Welcome to APL Quest 2014-8! Today's quest is Go The Distance:

> Write a dfn that returns the distance between two points in a space of any number of dimensions.

So, the formula for this is pretty simple: √((x₂−x₁)²+(y₂−y₁)²+(z₂−z₁)²+…)

0.5*⍨1⊥2*⍨-

From APLcart? ;-)

Oh, wait, it isn't exactly the same spelling.

Can't paste my solution but squared the input, added each element and took the square root.

There are two "inputs" (arguments).

The same I guess as RAbbit

@Adám No, no cheating :)

Yeah, sorry. I didn't notice that yours was unique.

I had {*∘.5+⌿*∘2⊢⍺-⍵}

obligatory +/⍢(×⍨)-, using Under (×⍨ → *∘2 to make it run in dzaima/APL)

@Adám yes, squaring the difference

@doug Interesting how you bind right args to create monadic square and squareroot fns.

@dzaima Yup. *∘2⊢ could also be ×⍨

This one is fun: +.×⍨⍤-*.5⍨

@Adám (misedit?)

ngn/k, 11 bytes: %+/*/1{x}\-

I’m not sure what I was thinking at the time. : )

@dzaima Yeah, that additional comment should have been for @doug

My fun one isn't shorter than rabbitgrowth's, though.

@PyGamer0 How does that work?

@Adám subtract -> square (basically catanate to self and reduce by multiplication) -> sum -> sqrt

Now I get how rabbigr’s works. It’s essentially the same but fancier. I still don’t have a great feel for how performance is impacted.

@Adám nice. I always forget outer product

Inner.

@PyGamer0 {x} looks funny..

Yes

@PyGamer0 I don't understand what the 1{x} is doing (as operand to \?)

@Adám it's like a cumulative ⍣, repeating {x} once, and giving the original and new results as an array

Ah.

@PyGamer0 Why does %+/{x*x}- not work?

@ovs {x*x} is a noun. You need %+/{x*x}@-

@PyGamer0 Isn’t 1{x}\ just 2#

@doug yea

@doug Use doubled backticks. (Not too late to press UpArrow and edit.)

@doug hmm true

@doug no that doesnt work

for some reason

@Adám tx

2#, did the trick

needs to be 2#,

⍢ is interesting. Also it looks so happy

Speed more or less the same for all of them I assume?

@rabbitgrowth The one you need here can be defined as {⍺←{⍵ ⋄ ⍺⍺} ⋄ ⍵⍵⍣¯1⊢(⍵⍵ ⍺)⍺⍺(⍵⍵ ⍵)}

@Richard I think so.

I can't really think of anything else to add here.

Yes, but though multi-dimensional complex vectors are maybe not so meaningful.

I gues it should

OK, then I'll run. See you next week.

,l👍

@PyGamer0 FWIW: variants

@Adám Would you mind explaining the ⍺←{⍵ ⋄ ⍺⍺} part?