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rabbitgrowth

04:09

Would ⊢⊂⍤⊢⌸⍨⊢|⍤-⍳⍤≢ be a good overall solution?

⋄ (⊢⊂⍤⊢⌸⍨⊢|⍤-⍳⍤≢) 2 5 3 1 4 6

TryAPL

@rabbitgrowth
┌───┬───┬───┐
│2 4│5 1│3 6│
└───┴───┴───┘

rabbitgrowth

The top J solution is 3 bytes shorter, but I'm not sure how comparable that is

LdBeth

04:41

guys what is the shortest method to generate all permutations of ⍳3?

LdBeth

04:52

nvm found that in APLcart

3 hours later…

Fawn Locke

07:40

⌽⍤0 1⍨ ⍳3 technically contains all permutations :)

3 hours later…

Adám

10:45

@compiledweird Hi there. Interested in APL?

4 hours later…

LdBeth

15:02

⋄ {(⍳⍵)(,[⍳2](⊢,⍤1 0~)⍤1)⍣⍵⍉⍪⍬}3

TryAPL

6 hours later…

LdBeth

21:17

Is there a better method to matching parentheses than ⋄ ((0<|)×¯1∘=++\)0 1 1 ¯1 ¯1?

TryAPL

@LdBeth 0 1 2 2 1

dzaima

(0<|) can be 0∘≠

LdBeth

21:40

thanks.

but how do I have 1 1 2 2 instead of 1 1 1 1 from 1 ¯1 1 ¯1?

dzaima

so you want a separate number for each matching parenthesis pair, not depth?

LdBeth

yes, and if possible get the numbering in ascending order.

dzaima

aight that'll probably be a decent bit more complicated

LdBeth

⋄ ((0∘≠)×¯1∘=++⍀) 1 ¯1 2 ¯1 3 4 ¯1 ¯1

TryAPL

@LdBeth 1 1 2 2 4 8 8 7

dzaima

21:50

what does a number not in ¯1 0 1 mean?

LdBeth

I'm trying to make first 1 a 1, second 1 to 2, third to 3 etc.

dzaima

i'm working on a way using progressive index of

doug

Like ranking? Double grade up?

dzaima

@dzaima actually that's way overkill

LdBeth

my problem context is, for a bit string of even length like 101001, and for every possible arrangement of parentheses of the same length, find if the the two bits on matching parentheses are distinct.

doug

21:56

⋄⍋⍋1 ¯1 2 ¯1 3 4 ¯1 ¯1

TryAPL

@doug 5 1 6 2 7 8 3 4

dzaima

aight this works

LdBeth

coool!

doug

Couldn't you just pick parens for the 1's and then fill in the 0's with the opposite ones? Or are you looking for a predicate?

LdBeth

oh it is more like "two bit strings can be matched together if for any one of possible parentheses arrangement, the parity on each pairing parentheses are same"

so 1111 matches. 1100 with parentheses {}{}

0101 matches 1010 for {{}} or {}{}

doug

22:15

Oh, opposite braces for the same bits on both?

Oh, no. Got it. Matching braces, but the lefts occur on the same bits for both.

lefts and rights. I thought the one implied the other but the first case shows that's not the case.

LdBeth

no, it's more complex than that, the full problem context is this challenge codegolf.stackexchange.com/questions/229334/…

doug

Thanks for the reference. I was still confused...

LdBeth

> Brackets corresponds to 2's and 3's. For matching brackets, if they are [ ]0 then they correspond to the same color, and different colors if they are [ ]1.

doug

But the left brackets are the same for both types...

You could number them but only after finding the matching brackets.

LdBeth

22:32

yes that's why I ask for a method to match parentheses

doug

The presentation is a little confusing. He sets it up describing how to get a coloring from a chromogram and then immediately changes the frame of reference to getting a chromogram from a coloring. I think I get the problem now, though.

olabaz

22:46

Hi, if I do something like: (⍳ 5) + ⍳ 5
does APL make the two arrays and then add them or does it actually just work with one pair of elements at a time?

I'm asking because if instead of 5 we have a large number N than memory scales with N

LdBeth

It makes two arrays.if you want the later, do +⍨⍤0⊢⍳5

And if you are working on dataset larger than memory, there is component file feature that can operate on arrays saved on harddisk.

dzaima

@olabaz memory will always scale with array size (this technically depends on the implementation, but most work like this, including Dyalog)

olabaz

I see, thanks

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