I think I have a fairly good golfed version of the Padovan sequence for this challenge: codegolf.stackexchange.com/questions/182797/…

⎕←2⌷{(2⌷2+/⍵),⍵}⍣6⊢1 1 0

@Sherlock9
4

Is there any way I could golf this further?

@Sherlock9 1 1 0 is part of the bytecount?

Yep. It's the starting array

2+/ not easy to train-ize :(

@Sherlock9 (…),⍵ → ⍵,⍨…

⎕← 2⌷(2∘⌷∘(2∘+/),⊢)⍣6 ⍝ doesn't even work :-(

⎕←2⌷2(⊢,⍨2⌷+/)⍣6⊢1 1 0

@Adám
4

Oh.

Didn't realize ⍣ would quietly pass along its ⍺

@Adám Oh much belated thanks!

@Sherlock9 Sorry, I should have pinged you.

I didn't see the first ping either, so that's alright :D

Is it just me, or do you sometimes wish there was a neat way to do A f A g A h B ?

It isn't just you

Darned if I have a good solution though. Or at least one that I'm sure won't have run into some language issue I don't yet know about

It would require a different and incompatible definition of trains.

Hm, actually…

(f g A h)B ←→ A f A g A h B but there's no obvious dyadic form of that.

And it isn't a very self-explanatory definition, nor is it nice that the left end of the train depends on the rightmost end.

So f g h, except stacked? In which case do you need that one?

@Ven I'm not sure what you mean. (f g h)B is (f B) g (h b). (f g A h)B would be equivalent to A(⊣f⊣g h)B

Right. I meant that it'd be f(g(h..)) instead of g(f, h)

@Adám I wondered why @ doesn't pass its ⍺ to its ⍵⍵

@Ven What would it then pass if called monadically?

It wouldn't pass it then.

@Ven I'm not sure (that we can ever find out, now that its author isn't with us any more). It does seem a bit odd to mix levels (main data vs replacement/modifier) like that, and forces writing ⍵⍵ to ignore left arg (just as annoying as with e.g. ⌺ and ⍣), but I can see the occasional use. Maybe if you think of @ as a special case of ⍢ it makes more sense that it shouldn't pass in ⍺, as ⍺ is really part of ⍺⍺ rather than a proper argument.

What do I mean by ⍺ being a part of ⍺⍺? Note that A f@g B is exactly the same as A∘f@g B and A@g B is exactly the same as A∘⊣@g B so you can see that ⍺ for @ isn't really a proper argument, but just an operand shorthand.

no I see what you mean

I'd just have a use for it sometimes.

@Ven It isn't hard to define or inline it when needed. APL isn't a golfing language.

Oh, for sure. I'm just sayin' I'd have a use for it when golfing

@ is a special case of ⍢ in that A f@g B and A∘f@g B are the same as A∘f⍢{(,g ⍵)/,⍵}B but not the same as A f⍢{(,g ⍵)/,⍵}B which would be (,g A)/,A)f@g B so you can see here too that passing A into g makes no sense for ⍢ and therefore no sense for @ either, the latter being a special case of the former.

(,g A)/,A)f@g B is probably missing a ( at the start

@Ven Indeed.

What could be consistent, is that the A f⍢g B would supply the "opposite" argument as left argument to g, but then what would the inverse application look like‽

And honestly, in a non-golfing-related tangent, I sometimes would like a better idiom for (g A)/A

Well, f/⊢ is not select, so I can't write that.

@Ven That is idiomatic, although I tend to write A/⍨g A. Yeah, I know that a few people have wanted the "filter" operator {⍵⌿⍨⍺⍺ ⍵} but @arcfide argues fiercely that it constitutes unnecessary abstaction.

Right, "idiom" is not the right word for that. I just want a better way :).

@Ven Right, your real problem is not the lack of a filter operator, but rather that slashes are schizophrenic.

Yup, pretty much.

@Ven I'm very confident that f⊢⍤⌿⊢ will work in 18.0 (although maybe not spelled like that). It isn't as neat as a dedicated primitive, but we'd need four of them, and you'll quickly learn to read ⊢⍤⌿ as "filters".

Explanation, please?

@Ven ⊢⍤⌿ is simply the function ⌿ (as opposed to the possibly-but-actually-unfortunately-in-this-case-operator)

@Ven I'm proposing that ⍤ is "overloaded" so that f⍤g means (f g) which cures schizophrenia because an operator cannot be an operand, so ⊢⍤/ forces / to be a function and (⊢ g) is the same as g so ⊢⍤/ is the function /

Is J's " overloaded in some other way?

@Ven Yes, f"g means f applied with rank of whatever the built-in default rank of the g function is.

ah alright. Thank you.

@Ven But J already has "atop" as @ even though it doesn't have mentally ill primitives ;-) however, atop is essential in J because (f g)B isn't f g B and "dfns" are very awkward.

And K doesn't need "atop" because (f g h)B is f g h B

I remember little of the J I did, but I do remember dfns being a pain point

@Ven Yeah, I really like how dfns resemble a formalised inline version of the curly-left-brace in TMN.

arthur whitney: "might be time to revive the rank operator." groups.google.com/d/msg/shaktidb/bAKz1ikMcjs/qYDV8DHaBwAJ

@Adám you called it chat.stackexchange.com/transcript/52405?m=47048015#47048015

@Sherlock9 1 1 0 -> 3≠⍳3 (⎕io=1)

@ngn Grr, I was looking into something like that. How did I miss it‽

actually, can't we start with 3⍴1?

Oh I could also use 1 1 1, 3⍴1

@ngn Jinx :P

and it might be easier to return the whole sequence instead of the n-th term. i'm not sure

2 bytes shorter :) {{⍵,+/2↑¯3↑⍵}⍣⍵×⍳3} (⎕io←1)

Oh nice

the length of the result is longer than ⍵ though. i don't know if that's allowed

Hm and just the 3⍴1 gets me to 18 bytes which is already shorter

@Sherlock9 ⊢3⍴1 -> ×⍳3

@ngn Edited it in. Thanks muchly, it's 17 bytes now

⎕←2⊤152

@Ven
0

Huh. I don't think I understand how ⊤ works.

@Ven Most often, you want ⊥⍣¯1 instead of ⊤:

⍞←2(⊥⍣¯1)152

@Adám 1 0 0 1 1 0 0 0

@Ven ⊤ treats a scalar left argument as a 1-element list, so you only get the trailing bit/digit.

Ah yeah, thanks

⍞←1 0 0 1 1 0 0 0 (1=-⍥(+/))1 0 0 1 1 0 0 1

@Ven SYNTAX ERROR

Oh duh, there's no Extended there.

^ This works if I'm already working with binary representations, but I want (f A) .. (f B). Maybe I can abuse ⍢, or maybe I'm better off with a pair

⋄ Ö←{(⍵⍵ ⍺)⍺⍺(⍵⍵ ⍵)} ⋄ ⎕←1 0 0 1 1 0 0 0 (1=-Ö(+/))1 0 0 1 1 0 0 1

@Adám
0

@Ven Right, but that one is simple enough to model inline:

Fair enough.

@Ven Actually, you can use extended, but it is a bit more awkward.

⍞←↑2(⊥⍣¯1)¨152 24

@Ven 1 0 0 1 1 0 0 0

Whoops, wrong command. Anyways: mix pads to the end. Not what I want, I'll have to find another solution

@Ven What were you trying to do?

I am, however, not sure it matters for this challenge.

⎕←2⊥⍣¯1⊢152 24

@Adám
1 0
0 0
0 0
1 1
1 1
0 0
0 0
0 0

⍞←1=-/+⌿2(⊥⍣¯1)152 24

@Ven 1

I guess I can save a byte in Extended by replacing 1= with ≠

@Ven Uh, no, only if the result can never be 2 or ¯1

thinks of other schemes

#tio do apl-dyalog-extended ⎕←1=-/+⌿⊤152 24

@Adám 1

Dammit! I had seen that part of ⊤ last time I checked :(.

OK, now I need to write a checker that the code itself can have a single byte changed and not work

@Ven A single bit, that is, right?

@Ven That will be hard, e.g. ⊥ and ⊤

Yes. I need to check the codepage.

@Ven Then watch out for ⎕AV vs SBCS. Anyway, gotta go.

#tio do apl-dyalog-extended ⎕←(1=(-/)∘(+⌿)∘⊤)152 24

@Ven 1

⎕←(1=(-/)∘(+⌿)∘⊥)152 24

@Ven
SYNTAX ERROR

OK, well, ⊥ and ⊤ won't be an issue it seems.

Actually Extended is probably worst suited for this challenge

#tio do apl-dyalog-extended ⎕←(1=1-/⍤⊥⊤)152 24

@Adám 1

@Ven ^^

@Adám Issue is that one-bit-changes have to mean an error. I can replace / with + (at least in ASCII) and it won't error

@Ven *one-bit

Right indeed.

here's a program for SBCS mapping

0 1 1
1 0 0
0 1 0

cmc: produce this matrix or its transposition

(my cunning plan: ⊃+.×⍣⍵⍨ of that matrix is the ⍵th padovan number)

wow. primes. who would have thought

@ngn APL (Dyalog Extended), 8 bytes 1⍭3 3⍴⍳9

@ngn J, 9 bytes 1 p:i.3 3

in apl the best i have is 10: 2 5 4⊤⍨3⍴2 so the whole padovan solution ⊃+.×⍣⎕⍨2 5 4⊤⍨3⍴2 is 17 bytes - exactly as many as Ven's best

@ngn APL (Dyalog Extended), 6 bytes ⊤2 5 4

⊤ is a good one

@ngn a direct translation of Galen Ivanov's recursive solution is also 17 bytes: {⍵<3:1⋄+/∇¨⍵-2 3}

@Adám what does it mean when I see a function definition that looks like r←{something} FunctionName args

@nathanrogers Optional left arg

oh well that's wierd

@nathanrogers Dfns and tacit functions are always ambivalent, but in Dyalog APL tradsfns have to declare themselves so. Other APL's don't give you a choice for tradfns (i.e. they are also always ambivalent).

oh tacit

@nathanrogers fixed