@LeakyNun How does this look for my next coding task?

There's a 9-byte Jelly answer by ais523 (checked with the userscript), but I haven't seen the answer yet.

Or I could use the leaderboard. :P

Oh? Hm. I never noticed that before.

Oh okay. My first attempt will probably be really long but I may need some hints later on

@LeakyNun My idea is to get 4 values (probably an array) and add 1 for open and subtract 1 for close, cumsum, and take the first instance of 0. Is this a bit overcomplicating things?

Alright. I will try this.

Alright.

Oh more alphabet problems! :D

@LeakyNun Is there a way to repeat the alphabet multiple times? It's always giving me AAAAAAABBBBBBBCCCCCCC...etc.etc

Oh, okay. Thanks. I was using normal x ;_;

I don't think so because it vectorizes (?) (gives the first element y times, then the second y times, etc.)

Hm. Interesting. What does the ' do?

@LeakyNun 16 bytes (rather long, I will work on golfing this now):

ØAẋ26
51RḂÐfR¢ṁY

Oh okay. I'll check the specs and see if I can find it.

"For monads, flat. For dyad, spawn."

What does this mean?

Oh alright.

This and this appear to both work.

Yep.

@LeakyNun I'm looking for something to take two arrays and split the first array into slices of the sizes according to the second array. Is there anything like that?

Let's say I have two arrays. x = [1] * 10, for example. y = [3, 2, 4, 1]. Then, I want a dyad that does (x, y) -> [[1, 1, 1], [1, 1], [1, 1, 1, 1], [1]].

Equivalent to y -> Map by R (range) -> mold

Hm. Alright.

Yes. I'm just getting used to it so most of my code might look to Pythonic for this reason.

Oh hm, I read your TNB comment, that would have been useful to know in the docs.

ØA
51RḂÐfR¢ṁY

Quick trivial edit for 3 bytes

I think the helper link is not necessary

51RḂÐfRØAṁY

Yay :D

Hm. Dennis's answer is still 1 byte shorter according to the userscript's count

@LeakyNun Can you help me check how different Dennis's answer is from mine?

Alright Thanks.

No, I wanted to know how different it was, like quite similar, or very different approach?

alright. i will try to figure it out

@LeakyNun Is it this?

51Rm2RØAṁY

Using m2 to get every 2nd element instead of filtering by bit.

It is now the same length as Dennis's.

Same length?

Alright. Will do.

26RḤ’RØAṁY

26ḶḤ‘RØAṁY

aww :P

lol xD

Wait, I recall something from one of my professor's T-Shirts

The odd numbers are the differences between squares.

So that's the third way to generate the odd numbers.

27Ḷ²IRØAṁY

@LeakyNun I appear to have hit the output cache from when you wrote this :P

Thanks :D

Yes. It appears that the site remembers what it output for a given program and given arguments and it just gives the same output whenever anyone uses that setup again

Continue what?

Ah. Yes, I am working on that right now.

@LeakyNun Darn, all of the brackets are 2 codepoints away from each other except () :(

That's annoying. That could have simplified things quite a bit :( lel

Oh hm. Good idea.

I can't seem to see the pattern...

() 0101000 0101001
[] 1011011 1011101
{} 1111011 1111101
<> 0111100 0111110

Hm I noticed that the first three digits are unique.

Oh, the second and third.

Wait no.

@LeakyNun Which digits do I need? I can't seem to find two digit places that would be unique.

But () and <> then have the same 01.

I'm trying to distinguish between different types of brackets, though that may be unnecessary.

My idea is to assign each bracket type a unique value (probably just the codepoint or a modification on it), take the square root, negate the close brackets, cumulatively sum it, and find 0. Doesn't seem very Jelly-mindset but I can't think of a different method off the top of my head (other than the straightforward method which will be too long)

See, let's say I assign 2 to { and 3 to (. Then {{{)) would give a net of 0.

(I know that that's not balanced)

So by taking the square root, the ratio between the values is irrational so I won't get false 0s from it.

hm

Wait I think that can't happen because the brackets must be balanced.

@LeakyNun I don't think I actually need the square root thing because the brackets have to be balanced so that won't happen.

That's a good question.

I don't...

Whoops.

I just need to distinguish which way it faces...

@LeakyNun Okay so I have an idea where to go from here. It's late now so I'm going to go to bed. Thanks for the help today, cya!