@MilkyWay90 Makes sense?

what do you mean by monadic fold?

like +/3 would be 3+3?

@MilkyWay90 That the derived function is a monadic function. And what / does is to make a folding function of its operand.

No, +/3 1 4 is 3+1+4

Ohh

okay

¨ is a monadic operator that takes its operand and makes it into a mapping function. We call ¨ each.

So to sum integers from 1 to 100, you would do this:

⍞←+/⍳100

@MilkyWay90 5050

⎕←{⍵ ⍵}3 ⋄ ⎕←{⍵ ⍵}¨3 1 4

@Adám
3 3
┌───┬───┬───┐
│3 3│1 1│4 4│
└───┴───┴───┘

Okay

I think I get it

⎕←{((0 = 15 5 3 | ⍵) ⍳ 1) ⊃ 'FizzBuzz' 'Buzz' 'Fizz' ⍵}¨⍳20

@Adám
┌─┬─┬────┬─┬────┬────┬─┬─┬────┬────┬──┬────┬──┬──┬────────┬──┬──┬────┬──┬────┐
│1│2│Fizz│4│Buzz│Fizz│7│8│Fizz│Buzz│11│Fizz│13│14│FizzBuzz│16│17│Fizz│19│Buzz│
└─┴─┴────┴─┴────┴────┴─┴─┴────┴────┴──┴────┴──┴──┴────────┴──┴──┴────┴──┴────┘

ninja'ed

Sorry. I should have let you have the pleasure.

⋄

oh whoops

And the close brace is in the wrong spot again.

And you're back to post-fix ⍳. Copy-paste much?

⋄ D←{((0 = 15 5 3 | ⍵) ⍳ 1) ⊃ 'FizzBuzz' 'Buzz' 'Fizz' ⍵} ⋄ ⎕← D¨⍳100

@MilkyWay90
┌─┬─┬────┬─┬────┬────┬─┬─┬────┬────┬──┬────┬──┬──┬────────┬──┬──┬────┬──┬────┬────┬──┬──┬────┬────┬──┬────┬──┬──┬────────┬──┬──┬────┬──┬────┬────┬──┬──┬────┬────┬──┬────┬──┬──┬────────┬──┬──┬────┬──┬────┬────┬──┬──┬────┬────┬──┬────┬──┬──┬────────┬──┬──┬────┬──┬────┬────┬──┬──┬────┬────┬──┬────┬──┬──┬────────┬──┬──┬────┬──┬────┬────┬──┬──┬────┬────┬──┬────┬──┬──┬────────┬──┬──┬────┬──┬────┬────┬──┬──┬────┬────┐
│1│2│Fizz│4│Buzz│Fizz│7│8│Fizz│Buzz│11│Fizz│13│14│FizzBuzz│16│17│Fizz│19│Buzz│Fizz│22│23│Fizz│Buzz│26│Fizz│28│29│FizzBuzz│31│32│Fizz│34│Buzz│Fizz│37│38│Fizz│Buzz│41│Fizz│…

Copy-pasted the wrong program, whoops

That looks good.

Wow, I've learned a whole lot in one day

Yes, APL is very easy to learn. Up for more or have you had enough for today?

up for more

How about writing a function that tells me whether its argument is a prime or not?

ok

pco?

You actually know enough to do so already. Do you want to try by yourself, or do you want guidance?

@MilkyWay90 pco? You're cheating?

@Adám Wait, that's cheating?

ok, I'll try without pco

Yeah, try just using the things you've learned here. Do you know what a prime is?

yes, why not

Tell me.

A number that is divisible by only one and that number

Right. What does "is divisible by" mean?

if the remainder of that number is 0

Right. So you'd have to check divisibility of the target number and all numbers (until that number).

yes

And then you need to check if only two numbers divide it. I.e. 1 and that number itself.

mhm

Go for it. What would the first step look like in APL, given N?

@adelidris Hi. Long time no see. How goes it?

probably ⍳N

@MilkyWay90 Right, that gives us "all the numbers until that number". Now what?

then a list of 0=⍳N|100

unless the 100 and ⍳N are switched

Right, but don't forget your parens.

oh

Well, 100 would be N, no?

why?

Because you are checking if N can be divided by all the numbers until N, no?

0=100|⍳N

Where did 100 come from?

I think I just had a "brainfart"

0=⍳N|N

@MilkyWay90 That's almost right, but you do mean 0=(⍳N)|N, no?

Oh yeah

0=(⍳N)|N

OK, this gives us a Boolean list indicating which numbers divide N evenly. Now what?

and then

we need to count all 1s

@MilkyWay90 Right. How might we accomplish that?

hmm

Remember, Booleans are just numbers (0 and 1) in APL.

oh yeah sum it

and see if it's equal to 2

You got it! Write it out, please.

2=+/0=(⍳N)|N

As a function:

D←{2=+/0=(⍳⍵)|⍵}

Spot on.

yay

Hmm

I'll try seeing if I can solve a simple PPCG challenge

@MilkyWay90 For that, you need to know more about APL's array model, and another function too.

oh okay

@MilkyWay90 I suggest you look at the chat lessons.

I'll learn more before attempting a challenge

@MilkyWay90 This one is easy (and already solved in APL, so don't peek):

You know all you need.

okay

@Adám Are anonymous functions allowed?

sure

wait nvm, they are allowed by default

Ok, I found a solution

{÷+/÷⍵}

That's it.

Was just a few bytes more than the other answer:

÷1⊥÷

4 vs 7 bytes

Yes, and that answer uses a couple of more advanced golfing tricks.

⎕←2=+/0=(⍳|⊢) 7

@TessellatingHeckler
1

a train version, I think

@TessellatingHeckler Almost, but 0=+/ is problematic. However, it s safe to use 1⊥ instead of +/

2=1⊥0=⍳|⊢ is indeed a train.

Alternatively, you can combine +/ with =: 2=0+.=⍳|⊢

@TessellatingHeckler This latter version is the one used on APLcart.

@Adám I wasn't thinking the whole thing to be a train, only the (⍳|⊢) part "a version with a train"; is that what you mean by 0= being problematic in describing the whole thing as a train? Or what else might it cause as a problem?

⎕←0=(⍳|⊢) 7

@TessellatingHeckler
1 0 0 0 0 0 1

that's a binary number for 65

⎕←1⊥0=(⍳|⊢) 7

@TessellatingHeckler
2

2??