@Uriel If you can make it shorter without assignment, sure, go ahead. Unfortunately ⎕ can't go immediately before the { } because ∘.⎕{ } would parse as (∘.⎕){ } instead of ∘.(⎕{ })

@ngn Doesn't '\-/ '[⊖⎕∘.{0::4⋄2+×⍺⎕⍵}⎕] work?

@Adám that inputs f multiple times

@ngn Chat doesn't do markdown for multiline messages, btw.

@Adám @ngn hey guys, could someone give me a hand here? I'm trying to iterate over a vector while adding to it. I've currently got 2{n←1⋄z←⍵,⍨⍺⍴⍺⋄(≢z)<⊃⌽⍵:z,←⍵,⍨z[n]⍴⍺⋄n+1}5 20. I'd like to recurse (or loop) until ≢z≥20. The ⋄n+1 there is not relevant atm, I was just testing

@J.Sallé that code is wrong, it's supposed to be 2{n←1⋄z←(⊃⍵),⍨⍺⍴⍺⋄(≢z)<⊃⌽⍵:z,←(⊃⍵),⍨z[n]⍴⍺⋄n+1}5 20

@J.Sallé You might be interested in the power operator, ⍣

@Ven Yeah that's what I was thinking, but I couldn't get it to work

I'll fiddle with it a bit more

(that's a doc link btw, it just doesn't show)

@J.Sallé that or you can do recursion with ∇

@ngn that'd be difficult to do since I've got a lot of stuff going on, wouldn't it? I'd have to pass n and z to the function, right?

@J.Sallé you could either have them as arguments (left or right; separately or in a pair) or you can modify them in place

@J.Sallé note that modified assignment like A,←B modifies A in place if it already exists

mh, what?

⎕←({⍵+1}⍣{⍵=0})0

@Ven
1

@Ven ⍣ applies the left operand until the right operand returns true for two consecutive values

ahhh, right, it's for fixpointing...

@Ven yes, the classic use case is f⍣≡ which means apply f until the sequence converges

@ngn surprisingly, this doesn't error, but it's not doing what I want either

Do I really need to pass z to it too? I've no idea how to handle that

@J.Sallé it will take me some time to find out what it's doing

@J.Sallé which problem are you solving?

⎕←+∘1(=∘1{v←⍺ ⍵ ⋄ ⍺⍺ ⍵:v ⋄ ∇ v})0

@Ven

Rebuilding user command cache... done
Was OFF -trains=box -fns=off
VALUE ERROR
 __field_initialize_result_←(⎕NS ⍬).⍎'⎕CY''salt''⋄⎕SE.UCMD''box on -fns=on -trains=tree''⊣enableSALT' ⋄ ⎕←+∘1(=∘1{v←⍺ ⍵ ⋄ ⍺⍺ ⍵:v ⋄ ∇ v})0
                                                                                                                   ∧

Real time: 1.275 s
User time: 1.114 s
Sys. time: 0.056 s
CPU share: 91.71 %
Exit code: 0

@ngn this one

:\ ...bah.

I'm trying to create the sequence first, not actually solving the challenge

⎕←1÷∘+⍣≡1

@ngn
1

ah, I got it wrong...

⎕←1+∘÷⍣≡1

@ngn
1.618033989

there's the golden ratio (or its reciprocal)

how does ∇even handle ⍺⍺..

⎕←+∘1(=∘1{v←⍺ ⍵ ⋄ ⍺⍺ v:v ⋄ ⍺(⍺⍺∇)v})0

@Ven for operators there's ∇∇

@Ven

Rebuilding user command cache... done
Was OFF -trains=box -fns=off
VALUE ERROR
 __field_initialize_result_←(⎕NS ⍬).⍎'⎕CY''salt''⋄⎕SE.UCMD''box on -fns=on -trains=tree''⊣enableSALT' ⋄ ⎕←+∘1(=∘1{v←⍺ ⍵ ⋄ ⍺⍺ v:v ⋄ ⍺(⍺⍺ ∇)v})0
                                                                                                                   ∧

Real time: 1.313 s
User time: 1.166 s
Sys. time: 0.060 s
CPU share: 93.32 %
Exit code: 0

@ngn I know, but does ∇ keep its function operand?

@Ven to be honest, I don't know what ∇ would be in a user-defined operator

⎕←+∘1(=∘1{v←(⍺{⍺⍺w})⍵ ⋄ ⍺⍺ ⍵:v ⋄ ∇∇v})0

@Ven

Rebuilding user command cache... done
Was OFF -trains=box -fns=off
VALUE ERROR
 __field_initialize_result_←(⎕NS ⍬).⍎'⎕CY''salt''⋄⎕SE.UCMD''box on -fns=on -trains=tree''⊣enableSALT' ⋄ ⎕←+∘1(=∘1{v←(⍺{⍺⍺ w})⍵ ⋄ ⍺⍺ ⍵:v ⋄ ∇∇ v})0
                                                                                                                    ∧

Real time: 0.955 s
User time: 0.897 s
Sys. time: 0.031 s
CPU share: 97.15 %
Exit code: 0

I'm not good enough for that.

@Ven I think you need to pass ⍺⍺ to the left of ∇∇ if you're trying to call it recursively

@ngn well, that's not the issue (yet):

⎕←+∘1(=∘1{v←(⍺{⍺⍺w})⍵ ⋄ ⍺⍺ ⍵:v ⋄1})0

@Ven

Rebuilding user command cache... done
Was OFF -trains=box -fns=off
VALUE ERROR
 __field_initialize_result_←(⎕NS ⍬).⍎'⎕CY''salt''⋄⎕SE.UCMD''box on -fns=on -trains=tree''⊣enableSALT' ⋄ ⎕←+∘1(=∘1{v←(⍺{⍺⍺ w})⍵ ⋄ ⍺⍺ ⍵:v ⋄ 1})0
                                                                                                                    ∧

Real time: 0.990 s
User time: 0.905 s
Sys. time: 0.046 s
CPU share: 96.09 %
Exit code: 0

I think I'm going to have to solve that with 2 or 3 functions

@J.Sallé so, in the tio link, you're creating z in the dfn and you're modifying it in the same dfn?

@J.Sallé that doesn't look like it's going to accumulate anything

@J.Sallé maybe if you create z outside and then append with z,← multiple times?

ah oopsie..

@ngn yes. I ultimately wanted it to become a vector like 2 2 5 2 2 5 2 2 2 2 2 5

I got my jot precedence wrong >_>

@ngn is there a trick to apply a on w?

without having to do (⍺{⍺⍺⍵})⍵

is z←⍬ a valid way to create an empty vector?

@J.Sallé yes

@Ven you're asking for a way to apply ⍺ to ⍵?

@ngn yes

@Ven well, ⍺ is supposed to be an array, not a function

I know it is :c

@Ven in an operator the operand is ⍺⍺

@ngn but I want to pass two operators. Guess I'll use ⍵⍵

@Ven you mean pass two functions to the operator?

@ngn yeah. I'll just use ⍺⍺ and ⍵⍵

ah, I think it's better now!

⎕←({⍵=5}{v←⍵⍵ ⍵⋄⍺⍺v:v⋄(⍺⍺∇⍵⍵)v}{⍵+1})0

@Ven
5

applies ⍵⍵ until the value matches ⍺⍺ :D.

Welp, I'll be back in a few

@Ven that's a very elaborate way for working around ⍣ :)

@ngn well, sometimes you don't want that "stable" behavior that ⍣ has.

@Ven I agree. In k there is a proper "until": f g/ a (g gets applied multiple times to a until f returns true)

@ngn I did try to use / in dyalog as well, but to no avail.

That language bar is really nice :).

@Ven I hope it's not too off-topic that I mention k so often in an APL room.

@Ven Yeah, you could use reduction, but you have to know in advance how many iterations you want to make.

@ngn if you (or @Adám) have a simpler version than my until, I'd gladly take it. I think though, wrt K, that it was fine in this room.

@Ven ⍣ is simpler

It doesn't do the same thing, though...

@Ven to save the extra iteration, you could use ⍺ instead of ⍵

⎕←{⍵+1}⍣{⍺=5} 0

@ngn
5

@ngn it's still iterating an extra time, right?

just discarding the result

@Ven f⍣g is still calling g an extra time, yes

@Ven but f is called the correct number of times

@Ven which is probably what you care about

well, it calls prev ⍵⍵ next right..?

so next is calculated even though it's discarded

since it calls 5{⍺=5}6

@Ven oops, I was wrong. It's actually next ⍵⍵ prev

⎕←{⍵+1}⍣{⎕←'⍺='⍺'⍵='⍵⋄⍺=5} 0

@ngn
┌──┬─┬──┬─┐
│⍺=│1│⍵=│0│
└──┴─┴──┴─┘
┌──┬─┬──┬─┐
│⍺=│2│⍵=│1│
└──┴─┴──┴─┘
┌──┬─┬──┬─┐
│⍺=│3│⍵=│2│
└──┴─┴──┴─┘
┌──┬─┬──┬─┐
│⍺=│4│⍵=│3│
└──┴─┴──┴─┘
┌──┬─┬──┬─┐
│⍺=│5│⍵=│4│
└──┴─┴──┴─┘
5

wait, what?

oh.

oh..

ok, so my think is really overcomplicated just because I switched ⍺ and ⍵. oops.

@Ven it's still a good exercise in operator syntax :)

s/think/thing

@Ven @ngn ∇ in a monadic dop is equivalent to ⍺⍺∇∇ and in a dyadic dop it is equivalent to ⍺⍺∇∇⍵⍵.

Why is ⍬ f 2 5 returning 5 instead of 2 2 5 for f←{⍺,⊃⌽⍵,⍨⊃⍵⍴⊃⍵}?

nvm I figured it out

I'm getting a value error now though D:

I also figured it out, but why does ⍬f 2 5 return a value error and ⍬ f 2 5 works fine?

@Adám ^

@J.Sallé your expression, a little simpler: n←2 ⋄ k←5 ⋄ z←(n⍴n),k ⋄ {20<≢z:z ⋄ z,←(z[⍵]⍴n),k ⋄ ∇⍵+1}1

@ngn oh, that's good! Thanks!

@J.Sallé likely it can be solved more efficiently with other techniques than recursion

@ngn I was trying to make a dop to work it out, but I got stuck >.>

@J.Sallé I forgot to mention: ⎕IO←0 above

Nice, thanks!

@J.Sallé It doesn't by me. TIO link?

@Adám hm, it's working properly after a )clear, but I took a screenshot

@J.Sallé You had f2 with no space between f and 2.

yeah I jsut noticed that too

by the way, does ⍬≢⍬=0?

@J.Sallé No, ⍬≡⍬=0 because ⍬=0 is ⍬ and ⍬≡⍬.

Oh, so that's why my dfn wasn't working, I assumed the contrary

@J.Sallé Well, ⍬=0 will extend the scalar (0) to be paired up with every element of the vector (⍬), i.e. none, so the result is 0⍴something boolean, i.e. 0

@Adám I see. Could you help me debug this? With that input it works fine, but when I input any other vector instead of ⍬ I get a rank error

@J.Sallé First, what is it supposed to do?

If the first element of the left argument is ⍬, it should create a (⊃⍵⍴⊃⍵),⊃⌽⍵ vector

@J.Sallé Let's straighten some things out. You have g←(⊃⍺)≡⍬:g, (which could be shortened to g←⍬≡⊃⍺:g,) but the guard only fires if the condition (and therefore g) is true, so you might as well write :1,

@J.Sallé And then on the next line, we know that g is 0.

It shouldn't be adding that 1 before the 1 2

@J.Sallé So, you don't need g at all.

Okay, I think I'm following

@J.Sallé In general, (A)f⍨B should always be changed to B f A

I ultimately want to create a vector following this sequence codegolf.stackexchange.com/questions/150262/…

Not trying to solve the challenge yet, just want to create the vector

And I'm trying to solve it without having to fall back to :For in the tradfns

Although I realize I'll probably need two or three functions for it to work out

@J.Sallé Hm, that's complicated. I think I have to get some dinner first.

Hahahah no problem, I'll need to leave in a few anyway

I'll get on that tomorrow, hopefully