Timeline for Uncollapse digits
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 22, 2018 at 17:31 | comment | added | Jonathan Allan | Heh, thanks Lynn, that allowed two more bytes to be golfed off too! | |
| Jun 22, 2018 at 17:30 | history | edited | Jonathan Allan | CC BY-SA 4.0 | added 162 characters in body |
| Jun 22, 2018 at 16:59 | comment | added | lynn | I'm not sure how I keep coming back to this question and noticing new things, but h=(ord(s[0])*ord(s[1])%83%7+1)%3+3 is 65 bytes! :) | |
| Apr 9, 2018 at 7:15 | history | edited | Jonathan Allan | CC BY-SA 3.0 | added 43 characters in body |
| Apr 9, 2018 at 7:11 | comment | added | Jonathan Allan | Oh wow, h(s) and h(s) how did I not notice?! Thanks Lynn! | |
| Apr 8, 2018 at 6:34 | comment | added | lynn | I'm revisiting this a lot later and realizing this could be 68 bytes: def f(s):h=[4,5,3][ord(s[0])*ord(s[1])%83%7%3];print(s[:h]);f(s[h:]) | |
| Nov 25, 2017 at 15:30 | history | answered | Jonathan Allan | CC BY-SA 3.0 |