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Brian

3:21 AM

The following will also work
)wsid newname
)save

10 hours later…

andycraig

1:45 PM

Hi, I'm having a go at this golfing question: https://codegolf.stackexchange.com/questions/239999/group-elements-by-their-displacement
Here's my solution, which seems to work:
{dist←|⍵-⍳⍴⍵ ⋄ dist ⊆⍥((⊂⍋dist)∘⌷) ⍵}t
The repeated dist makes me think there's an opportunity to shorten things with a fork or operator, but I can't quite manage to get there. Any suggestions welcome!

Fawn Locke

I don't think so. You'd want {(|⍵-⍳⍴⍵) ⊆{(⍺ ⍵⍵ ⍺) ⍺⍺ (⍺ ⍵⍵ ⍵)}(⌷⍨∘(⊂⍋)⍨) ⍵} which can't be expressed neatly afaik

{(|⍵-⍳⍴⍵) (⊆/⊂⍤⊣(⌷⍨∘(⊂⍋)⍨)¨,⍥⊂) ⍵} is an option but it isn't pretty

(⊃ ⊢(⊆/⊂⍤⊣⌷⍨∘(⊂⍋)⍨¨,⍥⊂)⍨∘|⊢-⍳⍤⍴)

Kamila Szewczyk

2:03 PM

how about ⋄f←⊢∘⊂⌸|∘(⊢-⍳∘≢)⋄f 2, 5, 3, 1, 4, 6

TryAPL

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

andycraig

2:18 PM

@FawnLocke Thank you! I'm going to have to spend a little while puzzling that out but it does seem to be what I was after!
@KamilaSzewczyk I think that does something slightly different, I don't get the same answer when I apply it to the test case.

(I've realised that my solution fails if the distance between an element an its index is 0, because then it's dropped by ⊆. But it can be fixed by adding 1 to all elements of dist.)

Thanks both for your help, have a good day!

Fawn Locke

You too :)

Kamila Szewczyk

@andycraig the order can be different

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