandy First convince yourself that it's true for points in the first quadrant, using a combinatorial argument. Then, for points elsewhere in the plane, use the fact that the value for that point is the same as the value for the three other points symmetrically located about the origin. If \$a+b\omega\$ is the first point, those other three points are \$-a-b\omega, b-a+b\omega,\$ and \$a-b-b\omega.\$ | |
| Apr 12, 2020 at 10:51 | answer | added | Uriel | timeline score: 2 | |
| Apr 12, 2020 at 10:48 | answer | added | Mr. Xcoder | timeline score: 1 | |
| Apr 12, 2020 at 10:13 | comment | added | jonrandy | Can you point me to an explanation of this? Take the absolute values of [a, b, a-b], and call it L. Calculate binomial(max(L), any other value in L) | |
| Apr 12, 2020 at 9:59 | answer | added | Neil | timeline score: 0 | |
| Apr 12, 2020 at 8:11 | answer | added | Noodle9 | timeline score: 2 | |
| Apr 12, 2020 at 7:33 | answer | added | ZaMoC | timeline score: 3 | |
| Apr 12, 2020 at 7:20 | history | asked | Bubbler | CC BY-SA 4.0 |