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B. Wilson
B. Wilson
02:24
Happy Friday, APL!
So, I didn't make much concrete progress on getting Dyalog running in Plan 9.
Focused mainly on the CI for now, so the RIDE efforts might be orthogonal.
Two issues: 1. I'm mostly a Plan 9 noob, and 2. linuxemu isn't compiling on my amd64 9front install.
That said, I did stumble across /lib/keyboard which basically configures a system-wide compose key-like mechanism, so getting the squiggles should be easy enough.
 
4 hours later…
sloorush
sloorush
06:29
sorry, ignore me, my brain sometimes just forgets that APL operators can be dyadic ⍨
Adám
Adám
07:02
@B.Wilson Don't forget to add a Plan 9 section to apl.wiki/Typing_glyphs when you have it all working :-)
 
2 hours later…
pmikkelsen
pmikkelsen
08:42
@B.Wilson If you just want something to play with, you can always try git.sr.ht/~pmikkelsen/APL9 It is pretty limited in may ways, but I always considered it a proof of concept thing. There is a script that can extend you keyboard map with apl glyphs included in the repository :)
Trying to run linux software on plan9 is missing the point a bit in my opinion, as you don't really get anything out of the operating system in that case. It would be better to just run it on linux and interface with it from plan9 if you want to.
 
3 hours later…
essie lovett
essie lovett
12:04
Is there a way to do a TAO comparison without grading?
dzaima
dzaima
there's also ⍺⍸⍵ but that & grade are the only things that use TAO iirc
Adám
Adám
13:00
Welcome to APL Quest 2018-6! Today's quest is What’s Your Angle?:
> Write an APL expression that, given a character scalar or vector representing some XML, returns 1 if the angle brackets are properly balanced and 0 if not; all left angle brackets < must be “closed” with right angle brackets > before another occurrence of a left angle bracket.
Richard
Richard
Did not spend much time on it so expect something better :) 
{{(∧/0≤+\⍵)∧0=+/⍵}-⌿'<>'∘.=⍵}
Adám
Adám
That looks very traditional.
Richard
Richard
yes perhaps. We used this already once before more or less
at least the first part
Adám
Adám
You could halve the amount of summing by noticing that +/ is the last element of +\.
Richard
Richard
ah yes :)
Adám
Adám
13:06
Huh, is the reference solution on the site really correct?
> Passed all basic tests – for extra points, consider cases like ('<name<>first>') as right argument which should give (1)
Looks invalid to me.
@Richard Your solution says it is valid!
Richard
Richard
ow...
can be fixed by also checking wether the sum is never larger then 1
but that makes it even more ugly
Adám
Adám
I have ((⊢≡≢⍴1 0⍨)⊢⍷∩)∘'<>'
@Richard So you agree that it is wrong, right?
Richard
Richard
yes I do
It probably needed another testcase
Adám
Adám
The reference solution is {0(∧.≤∧=∘⊃∘⌽)+\-⌿'<>'∘.=⍵} which is effectively yours with my optimisation.
Richard
Richard
What does ∧.≤∧ do? Inner product for testing both conditions?
Adám
Adám
13:15
(∧.≤∧=∘⊃∘⌽) is the same as ((∧/≤)∧=∘⊃∘⌽)
Richard
Richard
Also still looking at your solution. I don't think I understand it
Adám
Adám
The intersection of the argument and '<>' removes all other characters.
Richard
Richard
yes
Adám
Adám
If valid, the remaining characters must be <><><><> etc.
So we look for <> and check that it is found to begin at every other position, starting with the first.
Neater? ((⊢≡≠\⍤=⍨)⊢⍷∩)∘'<>'
Richard
Richard
and (1 0 ⍨) is the constant operator again?
Adám
Adám
13:22
Yes, indeed.
OK, I've updated the site.
Richard
Richard
yes, now my soluition isn't accepted anymore
Adám
Adám
Do you want to fix your solution?
@Richard It should be possible to check that the sum is always between 0 and 1.
Richard
Richard
yes, but therefore another comparison is needed. Or another logic trick
or a small train
Adám
Adám
Using interval index: {((0=⊃∘⌽)∧1∧.=0 2∘⍸)+\-⌿'<>'∘.=⍵}
Richard
Richard 
{{(∧/(<∘2∧≥∘0)+\⍵)∧0=+/⍵}-⌿'<>'∘.=⍵}
Adám
Adám
13:32
Elegant, if hard to follow: {0 2(=⍥⊃∘⌽∧1∧.=⍸)+\-⌿'<>'∘.=⍵}
Shall we call it a day, then?
We didn't do an "industrial" regex solution!
Richard
Richard
Still breaking my brains on your last one. I'll take that with me. Thanks for this week
Adám
Adám
Sure, see you next week for 2018-7: Unconditionally Shifty.
 
3 hours later…
xpqz
xpqz
16:57
      (a b)c d e(f g)←1 2 3 4 5
      a b c d e f g
┌→────────────┐
│1 1 2 3 4 5 5│
└~────────────┘
Thanks, APL.
Even if I understand it, I don't have to like it.
LdBeth
LdBeth
What's the problem with it?
xpqz
xpqz
17:13
I think my gut reaction was that it ought to error. It took me a while to see that the two parenthesised expressions will be filled with extension if there aren't enough values.
Too many, and you get the LENGTH ERROR I'd expect: 
      (a b)c d e(f g)←1 2 3 4 5 6 7 8
LENGTH ERROR
      a b←1
      a b
┌→──┐
│1 1│
└~──┘
is obviously fine.
Scalar extension ftw.
LdBeth
LdBeth
17:59
try (a a)c d e(a g)←1 2 3 4 5
I'm a little disappointed it isn't assigned from right to left :D
The Thonnu
The Thonnu
18:20
@Adám sorry, I'm a complete beginner, so this is probably obvious, but this should print [0, 1, 2, 3, 4, 5], right?
Or am I missing something?
dzaima
dzaima
you want ⎕←f 5
(or put f 5 in the "Input" field)
 
3 hours later…
The Thonnu
The Thonnu
21:12
@dzaima oh, thanks

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