Julia

If I'm not mistaken, [x y] actually constructs a 2-dimensional array (Array{T,2} where T = eltype(array)) whereas [x,y] constructs a 1-dimensional array.

^ this is true

Dennis

I didn't even know there was a difference...

Um, I restarted the REPL and my answer doesn't work anymore. How is that possible?

Ah, crap. generic function with 3 methods means I overloaded f, right?

Needed more .... Now I understand why you cannot redefine generic functions as anonymous functions and vice versa.

@AlexA. Switched to comma. I still need spaces in 2D arrays though, right?

Alex A.

7:57 PM

@Dennis That depends. It'll tell you that you have more methods if you have multiple function arguments

@Dennis You may have had things defined that it was reading that it shouldn't have been

In a fresh REPL it says that f has one method

Dennis

No, it was the overloaded f. I got away without the ... since it actually called previous definition when recursing.

Alex A.

Btw it works for me in a fresh REPL. The last test case is pretty slow though.

Dennis

Yeah, I posted a fix 11 minutes ago. The previous version shouldn't work.

Alex A.

o

@Dennis Forgot to respond to this. Yes.

Dennis

Does Julia automatically memoize function calls?

Alex A.

8:07 PM

I think so.

wafflemao

Dennis

Try executing the large test case twice.

Alex A.

Oh damn. Instantaneous

Dennis

Yup.

This is also very fast:

h(x)=(r=[];while x∉r push!(r,x);x=sum(j->circshift(x,[j÷3,j%3]-1),0:8)-x|x.==3end;(r[k=end]==x)k-1)

@AlexA. hahahaha

Alex A.

@Dennis r=[r;x] is shorter than push!

You also have a space after the comma

Dennis

That gives a DimensionMismatch.

Alex A.

8:10 PM

Oh

Ohhh I see why

Yeah

Is divrem(j,3) shorter than [j÷3,j%3]?

Nope

Even though ÷ is 2 bytes

Dennis

@AlexA. I don't. :P The whole array notation business is pretty obscure to me.

divrem wouldn't work anyway, since it returns a Tuple and I can't subtract 1.

Alex A.

oh right

You'd need [divrem(j,3)...] which is absurd

@Dennis What's the type of x as you're doing this?

It's a multidimensional array, right?

r=[] initializes r as Array{Any,1}

So trying to vcat a multidimensional array onto it is no bueno, but you can push! an array into it because its eltype is Any

@Dennis Since this isn't recursive you don't need to name it, which saves a byte or two.

two

Also, nice trick when golfing in Julia: You can score your submission by wrapping it in sizeof("")

You just need to escape quotes and $

Dennis

Still way longer than the recursive solution though...

But the speed difference is impressive.

Alex A.

8:33 PM

Which is faster?

Dennis

The iterative solution finishes instantly.

Alex A.

By iterative do you mean recursive?

Dennis

No, the while loop.

Alex A.

Oh

I realized I misspoke in my explanation of why r=[r;x] doesn't work. It's been ~~elevened~~ corrected. Does it make sense?

The syntax [array; array] is actually doing vcat(array, array)

Dennis

Sort of. I'm still fuzzy on the differences between spaces, commas and teary eyes...

Alex A.

8:36 PM

haha

Dennis

And severely annoyed by the fact that [x] doesn't wrap x in an array.

Alex A.

It doesn't? Huh. I guess I've never tried that.

Dennis

If x is an array [x]==x give a deprecation warning and true.

Alex A.

huh

> oldstyle_vcat_warning

Right

You used to be able to do [x,y] where x and y are arrays

That used to be vcat but now it's [x;y]

It's one of those readability things

I don't know what it thinks you're trying to vcat with [x] though

And the recommendation of collect is equally mysterious

Dennis

The weird thing is that it works if you prefix Any. All this array type business is very confusing.

Alex A.

8:44 PM

Yeah, because if it's Any then you're just listing elements in an array literal rather than vcating

Dennis

Should we create a Julia room? We discuss things rather frequently, and it's not always a fit for CGC.

Alex A.

I wonder if there's a Julia room in Chat.SO

brb looking

There is not

Oh but we talk about golfing so it would make sense for PPCG to be its parent

Okay I'll make room

Discussion about the Julia language and golfing in it — julial...

Dennis

\o/

Alex A.

Julia is objectively the best language ever to exist.

3

Dennis

It is realle great and does all things.

Should we migrate this conversation so we can continue it there?

Alex A.

8:52 PM

waffles are nice

Dennis

Disappointingly, this is longer than what I had before:

f(x,r...)=x∈r?(r[k=end]==x)k-1:f((g=x->sum(i->circshift(x',i),-1:1))(g(x))-x|x.==3,r...,x)

Feeds

Alex A. has added Dennis to the list of this room's owners.

Alex A.

@Dennis That's some mysterious code you have there

Dennis

g(x)=sum(i->circshift(x',i),-1:1)
f(x,r...)=x∈r?(r[k=end]==x)k-1:f(g(g(x))-x|x.==3,r...,x)

That's better.

Alex A.

Would it be shorter to define g as a function argument?

Actually I think it'd be the same

Dennis

9:00 PM

Too bad there's no Cartesian product. 2D ranges would also help.

Alex A.

Yeah. :/ Something like R's expand.grid would be nice.

Oh

[(i,j)for i=1:n,j=1:m] would do it

julia> [(i,j)for i=1:4,j=1:5]
4x5 Array{Tuple{Int64,Int64},2}:
 (1,1)  (1,2)  (1,3)  (1,4)  (1,5)
 (2,1)  (2,2)  (2,3)  (2,4)  (2,5)
 (3,1)  (3,2)  (3,3)  (3,4)  (3,5)
 (4,1)  (4,2)  (4,3)  (4,4)  (4,5)

Dennis

That's a lot longer than the div-rem approach though.

Alex A.

Yes

Damn I love the , syntax for nested for loops

Dennis

Wait, so if x is 2D, x[1] gets the first entry rather than the first row?

Alex A.

Yes

x[1,:] gets the first row

Dennis

9:12 PM