Jelly Hypertraining

11:43 AM

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Erik the Outgolfer has added Leaky Nun to the list of this room's owners.

Leaky Nun

Currently I've given him 3 challenges (I know it's many):

1. reverse and add degeneracy

2. reduce square root

3. Given e.g. 4, output:

 1  2  3  4
 8  7  6  5
 9 10 11 12
16 15 14 13

Erik the Outgolfer

@LeakyNun How can you reduce a monadic function?

Leaky Nun

@EriktheOutgolfer you took it too technically

I mean "simplify a square root".

Erik the Outgolfer

11:59 AM

Oh, so, for example, for input 8 you print 2 2 or something?

Leaky Nun

yes, that is what I mean

Erik the Outgolfer

Basically I think that the title of the challenge is "Simplify a square root", and a very important keyword is in the title itself :)

@LeakyNun And the third one is to output a square snake of a number, right?

Leaky Nun

Yes.

Erik the Outgolfer

Good. As a teacher, I will be preparing/copying golfed solutions now

user165474

Yay :) Thanks for making this room

Leaky Nun

12:13 PM

@HyperNeutrino yay you're back

user165474

Yes. I'd gone to bed when you pinged me.

user165474

Is the other challenge you said you had item 3 on the current list?

Leaky Nun

Yes

user165474

I see.

user165474

I kept working on item 1 for a while last night and then eventually the code just stopped making sense because I couldn't understand what the code I wrote before it meant :/

Leaky Nun

12:15 PM

I see.

user165474

For item 2, I figured out how to get b, but I just need to get a by taking [argument]/(b**2)

user165474

12:26 PM

Is there a built-in to iteratively take each element of an array until a certain condition is reached? (That is, if we make the condition x -> x != 0, it should give [1, 2, 3, 0, 4, 5, 6] -> [1, 2, 3])

user165474

Wait, jelly's i for array indices is 1-indexed?

Leaky Nun

@HyperNeutrino yes it is

Erik the Outgolfer

Everything in Jelly is 1-indexed.

user165474

Ohh...that would explain why my code was broken yesterday. Wow. :P

Leaky Nun

@HyperNeutrino TḢ’ḣ@ is the best I can do (beware the unterminated dyad)

user165474

12:31 PM

Okay. Thanks.

Erik the Outgolfer

I think that's the best possible.

user165474

Alright. I found an alternate way for my specific use for it which is a bit longer but doesn't require the extra helper link, but thanks.

Leaky Nun

I'm having a hard time doing 2 lol

user165474

lol

user165474

I figured out how to get the part under the new square root

user165474

12:39 PM

but every attempt at getting the other one gives me weird things which make no sense at all

user165474

For 1, is truthy/falsy fine as opposed to True/False?

Leaky Nun

my current solution for 2 has 12 bytes.

@HyperNeutrino sure

Erik the Outgolfer

@LeakyNun Well, I have a hard time doing 1.

Leaky Nun

@EriktheOutgolfer let me try

Erik the Outgolfer

Nah, concentrate on your 2 ;)

Leaky Nun

12:44 PM

@EriktheOutgolfer nah I'll do 1

user165474

I got a solution

user165474

For 1

Leaky Nun

that's ... long (but I haven't tried it so I can't comment)

user165474

spoiler

user165474

Yes, I agree.

user165474

12:45 PM

But <shrug>

user165474

Should I golf it or start on 2 to keep learning new stuff?

Leaky Nun

why is it in a spoiler? who are you keeping that secret to?

Erik the Outgolfer

For newcoming students I guess?

You know, students can come at any time.

user165474

Also in case Erik's trying I don't want to openly post the code

user165474

But yes, also for new students

Leaky Nun

12:46 PM

@EriktheOutgolfer not if you set this chatroom as private

Erik the Outgolfer

This is a gallery not private. There is a request access button.

Leaky Nun

alright

Erik the Outgolfer

@HyperNeutrino You should never use TIO v2 for production.

user165474

Oh right.

user165474

@EriktheOutgolfer Nexus then?

Erik the Outgolfer

12:49 PM

Yes, there isn't any other TIO.

user165474

Did v1 get taken down?

Erik the Outgolfer

Yes.

TIO Nexus link: Try it online!

user165474

Okay.

user165474

Thanks.

Leaky Nun

For the record, my code for 1 is 14 bytes

user165474

12:50 PM

Nice.

Leaky Nun

@HyperNeutrino so I guess you should golf it to keep learning new stuff

user165474

Yeah.

user165474

I'll share my method first I guess, and along the way I might see (code-length) optimizations.

Leaky Nun

I can read.

I know what your method is.

user165474

Alright. Yeah.

user165474

12:53 PM

I'll write out an explanation of my own code for myself to read and see if I see anything that can be shortened.

Leaky Nun

@EriktheOutgolfer have you tried 1/2?

Erik the Outgolfer

I'm trying 1 right now, and I think I've found a better method than Hyper's.

user165474

@LeakyNun does this explanation look correct? I think this is how my code works:

user165474

ṚḌ+                Helper Link; adds a number to the number itself backwards
  +                Addition
ṚḌ                 Reverses integer into decimal and undecimals

Ç6Ð¡µọ11TḢ’ḣ@ÆPÐfȦ Main Link
Ç6Ð¡               Calls the last link on the input 6 times, collecting all intermediate results
    µ              Starts new monadic chain
        TḢ’ḣ@      Takes all elements up until the condition is first satisfied
     ọ11           Checking for divisiblity by 11
             ÆPÐf  Filters to only keep primes

Leaky Nun

@EriktheOutgolfer but have you outgolfed me?

Erik the Outgolfer

1:01 PM

Well, not, but I think I've tied you, if my code is valid ofc.

Leaky Nun

@EriktheOutgolfer have you tried your code then

Erik the Outgolfer

Still implementing it.

Leaky Nun

@HyperNeutrino yes it is correct

@EriktheOutgolfer do you need a template?

user165474

Okay, thanks.

Erik the Outgolfer

@LeakyNun No. I think I've outgolfed you.

Leaky Nun

1:03 PM

@EriktheOutgolfer congratulations

Erik the Outgolfer

Running test cases now to confirm.

user165474

Does your code start with:

user165474

ṚḌ+
Ç6Ð¡

Leaky Nun

no it doesn't

user165474

Alright. Is there a better method to that too?

Leaky Nun

1:04 PM

Yes there is

Erik the Outgolfer

My code does but it's a one-liner, although that doesn't save any bytes.

user165474

Okay. So wait, does the better method completely change the way the code works or does it just have a shorter way to get the first 6 iterations?

Erik the Outgolfer

Mine completely changes the way it works.

Leaky Nun

mine changes partially the way it works.

Erik the Outgolfer

@LeakyNun Testcases seem correct, although I think there would be an issue when hitting more than 1 prime in the first 6 iterations.

Leaky Nun

1:07 PM

@EriktheOutgolfer truthy is truthy

and I was stupid, I'm at 11 bytes now.

@EriktheOutgolfer have you out-golfed me?

user165474

Oh, I think I found an optimization.

Erik the Outgolfer

Mine was 11 bytes too.

Leaky Nun

@HyperNeutrino congratulations

Erik the Outgolfer

I don't think it can get shorter than that.

Leaky Nun

@EriktheOutgolfer my md5 hash is 1f938eb9edd736e6cebd4e0cef09e8c9. yours?

Erik the Outgolfer

1:09 PM

050c7e6ee8c8923aa811c294bc90e1c2

Maybe similar algorithm?

Leaky Nun

oh, lol

user165474

Once it's hit a multiple of 11, it will always be a multiple of 11.

user165474

(I think)

Leaky Nun

Yes

because a reverse of a multiple of 11 must also be a multiple of 11

(corollary of the divisibility algorithm for 11)

user165474

Right.

user165474

1:10 PM

Yes.

user165474

Alright. That makes sense. Might be able to use that for something.

user165474

Wait, I just realized that prime numbers can't be multiples of 11 (excluding 11 I mean)

user165474

lol that was obvious >_>

Erik the Outgolfer

YEAH!!!11

OK golfing 2 now.

user165474

haha >_>

Leaky Nun

1:12 PM

@EriktheOutgolfer that's fast

Erik the Outgolfer

@LeakyNun May I link a spoiler?

Leaky Nun

Alright

Erik the Outgolfer

Spoiler: non-quitters don't click!

user165474

Alright. Won't click :)

Leaky Nun

Me neither

but I don't really see a way to optimize any part of my code

Erik the Outgolfer

1:14 PM

Mine either.

I'd say golfing 3 instead of 2.

Leaky Nun

Do you mean that your answer is now 8 bytes?

Erik the Outgolfer

What?

Leaky Nun

...

user165474

I think he means golfing the snake-matrix problem.

user165474

(item 3)

Leaky Nun

1:17 PM

@EriktheOutgolfer I completely misinterpreted this

I thought you golfed 2 bytes

@EriktheOutgolfer what does this contain?

Erik the Outgolfer

I won't spoil it here ;)

Leaky Nun

I mean, which item

Erik the Outgolfer

1

Leaky Nun

the 11-byte solution?

Erik the Outgolfer

Yep.

Leaky Nun

1:18 PM

lol, I can click it then

Erik the Outgolfer

NO! You will be a quitter then!

Leaky Nun

lol, our solutions are the same

almost the same

Erik the Outgolfer

OK now you can put your spoiler.

Leaky Nun

we just differ on the first 4 bytes

spoiler for item 1 (added 3 bytes to verify all test cases at once)

user165474

Do you have a hint that might help me shorten it a bit?

Leaky Nun

1:21 PM

@HyperNeutrino hint: you don't need to care about 11 at all

user165474

Oh really. Hm. I'm thinking of this the wrong way then. Thanks, I'll rethink the algorithm.

user165474

Wait I can just completely delete part of my program xD

user165474

14 bytes:

Leaky Nun

the last part can be golfed.

Erik the Outgolfer

@HyperNeutrino Please remove and repost (don't edit) this as a spoiler link :)

user165474

1:23 PM

Okay. :)

user165474

14 bytes SPOILER

user165474

And I'll work on the last part.

user165474

13 bytes SPOILER

Leaky Nun

I'll give you a hint: 2 is also truthy

user165474

Hm.

user165474

1:26 PM

Oh...

user165474

Well I don't seem to need the link separator at least, so 12 bytes

Leaky Nun

good, one byte to go

user165474

Yep :)

user165474

Is empty falsy or invalid?

Leaky Nun

falsey

Erik the Outgolfer

1:28 PM

[] is falsy.

user165474

11 bytes?

Leaky Nun

@HyperNeutrino that would work

user165474

Yay

Erik the Outgolfer

Congratulations!

Leaky Nun

you can click our spoilers then

user165474

1:36 PM

Yay :)

Erik the Outgolfer

@LeakyNun Does 3 need to have formatted output?

Leaky Nun

@EriktheOutgolfer yes it does

Erik the Outgolfer

Then I have a looong 11-byte solution ready. How much do you have?

Leaky Nun

@EriktheOutgolfer the official answer is 9 bytes, and I cannot golf it down any further

Erik the Outgolfer

Got to work!

user165474

1:49 PM

Can you explain what $ does? I don't fully understand it.

Leaky Nun

@HyperNeutrino combines the two links before it to form a single monad

e.g. ḅ10$ is equivalent to Ḍ

user165474

Oh. Hm. But in your program, it looks you only have 1 link, or am I reading it incorrectly?

Leaky Nun

where?

user165474

+ṚḌ$$

Leaky Nun

ṚḌ$ is one monad

user165474

1:52 PM

Or ṚḌ$+$

user165474

Oh? Hm.

Erik the Outgolfer

That's from my solution, right?

user165474

Okay, but what's wrong with just ṚḌ?

Erik the Outgolfer

ṚḌ is two links, not one.

user165474

Yes. The two are very similar but you have the plus sign in different places :P

user165474

1:53 PM

Okay. I think I might not understand what a link is then.

Erik the Outgolfer

a + b is a commutative math operation.

user165474

Yes.

Leaky Nun

@HyperNeutrino a monad is a link

a dyad is also a link

user165474

Oh. I thought a link was like a line in the code or something like that.

Erik the Outgolfer

The docs are a bit confusing about that.

Leaky Nun

1:55 PM

I agree.

user165474

Alright.

user165474

That makes some sense now.

user165474

Oh, that would make a few of my coding attempts easier.

user165474

That gives a fourth 11-byte answer which is pretty much identical to my first solution except using the same first 5 bytes as Leaky Nun

user165474

(Or a fifth using Erik's method but I won't bother)

Erik the Outgolfer

2:00 PM

Or a sixth using your multi-line method but with S instead of T at the end.

user165474

So there are 3 methods for the start and 2 for the end, giving 6 total solutions that are pretty much identical?

user165474

17 bytes for item 2 This is horribly long. My first successful attempt.

Erik the Outgolfer

@LeakyNun Damn item 3 is so difficult to golf to 9.

Leaky Nun

@EriktheOutgolfer I agree

Erik the Outgolfer

The lowest I've been to is 12. Horrible.

Leaky Nun

2:14 PM

not very horrible really

Erik the Outgolfer

What challenge has the 9-byte solution? (spoiler link again)

Leaky Nun

The challenge I linked to you on TNB

user165474

How short can item 2 be?

Leaky Nun

@HyperNeutrino my solution is 12 bytes

user165474

Alright. I'll try to get down close to that.

Erik the Outgolfer

2:17 PM

@LeakyNun That's a CnR.

Anyways I'll look into it.

Wow that's... clever.

user165474

@LeakyNun In your code for item 2, are you using ÆF?

Leaky Nun

yes

user165474

Alright.

user165474

So I'm not sure about the Ḣ*HḢḞ helper link; is that a good idea, or is it longer than necessary?

Leaky Nun

don't know what it does

user165474

2:22 PM

[x, y] -> x ** (y / 2) (integer division) (monadic)

Leaky Nun

oh I didn't use ÆF

I used Æf

user165474

Oh.

user165474

I see.

user165474

That might help shorten it. I'll see if I can change my algorithm to use the pure prime factorization instead.

user165474

Do you use ċ anywhere?

Leaky Nun

2:29 PM

not really

user165474

Hm? What do you mean by not really?

Leaky Nun

I didn't use that anywhere.

user165474

Okay.

user165474

I'm trying to use it but it doesn't seem to be working the way I want it to.

user165474

Darn, I can't figure out where to go using the Æf atom. Can you give me a hint?

Leaky Nun

2:42 PM

œ^/

user165474

Multiset symmetric difference?

user165474

Hm.

user165474

Can you explain that?

user165474

Also, can I add an answer to the reverse and add degeneracy problem, or will you because you got it first?

Leaky Nun

@HyperNeutrino I won't add

user165474

2:45 PM

Alright.

Leaky Nun

@HyperNeutrino well, it removes pairs

user165474

@EriktheOutgolfer Will you add your Jelly answer to the degeneracy problem?

Erik the Outgolfer

Yeah.

user165474

Hm. I'll play around with it on TIO and see if I can figure it out.

user165474

Alright.

Leaky Nun

2:46 PM

@EriktheOutgolfer can you explain how œ^/ works?

user165474

I'm pretty sure ours are too similar so I won't add my own.

Erik the Outgolfer

Yeah they are.

@LeakyNun It means "reduce by multiset symmetric difference".

Leaky Nun

@EriktheOutgolfer please explain to @HyperNeutrino how it works

Erik the Outgolfer

I'm not really sure how his code works...

user165474

After playing around with it, I think it does something along the lines of "for every element, if it's not in the list, add it to the list, otherwise, remove it from the list"

Erik the Outgolfer

2:49 PM

...anyways I'll explain.

user165474

Alright. Thanks.

user165474

Oh well sorry, I have to go now. I'll load back to here later and read the explanation. Thanks! :D

Erik the Outgolfer

Let's say you have the list [[1, 2], [1, 3], [3, 4]].

Since we are reducing, we first examine the first two elements, [1, 2] and [1, 3].

Since 1 is contained in both sets, we remove it, so we get back [2, 3].

Then, we take the result and the next element and do the same (2,3œ^3,4).

Since 3 is contained in both sets, we again remove it.

The final result is [2, 4].

Leaky Nun

@EriktheOutgolfer what's the significance of that?

Erik the Outgolfer

I am explaining how œ^/ works here.

user165474

4:14 PM

@EriktheOutgolfer Ohhh That makes a lot of sense now. Thanks! :)

user165474

4:46 PM

@LeakyNun I've figured out this much: 10 bytes, which gives the part under the square root. I can't seem to figure out how to get the other value concatenated properly in 2 bytes.

Leaky Nun

4:59 PM

@HyperNeutrino use µ instead

user165474

@LeakyNun You mean in place of the start_dyadic_link?

Leaky Nun

yes

well no

@HyperNeutrino your method should work

but if you had used the proper square root, you would need one more byte

(Æ½)

user165474

Using µ causes it to behave weirdly.

Leaky Nun

well you would need to change the logic as well

monadic chains and dyadic chains are drastically different

user165474

Yeah

user165474

5:02 PM

Oh okay.

user165474

Whoops, I have to go now (class). Thanks, I'll ping you if I figure it out. TTYL!

Leaky Nun

I'll post a spoiler here.

user165474

7:26 PM

@LeakyNun I got this which does the same thing as before for the same number of bytes, and it uses a monadic chain instead of a dyadic chain.

user165474

9:19 PM

@LeakyNun Figured it out!

user165474

It does k -> [a, b] where a**2 * b == k for maximum a.

user165474

Our solutions turned out almost identical, except the last byte. :P

user165474

@LeakyNun Can you quickly just scan over this to make sure I understand the code correctly? Thanks!

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