The APL Orchard

TessellatingHeckler

03:09

@ThePuzzlerThree How did you end up wanting to learn code golf but not knowing what code golf is? 🤔

H.PWiz

08:19

I was under the impression that all optimisations at dyalog were supposed to preserve behaviour with floating point numbers

Having just watched this. I still have the impression that +/≡{⍺+⍵}/ should always be true

Adám

@H.PWiz Yes indeed it should. The odd behaviour of +/ is very old, and we are reluctant to change it as that would surprise people with new values. It isn't an optimisation at all — quite the opposite, in fact — as it prevents us from reordering the summation to speed it up. That's why we recommend using 1⊥ which has no guarantee of order.

H.PWiz

So what order does +/ use?

Marshall says it goes from right to left in the talk. But surely that's what {⍺+⍵}/ does

Adám

@H.PWiz Strict left-to-right.

@H.PWiz I know, but that was just in general for f/ even though it wasn't technically true for the specific instance using +. I also noticed.

H.PWiz

That's funny

Bubbler

⎕←(+/-{⍺+⍵}/)1,9/1e¯13

Dyalog APL

08:27

@Bubbler
¯6.661338148E¯16

Bubbler

⎕←⎕CT

Dyalog APL

@Bubbler
1E¯14

Adám

      2⊃⎕WG'APLVersion' ⋄ 'cmpx'⎕CY'dfns' ⋄ f←?1e5⍴0 ⋄ cmpx '+/f' '1⊥f'
17.1.36847.0
  +/f → 9.4E¯5 |    0% ⎕⎕⎕⎕⎕⎕⎕⎕⎕
  1⊥f → 4.3E¯4 | +354% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

      2⊃⎕WG'APLVersion' ⋄ 'cmpx'⎕CY'dfns' ⋄ f←?1e5⍴0 ⋄ cmpx '+/f' '1⊥f'
18.0.37436.0
  +/f → 8.7E¯5 |   0% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕
  1⊥f → 3.4E¯5 | -62% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

cmpx didn't mark the results as different due to ⎕CT, but if I set ⎕CT←0 it will.

Bubbler

If I want to sum over an axis other than 1st, is 1∘⊥⍤k recommended?

Adám

@Bubbler From 18.0, yes, since 1⊥ is (luckily) like +⌿ and not +/. Btw, you don't need ∘ in most cases.

Bubbler

08:39

@Adám btw, have you checked a new APLcart entry proposal?

Adám

@Bubbler Yes. There are two reasons why I didn't take any (visible) action. One is that I'm on vacation this week. The other is that I was wondering if a shortcut was possible using ⎕DR (which is super fast):

⎕←5∧{⍉2⊥⍺⍺/{(-⍺⍺ 0>⍵)⍪⍺⍺ 2⊥⍣¯1⊢⍵}⍉⍺,[0.5]⍵}¯4 ⋄
⎕←5∧{2⊥(11⎕DR ⍺)⍺⍺(11⎕DR ⍵)}¯4

Dyalog APL

@Adám
4
4

Bubbler

Interesting idea, though that would break if the two values have different bit width

Adám

@Bubbler True, but this was just a prototype. One could combine the arrays with something like ⍺,[0.5]⍵ and use a single 11⎕DR.

Bubbler

(And it also breaks if the arguments go out of signed 32-bit range)

Adám

08:51

It would switch to floats.

RosLuP

09:33

@Adám Possibly i not understand well APL have in my mind one meaning
of each operator (¨) different... but
some APL primitives as "+" little seems to me a little or no,
thinked because it is easier doing
a+b when a and b are numeric array even different lenght
define + in the way

1 + 1 2 3 return 1 2 4
1 1 + 1 2 3 return 1 3 4

(so padding both array with necessary 0 in the left, for doing +)
and it is easy too doing the actual 1+1 2 3 in apl as

1+¨1 2 3

this is not only for + but for all operator as +

Adám

@RosLuP I already countered about the padding. You've gotten + wrong. You are thinking of it as ∘+¨ or +¨⊂:

⎕←1 2 4∘+¨1 2 3

Dyalog APL

@Adám
┌─────┬─────┬─────┐
│2 3 5│3 4 6│4 5 7│
└─────┴─────┴─────┘

Adám

⎕←1 2 4+¨⊂1 2 3

Dyalog APL

@Adám
┌─────┬─────┬─────┐
│2 3 4│3 4 5│5 6 7│
└─────┴─────┴─────┘

RosLuP

so you say that +¨ is ok for scalar in the right?

Adám

09:38

@RosLuP Yes, both left and right. M∘+¨N is the same as (⊂M)+¨N.

RosLuP

possible it is better each has as right argument only one list (even if contain only 1 element or 0 element too, but one list)

Adám

@RosLuP Or as left argument, yes. Or any non-scalar array, for that matter. Indeed, such are the rules of ¨ (and scalar functions): either the arguments must have matching shapes, or one has to be a scalar (well, singleton, but don't rely on that).

Ugh. We should disallow multiple assignment without parens!

Bubbler

@Adám How about {(⎕DR ⎕DR(⍺⍺⌿11∘⎕DR))⍺,[0.5]⍵}?

I noticed ⎕DR doesn't add a new axis, so the proper way to convert back should be towards the original format.

Adám

@Bubbler Just use ⎕DR⍤0 to add an axis.

⍞←10∧{(⎕DR ⎕DR(⍺⍺⌿11∘⎕DR⍤0))⍺,[0.5]⍵}¯4

Dyalog APL

09:54

@Adám 8

Adám

⍞←10∧{⍉2⊥⍺⍺/{(-⍺⍺0>⍵)⍪⍺⍺2⊥⍣¯1⊢⍵}⍉⍺,[0.5]⍵}¯4

@DyalogAPL Good morning.

⍞←10∧{⍉2⊥⍺⍺/{(-⍺⍺0>⍵)⍪⍺⍺2⊥⍣¯1⊢⍵}⍉⍺,[0.5]⍵}¯4

Dyalog APL

@Adám 8

Bubbler

2⊥ doesn't handle sign bit by itself anyway, and the new function above is blazing fast: (f1 is my original submission, f3 is the new one)

RosLuP

@Adám "Sure, filling with zeros can make sense for + and -, because that's their identity element. " even for max, min, x, division, etc

Bubbler

      arr←(2*31)-?100 100⍴2*32
      arr2←(2*31)-?100 100⍴2*32
      dfns.cmpx '#.arr ∧#.f1 #.arr2' '#.arr ∧#.f3 #.arr2'
  #.arr ∧#.f1 #.arr2 → 3.5E¯3 |   0% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕
  #.arr ∧#.f3 #.arr2 → 4.6E¯5 | -99% ⎕

Adám

09:57

@Bubbler I told you ⎕DR is super fast :-)

Bubbler

Yeah, it's really impressive.

Adám

@Bubbler No it isn't. Think about what the interpreter needs to do to convert something to its binary representation!

@RosLuP Uh, no. What would you say 10 6 3÷2 3 should give?

RosLuP

10 6 3÷2 3 ->> 10 6 3÷0 2 3 ok not for division

because would be one error division by zero

but the programmer has to write in the way never division by 0

Adám

Right. While this would be a possible extension (although the identity element makes more sense than constant 0), in bar far most cases, what would be a length error is actually indicative of a problem rather than the programmer wanting this.

RosLuP

the problem is division by zero, so to know that in the right there is no 0, and not have some positive extension with fill some 0

Adám

10:08

@RosLuP Why 0?

RosLuP

⎕←¨⊂1 2 3

Dyalog APL

@RosLuP
SYNTAX ERROR

RosLuP

@Adám because n ÷0 is one error

Adám

@RosLuP No, I mean why fill with 0 specifically?

RosLuP

because when one has to compute something using digits, the number has less digit 90% of time is fill in the left with 0

⎕←,¨⊂1 2 3

Dyalog APL

10:14

@RosLuP
┌─────┐
│1 2 3│
└─────┘

RosLuP

⎕←⊂1 2 3

Dyalog APL

@RosLuP
┌─────┐
│1 2 3│
└─────┘

Adám

@RosLuP Oh, you want to fill on the left. Hm. I was thinking that 1 1 1+ would increment the first 3 elements. And then 2 2× would double the first two element. But that'd require × to fill with 1s.

RosLuP

⎕←⊂,⊂1 2 3

Dyalog APL

@RosLuP
┌───────┐
│┌─────┐│
││1 2 3││
│└─────┘│
└───────┘

RosLuP

10:15

⎕←⊂,⊂1 2 3

Dyalog APL

@RosLuP
┌───────┐
│┌─────┐│
││1 2 3││
│└─────┘│
└───────┘

RosLuP

i don't like fill with 1... i like fill 0

Adám

@RosLuP I didn't mean always fill with 1. I meant fill with the identity element, or alternatively, do nothing on the extra elements.

RosLuP

⎕←≢⊂⊂1 2 3

Dyalog APL

@RosLuP
1

RosLuP

10:20

⎕←≢⊂1 2 3

Dyalog APL

@RosLuP
1

RosLuP

⎕←≢⍴⊂1 2 3

Dyalog APL

@RosLuP
0

RosLuP

⎕←⍴⊂1 2 3

Dyalog APL

@RosLuP

RosLuP

10:24

⊂1 2 3 is a scalar right? it is not one list… why ≢⊂1 2 3 return 1?

possible all is one list?

@Adám "Uh, no. What would you say 10 6 3÷2 3 should give? " it should give for me 10 6 3÷0 2 3 so (10÷0)(6÷2)(3÷3) so one division by zero error...

⎕←1999{k←⌈/≢¨(a b)←⍎¨∘⍕¨⍺⍵⋄f←{((k-≢⍵)⍴0),⍵}⋄10⊥(f a)⌈f b}2000

Dyalog APL

@RosLuP
2999

RosLuP

10:39

⎕←1990{k←⌈/≢¨(a b)←⍎¨∘⍕¨⍺⍵⋄f←{((k-≢⍵)⍴0),⍵}⋄10⊥(f a)⌈f b}0

Dyalog APL

@RosLuP
1990

RosLuP

⎕←1998{k←⌈/≢¨(a b)←⍎¨∘⍕¨⍺⍵⋄f←{((k-≢⍵)⍴0),⍵}⋄10⊥(f a)⌈f b}9

Dyalog APL

@RosLuP
1999

RosLuP

⎕←1{k←⌈/≢¨(a b)←⍎¨∘⍕¨⍺⍵⋄f←{((k-≢⍵)⍴0),⍵}⋄10⊥(f a)⌈f b}9

Dyalog APL

@RosLuP
9

RosLuP

10:42

⎕←9{k←⌈/≢¨(a b)←⍎¨∘⍕¨⍺⍵⋄f←{((k-≢⍵)⍴0),⍵}⋄10⊥(f a)⌈f b}11

Dyalog APL

@RosLuP
19

Adám

@RosLuP ≢ counts how many major cells there are. It doesn't say anything about the rank of its argument. A scalar has one major cell. ≢1⍴3 and ≢1 2⍴3 also both give 1.

@RosLuP I get why you would want to fill with 0 for + and -, but why would you want to fill with 0 for × and ÷?

RosLuP