The APL Orchard

@Adám, in this video the approximation for ∏ is completely wrong. I tried on TryAPL and it is still wrong. Is this a known problem? I think it has to do with floating point precision in internal calculations.

⋄ ((÷/1↓⊢)+⊃)(1∧0 1∘⊤,÷)○1

TryAPL

@AlexB 5.093937333

Alex B

⋄ ((÷/1↓⊢)+⊃)(1∧0 1∘⊤,÷)3.141593

TryAPL

@AlexB 3.141593

Alex B

⋄ ((÷/1↓⊢)+⊃)(1∧0 1∘⊤,÷)3.1415927

TryAPL

@AlexB 3.400642929

Adám

11:08 AM

That doesn't look right.

Alex B

You can see that it starts misbehaving at 3.1415927

The more decimal digits, the more it diverges.

⋄ ((÷/1↓⊢)+⊃)(1∧0 1∘⊤,÷)3.14159265358979

TryAPL

@AlexB 4.810752026

Adám

Even worse if you ask for more precision: ⋄ ⎕CT←0 ⋄ ((÷/1↓⊢)+⊃)(1∧0 1∘⊤,÷)3.1415927

TryAPL

@Adám 17.6756

Alex B

I think the algorithm is correct, but the internal representation is a problem. Either the reciprocal or the lcm or both.

Either way, you might want to not advertise this algorithm as good for ∏ approximation.

Adám

11:11 AM

We need to look into this. Thanks!

Alex B

I doubt it's a Dyalog interpreter problem. It looks like the limitations of a numeric algorithm.

Kamila Szewczyk

pi approximation

the simplest one i can come up with is using arctan(1)

Alex B

@KamilaSzewczyk I was not trying to compute PI. I was watching Richard's video and I noticed something was wrong.

Kamila Szewczyk

I know, I was just thinking how i'd approach it.

Adám

This seems to work: ⋄ ÷/((,÷∨)∘1)○1

TryAPL

11:14 AM

@Adám 3.141592654

Kamila Szewczyk

this is extra lame though.

it turns a pi constant into a rational and then back to a float.

Alex B

I think the train proposed by Richard in the video for a fraction equivalent to a rational number is nice, but it works only up to about 7 significant digits.

Adám

@KamilaSzewczyk We're trying to look at the validity of the generated rational, hence turning it back into float.

Kamila Szewczyk

I remember stumbling upon the exact same issue in kamilalisp funnily enough

with the precision making the rational approximation worse

i didn't find a solution to it sadly

i think it might just be floating point inaccurancies?

Alex B

@KamilaSzewczyk Typically one resorts to double precision for internal calculations. In Dyalog it's not available.

Kamila Szewczyk

11:16 AM

it's not really a matter of precision, some numbers are not representable as floats in the binary representation.

and the error creeps

i have zero idea why it creeps more the larger the algorithm's precision is, though.

Alex B

Quadruple-precision floating-point format

In computing, quadruple precision (or quad precision) is a binary floating point–based computer number format that occupies 16 bytes (128 bits) with precision at least twice the 53-bit double precision. This 128-bit quadruple precision is designed not only for applications requiring results in higher than double precision, but also, as a primary function, to allow the computation of double precision results more reliably and accurately by minimising overflow and round-off errors in intermediate calculations and scratch variables. William Kahan, primary architect of the original IEEE-754 floating...

Kamila Szewczyk

yeah, heard of that. kamilalisp worked on MPFR floats which funnily enough suffer from the same thing.

Alex B

@KamilaSzewczyk Yes, but this doesn't look like it's a representation problem. PI digits are random and they cannot all be idiosyncratic for binary representation.

Kamila Szewczyk

⎕←{4×1++/(¯1*⍵)÷(1+2×⍵)}∘⍳¨1E2 1E4 1E6

TryAPL

@KamilaSzewczyk 3.151493401 3.141692644 3.141593654

Kamila Szewczyk

11:21 AM

could use Shanks transformation to make it converge quicker. I always wanted to write an APL blog post about Shanks transformation anyway.

Adám

OK, I've sent an email to our internal language group, but don't expect a response until tomorrow.

dzaima

@Adám i think it'd probably be 1∧ choosing different multiples from the two arguments its given

Adám

Oh, that might make sense. It can't find a "correct" value, so it just chooses something.

Alex B

@KamilaSzewczyk it's your chance to do it! :)

dzaima

so unless (1∘|,÷) always keeps precise info about what fraction 1∧ would think each element there is, 1∧ will fail sometimes

Kamila Szewczyk

11:24 AM

too busy with cooler stuff for now :3

but i can reveal it only in approximately two weeks

dzaima

@dzaima actually, the issue is more in that ÷s argument is the raw un1|ed thing, (1∧⊢,÷)1|x is pretty good. But that "will fail sometimes" still holds, it's just that ÷ getting 3.141 instead of 0.141 makes it lose a lot of precision and makes that "sometimes" more likely

LdBeth

With rational arith I get 1E4 => 3.141692643590543, still computing 1E6...

Adám

Precise 128-bit decimal floats:

TryAPL

@Adám
NOT PERMITTED: Illegal token
      ⎕FR←1287 ⋄ ⎕DCT←0 ⋄ ((÷/1↓⊢)+⊃)(1∧0 1∘⊤,÷)○1
                ^

Adám

11:40 AM

@hyper-neutrino If I remove the triggering code, the bot should treat it as it does removing the entire message, no?

      ⎕FR←1287 ⋄ ⎕DCT←0
      ((÷/1↓⊢)+⊃)(1∧0 1∘⊤,÷)○1
3.141592654
      (○1) = ((÷/1↓⊢)+⊃)(1∧0 1∘⊤,÷)○1
1

LdBeth

11:57 AM

hmm, I give up after 10 minutes

Alex B

@Adám so, with 128 bits it seems to work?

Adám

Only if you also set the decimal comparison tolerance to 0.

Alex B

@Adám I am not talking about the = in the 2nd statement. I am talking about the visible digits in the 1st statement. They don't diverge. The result looks converged.

dzaima

..and 3.01234567890123456789 fails

Adám

Yeah, this is probably an exercise in futility. We're trying to exude exactness where none is to be found.

dzaima

12:06 PM

well i did give you a logical reason for why you won't get this to work

Alex B

@Adám I agree. Sorry for triggering this. All I think needs to be done is note that Richard's train cannot be used to approximate π.

lyxal

so uh this is probably a big no-no, but I just used ⍨ to treat a scalar as a single unit in a train: ⌊(⍴÷3⍨)

that should tell you how many times any number can be divided by 3 ignoring remainder

dzaima

@lyxal that's what it's kinda meant for actually

lyxal

oh okay good

I thought that wasn't something you were supposed to do with selfie

TIL.

Adám

@lyxal "a number"? You mean a string/list/vector?

@lyxal It is called Constant when used with an array operand.

lyxal

12:11 PM

@Adám ah yeah that's right - the length of a string

Adám

Consider using ≢ instead of ⍴ as the former will give you a scalar rather than a vector result.

RikedyP

@AlexB Is that also this? aplcart.info/?q=convert%20fraction#

I'll put a link to this conversation as a YT comment

Alex B

12:40 PM

@RikedyP This seems to work

⋄ ÷/((,÷∨)∘1)○1

TryAPL

@AlexB 3.141592654

dzaima

1 hour ago, by Adám

This seems to work: ⋄ ÷/((,÷∨)∘1)○1

TryAPL

@dzaima 3.141592654

Adám

CMP: Should the bot react to quoted commands?

⎕←'⋄ ⍳3'

TryAPL

@Adám ⋄ ⍳3

Adám

12:45 PM

It doesn't react to itself.

PyGamer0

@Adám do you mean it should react to itself?

Alex B

@dzaima so maybe this can be used to find the fraction in the form a + b/c

⋄ (2⌽((,÷∨)∘1(⊢-⌊)),⌊)○1

TryAPL

@AlexB 3 3612111 25510582

Adám

@PyGamer0 No, I meant that just like it doesn't react to itself, it shouldn't react to quotes either.

Alex B

(I am sure some of the train wizards can improve on my poor train)

PyGamer0

12:52 PM

@AlexB i am not a train wizard, i am a train beginner

Alex B

@PyGamer0 I was replying to @dzaima and @Adám

dzaima

@Adám the code has already been evaluated, it's just wasted space. And if it's from a long time ago, you can just click it

Alex B

(the threads got mixed up)

Adám

Someone who teaches trains can be described as a "trainer", right?

PyGamer0

@AlexB i was just saying ;)

Alex B

12:53 PM

@Adám or a train conductor...

Adám

@AlexB 2⌽(,÷∨)∘1⍤-∘⌊⍨,⌊

dzaima

@AlexB (⌊,1(,÷∨)⍨1|⊢)○1

Alex B

Great. One of these can replace the train that derailed with π.

(BTW, the more I learn APL, the less I know — typical Dunning-Kruger effect)

Learning APL has a very humbling effect. I think it should be mandatory learning for politicians of all colors. :-)

Kamila Szewczyk

@AlexB wait until you try C++ ;)

Fawn Locke

1:14 PM

@Adám You're welcome! I've wanted to contribute more, I'm just insecure about my writing style ^^'

Alex B

@KamilaSzewczyk I tried C++ in 1989 and I didn't like it. I sympathize with Alan Kay on this.

(to me C++ is in the same group with JS and PHP)

Adám

@FawnLocke It is fine, and prompts others (i.e. me) to contribute more.

Razetime

1:43 PM

@AlexB C++ is actually quite fine to program in if you just ignore everything and write it like C

still a very usable, feature filled but bloated, fast language which i wouldn't put anywhere near js or php

dzaima

@Razetime and JS is quite fine to program in if you ignore 97% of the language too. PHP, though..

Fawn Locke

lol

PyGamer0

Malbolge is also quite fine to program in if you ignore 100% of the language.

Kamila Szewczyk

Malbolge is actually fine to program in :)

dzaima

@KamilaSzewczyk so says, give or take, the single person in the world who actually can

Alex B

2:00 PM

@Razetime of course. That’s what I did. (BTW C++ was a C preprocessor, so you could just write K&R C)

Razetime

hm C++ has C inside of it so at least i know what i can use without banging my head, but js has some basic things that are quite bees

Master Quiz

Tacit version of 2 1 1 1 2{⍺⌿⍺/⍵}5 5⍴1?
Also, is there a better way to generate ⍺ than (1+⍳∊1∘,)⍤≢?

PyGamer0

@KamilaSzewczyk you use a transpiler that transpiles to malbolge right?

Kamila Szewczyk

who knows

Master Quiz

I've just thought there there is no reason to tacify it, It's already short, and I can do something like {mask⌿⍵/⍨mask←(1+⍳∊1∘,)⍤≢ ⍵}5 5⍴⍳1.
But I'm still curious

PyGamer0

2:14 PM

@KamilaSzewczyk idk, you?

Fawn Locke

2:28 PM

@MasterQuiz replicate sucks in trains, ⊣⊢⍤⌿⊢⍤/ is my attempt

In an ideal world ⊣⌿/ would work

Master Quiz

@FawnLocke I was trying (⊢⍤⌿⊢⍤/)

Fawn Locke

Doesn't work , a dyadic atop is g X f Y, i.e ⌿ doesn't get the left argument it needs

Master Quiz

@FawnLocke In ideal world there would exist this operator 2 1 1 1 2(⌿{⍺ ⍺⍺ ⍺ ⍵⍵ ⍵}/)5 5⍴1

Fawn Locke

Well that's just the fork (⊣⍺⍺⍵⍵) so I don't see much utility in having an operator for it

Master Quiz

@FawnLocke Yes, it's just that I've always thought that it recognise itself when a function should be used dyadically; but I guess also this doesn't make much sense

hyper-neutrino

3:40 PM

@Adám ... I thought it was supposed to ¯\_(ツ)_/¯

Adám

4:23 PM

@hyper-neutrino It removes its message if I remove mine, but it doesn't remove its message if I remove the triggering code from my message, leaving the rest of the message there.

Alex B

4:59 PM

@FawnLocke but this doesn't work in the way that @MasterQuiz needs. I am not sure how common its use would be, but I tend to agree with @MasterQuiz that the operator that he suggests is different and is exactly what he needs in his case.

@MasterQuiz, if you don't want to define an operator, you might just create a dedicated dfn:

⋄ 2 1 1 1 2{⍺⌿⍺/⍵}5 5⍴⍳25

TryAPL

@AlexB
 1  1  2  3  4  5  5
 1  1  2  3  4  5  5
 6  6  7  8  9 10 10
11 11 12 13 14 15 15
16 16 17 18 19 20 20
21 21 22 23 24 25 25
21 21 22 23 24 25 25

Fawn Locke

How is it different?

dzaima

@AlexB (⊣ f g) works precisely the same as f{⍺ ⍺⍺ ⍺ ⍵⍵ ⍵}g (if ignoring slashes sucking, which is mostly unrelated)

but there are at least 4 such useful operations - (⊣ f g), (⊢ f g), (f g ⊣), (f g ⊢) - and having an operator for each is a lot of characters. And there are quite a lot more frequent useful forks. You're not really even saving any characters if there were operators for those (the only exception being said already-awful slashes)

Alex B

@dzaima I never suggested adding an operator to the language. I was trying to analyze @MasterQuiz's problem and suggest solutions dedicated for him.

@FawnLocke Not different in principle but, as @dzaima pointed out (and you as well), it doesn't work with slash. Thus, @MasterQuiz has to use something else. Of course that's what he started with, so in a sense I went full circle to confirm (to myself) what he already knew.

Fawn Locke

It does, it's just ugly

(⊣⊢⍤⌿⊢⍤/)

Alex B

5:08 PM

@FawnLocke It's a bit less readable, IMO.

Adám

5:20 PM

If it wasn't for the stupidly working (,1)⌿1 2 3 4 I'd extend / so you could write ⍺ ⍺⌿⍵ meaning ⍺⌿⍺/⍵ (for matrices).

Wait, that's nonsense.

It if wasn't for 1≡⊂1…

But (,1)(,1) is distinguishable from 1 1 so it could work to extend ⌿ to map nested ⍺s over leading dimensions.

Fawn Locke

Huh how does (,1)⌿1 2 3 4 work

Adám

Sillily.

In many places, many APLs allow 1-element vectors (or even any singleton) where a scalar should be required.

It is largely because ≢ was missing, so (ab)use of ⍴ lead to 1-element vectors where a scalar should have been.

Actually, in this particular case, it is a good thing that ,1 is treated as a scalar, because it allows writing and enclosed "scalar" as ⊂,1

dzaima

@Adám of course, that's just ⍮ being missing and so (ab)using ⊂ for it

Adám

No, ⍮ wouldn't help. You still can't enclose a simple scalar.

dzaima

ah i misread (,1)(,1) as being the proper per-axis matched item replicate, which it isn't as there'd a constant length for each axis there

@Adám 1 2/2 2⍴⍳4 also works, and is a lot more important

Adám

5:32 PM

I don't follow.

I'd want (1 2)(3 4)⌿2 2⍴⍳4 to mean 1 2⌿3 4/2 2⍴⍳4

So how would you write 2⌿3/2 2⍴⍳4? — (,2)(,3)⌿2 2⍴⍳4

dzaima

@Adám you complained about singleton extension blocking just 2 3⌿2 2⍴⍳4, but i'm pointing out that regular replication also doesn't allow extending that

ah right after that there's "wait, that's nonsense" :| ignore me

Adám

Yeah, my singleton complain was a mistake. What blocks it is the lack of enclosed simple scalars. What rescues it is singleton extension.

Master Quiz

6:11 PM

How to take the vector of greater length between two vectors that has some scalar in common?
An example: (≢2 4)>(≢4), (≢7 10 11)>(≢10 11)>≢(,11), (≢3 5 8)>(≢,3)

⋄(,1)(,4)(6 9)(,11)(2 4)(7 10 11)(,3)(3 5 8)(10 11)
⋄(,1)(6 9)(2 4)(7 10 11)(3 5 8)

TryAPL

@MasterQuiz
┌─┬─┬───┬──┬───┬───────┬─┬─────┬─────┐
│1│4│6 9│11│2 4│7 10 11│3│3 5 8│10 11│
└─┴─┴───┴──┴───┴───────┴─┴─────┴─────┘
┌─┬───┬───┬───────┬─────┐
│1│6 9│2 4│7 10 11│3 5 8│
└─┴───┴───┴───────┴─────┘

Fawn Locke

"scalar in common" do you mean 1-element vector?

Adám

@MasterQuiz Is the common scalar actually relevant, or do you just want the longest of two vectors?

Master Quiz

@FawnLocke No, each item of the vectors

@Adám The longest of multiple vectors that have a common element

I can also have this matrix, maybe better cause I have to use the result as a mask after, and because I obtained the vector from this matrix

⋄ ↑(,1)(,4)(6 9)(,11)(2 4)(7 10 11)(,3)(3 5 8)(10 11)∊⍨¨⊂⍳11

TryAPL

@MasterQuiz
1 0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0 1
0 1 0 1 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 1 1
0 0 1 0 0 0 0 0 0 0 0
0 0 1 0 1 0 0 1 0 0 0
0 0 0 0 0 0 0 0 0 1 1

Alex B

6:50 PM

@Adám, I think there is a slightly unclear statement in the [Expand wiki page](https://aplwiki.com/wiki/Expand#Examples). It reads "For vector arguments, the number of positive values in the left argument must match the length of the right argument." but I think it should read "For vector **right** arguments..."
If you agree, I could edit the page.

Adám

Yes, that seems right. Go ahead.

Alex B

7:25 PM

@Adám Done. Thanks.

Alex B

7:47 PM

I don't understand why the first example succeeds and the second fails.

⋄ (⍳2) {⍺ (,⍤0) ⍵} 10×⍳2

TryAPL

@AlexB
1 10
2 20

Alex B

⋄ (⍳2) ({⍺ (⍵⍵⍤0) ⍵},) 10×⍳2

TryAPL

@AlexB (1 trailing line)
VALUE ERROR
      (⍳2)({⍺(⍵⍵⍤0)⍵},)10×⍳2

␄

Alex B

They should be equivalent, shouldn't they?

Fawn Locke

⋄ (⍳2)(,{⍺(⍺⍺⍤0)⍵})10×⍳2

TryAPL

7:49 PM

@FawnLocke
1 10
2 20

Fawn Locke

A monadic operator has a left function operand, not a right one

Also btw you don't need brackets around ,{⍺(⍺⍺⍤0)⍵} operators are parsed "first"

Alex B

@FawnLocke But doesn't depend on whether I specify double-alpha or double-omega? It seems that this rule is cast in stone, regardless of how I specify the operator.

This works:

⋄ (⍳2) (,{⍺ (⍺⍺⍤0) ⍵}) 10×⍳2

TryAPL

@AlexB
1 10
2 20

Alex B

But it's different from what I wanted. I wanted the operator to associate on the right. Apparently this is not allowed?

Master Quiz

@AlexB You can, but it has to have also a left operator

⋄ (⍳2) (⊢{⍺ (⍵⍵⍤0) ⍵},) 10×⍳2

TryAPL

7:58 PM

@MasterQuiz
1 10
2 20

Alex B

@MasterQuiz I see. There are some weird rules cast in stone in Dyalog APL, as I can see. Trying to learn them...

Thanks, @FawnLocke and @MasterQuiz!

Fawn Locke

It's the same with array operands, just the other way around

⋄ 1{⍺}0

Master Quiz

@AlexB Notice also that it's +/ and not /+

TryAPL

@FawnLocke 1

Master Quiz

@FawnLocke Did you mean functions?

Fawn Locke

8:03 PM

No, it's hard to clarify in a short sentence. Not related to this misunderstanding, (but related to the larger conversation) you can read about operator syntax here: help.dyalog.com/latest/Content/Language/…

You can use operands of either association but a monadic operand must always be present. Although I don't see much purpose in doing it this way

Alex B

I understand all this, but I did not understand until now that the associativity of a user-defined operator cannot be changed by playing with double-alpha and double-omega. It probably has to do with ambiguity during parsing. (If operator associativity sometimes were on the right, there could be situations that the parser cannot disambiguate.)

Master Quiz

@FawnLocke What are you replying to?

Fawn Locke

@MasterQuiz This

Master Quiz

@FawnLocke Isn't {⍺} a function?

Fawn Locke

Yes, I was trying to get across that you can create a function only referencing the dyadic operand, similarly to how you can with operators

but that a monadic operand must be present in both cases

Master Quiz

8:16 PM

@FawnLocke I didn't get "array operand" before, but I understood the simile