The beginnings of an implementation of Relational Algebra in APL: github.com/romilly/raapl

@RomillyCocking Use can be even easier: Start APL (e.g. via RIDE) with the configuration parameter LOAD=…/src/apl

Thanks

Btw, if the user does that, then Dyalog will call the function #.Run if it exists. You can use that as a replacement for good ol' ⎕LX

@RomillyCocking Consider adding a license.

Oops - yes. It started a s a private project, and I forgot to add one.

Now all three of your suggestions are implemented.

> Given a vector representing monthly sales figures, write a dfn that returns the greatest percent month to month increase.

 {∨/0=v:999 ⋄ 100×⌈/(¯1↓⍵)÷⍨-2-/⍵}

I found it a bit hard to understand what the percentage would relative to.

Huh, the reference solution is bad?

What do you mean?

Basically, I mean the question if you need ¯1↓ or 1↓

I had {100-⍨100×⌈/¯2÷/⍵}, which would fail like the reference solution on cases with 0s

Same issue with my 100⌈.ׯ2(-÷⊢)/⊢

 (⌈/100×(-2-/⊢)÷¯1↓⊢)

I vote for removing test cases with 0s.

@Richard Btw, you can use ¯2-/ to reverse the the segments before their reductions.

(or use -⍨)

ah, ok. like I did with the dfn?

Does it really make sense to have negative sales?

@Richard Yeah.

@Adám I didn't manage to make it tacit, and swapped it again.

@Adám I don't think so. 0 sales would make more sense, but cause more inconvenience :/

@Richard But that is tacit.

(was home work from @xpqz ;))

@ovs The text doesn't specify what should happen when going from 0 to something.

@Adám depends - could be you've bought something. That's how I first thought of negative numbers, that you owe someone something

Then what is the correct solution according to the practice problems page? Also catching the 0 gives the same error, instead of telling what the result should have been

Sure, but if so, then it doesn't make sense to month to month increase, does it?

@Richard Yeah, even the internal reference solution doesn't handle it.

APL (Dyalog APL), 35 bytes: f←⌈/2{100×⍵÷⍨⍵-⍺}/⊢

why is the precision trash?

OK, I've updated the test cases to be strictly positive.

yes is working now

@PyGamer0 For golfing purposes, that's 17 bytes.

@Adám oh

APL (Dyalog APL), 19 bytes: f←⌈/2{100×⍺÷⍨⍵-⍺}/⊢

@Adám forgot to change the encoding :p

You don't need to count the assignment, per CGCC rules.

yea

i know

Can anyone beat my 15?

It should really be phrased as 100⌈.ׯ2(-÷⊢)⌿⊢ or ⌈⌿100ׯ2(-÷⊢)⌿⊢ to be leading-axis oriented.

ngn/k, 20 bytes: |/{100*(y-x)%x}/'2':

You can save one byte in dzaima/APL as it allows omitting the trailing ⊢

@Adám 14: 100×1⌈.-⍨¯2÷⌿⊢

Wow.

How does this work? (Simpler to explain ⌈/100×1-⍨¯2÷⌿⊢)

Same principle as your earlier {100-⍨100×⌈/¯2÷/⍵} but I don't get it.

Adeline gives: (100×(( ¯2÷⌿⍵)⌈.-1))

(a÷b)-1 is the same as (a-b)÷b

Then its just the same as yours

@awagga It doesn't explain‽ ;-)

Heh..

@ovs I see, but I must have forgotten basic arithmetic or something, because I still don't understand.

¯2(-÷⊢)⌿⊢ → ¯2{(⍺-⍵)÷⍵}⌿⊢ → 2{(⍺÷⍵)-⍵÷⍵}⌿⊢ → 2{(⍺÷⍵)-1}⌿⊢ → 1-⍨2{⍺÷⍵}⌿⊢ → 1-⍨2÷⌿⊢

This would be how you could derive it from yours, but ÷/ was just my first thought when I read the problem

D'oh. Of course.

Leaving golfing, and thinking about efficiency…

I would guess that 100ׯ2(⌈/-/÷⊢/)⊢ would be fastest.

Whoa:

         n←?⍴⍨1e6
  (100ׯ2(⌈/-/÷⊢/)⊢)n → 4.3E¯3 |     0% ⎕
  (100⌈.ׯ2(-÷⊢)/⊢)n  → 2.5E¯1 | +5738% ⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕⎕

@Richard You win the performance price.

well... , :)

(100×1-⍨(⌈/¯2÷/⊢))n is 34% faster for me

Oh wow, by me too. Why is that?

Oh, what's with me today‽ It only does a single pass.

I like {100ׯ1+⌈/¯2÷/⍵} the most, then.

@Adám yes, agreed

Anything else we can say about this?

OK, then that concludes the 2014 problem set. See you next week for 2015-1: Nag A Ram.

@Adám what?

Hello everyone! I have a stupid question about the power operator ⍣. I would like to run my fun ← {?10} 10 times and I try fun⍣10 which results in nested vector of function, operator and 10. Not sure why it doesn't work.

@PyGamer0 (a-b)÷b = ((a-b)÷b)÷(b÷b) = ((a÷b)-(b÷b))÷1 = (a÷b)-1

@minerals It just shows you the derived function. You have to give it an argument to run it (even if the argument isn't used).

Btw, (fun⍣10)'dummy arg' does generate 10 random numbers, but you'll never see any of them other than the last generated one.

ah, I see. Yes, it gives me the last one (which makes sense), aand I forgot about 10?10 :)

@minerals They are not the same. ?10⍴10 generates ten independent 10-sided dice rolls, while 10?10 deals ten unique cards from a 10-card deck.

@Adám ok, that's good to know. I guess ?10⍴10 and ?⍳10 are equivalent.

For some reason, ¯2⊢/80 100 120 140 is returning 100 120 140 instead of 80 100 120 for me.

Is this an old bug? I'm on version 18.0.40684.

@minerals they're not. The latter's first item is always 1, the second is either 1 or 2, the third is 1, 2, or 3, etc

@rabbitgrowth yes, that does appear to be a bug; It's fixed in 18.2

Thanks, I was baffled as to why Adám's solution wasn't giving the right result for me