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9:00 PM
@cairdcoinheringaahing ah yeah good point
@Mr.Xcoder My pleasure!
@EriktheOutgolfer ah, thanks
@WheatWizard admittedly, the Add a language to a polyglot challenge doesn't have over 250 non-deleted answers, which is probably more of a problem than simply how long it's been going
What is the least permissive type of licence?
@Mr.Xcoder the "don't give your software to others" kind
"Nobody is allowed to use this, ever."
9:05 PM
CMR: How many integers from [1, n] have an odd number of divisors?
(The answer isn't that difficult, so try not to spoil it for other users)
Yeah, but what is it called?
@DJMcMayhem huh
what is CMR
Chat mini riddle lol
And bonus points if you can solve it without code
how do we answer without spoiling
@Poke Put any answers in a TIO link
9:06 PM
I add a link, but do not include the code
@EriktheOutgolfer It's the ISO country code for Cameroon
@DJMcMayhem Spoiler
not what is meant here
@DJMcMayhem test case?
Pretty much
9:07 PM
@Mr.Xcoder That's a given, but how many is that?
that was my answer too
Oh, were we not supposed to go further than that? ಠ_ಠ
Ok, I'll come up with a more rigorous definition
@cairdcoinheringaahing No, you can
But it's not very hard once you're there
9:09 PM
@KSmarts Correct
@KSmarts Yes, I think that's correct (it's what I had as well :P)
ideas for advent 7?
Honestly, the bulk of the riddle is finding out that spoiler
The rest is pretty straightforward after that
A more rigorous definition than my previous one
9:10 PM
@HyperNeutrino looks like there are no ideas currently, maybe ask later, but asking repeatedly without new people coming in won't help much :p
@Mr.Xcoder it was 5 until I realized something
true :P the thing is i need to finish it by 5 (that's in 50 minutes in this timezone) or i won't be able to do it (utc) today :P
@DJMcMayhem oh that's clever +1
@EriktheOutgolfer Jelly 2 bytes: Try it online!
Just take your time and think :P
@EriktheOutgolfer That's incorrect, as shown by the fact that this is exactly the same
9:13 PM
That must be correct...
@EriktheOutgolfer :)
I'm glad
An even more rigorous definition
but I don't think you can do it in 2 bytes
This is a riddle I learned growing up, but it was phrased like this:
> You're in a hallway with 100 lightbulbs, each one connected to a single switch. They are all off. On the first pass through, you turn all of them on, then you go back to the start. Then, you toggle every other switch. (2, 4, 6, etc.), and go back to the beginning. Then, you toggle every third switch. Then every fourth switch, every fifth, etc. up until the hundredth. After 100 passes, how many light bulbs are on?
9:13 PM
@Mr.Xcoder ÆD gets the divisors of the input, L€ maps each to 1, then ḂS is the same as S
for me that's 4 bytes
@cairdcoinheringaahing I know...
7 mins ago, by Mr. Xcoder
@DJMcMayhem Spoiler
It's a great riddle because most people can figure it out with enough work, but then the answer seems so obvious
And it seems very intimidating at first
@Mr.Xcoder and how do you do it in 2 bytes? that's 4 bytes for me
@DJMcMayhem I don't have any proof, but isn't it this?
9:15 PM
@EriktheOutgolfer Try square root - floor or see this
@cairdcoinheringaahing For n = 100?
@DJMcMayhem Yes (I may be wrong :P)
No, it's definitely this for 100
@Mr.Xcoder hm, the fact that it's 23:15 here sure helps
and I have exam tomorrow so my brain is filled with Biology material...
I see :) and understand you completely. Caird can confirm :P
9:16 PM
@Mr.Xcoder What can I confirm?
@DJMcMayhem not looking at your answer, but seeing as you're talking depending on the value of n I'm guessing my answer of "only the first bulb" is incorrect
@EriktheOutgolfer Good luck!
@cairdcoinheringaahing That I understand him when he says he's tired
9:17 PM
@Mr.Xcoder Ah, yes :P
I thought, "every light toggled an odd number of times stays on", which means it must have an odd number of divisors to stay on, so only the first bulb will stay on. Am I wrong? @DJMcMayhem
12 mins ago, by DJMcMayhem
CMR: How many integers from [1, n] have an odd number of divisors?
You are wrong. 1 isn't the only integer with an odd number of divisors
sya all
@Mr.Xcoder o/
@user202729 I was falling asleep when I realised that your nickname number is probably a date
9:18 PM
@DJMcMayhem oh, of course. I was thinking in pairs of factors. But squares will have odd numbers of divisors
So it is just the number of squares?
Pretty much
And before bed, I reached 10 answers on the OEIS challenge \o/
You're right about the riddle, it seemed daunting at first, but once I realised it's just looking for numbers with an odd number of divisors it doesn't seem like a massive task
Most non-mathematicians/programmers I've asked take a longer time to connect the odd number of divisors == on and then squares == odd number of divisors
It's a great riddle to tell IRL friends
Of course, I forgot to make the connection with squares
9:20 PM
@DJMcMayhem the power of the community
I made the connection quite rapidly because we're discussing divisibility in maths in school lately
@EriktheOutgolfer Needs an R
e.g. RÆDL€ḂS
@DJMcMayhem ik
btw I can't think about it very much, gotta keep all that biology till tomorrow
and that's why I gtg now
9:23 PM
No problem lol
@DJMcMayhem ಠ_ಠ I took the difficult way Who needs to count squares :P
I don't know the answer to this one, but what if you went through the hallway of lights and only flipped the switches of the prime-th lights
Start with everything off, go through every second, every third, every fifth, until every 97th
Well then every prime would be on
@Sherlock9 IS the question how many lights are on?
Yes, as well as which ones
@DJMcMayhem I think it would be every number with an odd number of prime divisors (without multiplicity)
Sorry I just realized how I was being unclear
Every second (2, 4, 6, ...), every third (3, 6, 9, ...) up until every 97th
So something like this, I suppose: tio.run/##y0rNyan8/z/ocJubz6OmNQ93NIX8///f0MAAAA
That seems correct
@Sherlock9 I know its not a challenge/code golf, but 5 bytes
Brilliant :D
I wonder how many other variations we could come up with
In the meantime, I'm heading to bed. Goodnight
9:46 PM
@Sherlock9 Implemented the actual corridor-based algorithm, got the same answer. TIO
is anyone willing to post advent 7? I didn't have time today and by the time I do it will probably be past UTC midnight
My /var/lib/pgsql is haunted. ls -a doesn't even list anything, even . and ...
Oh, nevermind, it's because I don't have permission over it and nothing was showing an error ._.
10:01 PM
Q: Command-line flags on front ends

Erik the OutgolferOur standard policy regarding command-line flags states that one should count the space before the dash (-) too, when there aren't any "free" options, since it doesn't exist in the shortest possible invocation. For example this: language -option program.p is 8 bytes longer than the shortest po...

10:12 PM
A: Create the slowest growing function you can in under 100 bytes

Anders KaseorgHaskell, 98 bytes, score=fε0−1(n) _#[]=0 k#(0:a)=k#a k#(a:b)=1+(k#(([1..k]>>fst(span(>=a)b)++[a-1])++b)) f n=[k|k<-[0..],k#[k]>n]!!0 How it works This computes the inverse of a very fast growing function related to Beklemishev’s worm game. Its growth rate is comparable to fε0, where fα is th...

Could anyone explain this answer to me? I'm not good with Haskell
@LeakyNun I'll try.
A: Sandbox for Proposed Challenges

Steven H.A Fine-Grained Mesh code-golf If you've used Matlab before, it's highly likely that you've heard of meshgrid. It's a function that has since mostly been obsoleted by broadcasting, but it still has its uses sometimes. The function itself is relatively simple. Given two vectors x and y of le...

@Zgarb thanks
The function f takes a number n, and finds the lowest number k such that k#[k] > n. The idea is that k # [k] grows very fast, so f grows very slowly.
k # x, where x is a list, can be understood as the number of steps in a process that ends when x is empty. k stays fixed, x is modified step by step.
_#[]=0 means that when x is empty, the number of steps is 0 (since it's the end).
k#(0:a)=k#a means that if the list begins with 0, we remove it and it doesn't cost a step.
I know no Haskell at all
10:23 PM
Ok. Are you with me so far?
f and # are just user-defined functions.
go on
k#(a:b)=1+(k#..) is the last case of #. It means that x has first element a and tail b (so it's not empty), and a is not 0 (otherwise we'd be in the second case).
The number of steps from x is the number of steps from .. which is the next iteration, plus one.
.. is this list: ([1..k]>>fst(span(>=a)b)++[a-1])++b
What happens here is that we take the longest prefix of b whose elements are all at least a (this is fst(span(>=a)b)), tack a-1 to its end (this is ++[a-1]), and repeat the result k times (this is [1..k]>>). We concatenate this with b, which means that we essentially replace a with this long list.
user image
For example, if k = 3 and x = [2,3,3,2,1,6], then a = 2 and b = [3,3,2,1,6]. The prefix is [3,3,2,1,6]. We tack a-1 to it and get [3,3,2,1,6,1]. Repeat k times to get [3,3,2,1,6,1,3,3,2,1,6,1,3,3,2,1,6,1]. Concatenate with b to get [3,3,2,1,6,1,3,3,2,1,6,1,3,3,2,1,6,1,3,3,2,1,6].
10:34 PM
@Pavel wtf
That's horrible, yet amazing
10:44 PM
I'm pretty hyped about the new years challenge. Threead will rise again.
@ATaco >tfw you did nothing all year
11:07 PM
Just recovered an old BTC wallet. Christmas came early this year.
How old? And do you have more than one BTC?
Even got a sizable amount of BCH from the split. Didn't even know this exchange did BCH deposits.
Well, pop the campaign.
@ATaco I have more than .1 BTC.
Ages ago, I used this wallet to pay server bills.
Much more than me :P
11:20 PM
I have ~$100 in cryptocurrencies now, thanks to that one user who was handing it out as a challenge prize.
EU exchanges also seem to be much more pessimistic about price, and have bigger, less frequent sales/buys. Let's see if it hits 15k EUR this year.
Anyone familiar with this project: tenso.rs? All the promised demos are online, but it's still not released. I begin to doubt it ever will...
11:49 PM
Yay, I got Yearling on WorldBuilding.
@mınxomaτ dang that's actually a good bit
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