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7:10 PM
CMCs or old challenges?
 
Let's see if we can find a good, old challenge
 
Ok, let the challenge searcher do its job :P
 
I found a / challenge that doesn't have a Jelly answer
 
7:13 PM
oh I remember that challenge, i have seen it before
I remember it as K Zhang's profile icon :P
But it is subjective
 
The J answer seems to be a good start
 
How?
 
J is tacit and Jelly's father language
 
Yeah but can you read that?
 
With help from the J docs :P
 
7:18 PM
i am fine making my own :P
 
I have a better challenge, but it has a Jelly answer
 
Ok lemme see it
 
What do you think of this?
 
ha... yeah cool
Brb making an answer in 05ab1e first, it's perfect for that
:P 7 bytes: AƶJQ42*
Now Jelly
 
Damn, I have 13 bytes (equal to ais532)
12 bytes \o/
11 bytes
 
7:25 PM
I was thinking of smth like ØaxJ
 
11 bytes
10 bytes
@cairdcoinheringaahing This is my 10-byter.
Same?
 
@Mr.Xcoder Nope :P
 
Very different
?
 
Kinda. Hold on, golfing mine a bit :P
9 bytes
 
7:29 PM
Ah, I think I know what you are golfing :P
Ok I give up
 
:P
I am having a hard time believing this is golfable :D
 
9 bytes seems optimal for Jelly given that 05ab1e has 7
 
7:33 PM
@cairdcoinheringaahing 05ab1e has a built-in for xJ¤
 
@Mr.Xcoder ƶ?
 
yes
 
Hehe, we beat one 05ab1e answer :P
 
CMC: Output all "product partitions" of an integer (i.e all consecutive integers which, when multiplied give the input).
Can't be bothered to make a test case rn, but I know a few: 24 -> [[1,2,3,4], [2,3,4], probably more...]
 
24 -> [[1, 2, 3, 4], [2, 3, 4], [24]]
 
7:36 PM
._.
 
ლ(ლಠ益ಠლ)ლ
 
@Mr.Xcoder Should we post as an answer, or comment on the other Jelly answer (by old ais532)?
 
I won't post any answer, but as it is entirely different, you can post it
 
@Mr.Xcoder What happened? :P
 
7:42 PM
I can't wrap my head around this, as I am quite fired tired for some reason
@cairdcoinheringaahing Can I just see your 6 byter?
Also I want to take a look at the next OEIS
 
Same algorithm as the beginner challenge :P
 

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