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13:35
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1 hour later…
14:50
Hey @Adám, if/when you have some time later today, could you lend me a hand in solving this challenge? I want to learn how to properly use the thingy and it seems like a good challenge for such
@J.Sallé OK, I'm a bit busy atm, but I'll try to get to it. Are you available later?
@Adám yup, no problem. I’ll be here for about 4 and a half hours
Can anyone enlighten me on using in place of index notation for ? ↑[0.5] and ↑[1.5] feels weird, and apparently there's a way to use instead, but I haven't found any clear documentation on the topic.
For instance:
  ⎕ML←3
  ⊃[1](1 2 3 4)(5 6 7 8)(9 10 11 12)
Is there an equivalent for ⎕ML≤1 using ?
ngn
ngn
15:15
@JackHenahan what is your ⎕IO?
if it's 1, you can achieve the same with ⎕ML←1 ⋄ ⍉↑(1 2 3 4)(5 6 7 8)(9 10 11 12)
if it's 0, just ⎕ML←1 ⋄ ↑(1 2 3 4)(5 6 7 8)(9 10 11 12)
⎕←⎕ML←1 ⋄ ⍉↑(1 2 3 4)(5 6 7 8)(9 10 11 12) ⋄ ⎕IO←0 ⋄ ↑(1 2 3 4)(5 6 7 8)(9 10 11 12)
@J.Sallé
1
rip bot
you need ⎕←...
oh, and the ⎕← prefix can be turned into to avoid confusion
@EriktheOutgolfer ah yeah, I forgot that
15:22
@J.Sallé maybe because it's something Adám added 2 days ago
⋄ ⎕ML←1 ⋄ ⎕←⍉↑(1 2 3 4)(5 6 7 8)(9 10 11 12) ⋄ ⎕IO←0 ⋄ ⎕←↑(1 2 3 4)(5 6 7 8)(9 10 11 12)
@J.Sallé
1 5  9
2 6 10
3 7 11
4 8 12
1  2  3  4
5  6  7  8
9 10 11 12
That's what I was going for.
@EriktheOutgolfer yeah but I saw him use it already, I just forgot about it :p
ngn
ngn
I don't think there's a way to do this with ⍤ and it wouldn't be shorter anyway.
@J.Sallé it's easy to forget something if it's recent ;)
15:24
Indeed
15:38
@JackHenahan ⎕ML 3's monadic is simply monadic in ⎕ML≤1.
@Adám I challenge you to (whenever you're not that busy) reduce ($₁%₁$₁)%₁$₁%₁$₁ to 5 tokens or less, given that $ₓ represents a monadic function and %ₓ represents a dyadic function (both single tokens), here is mine with ⍺⍺ being monadic and ⍵⍵ being dyadic
(that's a spoiler)
@JackHenahan Actually, I just tried and it doesn't work like I expected, so I've sent an email to our internal language group to ask about it, and I'll get back to you.
@EriktheOutgolfer Are all the $₁s the same function?
@EriktheOutgolfer If so, %₁⍨⍣2$₁ and %₁⍨$₁%₁$₁ are both 5 tokens. (Didn't look at spoiler.)
16:09
@Adám quick question, ⍺-1∇⍵-1 will be interpreted as ⍺-(1∇⍵-1) right?
16:23
@J.Sallé Correct.
16:51
@Adám that's what the ₁ means, yeah
@Adám the first one is exactly my spoiler
17:50
@J.Sallé I don't think is a good fit for that. is good when you need to make a decision for each cell according to consistent rules. Here, each cell needs to be treated differently. Have a good weekend!
@Adám ah, I see. I'll try something else then. Thanks a lot, have a good weekend!
@J.Sallé Maybe this is useful:
⎕←∘.≤⍨⍳5
@Adám
1 1 1 1 1
0 1 1 1 1
0 0 1 1 1
0 0 0 1 1
0 0 0 0 1
@Adám oh, indeed it is!
does this ∘.<something> trick work with every fn/op of the language bar?
Well maybe that's a stupid question, I didn't mean every fn/op. Maybe all the ones on the = group?
18:24
@ngn @J.Sallé @Adám Thanks for the pointers. I was just going through the APL tutor with Dyalog and got curious about ⍤. Looking forward to hearing what @Adám discovers.
@JackHenahan on the meanwhile, you might want to take a look at the weekly lessons that Adám teaches in this chatroom :)
18:52
@J.Sallé Thanks for the link. Looks like fantastic material :)

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