@EriktheOutgolfer Here is a wrapper that allows you to assign a one-liner dfn or tacit fn to f without counting f←.

@Adám hm, that looks useful, thanks

@Adám wait, would f be global?

@EriktheOutgolfer Yes; it is a tradfn.

@Adám holy moley. I had to leave early yesterday and only got to see this now

@J.Sallé Notice that it is your exact program, except using indexing instead of the inner :Ifs and collapsing :For-loops to {_}¨ while giving a left argument to replace the global.

@J.Sallé Of course, one should never write like this in production code. Yours was perfectly good and legible.

@Adám hahahahah I'll keep that in mind. I'm trying to use it on the simplest case '>' but it throws a value error on the last ⍳⍺, do I need to call it with parens?

@J.Sallé Yes, it is a train. Also remember ⎕IO←0 as before.

@Adám What am I doing wrong?

@J.Sallé It needs a character vector (not scalar) as argument.

@Adám like so?

@J.Sallé Yes. If that result format isn't acceptable, stick a ∊ on the far left.

@Adám will do.

@J.Sallé Here it is "ungolfed". I'm sure you can explain it like this:

⊂{
      i←⊃t←⍬⍳⍺
      t/⍵⊣⍵{
          i+←¯1 1 0⊃⍨'<>'⍳⍵
          e←t[m←⍺|i]
          t[m]←('01!'⍳⍵)⊃0 1,e,⍨~e
      }¨⍺
  }¨1+⍳∘≢

@J.Sallé Actually, if you want, you can change i⊢←i+ into i⊣i+← for the same byte count but easier to explain and ungolfed. This is what I used in above "ungolfed" version.

then maybe you haven't seen ngn'd code

@EriktheOutgolfer But he's right, I just golfed his code with not understanding of what it was supposed to do. ngn would use a (cumulative) reduction like you did.

@Adám uh, what did I do?

I sincerely hope I didn't sleep-edit somebody else's answer

@EriktheOutgolfer {⍎∊'0@1' '1@1' '~@1' '1⌽' '¯1⌽'['01!<'⍳⍺],'⊢,',⍕⍵}/

@Adám ah, nice! I'll edit it in

@Adám oh haha if it was cumulative it would've been impractical

@EriktheOutgolfer that was a reference to this message actually hahahahah

@Adám What is i←⊃t←⍬⍳⍺ doing exactly? Is that a dyadic ⍳?

@J.Sallé It finds the locations of each element of ⍺ in ⍬, but since there are no elements in ⍬, each becomes the first index beyond the end, i.e. 0 in ⎕IO←0. So it makes one zero for each element in ⍺, and since that's all zeros of course, we can take the first one and use for i.

@Adám ah, I see. That's clever, I don't think I've seen that before

@Adám and then the t/⍵⊣⍵ replicates the vector of 0's in t to ⍵ and uses that as ⍺ for the nested function?

@J.Sallé No, notice the ⊣. It is ⍵ which becomes the left argument, but the result isn't needed, so we discard it in favour of ⍵ using ⊣. Then t, which is the final memory state is used to compress ⍵ (the length), because if there are is a 1 in the final memory state, it will give ,⍵, while if there are none, it gives ⍬.

@Adám ah, I see. I'm also having trouble with explaining i+←¯1 1 0⊃⍨'<>'⍳⍵. I think it's 'mapping' (perhaps not the best word, but anyways) '<>' in ⍵, but where does the 0 comes from?

If it's anything other than '<>', it becomes a 0?

@J.Sallé Yes, remember that dyadic ⍳ will give an index one higher than the last index if the sought element isn't found. That's exactly the same principle as used with ⍬⍳.

@J.Sallé Remember that all functions are right-associative, so the dfn can only "catch" ⍵ while the ⊣ must use the dfn's result as its right argument.

@Adám yeah, I think I got that.

I'll finish the edits later, need to go get some lunch now :p brb

@J.Sallé "And do that n times, where n is the length of the input." → "And do that for each possible tape length from 1 to the length of the input."

@Adám sounds good, gonna change it. Thanks :)

@J.Sallé Nice stuff, I had (4∨?2)⌽(⌽,1∘↓)3?10