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Dyalog are attending the FinnAPL Forest Seminar - taking place at Hämeenkylä Manor, Vantaa, Finland on 15-16 March. For details go to http://www.finnapl.fi/tapahtu.htm

Tweeted by dyalogapl on February 26, 2018 at 11:00 AM

J. Sallé

13:33

@Adám when shaping a matrix as 3 3⍴values, is 3 3 = ≡mat ≢mat? Or is it the other way around? Or is it not that at all? :p

Adám

@J.Sallé 3 3 would be ⍴mat. ≢mat is the number of major cells (3). ≡mat is the depth (1):

J. Sallé

@Adám ah, naturally. Makes sense. Thanks

Adám

⎕←mat←3 3⍴⎕A ⋄ ⎕←⍴mat ⋄ ⎕←≢mat ⋄ ⎕←≡mat

Dyalog APL

@Adám
ABC
DEF
GHI
3 3
3
1

J. Sallé

Yup, that solves it

@Adám no idea what your comment meant >.>

The function doesn't work without the (⍴⍵)⍴

Adám

13:45

@J.Sallé Really‽

J. Sallé

@Adám hur dur, I left the last ⍴ there

I used it because I was getting a rank error when doing {^/∊⍵=1 ¯1×⍉⍵}, and I didn't try using (⊂1 ¯1) by itself

Just thought I had to shape it before using or I'd always get a rank error

Adám

@J.Sallé You want to multiply the entire array 1 ¯1 with each element of ⍉⍵ so write that! (∘×¨)

J. Sallé

@Adám aaaaah, cool! I actually tried to use ×.¨ and it didn't work, obviously

Adám

@J.Sallé No, you can't have DyadicOp DyadicOp (except ∘.).

Erik the Outgolfer

@Adám ...or DyadicOp Op in general

Adám

13:52

@EriktheOutgolfer Yeah, sorry, that's what I meant (as ¨ is a monadic operator).

J. Sallé

Yeah I still get mixed up with using . and ∘

Adám

@J.Sallé . is just for inner/outer product (and namespaces and decimals — ugh, it's so overloaded).

J. Sallé

@Adám Hahahah the magic dot!

Erik the Outgolfer

@J.Sallé ∘. is a different syntax element altogether, in that it only takes a right operand, but otherwise it acts like an operator...yeah, that's confusing at first, but you can also think of . taking a left operand that can either be a function or ∘, where in the latter case it acts as outer product using the right operand

Adám

@EriktheOutgolfer Or just think of ∘.f as syntactic sugar for f⍤0 15.

J. Sallé

13:57

Okay I'll try going tacit now, wish me luck :p

Erik the Outgolfer

@Adám yeah, like prefixing a function with ∘. being the same as prepending ( and appending ⍤0 15)

oh, and one last thing, +.∘.- is a syntax error >_>

Adám

@EriktheOutgolfer Of course. Operators are left-associative, so +.∘.- is (+.∘).-:

⎕←+.∘.-

Dyalog APL

@Adám
  .
 ┌┴┐
 . -
┌┴┐
+ ∘

Erik the Outgolfer

@Adám I feel like ∘. syntax should have a higher priority than operator syntax

Adám

@EriktheOutgolfer Yeah, I've suggested that too.

@EriktheOutgolfer +.-⍤0 15 doesn't give you what you want either. You'd need +.(-⍤0 15) just like +.(∘.-).

J. Sallé

14:07

@Adám I recall having seen somewhere a little "conversion table" from Dfns to Tacit functions, do you know what I'm talking about?

Erik the Outgolfer

@Adám but I feel like ∘. should do the parenthesizing automatically

J. Sallé

I don't know if I've seen it here or in one of the resources you posted. Can't seem to find it

Adám

@J.Sallé Translation of D-functions into tacit form.

J. Sallé

@Adám Yes, that! Thanks

Adám

14:20

@EriktheOutgolfer If ∘.f was f\ would you expect +.-\ to be +.(∘.-) or ∘.(+.-) ?

Erik the Outgolfer

@Adám what? (markdown issues)

J. Sallé

The problem I'm having now is that I can't seem to use the original matrix as the left argument for the ≡, should I assign it to a variable or can I do without it if I'm smart enough?

Erik the Outgolfer

@Adám the latter, because now the monadic operator is postfix

Adám

@J.Sallé How about ⊢≡… ?

J. Sallé

@Adám That yields 2 for some reason. The fn I have is ⊢≡1 ¯1∘×¨⍉⊢

1 ¯1∘×¨⍉⊢ gives me the transposed and negated matrix

I just need to match that to the original one now

Erik the Outgolfer

14:24

@J.Sallé why do you use dyadic ⍉?

J. Sallé

@EriktheOutgolfer I'm not? I think >.>

Erik the Outgolfer

@J.Sallé then why are you forking with it in the center?

J. Sallé

That gives me the exact same result as the one in my Dfn without the ≡

Erik the Outgolfer

1 ¯1∘×¨⍉⊢ means (1 ¯1∘×¨)(⍉)(⊢)

@J.Sallé X(f g h)Y is the same as (X f Y)g X h Y

(f g h)Y is the same as (f Y)g(h Y)

so (1 ¯1∘×¨⍉⊢)X is the same as (1 ¯1∘×¨X)⍉⊢X

J. Sallé

I think that might be enough to help me

I suck at visualizing these kinds of things tbqh, that's probably why I couldn't learn calculus for the life of me

Adám

14:34

@J.Sallé Use the boxed display to visualise for you:

⎕←⊢≡1 ¯1∘×¨⍉⊢

Dyalog APL

@Adám
┌─┼───┐
⊢ ≡ ┌─┼─┐
    ¨ ⍉ ⊢
  ┌─┘
  ∘
┌─┴──┐
1·¯1 ×

J. Sallé

14:49

bangs head on wall

I got as far as ⊢≡⍨1 ¯1∘×¨(⍉⊢)

But that seems to use the transposed/negated matrix as the argument, not the original one

⎕←⊢≡⍨1 ¯1∘×¨(⍉⊢)

Dyalog APL

@J.Sallé
┌──┴──┐
⊢ ┌───┼──┐
  ⍨   ¨ ┌┴┐
┌─┘ ┌─┘ ⍉ ⊢
≡   ∘
  ┌─┴──┐
  1·¯1 ×

Adám

15:00

@J.Sallé Hint: Note that the order of flipping and negating is irrelevant, so you may negate first, and flip afterwards.

@J.Sallé Alternative hint: Operators are left-associative.

@J.Sallé Btw, ≡⍨ is the same as ≡.

J. Sallé

Okay I'm getting somewhere I think

Adám

@J.Sallé One more hint. A≡⍉B ⇔ (⍉A)≡B.

@ngn OK, fine, ≡⍨ isn't always the same as ≡, but for J.Sallé's purposes, it is.

Erik the Outgolfer

@Adám wrong ping?

J. Sallé

@Adám see my problem is that I only see B in this case ⍨

Adám

@J.Sallé No, A is ⊢ and B is 1 ¯1∘×¨

ngn

15:08

@EriktheOutgolfer or he might have been trying to predict my thoughts :) though I wouldn't argue over that

Adám

@EriktheOutgolfer ^

Erik the Outgolfer

@Adám ...that's like saying "you'll be pedantic now and I know it", no?

Adám

CMC: Shortest possible expression showing that ≡ is not the same as ≡⍨.

@EriktheOutgolfer Yup.

ngn

@Adám ngn being ngn now: why is that not obvious, why is it important and what are you trying to divert our attention from? :)

Adám

@ngn Not obvious to someone who thinks symbolically instead of numerically (with limited precision), not important (just a fun challenge made for you), nothing.

Erik the Outgolfer

15:15

lol that would be very trivial since lesson 1

ngn

@Adám should we assume it's a dyadic ≡?

Adám

@ngn Yes, of course.

ngn

@Adám ok, it's not obvious (to me) then

Erik the Outgolfer

@Adám that changes like the whole CMC a few levels of magnitude

J. Sallé

I think my brain is gonna shut down soon if I can't solve this ⍨

Only way I managed to use dyadic ≡ was (⊢⍉)≡1 ¯1∘×¨

Adám

15:19

@EriktheOutgolfer The whole thing started with J.Sallé needing to compare two matrices.

J. Sallé

and that still gives me 2

Adám

@J.Sallé Ah, you're forgetting to apply it with parens!

J. Sallé

@Adám I literally just did that facepalm

Erik the Outgolfer

@Adám but...but I could've abused something like (≡1)≢(≡⍨1)

but since we must assume the ≡ is not monadic ≡, oh well

Adám

@J.Sallé Now you just need to simplify one more thing.

J. Sallé

15:22

@Adám I assume I can remove those parens at (⊢⍉)

Adám

@J.Sallé Yes, and ⊢A ⇔ A.

J. Sallé

Good god that was hard

Finally got (⍉≡1 ¯1∘×¨) though

Erik the Outgolfer

@J.Sallé practice and you'll get it sometime :)

Adám

@J.Sallé I have not come up with a shorter one.

If APL had J's j. function Ⓙ←⊣+0j1×⊢ it would be just (+≡⍉)Ⓙ/¨

ngn

@Adám maybe something like ⊢∘-\ instead of 1 ¯1∘×?

Erik the Outgolfer

15:30

(and of course ngn's strongest weapon is cumulative reduce, as we can see here)

Adám

@ngn Wait, does that always hold?

ngn

@Adám I don't know, I haven't even copied the tests yet

Adám

@ngn Obscure (and longer than J.Sallé's): ⊂¨∘⍉≡,∘-/¨

Erik the Outgolfer

@Adám well, 1 ¯1×A B is the same as (1×A)(¯1×B) -> A(-B) and ⊢∘-\A B is the same as A(A⊢-B) and therefore A(-B), so...

Adám

@EriktheOutgolfer Sure, but you need to do that on every element.

J. Sallé

15:33

@EriktheOutgolfer f←⍉≡ ⊢∘-\¨ fails the last test case apparently

Adám

@J.Sallé No, you're missing a ⊂

J. Sallé

(There's a space there I hadn't noticed but it makes no difference)

@Adám oh duh. It wasn't failing for the other function so I didn't notice

Adám

@ngn So yes, ⍉≡⊢∘-\¨ is right.

J. Sallé

⎕←⍉≡⊢∘-\¨

Dyalog APL

@J.Sallé
 ┌─┼─┐
 ⍉ ≡ ¨
   ┌─┘
   \
 ┌─┘
 ∘
┌┴┐
⊢ -

J. Sallé

15:39

Just to clarify here, the \ is replicating ⊢∘- into the argument?

Adám

@J.Sallé No, it is a cumulative ⊢∘-/. But since there are only two terms, Re Im, it is Re (Re⊢∘-Im).

J. Sallé

@Adám ah, I see.

Erik the Outgolfer

@J.Sallé that's the issue with hybrids :P

Adám

@EriktheOutgolfer Well, there's no issue here, other than slight obscurity.

J. Sallé

I'm just finding it hard to work out an explanation. I'll post it and you can check it for me @Adám?

Erik the Outgolfer

15:42

@Adám ⊣/∘⊢⌽ ≢ ⊣(/∘⊢)⌽

Adám

@J.Sallé Sure, or maybe @ngn?

J. Sallé

Yeah, anyone who knows better than me :p

Adám

@EriktheOutgolfer Right, schizos prefer to be operators if at all possible, so ⊣/∘⊢⌽ is (⊣/)∘⊢⌽

J. Sallé

Posted

Adám

@J.Sallé Repeat the first element and negate the second → Ignore the first element in favour of the negated second element or something.

ngn

15:47

@Adám that's what I was gonna say too, the only purpose of ⊢∘ is to force - to be monadic

Erik the Outgolfer

good luck explaining ngn's cumulative reduces, they're just so packed together :D

5

Adám

@J.Sallé Maybe you meant to imply that the original left is also returned?

J. Sallé

@Adám @ngn sounds good

@Adám yeah, that's what I meant actually, but I see how the wording is off

Adám

@J.Sallé Maybe "scan" isn't descriptive enough. Cumulative reduction using… ?

J. Sallé

⍉≡⊢∘-\¨ ⍝ Anonymous tacit function.
      ¨ ⍝ To each element of the argument:
     \  ⍝ Cumulative reduction, using
  ⊢∘-   ⍝ Ignore the first element, then negate the second
 ≡      ⍝ And match
⍉       ⍝ To the argument's transposition.

ngn

15:50

this is so boring, someone should create an auto-explainer for apl

Ven

seems like a np-complete problem

Adám

@ngn Not really possible, as explanations include references to what the data stands for.

Erik the Outgolfer

sure, it's just that we need to train it to recognize ngn's cumulative reduce ab clever uses :P

J. Sallé

would probably be possible if every squiggle was only a single operation every time

ngn

@EriktheOutgolfer there's nothing wrong with using scan :)

Erik the Outgolfer

15:52

@ngn lol where did I say that

ngn

@EriktheOutgolfer reduce in apl is bad, but scan is ok

@EriktheOutgolfer you didn't say it but you thought it :)

J. Sallé

@ngn there is when you use it to do sorcery such as that ಠ__ಠ

Erik the Outgolfer

@ngn well yeah, since most challenges asking for specific reduces reduce from left-to-right

ngn

@EriktheOutgolfer I think it should have been left-to-right, like array indices go, not right-to-left like functions do

@EriktheOutgolfer and the thing with the required ⊃ after reduction is a bit annoying, but at least it's consistent with the rest of apl

Erik the Outgolfer

@ngn maybe in the early 60s people thought differently :P

@J.Sallé remember this is code-golf, where even the worst abuses ever are allowed and even encouraged ;)

Adám

15:56

@ngn -/ and ÷/ are only interesting RTL.

J. Sallé

@EriktheOutgolfer black magic though D:

Adám

16:31

→ 39 messages moved to The Waiting Room

ngn

16:54

@Adám can your golfing lib be used with tio?

Adám

@ngn Yes (opt/agl/AGL.dyalog) but it is in a broken state, and I've basically abandoned it due to the cost of inclusion.

ngn

@Adám so, if that is allowed, then presumably adding a library as "header" is also allowed?

Adám

@ngn Yeah, I guess with the latest meta developments, it could be considered a separate language. Maybe I should take up the work again?

ngn

@Adám I think that's one way to fix some of APL's golfing shortcomings.

Adám

@ngn Yes. I think most of the code is ok, btw, the lines just need reordering so identifiers are not used before they are defined. Feel free to submit a PR.

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