« first day (1764 days earlier)      last day (1187 days later) » 

Fawn Locke
Fawn Locke
00:01
⋄ (2÷1+0.5*⍨5)(1-⍨+∘÷⍣≡⍨1)(⊃⌽2÷/{⊃+\∘⌽⍣⍵⊢0 1}¨⍳100)
TryAPL
TryAPL
@FawnLocke 0.6180339887 0.6180339887 0.6180339887
milia
milia
00:33
@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.
Bubbler
Bubbler
The only connection with 23/11 (more like 11/23) and Fibonacci is that the sequence starts with 1, 1, 2, 3
 
2 hours later…
milia
milia
02:13
@Bubbler I feel 2x stupid now :|...
 
4 hours later…
milia
milia
06:29
Good morning APLers :) Today I had my 2nd APL dream and woke up with a happy feeling. Wish the same to you too. Cheers.
Adám
Adám
Thank you!
@milia Did your dream provide the solution to a problem?
milia
milia
@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. :)
Adám
Adám
Savour the moment!
2
milia
milia
07:05
@Adám I most certainly will :) Have a nice day Adám !
 
2 hours later…
elliptic00
elliptic00
09:23
norm ← {((⍺*2) + ⍵*2)*÷2}
any idea how to reduce those bracket from my function?
Adám
Adám
@elliptic00 What can you tell me about the arguments? Simple scalars?
elliptic00
elliptic00
@Adám, just a complex number
Adám
Adám
{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⍨
elliptic00
elliptic00
09:38
Nice.. I need time to understand it,
Adám
Adám
I wish we had the Under operator ⍢, as then we could write the dyadic function as +⍢(×⍨) and the monadic function as +/⍢(×⍨)
elliptic00
elliptic00 
¯1*÷2
0J1
      2*∘÷¯1
0.5
      2*∘÷⍨¯1
0J1
Adám
Adám
If you swap the arguments, you have to use ⍨
elliptic00
elliptic00
yep.. I just realize that.. thanks..
Adám
Adám
⋄ U←{⍺←{⍵ ⋄ ⍺⍺} ⋄ ⍵⍵⍣¯1⊢(⍵⍵ ⍺)⍺⍺(⍵⍵ ⍵)} ⋄ 3J3 +U(×⍨) 4J4
TryAPL
TryAPL
09:43
@Adám 5J5
elliptic00
elliptic00 
⍝ so ⍨ apply on (*∘÷),  not just ÷ ?`
Adám
Adám
Correct. Operators have long left scope.
elliptic00
elliptic00
I c, I'm always wondering how that possible because I always think the smile face apply on the division only
 
2 hours later…
Adám
Adám
11:50
CMC: Shortest expression that begins with #! and has no effect.
rak1507
rak1507
@Adám #!f where f is any fn?
Adám
Adám
That will print 
 # !+
rak1507
rak1507
oh
Adám
Adám
I've got 10, or 7 if boxing is not style=max.
 
2 hours later…
Fawn Locke
Fawn Locke
13:58
Are "flat-arrays", in regards to performance, non-nested rank 1 arrays or can they be of higher rank
dzaima
dzaima
14:11
can be higher rank
Fawn Locke
Fawn Locke
Thanks
 
4 hours later…
elliptic00
elliptic00
18:20
∘ is bind like function composition?
What is similar operation on other language like Python or Haskell?
Adám
Adám
@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
rak1507
rak1507
18:59
@elliptic00 in haskell it's similar to .
elliptic00
elliptic00 
⍝ Yep, but In APL, it is more complicated, three compose function operators: ⍤ ∘ ⍥
Adám
Adám
Not more complicated; more possibilities! They are all equivalent in the monadic f⎕g case.
Kamila Szewczyk
Kamila Szewczyk
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
Fawn Locke
Fawn Locke
I definitely miss "under" occasionally, and "before" often
Adám
Adám
Under isn't a true composition, but before – yes.
Kamila Szewczyk
Kamila Szewczyk
19:11
before is... reverse compose?
Fawn Locke
Fawn Locke
Yeah
@elliptic00 It's the B/Bluebird combinator in combinatory logic
Adám
Adám
I tend to define O̲←{⍵ ⍵⍵⍨⍺⍺ ⍺} when I need reverse-compose.
elliptic00
elliptic00
you mean like SKI combinator?
Fawn Locke
Fawn Locke
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?
Adám
Adám
@FawnLocke ⌿ is treated as a function when used as operand (although it can change back into an operator later: ⎕←⌿{+⍺⍺ ⍵}2 3
TryAPL
TryAPL
19:18
@Adám 5
Fawn Locke
Fawn Locke
Gotcha, thanks
Adám
Adám
However, ⌿⌿ does not guarantee that the lefthand ⌿ is a function: ⎕←+⌿⌿2(3 1 4 1 5)
TryAPL
TryAPL 
@Adám
┌───────┐
│4 5 5 6│
└───────┘
Fawn Locke
Fawn Locke
So would it be a hyperator (or round about hyperator in that case). The overloading can be confusing ^^'
Wait, nevermind.
I understand now, thanks.
Adám
Adám
The only thing in regular APL you could consider a hyperator is . because it "takes" ∘ as left hyperand, ∘ being an operator.

« first day (1764 days earlier)      last day (1187 days later) »