Timeline for finished Josephus-Problem in C#, how to make this more efficient?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 21, 2020 at 15:19 | comment | added | Jeremias T. | @HenrikHansen Thank you! That helped a lot :)! | |
| May 21, 2020 at 14:53 | vote | accept | Jeremias T. | ||
| May 21, 2020 at 14:53 | vote | accept | Jeremias T. | ||
| May 21, 2020 at 14:53 | |||||
| May 21, 2020 at 12:46 | comment | added | user73941 | @RickDavin: I havn't missed your comment, but as far as I can read, it's about eliminating every other person (k = 2) only - not an arbitrary interval - for instance k = 3, og maybe I have missed something? | |
| May 21, 2020 at 11:45 | comment | added | Rick Davin | You may have missed my comment to OP, but there is no to work through an elimination process. A formula may be used instead. See exploringbinary.com/powers-of-two-in-the-josephus-problem | |
| May 21, 2020 at 7:36 | history | edited | user73941 | CC BY-SA 4.0 | added 100 characters in body |
| May 21, 2020 at 7:12 | history | edited | user73941 | CC BY-SA 4.0 | added 29 characters in body |
| May 21, 2020 at 7:04 | history | answered | user73941 | CC BY-SA 4.0 |