⋄ (2÷1+0.5*⍨5)(1-⍨+∘÷⍣≡⍨1)(⊃⌽2÷/{⊃+\∘⌽⍣⍵⊢0 1}¨⍳100)

@FawnLocke 0.6180339887 0.6180339887 0.6180339887

@RGS I've posted with an alt account on r/math the proposal to keep 13/8 as the Fibonacci day as 1.625 is much closer to phi (~1.618) than 23/11 (~2.0909). Thanks for pointing this out, you were right. Good night room.

The only connection with 23/11 (more like 11/23) and Fibonacci is that the sequence starts with 1, 1, 2, 3

@Bubbler I feel 2x stupid now :|...

Good morning APLers :) Today I had my 2nd APL dream and woke up with a happy feeling. Wish the same to you too. Cheers.

Thank you!

@milia Did your dream provide the solution to a problem?

@Adám no, far from it. It just reassured me that I'm on the correct path, and all the 'pain' needed to master APL will in the end reward me with precious gifts, and perhaps dear friends :)

from there, I just did some work on the problem

experimenting and testing ideas on RIDE

it's very weird. I had a similar feeling 21 years ago, as a freshman at the physics school, when I felt that I was at the right place the right time. Second time I felt that, was at the math school (2015). And now. :)

Savour the moment!

@Adám I most certainly will :) Have a nice day Adám !

norm ← {((⍺*2) + ⍵*2)*÷2}

any idea how to reduce those bracket from my function?

@elliptic00 What can you tell me about the arguments? Simple scalars?

@Adám, just a complex number

{2*∘÷⍨+/⍺⍵*2}

2*∘÷⍨ can also be written 0.5*⍨

Tacit: 2*∘÷⍨,+.×, or 2*∘÷⍨1⊥,*2⍨

If you make the function monadic (taking a vector of two numbers), you can write 2*∘÷⍨+.×⍨ or +.×⍨*∘÷2⍨

Nice.. I need time to understand it,

I wish we had the Under operator ⍢, as then we could write the dyadic function as +⍢(×⍨) and the monadic function as +/⍢(×⍨)

¯1*÷2
0J1
      2*∘÷¯1
0.5
      2*∘÷⍨¯1
0J1

If you swap the arguments, you have to use ⍨

yep.. I just realize that.. thanks..

⋄ U←{⍺←{⍵ ⋄ ⍺⍺} ⋄ ⍵⍵⍣¯1⊢(⍵⍵ ⍺)⍺⍺(⍵⍵ ⍵)} ⋄ 3J3 +U(×⍨) 4J4

@Adám 5J5

⍝ so ⍨ apply on (*∘÷),  not just ÷ ?`

Correct. Operators have long left scope.

I c, I'm always wondering how that possible because I always think the smile face apply on the division only

@Adám #!f where f is any fn?

That will print

 # !+

oh

I've got 10, or 7 if boxing is not style=max.

Are "flat-arrays", in regards to performance, non-nested rank 1 arrays or can they be of higher rank

can be higher rank

Thanks

∘ is bind like function composition?

What is similar operation on other language like Python or Haskell?

@elliptic00 There are 4 cases, but in the f∘g x case it is like (f∘g)(x) of traditional mathematics.

∘ with one array operand is currying, but with two function operands, it is but one of many possible function compositions

@elliptic00 in haskell it's similar to .

⍝ Yep, but In APL, it is more complicated, three compose function operators: ⍤ ∘ ⍥

Not more complicated; more possibilities! They are all equivalent in the monadic f⎕g case.

to be entirely fair, the entire time i've used APL, i never lacked function compositions

there was a way to express everything i came up with

so i'd agree on the more possibilities part - it's fairly cool

I definitely miss "under" occasionally, and "before" often

Under isn't a true composition, but before – yes.

before is... reverse compose?

Yeah

@elliptic00 It's the B/Bluebird combinator in combinatory logic

I tend to define O̲←{⍵ ⍵⍵⍨⍺⍺ ⍺} when I need reverse-compose.

you mean like SKI combinator?

More broad than that en.wikipedia.org/wiki/Combinatory_logic

I don't know that much about SKI calculus, it could have the B combinator

@Adám Is ⌿ a hyperator since ⌿⌿ is valid. Or is ⌿ (or /) treated as a function when used as a left argument?

@FawnLocke ⌿ is treated as a function when used as operand (although it can change back into an operator later: ⎕←⌿{+⍺⍺ ⍵}2 3

@Adám 5

Gotcha, thanks

However, ⌿⌿ does not guarantee that the lefthand ⌿ is a function: ⎕←+⌿⌿2(3 1 4 1 5)

@Adám
┌───────┐
│4 5 5 6│
└───────┘

So would it be a hyperator (or round about hyperator in that case). The overloading can be confusing ^^'

Wait, nevermind.

I understand now, thanks.

The only thing in regular APL you could consider a hyperator is . because it "takes" ∘ as left hyperand, ∘ being an operator.