Timeline for Closest vertex in a 3D fcc lattice
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2017 at 12:50 | vote | accept | user3433489 | ||
| Aug 28, 2017 at 21:53 | answer | added | Adam P. Goucher | timeline score: 1 | |
| Aug 28, 2017 at 19:34 | comment | added | Gerhard Paseman | Then start with the central cube. You will find that deciding on the orthant, followed by looking at (but not necessarily calculating) the Voronoi cells will lead to a simple algorithm based on relative size of coordinates. You can use translation to do this in other cubes. Gerhard "First, Make The Problem Local" Paseman, 2017.08.28. | |
| Aug 28, 2017 at 19:05 | review | Close votes | |||
| Aug 29, 2017 at 6:08 | |||||
| Aug 28, 2017 at 18:31 | history | edited | user3433489 | CC BY-SA 3.0 |
fixed centered--> cubic
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| Aug 28, 2017 at 15:20 | comment | added | user3433489 | I'm fine with conditionals, which is why I said piecewise. I just want a simple method that does not require distance calculations to multiple points. | |
| Aug 28, 2017 at 14:38 | comment | added | Gerhard Paseman | An algorithmic solution is more easily expressible. If you are looking for a purely arithmetic formula without conditionals, you will have something that is very hard to implement. What are you going to do with such a formula if you get it? Gerhard "Thinks The Goal Is Wrong" Paseman, 2017.08.28. | |
| Aug 28, 2017 at 14:20 | history | edited | user3433489 |
eliminated geometry tag
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| Aug 28, 2017 at 14:15 | comment | added | Martin Sleziak | (geometry) tag is deprecated, see the tag-info. Please, try to choose other suitable tag. | |
| Aug 28, 2017 at 13:54 | history | asked | user3433489 | CC BY-SA 3.0 |