Jelly Hypertraining

btw i saw you in the contact (or maybe Spyfall, I can't remember now) and went "Caird!" but then thought it would be weird to reach out to you out of the blue :)

caird coinheringaahing

9:21 PM

Yeah Spyfall I think

I realised I hadn't posted anything this year

Mr. Xcoder

I posted some answers in july and august

but complete break between january and april and april and june

caird coinheringaahing

yep :/

Mr. Xcoder

I actually think this one is pretty challenging btw :)

It will surely take a while for me

caird coinheringaahing

Yeah, its definitely making me realise how rusty I am

Mr. Xcoder

Though

I remember I bountied leaky's answer

on this exact question on main

and IIRC it was at least 15 bytes

caird coinheringaahing

9:25 PM

Yeah but Leaky's nuts

Mr. Xcoder

Of course

I'm tempted to check if Dennis answered too but I don't want to look

Uhh..

@cairdcoinheringaahing I must confess that I don't really know how to do this in Jelly

caird coinheringaahing

I'm giving it a go, but I'm inclined to agree

at 21 bytes so far :/

Mr. Xcoder

Because the way I thought I'd solve this is using the method with column elimination (ie Laplace)

And this requires recursion

And I still don't know how to use ß in Jelly :)

caird coinheringaahing

I'm using recursion as well, but with multiple links, not ß

Mr. Xcoder

Ah well

Then

I may be able to do it

I'll give it a go actually

caird coinheringaahing

9:34 PM

Remind me how Ŀ gets indexes again?

Is 1 the bottom link?

Mr. Xcoder

No

It's the top one

AFAIK

caird coinheringaahing

Hmmm

Mr. Xcoder

If you want the bottom link you can use Ñ if you're on the link immediately before the main link

caird coinheringaahing

Or Ç to cycle from the top to the bottom

Mr. Xcoder

My answer is starting to look like a program lol

Oh god

I think I'm done

But... 25 bytes

caird coinheringaahing

9:45 PM

I've had to restart :/

Mr. Xcoder

No rush anyway...

I'll have to golf this thing down

Mr. Xcoder

10:01 PM

So I realised my solution is faulty

I disregarded the fact that the head atom pops....

caird coinheringaahing

:/ Unlucky

Mr. Xcoder

10:16 PM

@cairdcoinheringaahing How about working together on it from scratch?

caird coinheringaahing

Yeah that seems better than anything I've got

Mr. Xcoder

Perfect

So

caird coinheringaahing

First thing is figuring out how to get the co-factor of (x, y)

Actually, that gives me an idea:

CMC: Given a matrix M and a tuple (x, y) return the co-factor of the corresponding element

Mr. Xcoder

That should be easy

caird coinheringaahing

[[1, 2, 3], [4, 5, 6], [7, 8, 9]] (3, 3) -> -3

Mr. Xcoder

10:22 PM

Wait a second

I think this is actually harder than needed

Can I show you my thoughts?

caird coinheringaahing

Yeah go ahead

Mr. Xcoder

So for example for [[1, 2, 3], [4, 5, 6], [7, 8, 9]], I was thinking about doing a first row expansion so 1*det([[5, 6], [8, 9]]) - 2*det([[4, 7], [8,9]]) + 3*det([[4, 5], [5, 8]])

Makes sense so far?

caird coinheringaahing

Yeah

Mr. Xcoder

Well for this task

I had something along the lines of: ḢNÐeṚḋZœcLÑƲ

So given the matrix, do ḢNÐe so [1, -2, 3]

Ooops, idk what that ḋ is doing there

But anyway

Since head pops, we're left with [[4,5,6],[7,8,9]].

Which is kinda what we want

We do ZœcL to get the right combinations of matrix-element pairs

So at this point we have the matrices and the factors

One just needs to recurse

And of course multiply and sum

But this is where I get stuck

caird coinheringaahing

I don't get how the last bit ZœcLÑƲ works

Mr. Xcoder

10:29 PM

So

We have [[4,5,6],[7,8,9]]. We want to associate 1 with the columns not containing it

We zip [[4,5,6],[7,8,9]] to get [[4,7],[5,8],[6,9]], the columns

and 1 needs to be paired with [5,8],[6,9], 2 with [[4,7], [6,9]] and 3 with [[4,7],[5,8]]

Wait a second...

Yeah it's fine

So

We need this association

Regarding ÑƲ, it's just a placeholder for now

But do you get the main point so far?

caird coinheringaahing

Oh, that's the bit that was confusing me

Mr. Xcoder

Good

We'll have to clarify that later

Do you agree that now all we need to do is multiply the elements we already have with the determinants of the matrices we already have and sum to get the answer?

caird coinheringaahing

Yep

Mr. Xcoder

Good

Setting up this recursion is where I get stuck though...

Pretty damn hard challenge for a warm up @cairdcoinheringaahing

caird coinheringaahing

Yeah, didn't exactly think it through :/

Mr. Xcoder

10:38 PM

Do you even want to proceed with this or pause?

caird coinheringaahing

I'm testing a few recursion methods

Mr. Xcoder

Good

caird coinheringaahing

I feel like a ? clause may work

Mr. Xcoder

Well, yes

Help me get some things straight

Should we recurse until we reach a 1 by 1 matrix?

caird coinheringaahing

I think that would be better than a 2x2

Mr. Xcoder

10:41 PM

Then Ḋ? might work

caird coinheringaahing

That's if deque right, not a single quick?

Mr. Xcoder

Yes

I wonder what I'm doing wrong...

I have ḢNÐeṚḋZœcLßƲµḢḢ$Ḋ? and I thought it may work

caird coinheringaahing

I don't think you need Ḣ$, because you're just doing Ḣ on a 1x1?

Mr. Xcoder

Hmm isn't that 1 by 1 of the form [[something]]?

caird coinheringaahing

Oh yeah :/

Mr. Xcoder

10:48 PM

So ḢNÐeṚ gets the first row, signed appropriately (presumably) and dot-products it with the (presumably computed through) recursion determinants

caird coinheringaahing

Got it, add a € after the ß, because it's currently recurring with all of the sub-matrices at once, rather than one by one

Mr. Xcoder

Ohhh

Yeah

Nice catch

It works!

caird coinheringaahing

Thanks

Mr. Xcoder

Thank you

So I'm down to 19: ḢNÐeṚḋZœcLß€ƲµḢḢ$Ḋ?

With your help...

caird coinheringaahing

Weird phenomenon: if you take an nxn matrix and fill it with 1, 2, 3..., 9, 0 (repeating to fill), the determinant is 0

Mr. Xcoder

10:51 PM

Yeah I always found that interesting

Any shorter way to generate the matrix without the nth column?

caird coinheringaahing

I can't think of any way

Mr. Xcoder

Oh there actually is!

caird coinheringaahing

How?

Mr. Xcoder

> Ƥ Map a link over prefixes, and if given a <nilad>, overlapping infixes if positive, or the overlapping outfixes of abs(<nilad>) if negative. <monad>Ƥ or <monad><nilad>Ƥ

Don't overlapping outfixes mean exactly that?

caird coinheringaahing

Umm

Mr. Xcoder

10:54 PM

Of course set to length 1 so -Ƥ?

caird coinheringaahing

I never quite understood how Ƥ works aside from prefixes

Worth a shot though

Mr. Xcoder

Let me test

Look: tio.run/##y0rNyan8///hjkV@hyekPtw56@GO7qijk5N9Ds9/…