The Nineteenth Byte

@trichoplax That's just insane. Like, the cheapest thing out and about here is usually soda. You can get 64oz gas station fountain soda for like $0.49 usually.

DJMcMayhem

@HelkaHomba since I can't see the timer, what's it at?

Geobits

44 secs ago, by Geobits

8 minutes and counting...

DJMcMayhem

6:22 PM

Oh. -_-

TimmyD

What's the timer?

DJMcMayhem

A YouTube live stream

Geobits

Unfortunately, I have to leave in the next couple minutes, so I won't be able to bask in the radiant light of Helka stream.

betseg

oh ---_-

Helka Homba

I'm trying a livestream in a few minutes. Nothing fancy. youtube.com/HelkaHomba/live

betseg

6:24 PM

@HelkaHomba CMC for stream: pronounce /y/ :p

trichoplax

@TimmyD You can buy 4lbs of soda??

Geobits

Don't know if serious...

trichoplax

Oh wait that's only about 2 litres

mınxomaτ

2.5l bottles of coke are pretty easy to get here.

Geobits

To be fair, 64oz of soda probably does weigh around 4lbs :)

trichoplax

6:25 PM

We do have 2 litre bottles I was just thinking of servings

mınxomaτ

Some weird duty-free airport store probably sells soda in barrels. To go along with their 15m toblerone.

I never got this oversize thing.

Helka Homba

6:42 PM

Can you hear the stream?

flawr

@El'endiaStarman youtube.com/watch?v=4k1jegU4Wb4

Nathan Merrill

in mathematics, is the statement infinity > 2 true?

DJMcMayhem

Yes?

Nathan Merrill

I mean, we don't consider infinity to be a number

DJMcMayhem

Oh. I'm not sure

PhiNotPi

6:47 PM

there's positive and negative infinity

and also a concept of "undefined"

+infinity > 2 is true, while 5/0 > 2 is undefined

Martin Ender

@NathanMerrill infinity is defined as a symbol such that for all real numbers x, infinity > x is true

ArtOfCode

@NathanMerrill TypeError: infinity is not a number

Martin Ender

(usually, anyway)

DJMcMayhem

Don't floats have a special value to represent +/- infinity and NaN?

mınxomaτ

Depends on the implementation.

I know that MSIL floats do have -Inf and +Inf separately.

Nathan Merrill

6:52 PM

@flawr this video really confuses me. the -1/12 makes no sense (and I even watched the extra footage)

Martin Ender

@DJMcMayhem IEEE 754 floats do

betseg

$  echo '#include <math.h>
1> main(){printf("%d",INFINITY>2);}'|gcc -x c - -lm -o inf 2>/dev/null && ./inf
1

Martin Ender

(which are the ones you'll most likely work with)

betseg

C says INF > 2

Nathan Merrill

if all of the elements of 1,2,3,4 are strictly greater than 1, 1/2, 1/3, 1/4, then the same proof should apply to both

DJMcMayhem

6:52 PM

@MartinEnder That's what C# uses, right? Do you know what C does?

mınxomaτ

@DJMcMayhem C# compiles to MSIL, so yes.

DJMcMayhem

I don't know what MSIL is...

mınxomaτ

@DJMcMayhem The bytecode that powers all DotNET languages.

DJMcMayhem

Ah

mınxomaτ

{C,J,F}#, VB.NET etc.

What C does depends on the settings the compiler uses. I.e. how strict it should adhere to the IEEE float std.

Yodle

6:55 PM

Isn't it CIL now?

mınxomaτ

Yes, now.

Yodle

Anyone golf in it?

mınxomaτ

You mean IL assembly, or DotNET things in general?

Yodle

IL assembly, I've seen people use x86.

mınxomaτ

Dunno, but in that case it might be a better idea to use IL bytecode and not IL assembly, that'll be shorter and equally inconvenient.

Quite powerful:

List of CIL instructions

Main article: Common Intermediate Language This is a list of the instructions in the instruction set of the Common Intermediate Language bytecode. Base instructions form a Turing-complete instruction set. Object model instructions provide an implementation for the Common Type System. == See also == Common Intermediate Language is the assembly language that uses the instruction set. Common Language Infrastructure is the standard in which the Common Intermediate Language is defined. .NET Framework is a platform and implementation of the Common Language Infrastructure. Mono is a cross-platform open...

El'endia Starman

7:03 PM

@flawr Man, the ant on the rubber band is so mind-blowing, but then it totally makes sense. :P

Nathan Merrill

after more research, it appears that 1 + 2 + 3 + 4 diverges but also has a value of -1/12

which is odd

aka, it approaches positive infinity, and then, after adding all of the terms, it hits -1/12

Mama Fun Roll

Hey guys!

github.com/molarmanful/DASH

I have a new language!!!

Poke

have you been watching numberphile, @NathanMerrill

PhiNotPi

CMC write a QFTASM-to-CIL compiler

Yodle

1 + 2 + 3 + 4 + ⋯

The infinite series whose terms are the natural numbers 1 + 2 + 3 + 4 + ⋯ is a divergent series. The nth partial sum of the series is the triangular number ∑ k = 1 n k = n ( n + 1 ) 2 , {\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}},} which increases without...

PhiNotPi

7:07 PM

@MamaFunRoll 10/10 logo

Nathan Merrill

@Poke yes, look at my earlier comments

@MamaFunRoll what does rng do, and why doesn't it generate a random number?

Poke

@NathanMerrill pretty cool channel

Wheat Wizard

If anyone wants some rep meta.codegolf.stackexchange.com/a/10109/56656

Mama Fun Roll

@NathanMerrill it's range

Helka Homba

@Geobits @DJMcMayhem Did you watch? It seemed to say 1 viewer the whole time ;_; Jk, I just wanted to make sure the audio was ok. (youtu.be/Pir3olQlfxc)

Nathan Merrill

7:11 PM

range is too many characters?

are you a golfing language or not?

Mama Fun Roll

lol yes

not rly

just a language for lazy people

Nathan Merrill

well, rng IMO means "random number generator", not "range"

Geobits

@HelkaHomba Umm, I had the tab with the timer open, but I was out until just now, so no idea.

DJMcMayhem

@WheatWizard If you're not careful there, you might end up giving Dennis (or someone else) all of your rep at this rate. :P

Wheat Wizard

@DJMcMayhem I make a 500 rep bounty every time I hit 1500. I don't care much for rep past 1000.

DJMcMayhem

7:16 PM

huh. I certainly do! A lot of the privileges are really nice

Helka Homba

@Geobits >.> (here's some substantive discourse so this message is not viewed as spam: Does anyone else like raw potatoes?)

Martin Ender

@WheatWizard you should also be aware that the privilege thresholds will increase once we get our design

Geobits

Well I tried to give warning :P

54 mins ago, by Geobits

Unfortunately, I have to leave in the next couple minutes, so I won't be able to bask in the radiant light of Helka stream.

DJMcMayhem

@WheatWizard Hey, would you mind if I did a more extensive edit on your tips post to provide some more information about what brain-flak is, and to fit with the general flow of most tips questions?

betseg

@DJMcMayhem removing vim

DJMcMayhem

7:18 PM

@betseg ???

Wheat Wizard

@DJMcMayhem ok that would be nice.

Geobits

@HelkaHomba Raw potatoes? No, I'm not a savage.

betseg

installing gvim (bcoz it has clipboard support)

Wheat Wizard

@MartinEnder How much will they increase?

DJMcMayhem

Why remove vim? Why not just have both? (Also, what OS are you on?)

betseg

7:19 PM

@DJMcMayhem arch, they conflict with each other

same package, different configs

DJMcMayhem

@MartinEnder Hey, could you faq this question? It seems like it would be a nice FAQ

Nathan Merrill

I still think we need a faq listing all of the standard types

"What counts as a boolean/integer/double/string/list"

betseg

normal vim in arch isnt built with +clipboard, gvim is, and both have /bin/vim

@DJMcMayhem any recommendated bind for :paste :nopaste?

Beta Decay

Ooh, there's been an update to the mobile site... You can now upload pictures for answers

betseg

yup

Beta Decay

7:32 PM

Has anything else changed?

DJMcMayhem

@betseg I've actually never used those options, so I'm not sure what they do. How about nnoremap <leader>p :se paste!<cr>:se paste?<cr>

betseg

@BetaDecay the code tag became actually monospace a while ago

@DJMcMayhem it says vim "this thing im gonna paste has indentations, dont mess with em"

DJMcMayhem

Oh. That sounds really nice. It really bugs me when I do (from insert mode) <C-r>* and it screws with the indentation.

Beta Decay

@betseg Oh, I thought it did but I thought I'd done something by messing around with Firefox

betseg

lol

oh i think +clipboard disables :paste

sad :(

wat

flawr

7:46 PM

@El'endiaStarman I didn't really get that one.

betseg

i think its removed from vim 8

DJMcMayhem

I no longer have 15k. Oh well, it was fun while it lasted

betseg

wat happened

DJMcMayhem

Dennis had to go ahead and break a record...

flawr

@NathanMerrill Well obviously with our epsilon-definition of limits, 1+2+..+n diverges. But then you can still try to assign a real value to a divergent series (and you can do that in very different ways) and in one of these ways you'll end up with -1/12

betseg

7:48 PM

wait I'm an idiot

it's :set paste and :set nopaste

DJMcMayhem

Or :se paste! to toggle

Nathan Merrill

@flawr right :) I eventually came to that conclusion. It makes me wonder what other divergent series have finite sums

Beta Decay

Is it possible to start a bounty on a question (by someone else) which already has an active bounty (also by someone else)?

DJMcMayhem

@BetaDecay No

flawr

Another way that does not work for 1+2+...+n, but for other divergent series is the Cesàro summation.

DJMcMayhem

7:49 PM

Chat mini-poll: What license do you use/recommend for open-source code? Why that one?

Beta Decay

@DJMcMayhem Ugh, that's annoying. I'll forget in 7 days

DJMcMayhem

What are you hoping to bounty-ize?

Beta Decay

I will give a bounty to the shortest code which colours the output using red and black to make the heart — Beta Decay Sep 27 at 21:56

flawr

@DJMcMayhem Why would anyone care about a license.

trichoplax

@BetaDecay Post to the open ended bounties meta question then someone can remind you if you forget

Beta Decay

7:51 PM

Good idea

DJMcMayhem

I either starred or posted every single message currently on the starboard. Interesting...

Nathan Merrill

wouldn't the sum of randomly chosen integers (out of all integers) be the same as 0 + 1 + -1 + 2 + -2 ...??

trichoplax

@BetaDecay Presumably that also has to be code that can't be shortened by removing the colour requirement, otherwise it'll be deleted as insufficient effort?

Nathan Merrill

because over an infinite number of "picks", you're going to get an even distribution across all numbers

trichoplax

I guess the mean remains zero, but the variance and standard deviation will grow without limit as you increase the range you choose from

Beta Decay

7:54 PM

@trichoplax Yeah, I would've thought so

flawr

The problem here again is, how do you choose "random" numbers? You need to assume some distribution first.

There is no uniform distribution over the integers (every integer equally likely)

Nathan Merrill

why not? every integer has a uniform 0 chance of getting picked

trichoplax

I must admit I assumed uniform over a range, then extended the range, even though uniform doesn't work for an infinite range...

The problem with uniformly picking an integer is that the expected length of the integer is infinite

flawr

@NathanMerrill But the probabilities should sum up to 1, in order to form a probability distribution.

trichoplax

Oh there's that too...

Nathan Merrill

7:58 PM

@trichoplax I don't understand this...expound?

betseg

youtube.com/watch?v=bMtLvFpBmI8 i did a video about :set paste :P

orlp

is it just me

trichoplax

flawr's is the serious mathematical point to be paying attention too. Mine was more a jokey "it'll have infinitely many digits"

Nathan Merrill

@flawr well, since 1/infinity isn't really 0, my initial "uniform 0 chance" isn't actually true

orlp

or did google chrome just update it's looks?

Nathan Merrill

7:59 PM

the fact that you can't define a probability to a particular integer doesn't really matter

@orlp I had this a couple weeks ago

I really like it

trichoplax

@NathanMerrill You can make the range as large as you like, and still have a small but finite probability for each integer in the range, but that breaks down if you let the size of the range tend to infinity

Nathan Merrill

I think the font is different

Anonymous

@orlp It did about a week ago

orlp

it just changed for me

'material' it's called

trichoplax

@NathanMerrill You end up with weird properties, like the probability of your randomly chosen integer being less than N from zero being zero, for arbitrarily large finite N

Nathan Merrill

8:01 PM

@trichoplax right, I agree. The probability of each integer is "undefined". That doesn't mean that the original "uniform distribution" is undefined though

Martin Ender

@WheatWizard Pro tip when you're offering someone a bounty: leave it up for the full 7 days before awarding it. That draws more attention to the answer which usually rewards the answerer with some additional upvotes on top of your bounty.

3

New Main Posts

1

Q: 2048-like array shift

Assume we want to shift an array like it is done in 2048 game: if we have two equal consecutive elements in array, merge them into twice the value element. Shift must return a new array, where every pair of consecutive equal elements is replaced with their sum, and pairs should not intersect. Sh...

Anonymous

Why do I get the feeling that, if anything with Chrome changes once I start working at Google, people here will be like "Mego what did you do?!?!?"

10

trichoplax

In other words, if you pick a uniformly random integer, it's infinite

Nathan Merrill

@Mego and now you wonder why I didn't accept the job

Martin Ender

8:03 PM

oh you didn't? I thought you did.

flawr

@trichoplax But you can show by induction that every integer is finite.

xD

Nathan Merrill

@MartinEnder I technically accepted, then went back

Martin Ender

oh okay

Nathan Merrill

I had only accepted for a couple of days

Anonymous

@NathanMerrill So really now you're an android pretending to be the late Nathan Merrill

trichoplax

8:03 PM

@flawr Hmm. True. Perhaps I should have said arbitrarily large?

DJMcMayhem

@Mego I'm actually giving up chrome. Every time I use it it's more buggy than the last time

Nathan Merrill

@trichoplax makes sense.

trichoplax

So adding two of them together could give you +/- infinity

flawr

@trichoplax Yes probably. But it is still fun ot see how many people do not get that every real number is finite=)

Nathan Merrill

wait..

this assumes that all integers are larger than 0

I didn't make that assumption

you can't say that the probability of the number < N for an arbitrarily large N is 0

trichoplax

8:05 PM

Well, not infinity. Again I should have said arbitrarily large. Is it obvious how far out of my depth I am? :P

@NathanMerrill I often make assumptions without spotting them. Where was it?

Nathan Merrill

nevermind, I proved it to myself

trichoplax

I meant the size of the number, so -5 would be 5 from zero. My wording was appalling though

Nathan Merrill

you can say that the probability of the number A, being between N<A<M is 0, for arbitrarily small N and arbitrarily large M

flawr

@NathanMerrill under what distribution?

trichoplax

Mine was intended to mean that, but with N = -M

Nathan Merrill

8:08 PM

right

DJMcMayhem

3

Q: 2048-like array shift

Assume we want to shift an array like it is done in 2048 game: if we have two equal consecutive elements in array, merge them into twice the value element. Shift must return a new array, where every pair of consecutive equal elements is replaced with their sum, and pairs should not intersect. Sh...

code-golf array-manipulation

Nathan Merrill

@flawr uniform

DJMcMayhem

What's unclear about this?

Nathan Merrill

It doesn't describe the merging process very well

flawr

@NathanMerrill But as I said, there is no uniform distribution.

Nathan Merrill

8:09 PM

right, but we were proving that.

trichoplax

Thanks to whoever's been browsing my network profile the last day or two. I've had votes and badges trickling in on ancient posts... :)

DJMcMayhem

@NathanMerrill It does now

Anonymous

I've upped my no-deadline bounty amount to 500, since it's been several months with no takers

trichoplax

Wow. There are a lot of 500s on that list...

flawr

@NathanMerrill Just note that the probability is sigma additive. So if there was an uniform distribution with probability 0<p<1 for each number, then you could have at most 1/p numbers, but the integers surely contain more than that =)

trichoplax

8:13 PM

I find it fascinating that you can distribute probabilities that sum to 1 over the integers with all being non-zero, you just can't do it evenly

Nathan Merrill

@flawr right, but I'm already ok saying that each number has an undefined probability, so that doesn't really hold

flawr

well undefined probabilities are especially cumbersome to work with^^

Nathan Merrill

right :)

trichoplax

Never tell me the odds. Especially if they're undefined

flawr

We recently made a petition in our math department to only allow finite things, as everytime infinity shows up, trouble arises.

Nathan Merrill

8:16 PM

@Mego have you gotten the massive "welcome to google" pdf packet?

Anonymous

@NathanMerrill Not yet. Still trying to get in touch with recruiter to finalize details

Nathan Merrill

well, I got super spammed with a gazillion "you're a Noogler!" emails

ok, it wasn't that bad

but still :)

El'endia Starman

@flawr An ant starts walking on a 1-meter rubber band, at the rate of 1 cm per second. Every second, the rubber band grows by 1 meter. Does the ant ever make it all the way around? Yes, because the rubber band grows behind the ant too.

Anonymous

I'm sure I'll get those emails too when it comes time :)

Anonymous

@El'endiaStarman My brain hurts

El'endia Starman

8:20 PM

@Mego You can prove it too by directly linking it to the harmonic series, which diverges.

flawr

@El'endiaStarman Ah now I see, we stretch the band. Somehow I had the image of adding one meter at the end in my mind.

PhiNotPi

@DJMcMayhem so... I have a program that can parse bflk into a tree and vice-versa... now I'm trying to figure out the best way to implement simplification rules.

El'endia Starman

@flawr Yes, that's the trick. That's an assumption we pretty much automatically make.

Anonymous

@Mego The best part will be the one time it really was me

ArtOfCode

Even with that, not seeing how that works. Assuming a uniform band, that still means it's growing by 0.5m/s in front of the ant

Nathan Merrill

8:22 PM

@ArtOfCode only when its halfway around

Martin Ender

@ArtOfCode only when the ant is in the centre

ArtOfCode

Are we talking about a regular, circular band, or what

Martin Ender

but since the ant is moving it don't be in the centre any more a little later.

you can consider it to be a rubber string if you like (not circular)

ArtOfCode

I can see it works on a line, but not on a circle

flawr

@ArtOfCode You'll expand the band uniformly.

Nathan Merrill

8:24 PM

@ArtOfCode yes. When the ant is 1/10 of the way around the band, its growing .9m/s in front of the ant, but when its 9/10s of the way around, its only growing .1m/s in front

ArtOfCode

@NathanMerrill no, that would only be true if there was a fixed point on the circle, which you can't assume. Otherwise, it's permanently growing 1m/s in front of the ant, because "in front" can be all the way around the circle.

It doesn't work on a circle.

Nathan Merrill

you define the "top" of the circle to be the start/finish

Martin Ender

on a circle the ant is still increasing its angle relative to some fixed point

ArtOfCode

@NathanMerrill the "top" has to be a fixed point, then, for that to work.

Nathan Merrill

well, the ant is moving relative to somthing

otherwise you could say that the ant moving 200 m/s isn't moving at all

Eᴀsᴛᴇʀʟʏ Iʀᴋ

8:27 PM

@DJMcMayhem but but no-internet-dinosaur game???

DJMcMayhem

@PhiNotPi Hmm. Well a start would probably be checking if any nodes whose children are constant can be joined together, like turning (()()())(()()()) into ((()()())) or (()()[()]) into (()). Only downside is that it won't drastically shorten things

You could also do single integer golfing on any (()()()()...) chunks

Wheat Wizard

@DJMcMayhem Remember you can only do this in a zeroed scope

Martin Ender

@PhiNotPi What I've been meaning to do for Stack Cats for a while is brute force short programs and sort them into equivalence classes. You could do the same for short expressions.

DJMcMayhem

Oh, the top scope. Good point

mınxomaτ

Leaked photo of next year's Google Pixel.

Nathan Merrill

8:29 PM

? it has a huge bar at the bottom?

Wheat Wizard

@DJMcMayhem or inside of <>

flawr

mınxomaτ

@NathanMerrill That's the joke. The current Pixel is a design fiasco, mostly because of the useless chin.

flawr

@ArtOfCode Do you agree with that?

ArtOfCode

Okay, I simulated it for a circular rubber band. The position and the circumference diverge - i.e. it doesn't work.

ant position:  1 band circumference:  200
VM337:6 ant position:  2 band circumference:  300
VM337:6 ant position:  3 band circumference:  400
VM337:6 ant position:  4 band circumference:  500
VM337:6 ant position:  5 band circumference:  600
VM337:6 ant position:  6 band circumference:  700
VM337:6 ant position:  7 band circumference:  800
VM337:6 ant position:  8 band circumference:  900
VM337:6 ant position:  9 band circumference:  1000
VM337:6 ant position:  10 band circumference:  1100
VM337:6 ant position:  11 band circumference:  1200

flawr

8:30 PM

@ArtOfCode Don't you agree witht ht the formula above?

Nathan Merrill

@mınxomaτ ah, I had no idea

flawr

@ArtOfCode You ignore the stretch behind the ant.

PhiNotPi

@DJMcMayhem we're talking some more in the brain-Flak room

DJMcMayhem

@WheatWizard well (((()()()){})) would still be shorter

ArtOfCode

@flawr If we're on a circle, behind is in front.

DJMcMayhem

8:31 PM

Wait, is that right?

ArtOfCode

That's our fundamental disagreement

flawr

@ArtOfCode Well you have to mark the starting point.

Nathan Merrill

@ArtOfCode ok. Define "ant position"

Martin Ender

@ArtOfCode every point on the circle is a fixed point. so there is always a reference point.

ArtOfCode

@NathanMerrill distance which ant has travelled

Nathan Merrill

8:32 PM

from where?

ArtOfCode

wherever it started?

ah

:)

Wheat Wizard

@DJMcMayhem Than what? I'm a bit confused

Martin Ender

yeah, but the part between where it started and where it currently is has stretched as well

ArtOfCode

mkay, in that case, CMC: calculate how far the ant has travelled at second n

DJMcMayhem

@WheatWizard if you have (()()())(()()()) but need the scope, I think you could do (((()()()))){}

Wheat Wizard

8:35 PM

@DJMcMayhem Oh yes that will always work

flawr

@ArtOfCode did you see my formula?

Nathan Merrill

are there are any convergent series that are infinite when evaluated?

Steven H.

I can't work on the numerical regex lang if I'm at school, does anyone have any good turing tarpit ideas?

flawr

@NathanMerrill isn't that a little bit contradictory?

Nathan Merrill

well, there's the inverse. We have divergent series that are finite when evaluated

aka, 1 + 2 + 3 + 4 + 5

Steven H.

8:39 PM

Finite when evaluated with a finite amount of elements

Nathan Merrill

nope, 1 + 2 + 3 + 4 +5 ... is equal to -1/12

Steven H.

You can't get infinity from a finite number of finite numbers

Oh yeah, that

Nathan Merrill

but it also diverges

flawr

@NathanMerrill Well most "ways of evaluating" are trying to extend the notion of the convergence, so not much interesting is happening there.

Steven H.

> Indeed, limits are very often used to find a meaningful surrogate for f(x) for those cases where f(x) is itself undefined. In the case of convergent sums, f(∞) is generally undefined (by virtue of infinity not being a number), but if the function converges to 0, we can treat the limit of the sum as the sum itself.

curiouscheetah.com/BlogMath/infinity-and-string-theory

If it converges, f(∞) = 0, and f(∞) is what makes sum(range[1,∞],n=>n) == -1/12

Martin Ender

8:45 PM

@WheatWizard stackoverflow.com/help/privileges

@DJMcMayhem status-completed

Nathan Merrill

huh. Is that really true? If a summation converges, is that summation always evaluated to the value the summation converges to?

flawr

@StevenH. Consider zeta(s) = sum( 1/n^s for n=1 upto infinity)

Then consider the limit zeta(s) for s to 1

Steven H.