@JeffZeitlin @dzaima Who should get the bounty here? The first or the correct solution?

In the RIDE interface, is there an option to make it behave a bit closer to a "traditional" REPL? Like up-arrowing through the executed expressions, rather than also going through any output? Or failing that, some sort of history command? Or am I just doing it wrong?

@xpqz Ctrl+Shift+Backspace/Enter. You can customise those.

oh man -- I could have saved myself so much scrolling :)

Is there also a setting to make all output be run through DISPLAY by default?

@xpqz ]box on -style=max

You guys have thought of everything.

@Adám - I question whether there can be any correct solution to the problem as stated. While dzaima's solution does provide a correct string of digits of arbitrary length, it's returning an integer (or BigInt), not the actual value (less than 1) requested. I leave the question of the bounty to your discretion; I have come to the conclusion that the problem as stated should have been considered invalid.

@JeffZeitlin Yeah, I noticed that others didn't follow the spec either, which is why I made the bounty to you.

@Adám - For what it's worth, I spotted a couple of other "solutions" that used exactly the same algorithm - the K solution given is essentially an exact translation of my APL solution. (Or vice-versa; it seems that the K solution was posted in 2017.)

Right.

@JeffZeitlin it does ask for precision as the input, which makes it 100% possible.. (given enough memory & time)

@JeffZeitlin that k solution is also invalid.

either way, your solution as-is in Dyalog is definitely invalid as is just plainly ignores a rule of the challenge, whether or not you call that rule possible to follow or not (spoiler: it is possible) is unrelated

@JeffZeitlin i could remove the trailing L and prepend 0. but i didn't feel like it was required. the author has been around, and hasn't called my answer invalid so ¯\_(ツ)_/¯

How can I pick out the last item of each element vector in

⎕←1 5 ⍴ (4 7) (5 6) (6 5) (7 4) (8 3)

So expecting 7 6 5 4 3

My thinking was to increase the rank, transpose and pick last row.

But that won't work because of the shape (I think).

This works: 1⌷⍉↑(4 7) (5 6) (6 5) (7 4) (8 3)

If the arg is a vector, not a 1-row matrix

not (⊃⌽)¨?

Yup, that works...

@xpqz 👍

So that's "pick first of the reverse of each"

Makes sense.

Thanks.

@user41805 I don't see what difference that makes.

@Adám i have a solution for which that makes a difference

@user41805 I'm down to 16 for the general case.

does your solution also work correctly for 13 and 15?

@user41805 Yes, gives 133 and 1515

@Adám you mean with that restriction?

@user41805 Now I'm beginning to doubt what you mean by "this restriction". Do you mean that it in principle has to work for all numbers?

@ksi Welcome. If you want explicit write access to this room, just email me: adam@ with the same domain as www.dyalog.com

by restriction, I mean the rule of supporting up to 10000, but my solution for 16 bytes cannot realistically support up to 10000, my solution assumes arbitrary precision integers

@user41805 dzaima/APL, 7 bytes: ⊤⌿≢∘⊤⍴⍕ Try it online!

i find it interesting that this doesn't work in Dyalog but works in my impl

hm, this does however work. ¯\_(ツ)_/¯

@dzaima ((≢∘b⍴⍕)(⌿⍨)(b←2⊥⍣¯1⊢))¨⍳16 this also works

It is truly odd. Cut down example: (⊢(⌿⍨)b←⊢)1 vs (⊢(⌿⍨)⍨⍨b←⊢)1

half-relatedly, i once wanted to brute-force compare dzaima/APL to Dyalog to find differences but quickly realized there'd be way too many. i wonder if there's an acceptable hopefully-equal subset that could be properly brute-forced

At least in 18.0 you can write (⊢⊢⍤⌿⍨b←⊢)1

@dzaima (wouldn't have caught this anyway, no schizos in dzaima/APL (though they might be implementable for purposes of comparison))

@user41805 My solution fails on some cases.

i realised my 16-byter was wrong for some cases

Fixed at 18

@Adám this makes it directly equivalent to the j answer

@user41805 It is. Pretty obvious solution really.

@Adám i have a 20b solution derived from the apl answer to the challenge, hmm

@user41805 Want to see mine?

i have 17 bytes in Dyalog

My 18: Spoiler

@Adám golfed that to 16

@dzaima Show it already!

two 17-byters and the 16

Nice.

«unprintable mumbles» maybe someday I'll internalize that f⍣¯1 is the inverse function of f...

@Adám oh ⊆, nice