@Adám suppose I have a vector of boolean pairs such as (0 0)(1 0)(0 1); is there a way to apply ⍸ (or some other op/fn) so it would return, for this example, 2 3?

@J.Sallé Why 2 3?

@Adám Well, the idea is that it would return the same as ⍸(0 1 0)(0 0 1)

@J.Sallé A domain error?

@Adám >.> stupid me, it's ⍸¨(0 1 0)(0 0 1)

⎕←⍸¨↓⍉↑(0 0)(1 0)(0 1)

@Adám
┌─┬─┐
│2│3│
└─┴─┘

@J.Sallé ↓⌽↑ is a Dyalog-style idiom, i.e. it is very fast and doesn't waste memory.

Ah, nice! That's what I needed!

@J.Sallé You may also be interested in:

⎕←⍳∘1¨↓⍉↑(0 0)(1 0)(0 1)

@Adám
2 3

Alternatives are 1⍳¨⍨↓⍉↑ and 1⍳⍨⍤1⍉↑

Okay, just a second

Ah, I needed a second ↑ there

I might need some help actually, my function's rapidly becoming an abomination

Yup, an abomination indeed: {⍵⊃¨⍸¨↓⍉↑↑{(>,<)/≢¨{(0=(⍳⍵)|⍵)/⍳⍵}¨⍵,⍎⌽⍕⍵}¨⍳1e4}

I wanted to get that into Extended to use the infinity symbol instead of 1eX (if that's even possible) since idk how big X should be

That's a solution for this challenge btw

Oh, I knew there was a shorter version of that

@J.Sallé Or tacit: ∪∨∘⍳⍨

Even better

@Adám is it possible to swap the 1eX for the infinity symbol in extended?

@J.Sallé No, that'd require too much memory.

Hm, I see

my problem is 1e4 only goes as far as 100odd numbers while 1e5 takes very long to run

@Adám is there an overhead difference between funcall and ⍎'funcall'

@Nathan yes. ⍎ is not "dereference" — a function we've contemplated adding.

thanks

@J.Sallé {⍵⊃¨⍸¨↓⍉↑↑{(>,<)/.. -> {⍵⌷⍉1↓⊢⌸{×-/..

@ngn oh damn

{⍵⌷⍉1↓⊢⌸{×-/≢¨(∪∨∘⍳)⍨¨⍵,⍎⌽⍕⍵}¨⍳1e4} still takes a while but not nearly as long

Very nice, thanks @ngn

@J.Sallé tacit isn't always shorter

≢¨(∪∨∘⍳)⍨¨ -> {≢∪⍵∨⍳⍵}¨

@ngn nice, I didn't see that

well, could've been (≢∘∪⊢∨⍳)¨ but dfns are easier :)

@ngn Look ma, no parens: ≢∘∪¨∨∘⍳⍨¨

@Adám are you still on?

@Nathan Yup. Laptop troubles made boot to safe mode, so the chat agent didn't auto-start.