Timeline for The 3x3 Hexa Prime Square Puzzle
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 17, 2020 at 9:04 | history | edited | CommunityBot |
Commonmark migration
|
|
| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
replaced http://codegolf.stackexchange.com/ with https://codegolf.stackexchange.com/
|
|
| Oct 15, 2016 at 22:36 | history | edited | Karl Napf | CC BY-SA 3.0 |
added 4 characters in body
|
| Oct 15, 2016 at 22:19 | comment | added | Jonathan Allan |
Couple more bytes by printing a list of strings: print[s for s in map(''.join,permutations(input()))if all(int(s[3*i:][:3],16)in P and int(s[i::3],16)in P for i in(0,1,2))]
|
|
| Oct 15, 2016 at 22:11 | history | edited | Karl Napf | CC BY-SA 3.0 |
added 1 character in body
|
| Oct 15, 2016 at 21:54 | comment | added | Karl Napf |
@JonathanAllan The import saved only 1 byte in this case, thought it helps only if stuff was used multiple times. Oh the efficiency, it is hard for me to drop it ;) That refactor with the slice for P is great.
|
|
| Oct 15, 2016 at 21:52 | history | edited | Karl Napf | CC BY-SA 3.0 |
added 44 characters in body
|
| Oct 15, 2016 at 21:43 | comment | added | Jonathan Allan |
Also from itertools import* is a neat golf trick that'll save a couple of bytes. Also forgo the efficiency and do k<5e3.
|
|
| Oct 15, 2016 at 21:39 | comment | added | Jonathan Allan |
How about P+=[k][:p%k] - when p%k is not 0 the slice will yield [k], when it is 0 it will yield [].
|
|
| Oct 15, 2016 at 21:34 | history | edited | Karl Napf | CC BY-SA 3.0 |
added 126 characters in body
|
| Oct 15, 2016 at 21:14 | history | edited | Karl Napf | CC BY-SA 3.0 |
edited body
|
| Oct 15, 2016 at 20:57 | history | edited | Karl Napf | CC BY-SA 3.0 |
added 49 characters in body
|
| Oct 15, 2016 at 20:51 | history | answered | Karl Napf | CC BY-SA 3.0 |