Sorry for the extremely basic question, but: how do you read in a text file in gnu apl? I'm trying to do this year's advent of code in APL, but I'm stuck at the first hurdle (loading a text file of numbers).

@Lapwing482 fopen just returns a file descriptor integer, which you generally pass around to other libc functions. You probably want fread instead.

AoC Day2: data←{' '~⍨{'ABC'['XYZ'⍳⍵]}@(∊∘'XYZ')⍵}¨⊃⎕NGET 1,⍨⊂'input.txt', and Part1 +/{(3 0 6)[1+3|-/i]+⊃⌽i←'ABC'⍳⍵}¨data

Part 2: +/{(1+3|+/i)+(0 3 6)[x←⊃⌽i←'ABC'⍳⍵]}¨data

(⎕io←1)

@LdBeth Maybe you want to add your code to apl.wiki/aoc?

I'm not sure if I'll go for speed running or give more meaningful programs, but I'll make an edit to aplwiki if I decide to collect my answers. (Still fighting with school's ARM assembly project tonight)

Here's my attempt:

⎕IO←0
m←3|'ABCXYZ'⍳1 0 1/↑⊃⎕NGET'input.txt'1
⎕←(+/(1+⊢/)+3×3|1+-⍨/)m
⎕←(+/(3×⊢/)+1+3|1-⍨+/)m

It was surprising to see how similar the solutions for the two parts turned out to be

Welcome to APL Quest 2017-4! Today's quest is Slice(s) of Pie(s):

> Write a function that calculates and returns the areas of 0 or more pie slices. The left argument is 0 or more angles (in degrees). The right argument is 0 or more pie diameters.

{360÷⍨4÷⍨⍺×○1×⍵*2}

1×‽

Also, you can combine the two divisions into 1440÷⍨

○1

1 times pi

○ is a normal function, so it takes everything to its right as its argument.

@Adám yes, just for clarity

○1×⍵*2 ←→ ○(1×⍵*2)

ah ok. Makes sence

How concise and efficient can we make it?

All the × and ÷ can be combined

True, but does that help?

Try going tacit.

Since there's only one monadic function, you should have that as the last step, the far left.

(1440 ÷⍨⊣ × ○∘1 × *∘2)

more or less, working on it

The ○∘1 can't work like that.

Oh, I see, that's WIP code.

yes.

Hi @Silas, you like tacit stuff, right? What can you make of {360÷⍨4÷⍨⍺×○1×⍵*2}?

I like {(⍺÷360)×○2*⍨⍵÷2} or even {(⍺÷360)×○(⍵÷2)*2} for clarity, personally

I will have a think about tacit

you mean, how would I make that tacit?

Yeah.

@RikedyP Literally spelling out the formulas, OK.

hehe

actually I think mine's wrong - oops

oh no it's fine

I'm misreading test cases

My shortest solution needs ⎕DIV←1

your right tine is an array

which is not a valid train

You can fix it with a trailing ⍨

But also, your ○ will be called dyadically.

how long's your shortest @Adám?

9 with ⎕DIV←1, 10 otherwise.

geez

I had 360÷⍨×∘(○2*⍨÷∘2)

Right, that's basically the tacit equivalent of @RikedyP's spelling out.

@Adám wow

Just need a bit of mathematics for moving the terms around.

@Adám does it have the literal 1440 in it?

Yes, both do.

1440÷⍨⊢××∘○ and ×∘○÷1440÷⊢ with ⎕DIV←1, 11 and 10 chars

Good. Now you just need to move ○ to the far left.

wild

Bravo. I had ○⊢××÷1440⍨

And ○⊢÷1440÷× with ⎕DIV←1

Richard, you're the winner…

See you next week for 2017-5: Slice(s) of Pie(s)!

Next week!

btw it's 'DNA?'

problems.tryapl.org/psets/2017.html?goto=P5_DNA

Did a quick runtime: unsurprisingly, doing less maths takes less time:

  ]runtime -c "a{(⍺÷360)×○(⍵÷2)*2}b" "a(1440÷⍨⊢××∘○)b" "a{1440÷⍨⍵×⍺×○⍵}b"

  a{(⍺÷360)×○(⍵÷2)*2}b → 3.9E¯5 |   0% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕
  a(1440÷⍨⊢××∘○)b      → 2.7E¯5 | -31% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕
  a{1440÷⍨⍵×⍺×○⍵}b     → 2.7E¯5 | -32% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

Hmm, still going but why doesn't (2*⍨)12 work?

gives missing right argument

it probably missed the right argument before swap is being done?

yeah, just confused as thought would be valid

(2∘×)4 is however, Which is a single function by ny understanding

But there are better people to explain it

(2∘x) is even not a train i guess

I think (2*⍨)12 is valid in some dialects?

aplwiki.com/wiki/Train

> Only dzaima/APL allows (A h), which it treats as A∘h.

Different approach to AoC day 2
(4 8 3 1 5 9 7 2 6)+.×w[⍋{⍺(w←≢⍵)}⌸⊃⎕nget 'input02.txt' 1;2]

works as long as every combination is used

and didn't use logic to calculate to victory points

@Richard interesting

(3 4 8 1 5 9 2 6 7)+.×w[⍋{⍺(w←≢⍵)}⌸⊃⎕nget 'input02.txt' 1;2]

and second one

but other solutions are smarter

I had:

i←⍉↑¯65 ¯88+↓⎕UCS⍉(↑⊃⎕NGET'd/2'1)[;0 2]
+⌿(1+3 6 0 0 3 6 6 0 3[x←3⊥⍤1⊢i])+u←⊢/i ⍝ part 1
+⌿(1+2 0 1 0 1 2 1 2 0[x])+3×u          ⍝ part 2

for ⎕IO←0

⊥3 is nice @xpqz :)

+/{⍵[2]+⍵⌷↑(3 6 0)(0 3 6)(6 0 3)}¨data ⍝ part 1

+/{(0 3 6[⍵[2]])+⍵⌷↑(3 1 2)(1 2 3)(2 3 1)}¨data ⍝ part 2

where data is a vector of vectors of the input, where A/X -> 1, B/Y -> 2, and C/Z -> 3

No idea how to do code blocks in SE chat lol

Oh, I guess I don't need that 3| if I'm doing 23| (was trying to golf a bit)

@voidhawk Ctrl-K does it