Timeline for Not Quite Roman Ternary
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 28, 2020 at 10:36 | history | edited | Kevin Cruijssen | CC BY-SA 4.0 |
added 74 characters in body
|
| Jun 17, 2020 at 9:04 | history | edited | CommunityBot |
Commonmark migration
|
|
| May 15, 2018 at 9:55 | vote | accept | FrownyFrog | ||
| Mar 29, 2018 at 10:56 | comment | added | Arnauld | I rely more on intuition than on rigorous maths to find a correct formula, so it's quite hard to write a tip. A multiply followed by one or several modulos works reasonably well most of the time, especially when combined with a lookup table. Here, we want to find the correct value directly, so it could indeed become more complex very quickly. Maybe someday I'll try again to work on a more clever and more generic brute-forcer that starts with quick and easy formulae and gradually increases the complexity. | |
| Mar 29, 2018 at 10:36 | comment | added | Kevin Cruijssen | @Arnauld Ah, that's more basic than I was expecting, thanks for sharing though. I guess it can quickly become more complex when more values are being asked, and where you end up needing more modulos, bitwise operations, base-conversions and/or standard arithmetic operations. Is there any general rule of thumb you can give as a tip to determine what is best to try next? For example, if your linked code wouldn't have given a result, would you add a third modulo? Just trying to understand it a bit more so I can hopefully do something myself with future challenges. :) (Thanks for your time, btw) | |
| Mar 29, 2018 at 10:30 | comment | added | Arnauld |
(Also, I really should test whether p=1 and don't include *1 in the code if it is -- even though it doesn't lead to a better formula in that case.)
|
|
| Mar 29, 2018 at 10:22 | comment | added | Arnauld | I've already thrown it away ... :-/ But here is the last one. (Very inefficient, but that's OK for such small values.) | |
| Mar 29, 2018 at 10:11 | comment | added | Kevin Cruijssen |
@Arnauld Out of curiosity, what does the brute-forcers look like you've used for these last two magic numbers (when you still used i, and when you re-use c)?
|
|
| Mar 29, 2018 at 10:06 | comment | added | Kevin Cruijssen |
@Arnauld Nice one! And -1 more byte by putting r in the loop body. Thanks!
|
|
| Mar 29, 2018 at 10:05 | history | edited | Kevin Cruijssen | CC BY-SA 3.0 |
deleted 198 characters in body
|
| Mar 29, 2018 at 9:59 | comment | added | Arnauld | I think this works: 103 bytes | |
| Mar 29, 2018 at 9:28 | history | edited | Kevin Cruijssen | CC BY-SA 3.0 |
forgot to update actual golfed code in my last edit..
|
| Mar 29, 2018 at 8:59 | history | edited | Kevin Cruijssen | CC BY-SA 3.0 |
added 42 characters in body
|
| Mar 29, 2018 at 8:29 | history | edited | Kevin Cruijssen | CC BY-SA 3.0 |
added 244 characters in body
|
| Mar 28, 2018 at 18:11 | history | edited | Kevin Cruijssen | CC BY-SA 3.0 |
added 166 characters in body
|
| Mar 28, 2018 at 18:03 | history | rollback | Kevin Cruijssen |
Rollback to Revision 2
|
|
| Mar 28, 2018 at 17:58 | history | edited | Kevin Cruijssen | CC BY-SA 3.0 |
added 96 characters in body
|
| Mar 28, 2018 at 16:00 | history | edited | Kevin Cruijssen | CC BY-SA 3.0 |
added 105 characters in body
|
| Mar 28, 2018 at 15:49 | history | answered | Kevin Cruijssen | CC BY-SA 3.0 |