@RubenVerg Haha. You've inspired me:

(5 5 5⍴⍳125){(+⌿÷≢)⌷∘⍺¨,⊃∘.,⌿↓(⌊⍵),⍪⌈⍵}0.9 2.2 3.1

Not quite four characters, though ⍨

Oh, wait. That's not right. I'm not actually lerping correctly there. Oh well.

@RubenVerg Uh oh, now we need to decide if satire/humour posts go on ∘.×

@Adám I think there's just enough actual APL stuff in there that it might fit on jdt

@B.Wilson I think (1⊥,⍤⊢×(×⌿¨⍤1∘|⍤|⊂⍤⊣-¨~⍤,⍤⍳⍤⍴⍤⊢))(2↑⌊⍤⊣↓⊢) works with io 0

@RubenVerg You're a train monster.

@B.Wilson I'll take that as a compliment :)

That does seem to break the idea of taking a[1.5] <- 4 as inserting between a[1] and a[2] though...

@Silas Stictly speaking, they don't conflict.

Welcome to APL Quest 2021-6! Today's quest is Fischer Random Chess:

> Write a function that:

> • has a character vector right argument that is a permutation of 'KQRRBBNN'

have two solutions, partly different

Yes, there are a lot of interesting approaches to this.

{(≠/2|⍸'B'=⍵)∧⊃1=⍸/⍸¨,/'RK'∘.=⍵}

{(≠/2|⍸'B'=⍵)∧1='R'⍸⍥(⍸⍤=∘⍵)'K'}

{(≠/2|⍸'B'=⍵)∧'RKR'≡⍵∩'KR'}

@rabbitgrowth That looks very familiar.

2(⊣|↑+.∧↑∘⊖<.<⊃⍤↓)⍋

@mitchelljohnstone If you edit your message (press UpArrow) and insert either 4 leading spaces, or backticks (`) around your code, it will look better.

Did that help?

Yes.

Thanks!

Now, that solution looks very interesting. I need to study how it works, unless you want to explain.

@mitchelljohnstone ⍨

@rabbitgrowth I have (≠/2|∘⍸'B'∘=)∧('RKR'≡∊∘'RK'⊢⍤/⊢) which is pretty much the same thing

@rabbitgrowth last part is smart!

@RubenVerg Yes, you just implemented ∩

ye, that intersection is really nice

@Adám The main idea was that if you use ⍋, Bishops will be the first 2 spots, Rooks will be the last 2, and the King will be the 3rd.

@Richard Where would I as the ⍨?

@mitchelljohnstone was just confused about your nice solution :)

@Richard Lol, thanks!

@mitchelljohnstone apl.wiki/APL_Orchard#Emoticons

ah, then ∘.○

The other solutions are obvious. I'm still trying to grasp the ⍋-based approach.

Comparing the position of the the King (⊃⍤↓) and the position of the Rooks (↑∘⊖), We can get a single result 1 or 0 from <.<. Then, grabbing the position of the bishops (↑), it get's "anded" by the prior result and added +.∧. If the rook/king comparison was 0, the ∧0 will result in 0. Otherwise, the ∧1 is an identity, so it moves onto the bishop position sum (mod 2)

@mitchelljohnstone Yeah, I was just finishing up the reasoning too. Amazing. Just amazing. Wow!

\○/

@Richard You're using a lot of ⍸…=⍵. Have you considered breaking that out to a small utility?

{I←⍸=∘⍵ ⋄ (≠/2|I'B')∧1='R'⍸⍥I'K'}

You may also want to fix the '' case.

no i did not. Just thought ⍸/⍸ looked nice. But yours is better indeed.

I quite like this variant: {I←⍸=∘⍵ ⋄ (2|I'B')∧⍥(≠/)'K'<⍥I'R'}

@Adám didn't manage without {⍵='': 0 ⋄ }

@rabbitgrowth You can make the right side of ∧ be tacit.

I think it actually looks better because it avoids nested enclosures:

{(≠/2|⍸'B'=⍵)∧'RKR'≡⍵∩'KR'}
 {≠/2|⍸'B'=⍵}∧'RKR'≡∩∘'KR'

We can even make the right side explicit for symmetry and ease of grouping:

 {≠/2|⍸'B'=⍵}∧{'RKR'≡⍵∩'KR'}

I like making small function that do one thing each, and then chain them together in conditions using logical functions.

Then my code can become almost (?) a DSL:

    Char←0 2∊⍨10|⎕DR
    Num←2|⎕DR
    Null←∧/⎕NULL≡¨⊢
    Simple←{1=≡,⍵}
    Scal←⍬≡⍴
    Vec←{1=≢⍴⍵}
    Mat←{2=≢⍴⍵}
    String←Simple∧Char∧Vec∨Scal
    Table←Mat>0∊(String∨Simple∧Scal∧Num∨Null)¨

(from here)

Anyway, I think that'll be enough for today.

@Adám you say can't do ^.= but could you do ^.≡ ?

Yes, but that's not clearer.

@Adám true, just thought might be faster, although probably not with the each

@Adám I think I like this one best. The other one feels like an odd mix of dfn and tacit

Thanks Adám!

∩'KR' is very clever

Why stencil? Does windowed reduction not work?

I just wanted to note down the first solutions that came to mind before scrolling down to answers