Hey @Adám, I remember a while ago someone saying something about a syntax that would work like a :While loop in a dfn, am I mistaken or does such syntax exist?

@J.Sallé Tail recursion is free, so you can do {while_true:∇ something ⋄ finally}

@J.Sallé Another way is to use the power operator: {action}⍣{~while_true}⍬

Can someone tell me why is this working perfectly up to 13 and erroring for 14+?

The error message is weird, too

⎕←{¯1↑⍕∊(∊0≠⍎¨⍕!⍵)⊆⍕!⍵}¨⍳14

oh the bot's not here facepalm

@J.Sallé I think because !14 is represented as 8.71782912E10

⍎SYNTAX ERROR
 .
∧

@EriktheOutgolfer oh, I need a larger ⎕FR then

@J.Sallé nope, that will not work either

@EriktheOutgolfer yeah, just tested. Worked when I did ⎕PP←34

⎕FR is for floating-point size

Yeah, I mixed it up

actually doing both ⎕PP←34 and ⎕FR←1287 you can extend it up to !31

@EriktheOutgolfer yeah I tested up to !30 and it worked fine

So, I have this tacit fn 0≠∘⍎¨⍕!5, but I needed to replicate (or enlist/partition) the resulting vector using ⍕!5 as argument again. Any tips?

What I mean is 0≠∘⍎¨⍕!5 results into a vector 1 1 0 and I wanted do 1 1 0/⍕!5

@J.Sallé {⍵/⍨0≠∘⍎¨⍵}⍕!5?

@EriktheOutgolfer I have a Dfn for the challenge already made, now I wanted to find a tacit solution

{¯1↑⊃(∊0≠⍎¨f)⊆f←⍕!⍵} it's this, for reference.

@J.Sallé first of all I think the tacit function is 0≠∘⍎¨⍕∘!, right?

and you want to use the same argument again

@EriktheOutgolfer yeah, the 5 is the argument

@J.Sallé you'd need parens to clarify

@EriktheOutgolfer Yeah, that's what I imagined. I can't seem to make it work though

@J.Sallé first of all what's the tacit function on its own?

like, before the / replication

(0≠∘⍎¨⊢)⍕!5 this also returns 1 1 0

so the argument is 5 or !5?

@EriktheOutgolfer Currently, that's it 0≠∘⍎¨⍕!

Argument is 5

@J.Sallé why are you using dyadic ⍕?

@EriktheOutgolfer it's not supposed to be dyadic >.>

It's formatting !5 into a string

@J.Sallé then I assume you want 0≠∘⍎¨ separate from ⍕!

@EriktheOutgolfer probably? I think it's being read separately

@J.Sallé assuming you want something like (0≠∘⍎¨⊢)⍕∘! then

@EriktheOutgolfer That also results in 1 1 0 though

@J.Sallé in your case the solution should then be as simple as (⊢/⍨0≠∘⍎¨⊢)⍕∘!

but this will make ⊢/ into one function

hm

@EriktheOutgolfer that results in a length error in my session

@J.Sallé because ⊢/ means "reduce by right"

@J.Sallé but using a dfn in there would be the same size! {⍵/⍨0≠∘⍎¨⍵}⍕∘!

that shouldn't make / into an operator anymore

@EriktheOutgolfer hm, I think that might work

@EriktheOutgolfer Worked well, I believe. The final function is 2 bytes shorter than my Dfn ¯1↑∘{⍵/⍨0≠∘⍎¨⍵}⍕∘!

@J.Sallé can't you do {¯1↑⍵/⍨0≠∘⍎¨⍵}⍕∘!?

@EriktheOutgolfer apparently I can hahahahah

I'll edit it all later though, I need to go home. Been at work for 12 hours