@ngn it wasn't for golfing, but .. why not? I was picking up on people here saying "each" is more imperative and less array focused, and challenging myself to rewrite it

using -2 rank-0 take, I see what it's doing; I really need to study rank and shape, I'm getting stuck there a lot

ok forget all that, I've just thought to ask what the shape of a scalar is, and they don't have any

@TessellatingHeckler Markdown doesn't work in multi-line messages, but if you write multiple messages, the interface will merge them visually.

@TessellatingHeckler The shape of a scalar is ⍬ as it doesn't have any dimensions.

@Adám ok to both those 👍

Is there a way to promote a thing to have +1 dimensions?

@TessellatingHeckler Yes, but where do you want the additional dimension?

I don't know that I can answer that.

    a←1 2 3 ⋄ b←4 5 6 ⋄ 2 3 ⍴ a,b

this makes a 2 3 matrix, but it feels laborious

a←1 2 3 ⋄ b←4 5 6 ⋄ (1 3⍴a)⍪b

⋄ a←1 2 3 ⋄ b←4 5 6 ⋄ ⎕←c←↑a b ⋄ ⎕←--- ⋄ ⎕←⍴c

this, making a into a matrix but with only one *row, feels nicer

@Adám
1 2 3
4 5 6
┌─┼─┐
- - -
2 3

In general, you can insert a 1 in front of the shape vector of A using ↑,⊂A.

(If you use my Extended APL, that'd be just ↑⍮ doing exactly the same.)

@TessellatingHeckler Btw, you could also write ⍉⍪a for a vector-to-one-row-matrix conversion:

hmm, NARS2000 doesn't use uparrow and downarrow for mix and split, does it?

⋄ a←1 2 3 ⋄ b←4 5 6 ⋄ ⎕←a⍪⍉⍪b

@Adám
1 2 3
4 5 6

@Adám oh interesting; I did find a way to table them and then transpose

@TessellatingHeckler It has the monadic ⊃ and ↑ exchanged, so mix is ⊃ and first is ↑.

@TessellatingHeckler Yeah, you could merge them first too, and then transpose:

⋄ a←1 2 3 ⋄ b←4 5 6 ⋄ ⎕←⍉a,⍪b

@Adám
1 2 3
4 5 6

@Adám monadic right shoe, toolbar hint says "Disclose", which I imagine as being the opposite of box, but .. it does work as mix

@Adám what is that general approach doing; up arrow, catenate, box? Is one or more of those an operator?

@TessellatingHeckler Yeah, NARS's names and symbol choices (they stem from APL2s) are confusing, imho.

@TessellatingHeckler It encloses the entire array, then makes it into a vector of length one, then trades the outermost level of depth (i.e. the one-element vector of something) for increased outer rank, i.e. a leading axis of length 1.

@Adám comma is monadic ravel in this case, which I would have guessed to do nothing on a box, but apparently does something subtle. Trying things, I see the shape of a box is Zilde, so it's a scalar. Ravel seems to unroll all dimensions into a vector, but not change nesting, compared to monadic enlist which seems to completely flatten

@Adám "trades the outermost level of depth for increased outer rank" - I can't quite follow; is that a description of what mix does generally, or what mix is being used to do in this situation?

@TessellatingHeckler Monadic , ravels the elements into a vector, but any nested elements stay. A scalar (nested or not) becomes a 1-element vector. ⍬ is simply an empty numeric vector, i.e. ⍬≡0⍴0

@TessellatingHeckler What it does generally (and here, of course).

It is maybe easier to understand what ↑A does by looking at what ↓A does:

⎕←↓3 4⍴⍳12

@Adám
┌───────┬───────┬──────────┐
│1 2 3 4│5 6 7 8│9 10 11 12│
└───────┴───────┴──────────┘

It splits the major cells into elements of a vector. ↑ re-assembles the original array.

Anyway, gotta board my plane ⍋

@Adám Thank you :) Safe flight!

@TessellatingHeckler Thanks. Don't be afraid of experimenting!

Meh, then line of people boarding continues around the corner. Not worth it…

@TessellatingHeckler what helped me out of that problem was a function that showed the shapes trough all the depths. e.g. ⊂1 2⍴⊂⍳3 → [|1 2|3]. examples with it

visually what ↑ is it removes the first |, and ↓ inserts a | before the last number in the first group

@TessellatingHeckler @dzaima There's also displayr from dfns which shows dimension lengths along the top and left sides, and depth in the lower right corner:

⎕←displayr ⊂1 2⍴⊂1 1 2⍴⍳2 ⊣ ⎕CY'dfns'

@Adám
┌─────────────────────┐
│ ┌2────────────────┐ │
│ 1 ┌┌2───┐ ┌┌2───┐ │ │
│ │ 11 1 2│ 11 1 2│ │ │
│ │ └└~───┘ └└~───┘ │ │
│ └2────────────────┘ │
└3────────────────────┘

The outermost box is a scalar (no dimension lengths in top left corner) of depth 3, containing a 1-row (left border 1) 2-column (top border 2) matrix of depth 2 (bottom left 2) which has two identical elements, both being a 1-by-1-by-2 numeric (~) arrays of the numbers 1 2.

@Adám both have good uses - displayr displays more structure while my fds gives just the shapes linearly and no extraneous data

e.g. for the case of ↑, displayr, imo shows the transformations way worse than fds

@dzaima Fun exercise: Function that, given an array, gives a complete English description of it. E.g. ⊂1 2⍴⊂1 1 2⍴⍳2 gives Scalar containing a 1-row 2-column matrix containing [a 1-layer 1-row 2-column array containing the numbers 1 and 2] and [a 1-layer 1-row 2-column array containing the numbers 1 and 2].

@dzaima Maybe. I find displayr very clear: you see depth decrease from 3 to 2 to 1 (—) while additional axes (of length 2) are appearing.

@Adám the shapes 1 2 nor 1 1 2 in your example are displayed in a straight line, and imo ...||... is a way clearer way to show a scalar than an empty border

@dzaima True. Maybe it just makes me nervous that your function looses info on irregular structure, and of course the actual data.

@Adám different uses may require different representations. it may be useful to make the function error on irregular data but that'd make 1 2⍴⊂3 4⍴⊂'foo bar' 'baz' error even though it'd be helpful to at least see [1 2|3 4|3. but the displayr output of that is just useless for me

also big tables - if i have a 10000 row table, I definitely don't want to see the data white trying to figure out what mix of ↑ and ↓ arrows do i need to do before a ⍉

@dzaima Yeah, I agree. Maybe something like displayr that don't show that data, only the axes' lengths.

@dzaima hmm well cutting off at the place of irregularity like that is also an option

Why exactly does this test case fail: Try it online!. I'm assuming it's a precision issue but I don't see exactly how it hapening. Context is this problem

Note: All other test cases are passing.

@Jonah how does that work? it uses ln(x)*y instead of x^y?

note that x^y is equivalent to exp(ln(x)*y), not just ln(x)*y

@ngn Correct. It still works though bc a^b < c^d <==> b*ln(a) < d*ln(c). And we're only interested in comparing the two expressions.

@Jonah but then you use ln(x)*y as the exponent in the next step

and ln(x)*ln(y)*z is not equivalent to x^(y^z)

so you're saying the whole approach is just wrong? is it just coincidental that it works in all the other cases?

@Jonah i think so

so many downvotes for such a nice problem

downvoted bc it's a slight variation of the same problem with 3 elms. i think it's different enough to be new though, since it requires a different approach.

@Jonah i agree

 x^(y^z) < a^(b^c)
 y^z ln(x) < b^c ln(a)
 z*ln(y) + ln(x) < c*ln(b) + ln(a)

that should be valid, correct? (each line is an iff)

@Jonah shouldn't it be ln(ln(x)) and ln(ln(a)) in the last line?

@ngn yep, it should

thanks

i keep getting wrong results for that test case, even after trying to solve it "correctly" in different ways

@ngn interesting. could it be precision? (fwiw, i verified on wolfram alpha that the test case is correct)

pari/gp confirms it too:

? log(2)*(2^2^2^3) > log(10)*(4^3^2^2)
%1 = 1

did you try with the identity produced from the same transformation i did above but for 5 elms and taking ln 4 times?

@Jonah that can't work because of the 1s

ln(ln(1)) is undefined

i tried folding both sequences together and normalizing (i.e. dividing both numbers by the larger of them) after each step

⎕←{>/⊃{r÷⌈/r←⍺*⍵}/⍵}2 2 2 2 3,¨10 4 3 2 2

Interesting. I was expecting this to be much easier. I'll take a look tonight, need to take off for a bit.

my approach is probably wrong...