The APL Orchard

Bubbler

7:07 AM

This Haskell library sounds a lot like a backend data structure for APL

Time to build yet another APL derivative...

Adám

@Bubbler It does indeed, but interestingly, it is last-axis focused, as opposed to APL/J/K that are first-axis focused.

Feeds

9:19 AM

0

Q: Getting the type of a value in APL

I have a long vector which should be a character vector but when I print it using Dyalog's DISPLAY function it turns out to be a mixed vector. Now I need to find out which of the elements is not a character. How do I retrieve the type of a value in APL?

Adám

1:43 PM

@jordancurve Welcome. Interested in APL?

nathan rogers

3:58 PM

oops

Dyalog APL

@nathanrogers
{(⍴⍵)⍴(((0<⍺)×|⍺)⍴0),(-⍺)↓⍵,((0>⍺)×|⍺)⍴0}

nathan rogers

how do you do 2 left arguments with ¨?

I'm trying right f ⍨¨left doesn't seem to work in my case for some reaspon

and left¨ f right doesn't work either

J. Sallé

4:14 PM

@nathanrogers you need to do each of ⍺f ⍵?

If so, ⍺f¨⊂⍵ should work

nathan rogers

4:27 PM

why is that?

⎕← 3 ¯3{(⍴⍵)⍴(((0<⍺)×|⍺)⍴0),(-⍺)↓⍵,((0>⍺)×|⍺)⍴0}¨⊂1 0 1 1 1 0 1 1

Dyalog APL

@nathanrogers
┌───────────────┬───────────────┐
│0 0 0 1 0 1 1 1│1 1 0 1 1 0 0 0│
└───────────────┴───────────────┘

nathan rogers

for this I think I have a solution to problem 7 from 2018 challenge

but I'm not sure

Adám

@nathanrogers Sure, but I think you can make it much simpler.

nathan rogers

it doesn't use a guard, like the question suggests

H.PWiz

No, and it's flat, which is also good.

Adám

4:29 PM

@nathanrogers True, but it does a lot of extra work.

nathan rogers

@H.PWiz flat?

@J.Sallé why does that ⍺f¨⊂⍵ work to each ⍺?

H.PWiz

@nathanrogers No nested arrays

nathan rogers

oh, hmm. do many solutions have that?

H.PWiz

Yes

J. Sallé

@nathanrogers @Adám can explain that a lot better than me :p

nathan rogers

4:31 PM

@H.PWiz I can't imagine an example. could you show me?

I was thinking of trying this with ⊤⊥ but I couldn't seem to get it to work correctly

there's probably some kind of ...bits.../bits kind of solution

H.PWiz

Oh, hang on, I misunderstood your question. I was thinking of a different "that". I don't know if there are solutions that use nested arrays.

nathan rogers

Problem 7 - Unconditionally Shifty Logical bitwise shifting is a common operation where bits in a fixed width field are moved to the left or right by a specified amount, N, causing N bits to be "shifted out" at one end and N 0's to be filled in at the other end.

that's the problem from 2018

H.PWiz

If ↓⍣¯1 worked, a cute solution would be {⍺↓⍣¯1⊢(-⍺)↓⍵}. (I think this is correct)

However:

⎕←0↓⍣¯1⍳5

Dyalog APL

@H.PWiz

nathan rogers

eh

H.PWiz

4:40 PM

Drop ⍺ elements from one end. Append ⍺ 0s to the other end

Actually it doesn't work, when ⍺ is bigger than ⍴⍵

J. Sallé

⎕←3 ¯3{⍎(0>⍺)⊃'⌽(≢⍵)↑⍺↓⍵' '(≢⍵)↑⍺↓⍵'}¨⊂1 0 1 1 1 0 1 1

Dyalog APL

@J.Sallé
INDEX ERROR

J. Sallé

oh IO

⋄←⎕IO←0⋄⎕←3 ¯3{⍎(0>⍺)⊃'⌽(≢⍵)↑⍺↓⍵' '(≢⍵)↑⍺↓⍵'}¨⊂1 0 1 1 1 0 1 1

Dyalog APL

@J.Sallé
SYNTAX ERROR

Adám

I have a 12-byte solution.

J. Sallé

4:50 PM

I don't remember the correct syntax, mr bot.

Adám

@J.Sallé Hint: )about

nathan rogers

something like ⊣⎕IO←0

Adám

⋄⎕IO←0⋄⎕←3 ¯3{⍎(0>⍺)⊃'⌽(≢⍵)↑⍺↓⍵' '(≢⍵)↑⍺↓⍵'}¨⊂1 0 1 1 1 0 1 1

J. Sallé

@Adám I've been here a while, shouldn't need that D:

Dyalog APL

@Adám
┌───────────────┬───────────────┐
│0 0 0 1 1 0 1 1│1 0 1 1 1 0 0 0│
└───────────────┴───────────────┘

J. Sallé

4:51 PM

Ah it doesn't have the first ←

Cheesy solution btw

Adám

It is just like a normal session, except you must begin with ⎕← or ⍞← or ⋄ and you must use ⎕← for any further output.

J. Sallé

Yeah, I just didn't remember it was just ⋄ instead of ⋄←

nathan rogers

let's see those 12 bytes

Adám

@J.Sallé ⋄← is never valid APL.

nathan rogers

or at least give us a hint

Adám

4:53 PM

@nathanrogers @J.Sallé is well on the way. Just need to combine the two parts.

@nathanrogers Hint: A negative left argument to ↑ takes from the right and pads on the left.

nathan rogers

5:17 PM

oops

⎕←3 ¯3{(-⍺)↓((-×⍺)×(≢⍵)+|⍺)↑⍵}¨⊂1 0 1 1 1 0 1 1

Dyalog APL

@nathanrogers
┌───────────────┬───────────────┐
│0 0 0 1 0 1 1 1│1 1 0 1 1 0 0 0│
└───────────────┴───────────────┘

Adám

@nathanrogers The bot doesn't understand edits.

nathan rogers

⎕←3 {(-⍺)↓((-×⍺)×(≢⍵)+|⍺)↑⍵}1

Dyalog APL

@nathanrogers
0

nathan rogers

⎕←5{(-⍺)↓((-×⍺)×(≢⍵)+|⍺)↑⍵}⍬

Dyalog APL

5:19 PM

@nathanrogers

nathan rogers

is that more like what you had in mind @Adám

Adám

@nathanrogers A f¨B (by definition) pairs up each scalar of A with each scalar of B exactly like A+B. If any of A or B is a single scalar, then that scalar is paired up with all the scalars of the other side.

@nathanrogers Try first removing the unneeded bits, and only afterwards "over-taking" to restore original size.

nathan rogers

⎕←3 ¯3{((-×⍺)×≢⍵)↑(-⍺)↓⍵}¨⊂1 0 1 1 1 0 1 1

Dyalog APL

@nathanrogers
┌───────────────┬───────────────┐
│0 0 0 1 0 1 1 1│1 1 0 1 1 0 0 0│
└───────────────┴───────────────┘

Adám

@nathanrogers Perfect!

And I just noticed that my solution needs 14 bytes, not 12. (Because the problem uses -⍺ for some reason.)

nathan rogers

5:42 PM

:P

I guess you could just define it differently, for some reason I imagine ¯3 as shifting right not left

⎕←3 ¯3{((×⍺)×≢⍵)↑⍺↓⍵}¨⊂1 0 1 1 1 0 1 1

Dyalog APL

@nathanrogers
┌───────────────┬───────────────┐
│1 1 0 1 1 0 0 0│0 0 0 1 0 1 1 1│
└───────────────┴───────────────┘

nathan rogers

@Adám that makes more sense to me, but the negatives shifting left for negative seem to be more APL-ish?

Adám

@nathanrogers Right, which is why I suggested the author to swap it. But he disagreed.

@nathanrogers That one lends itself well to become a 3-train: {(×⍺)×≢⍵}↑↓

nathan rogers

5:57 PM

that doesn't work for me

Adám

Notice that you have (⍺ f ⍵) g ⍺ h ⍵. That's how trains work; it becomes f g h.

⎕←3 ¯3({(×⍺)×≢⍵}↑↓)¨⊂1 0 1 1 1 0 1 1

Dyalog APL

@Adám
┌───────────────┬───────────────┐
│1 1 0 1 1 0 0 0│0 0 0 1 0 1 1 1│
└───────────────┴───────────────┘

nathan rogers

why are the parens necessary?

what does the expression mean without them?

Adám

3 ¯3{(×⍺)×≢⍵}↑↓¨⊂1 0 1 1 1 0 1 1 is simply evaluated as it looks. First ↓¨⊂1 0… then monadic ↑ then dyadic dfn.

@nathanrogers An expression is a function if the rightmost token is a (derived) function. So you must isolate trains by parenthesising or naming.

H.PWiz

6:31 PM

@nathanrogers That doesn't work for 0 as a left argument

Adám

@H.PWiz Oh, I did't think of the edge case. It has been almost a year ¯\_(⍨)_/¯

nathan rogers

sign of zero

is 0?

that's annoying

sign of zero or 1 I guess

Adám

⍞←¯1*¯3 0 4

Dyalog APL

@Adám ¯1 1 1

nathan rogers

or that

H.PWiz

6:36 PM

You could add 0.5

Adám

@H.PWiz Ah yes, that'd work.

nathan rogers

{((1∨(×⍺))×≢⍵)↑⍺↓⍵}

why not that?

Adám

⍞←1∨¯3

Dyalog APL

@Adám 1

nathan rogers

thinking in js

undefined || 1
-3 || 1
3 || 1

H.PWiz

6:38 PM

What makes you think ∨ is or

⎕←∘.∨⍨⍳10

Dyalog APL

@H.PWiz
1 1 1 1 1 1 1 1 1  1
1 2 1 2 1 2 1 2 1  2
1 1 3 1 1 3 1 1 3  1
1 2 1 4 1 2 1 4 1  2
1 1 1 1 5 1 1 1 1  5
1 2 3 2 1 6 1 2 3  2
1 1 1 1 1 1 7 1 1  1
1 2 1 4 1 2 1 8 1  2
1 1 3 1 1 3 1 1 9  1
1 2 1 2 5 2 1 2 1 10

nathan rogers

? because it is?

Adám

@nathanrogers It is indeed, but (unfortunately, imho) extended to GCD

dzaima

@nathanrogers it is for booleans, but not beyond that

H.PWiz

It behaves like or on 0/1 but so does ⌈, you wouldn't call ⌈ or

@Adám I like having GCD, personally

nathan rogers

6:40 PM

{(((×.5+⍺))×≢⍵)↑⍺↓⍵}

so that then

H.PWiz

And since there wasn't another symbol for it already...

Adám

Me too, but I don't like it being paired with ∨

@nathanrogers Other than the double-paren, yes.

nathan rogers

oh, hah, I was adding on the wrong side

Adám

@H.PWiz I would have wanted to define ~ as 1+- and ∧ as × and all other Boolean functions based on those two.

However, I'd have used ¬ for NOT, as that shows the relationship with -.

@PaulMansour Long time no see. Welcome back!

nathan rogers

@Adám in any case there really should be a this or that operator

I suppose (bool)/this that is ok

but this || that would be welcome for sure

dzaima

6:47 PM

@nathanrogers imo that doesn't really fit the APL theme

Adám

@nathanrogers What does that do?

dzaima

for one APL doesn't remotely allow short-circuit

Adám

@dzaima Uh, not true:

⋄ ⍎⎕FX'f' ':if 1' ':orif 1⊣⎕←''boo''' '⎕←''yay''' ':endif' ⋄ ⎕←⎕VR'f'

Dyalog APL

@Adám
yay
⍎VALUE ERROR

dzaima

@Adám *not for regular functions

Adám

6:52 PM

⎕←⍎⎕FX'r←f' ':if 1' ':orif 1⊣⎕←''boo''' 'r←''yay''' ':endif' ⋄ ⎕←⎕VR'f'

Dyalog APL

@Adám
yay
     ∇ r←f
[1]    :If 1
[2]    :OrIf 1⊣⎕←'boo'
[3]        r←'yay'
[4]    :EndIf
     ∇

dzaima

obviously it can with special syntax, but any language can do anything with special syntax for it

Adám

@dzaima Tradfns are regular.

dzaima

@Adám I meant as making a regular function e.g. or that wouldn't evaluate ⎕←'hi' in 1 or ⎕←'hi..

Adám

⋄ or←{⊃⍺⍺⍬:1 ⋄ ⊃⍵⍵⍬} ⋄ ⎕←1 or {1⊣⎕←'hi'} ⍬ ⋄ ⎕←0 or {1⊣⎕←'hi'} ⍬

Dyalog APL

6:58 PM

@Adám
1
hi
1

Adám

You can use inline dfns as code blocks.

dzaima

@Adám right, but that's again very annoyingly long, and doesn't pass ⍺/⍵/⍺⍺/⍵⍵/∇/∇∇ down to the inner functions

nathan rogers

@dzaima what doesn't?

Adám

@dzaima How is it long? You define it once, and then you have it. I'm not sure what the problem is with passing is, but sure, you can define or←{⍺←⊢ ⋄ ⍺ ⍺⍺ ⍵:1 ⋄ ⍺ ⍵⍵ ⍵} and then use it as ⍺ {cond1} or {cond2} ⍵`

dzaima

@nathanrogers Adám's or, e.g. {0 or {1+⍵} ⍬}123 doesn't give the expected 124

@Adám that has 4 unnecessary chars in the calling of or - ⍺, {, }, ⍵ that all could be avoided with a syntactical function (I'm not saying that a thing like that should exist though, quite the opposite)

Adám

7:11 PM

@dzaima Fair enough. Btw, we are considering allowing multiple guards, but that will be short-circuiting "andif". E.g. {0:1⊣⎕←'hi':'yay' ⋄ 'nay'} will not print hi.

nathan rogers

10:41 PM

https://tio.run/##PU@9bsIwEN79FLdhq@LHToasTEw8AWKwBIqqUjsi7YAQEwglQCRegS5UVbe0S8c@yr1IuHMQ0vm@u@@@@@Sz2aI7W9mFT5umM7Yvzy4FCyPvZzBcpu@vc/fWEVmC@/PaUcLyA6s/PO6kw/Ly1OemOGBVY/UT4ldhcZWSdcUXTRXJqLHMWjZgtsZT0Q9D3mW4ulbXemyEzUkrI9BgFEjzwCigBh1yqHuEECnQwmtaYk0cbGoTC2@IehDaCM/GXsM9zL3A6lss@UqyhkF4nEXTEFte1hIPW5tP6G9TBVkC1Pq23fx/8kF0Yn4D
This is a solution for problem 8

except for the case where the left argument is ⍬

when maxnum≡⌈/⍬

how do I avoid this?

1≡⍬ p8 1 is the expected resuilt

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