Oh alright. I don't like the idea of using so many increment/decrements.

I think it could be much shorter.

Can you give me a hint for the array generation? I might be missing an atom that I can't find, perhaps.

Ohhhh...

Whoops.

Right, I made a 5-byte program to squarify a matrix >_>

Thanks!

@LeakyNun ²R$s lol

Hm. I forgot to test without it.

Even better :P

Oh okay, I see. Thanks.

lol 13 bytes -> 3 bytes what was I thinking? :P

though the arrays are now in the opposite direction. shouldn't cause issues though

Oh, wow! Beautiful.

Nice :)

@LeakyNun I'm guessing that I will need ¦.

Alright. Thanks. (I was looking for something like that but I missed it the first time ???)

@LeakyNun So, right now I have the following code:

RḂÐf
²Rs

The first link will give the indices that don't need to be reversed (though I can easily do it the other way around)

The second (main) link gives the square matrix

But I can't seem to figure out how to do the last part (before formatting)

RḂÐf
²RsµU³Ç$¦

This is what I have

but it gives an error saying that ints are not iterable

Oh hm. I need to look at what that does. Thanks.

Alright.

Perhaps.

Turns out I need the filter-in not filter-out. :P

Well this works (without the output formatting):

RḂÐḟ
²RsµU³Ç¤¦

I think I get how the square box thingy works, but I'll wait for Dennis to reply to you to see what the $ actually does

A single monad?

G

RḂÐḟ
²RsµU³Ç¤¦G

Yay :D Thanks!

Apparently it could be one byte shorter. Hm. (I think)

Actually 2 bytes shorter

Retrieved from this answer

Oh true. Sorry.

I'm going to do this challenge without looking at the Jelly answer (if one exists).

Oh yup, you have an answer. (userscript to tell me the winner)

Ooh, 5 bytes. Hm. Interesting.

here

No problem

@LeakyNun I got this for 10 bytes... badly golfed; this is my first attempt using a pretty straightforward method.

Oh. Hm.

Whoops, forgot to consider duplicate elements.

@LeakyNun So, my next method is to somehow get rid of one instance of the current number. I thought of an old method I used to get a desired value to the front, which was good for Python but maybe not for Jelly, was to slice the array up until that point and move the slice to the back (kind of like linking the array to itself in a loop and cutting it at a different point). I did this with Python using list splicing; I could theoretically do that with Jelly, but it might be too long.

Yes. I'm assuming there's a built-in to move a bunch of elements from one side to the other.

Well, let's see.

I'm going back to the loop thingy I described.

Rather than cutting it off at a different point...

What if I moved the elements?

Would that be called translating?

Hm.

Since it's on a loop.

So...

Rotate?

ṙ?

That looks promising.

@LeakyNun So just doing ṙ doesn't work because it's rotating by the current element, so I need to rotate by the index, but I feel like using enumerate would be too long.

Yes. I suppose that's probably mostly what I'm training for right now.

Oh, since the indices are just a list of increasing consecutive numbers, I could just use LR to get the length and get a range from that.

Or apparently I could use J because there's a single atom to get a range up to the length of another array.

ṙJ rotates it into the right place.

@LeakyNun my thought for removing the first element in every array was to zip, take out the first array altogether, and then zip again.

But it looks like the first zip is unnecessary, because it gives the same thing back (which makes sense)

Oh wait, because J starts at 1, I need to get rid of the last array.

@LeakyNun ṙJṖZ gives me the arrays, and I just take the minimum of each of them I guess.

@LeakyNun this appears to work for 6 bytes.

Thanks! :D Now I need to figure out where that one byte can come off.

It appears that operations seem to go column-wise between arrays.

So I don't need to zip.

But ṙJṖṂ won't work because Ṃ will just compare the sub-arrays individually...

Wait, lemme see what's equivalent to Ṃ but requires vectorization

Well I guess there are two ways of finding the minimum.

The first is to run through them all, replacing the variable value if the current value's smaller

but that doesn't sound very Jelly-ish

because it has variables and those don't exist per se in Jelly

Or, keep chucking out any element that is higher than another element.

So assuming there's a built-in to take the minimum of two values, just reduce over that? maybe?

@LeakyNun ṙJṖ«/

Alright.

Oh hey, they're identical. Hm.

Which one? The totient?

EDT (don't ask why i'm up this late; I typically don't sleep til 2)

Which means I will be going soon, in probably less than half an hour.

Alright.

Hm. I will probably need to think of this challenge from a mathematical perspective first.

Well obviously totient(n) < n

So if totient(x) == n then x > n.

I could map over a generator and stop when I get the right value in Python, but I don't think Jelly has generators, so I need to figure out what the maximum of x will be.

Well if we have x = n**2 + 1, then all multiples of n up to x are in the list of coprimes, so we have totient(x) >= n. I conjecture that we will never need to go over this number, but I don't have rigid proof.

@LeakyNun Assuming that my conjecture is correct (which I will attempt to prove when it's not 2am), then ²RÆṪi: should work; square, range, map using totient, find the index, and divide it by itself to get 1 or 0 because the value has to be consistent.

I tested it with a few of your cases, but I will add 3 bytes to run through all cases on the OEIS.

Alright. Thanks, see you later!

I ran through the tests and from what I can tell, it works perfectly. I checked the answer, and it's pretty much identical (square, range, map to totient), but Dennis uses the function to check if it's in the array and I divide the index by the argument which seems to work, but actually might not work for certain values.