@MasterQuiz ah I didn't know that thanks!

i don't think you noticed @xpqz but i am pretty sure there is a typo on iteration#del

CMC: for (2≤≢⍵)∧1=⍴⍴⍵, if ≢⍵ is odd, extract odd elements and even elements, e.g. for 1 0 2 3 4 return (1 2 4)(0 3), if ≢⍵ is even, the last element belongs to both sequence, e.g. for 1 0 2 3 return (1 2 3)(0 3)

@LdBeth Why doesn't 1 0 2 3 give (1 3)(0 2 3)?

oh, I mean odd and even position elements, determined by ⎕io←1 ⋄ ⍳≢⍵

so the parity of numbers in the sequence doesn't matter, the function could even accept characters, e.g. 'bananas' gives 'bnns' 'aaa'

@LdBeth Dyalog Extended, 20: 0 ¯1↓¨¯1.5 2,⌿⍤⍴⊢,⊢/ Try it online!

wow, that's a lot shorter that I expecting

Oh, I thought you'd have a much shorter, clever solution.

@LdBeth Also note that it uses the extended ⍴. The equivalent vanilla Dyalog solution would be 0 ¯1↓¨(⌈2,⍨2÷⍨1+≢),⌿⍤⍴⊢,⊢/

@user16722171 Hi there. If you want to participate here, please email access@apl.chat

I was using Replicate instead of reshaping which makes it much longer

^ my font with little support for apl

also how do i make that shorter ^ (i am talking about all those 2s)

@pVCaecidiosporeadduced (8⍴2)

Or shorten the whole thing to {⍵⊤⍨8⍴2}

⎕← 8 {⍵⊤⍨⍺⍴2} 231

@pVCaecidiosporeadduced 1 1 1 0 0 1 1 1

2∘|J <- what is the ∘|?

@pVCaecidiosporeadduced ∘ with an array operand is "Currying". It binds the 2 as constant left argument to |, thus deriving a monadic function you could call "parity". Technically speaking, ∘| can stand alone as a monadic operator, but that is probably not what you have in mind.

⎕← 4×.÷ 2 2⍴⍳4

@pVCaecidiosporeadduced 5.333333333 2

is there a way i can visualise these operations?

@pVCaecidiosporeadduced Like this?

      'tc'⎕CY'dfns'
      4(×tc).(÷tc) 2 2⍴⍳4
4  ÷  1  =>  4
4  ÷  2  =>  2
4  ÷  3  =>  1.333333333
4  ÷  4  =>  1
4  ×  1.333333333  =>  5.333333333
2  ×  1  =>  2
5.333333333 2

yes something like that

@pVCaecidiosporeadduced An alternative is ⎕←4{⍺'×'⍵}.{⍺'÷'⍵} 2 2⍴⍳4

@Adám
┌───────────────┬───────────────┐
│┌─────┬─┬─────┐│┌─────┬─┬─────┐│
││4 ÷ 1│×│4 ÷ 3│││4 ÷ 2│×│4 ÷ 4││
│└─────┴─┴─────┘│└─────┴─┴─────┘│
└───────────────┴───────────────┘

@TryAPL this looks better

Using a scalar left argument to f.g doesn't really illustrate what it does though. A better example is ⎕←(2 2⍴⎕A){⍺'×'⍵}.{⍺'÷'⍵}(2 2⍴⎕C⎕A)

@Adám
┌───────────┬───────────┐
│┌───┬─┬───┐│┌───┬─┬───┐│
││A÷a│×│B÷c│││A÷b│×│B÷d││
│└───┴─┴───┘│└───┴─┴───┘│
├───────────┼───────────┤
│┌───┬─┬───┐│┌───┬─┬───┐│
││C÷a│×│D÷c│││C÷b│×│D÷d││
│└───┴─┴───┘│└───┴─┴───┘│
└───────────┴───────────┘

@pVCaecidiosporeadduced You're not wrong. I shall fix. Many thanks for eagle-eyed reading.

@zeynel Welcome back. Long time no see!

@pVCaecidiosporeadduced Is there a reason why you want to put 8 twos? Cause you can do something like that 2⊥⍣¯1⊢

@MasterQuiz to get 8 bits

Announcement: BAA webinar in 15 minutes. Zoom 858 532 665, passcode: ×/1920 12 17

@Adám was it supposed to end?

i thought the audio got cut off abruptly

@Razetime Yes.

alright

@pVCaecidiosporeadduced Just asking, cause when I didn't know about the inversion trick, I used to put some twos to be sure I always get the first "valid" one...

@MasterQuiz Well, you can compute how many 2s you need by taking the 2-logarithm of the argument:

      a←1000?1e6
      ]runtime -c {⍵⊤⍨2⍴⍨⌈2⍟2⌈⌈/⍵}a 2⊥⍣¯1⊢a

  {⍵⊤⍨2⍴⍨⌈2⍟2⌈⌈/⍵}a → 4.0E¯6 |   0% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕
  2⊥⍣¯1⊢a           → 5.1E¯6 | +26% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

what!? how is that faster!!

I think it is because ⊥⍣¯1 has to do all kinds of extra checks. E.g. my dfn doesn't support non-positive numbers.

oh

still, pretty rare a pure(r) apl solution beats a 'primitive'

even though ⊥⍣¯1 isn't really a primitive (is it? idk)

¯\_(⍨)_/¯

@user922600 Hi Dan H. If you want to participate here, please email access@apl.chat

@rak1507 it at least must be implemented as a single thing

yeah