The Nineteenth Byte

Zgarb

19:04

@MartinEnder Map each even integer 2n to 2n+1 and vice versa (equivalently, flip least significant bit).

Lynn

@MartinEnder Doing the same with the bits is also fun!

flawr

You could alternatively just xor with any constant.

Lynn

Flipping them in pairs, I mean.

Martin Ender

@Zgarb yeah, I actually considered that (x XOR 1), but figured it might be redundant since I already have x XOR y

(then again I also have 0-x as a separate operator, so maybe I'll add it anyway)

Lynn

> (a, b) -> (a, a - b)

You meant b – a I guess?

Martin Ender

19:06

ummm, I don't think so?

Lynn

Ah, maybe I misunderstand the notation

I was thinking, f(x) = 5 – x is self-inverse but f(x) = x – 5 isn’t

Martin Ender

right, that's why I subtract the one that's changing from the one that isn't

Zgarb

For lists of lists, you could reverse the order of their lengths while keeping their concatenation constant.

Martin Ender

if you apply it twice you get (a, a - (a - b)) == (a, b)

Lynn

Ohh, I see, it really is a function on pairs of integers

Zgarb

19:10

So [[1],[2,3,4],[5,6]] -> [[1,2],[3,4,5],[6]] for example.

Lynn

I like @Zgarb’s

Martin Ender

@Zgarb make that a challenge? :)

Lynn

I was thinking my prime-y one is a cool potential challenge

It has some nice property that I can’t remember!!

It maps primes to primes, I guess?

flawr

If you consider the lists of fixed lengths of integers as a Z-module, you can find a base and consider the maps of involutory matrices. (Or alternatively idempotent matrices, with the trivial bijection)

Martin Ender

I know some of those words. Like "the".

flawr

19:15

A module is basically a vector space, but you replace the field with a ring.

In this case the integers.

Involutory matrices are just linear maps that are self inverse, as e.g. the reflections, or the permutations with cycles of order at most 2.

Idempotent matrices are just matrices A that satisfy A^2 = A

Lynn

Would a base be something like {[1, 0, 0], [0, 1, 0], [0, 0, 1]}, here?

(For length 3)

flawr

Exactly!

And a matrix A is idempotent if and only if 2A-I is involutory

Lynn

Hmm. For your function to be total, you’d have to make sure whatever involutory map you pick works for all lengths.

Nathan Merrill

@MartinEnder I'm lost as well

Lynn

Or, I mean, can be extended to all lengths

flawr

19:20

@Lynn You can trivially extend it by just "appending" identities.

Lynn

Oh, right, one nice property about my prime-exponent-flippy involution is that it nicely extends to the rationals!

Martin Ender

okay, maybe I should divulge a bit more background. what I've actually got is a tape of stacks (with implicit infinite zeros at the bottom of each of them).

so when I say "reverse list" what I've actually got is "reverse the longest non-zero prefix of the current stack"

flawr

And the tape is infinite too?

Martin Ender

yes

on both sides

flawr

Oh, I thought it was infinite in the middle but not at the ends

=)

Martin Ender

19:24

it could be semi-infinite...

Lynn

“Some permutation on the tops of all non-empty stacks” might be interesting?

El'endia Starman

@flawr Chat challenge: think of a situation where this makes sense. :P

Martin Ender

@Lynn there's a tape head, so local changes seem a bit more natural. one thing I'm considering is "swap stacks left and right of the tape head"

Lynn

That works nicely

Are these esolang-opcodes-to-be?

Martin Ender

indeed

hopefully ready to be released at some point next week

Lynn

19:30

Maybe left, center, right = reverse(right), reverse(center), reverse(left) would be funny

Turning the three stacks closest to the tape head upside-down

Zgarb

@MartinEnder Maybe later if I don't forget it.

Martin Ender

oh, so basically rotate

(although if I add "swap" stacks that can be done quite easily manually since I do have "reverse" already)

Nathan Merrill

@Zgarb you could even make it more difficult: [[4], [5,6,[7]],[]] -> [[], [[4],5,6], [7]]

better :)

flawr

@El'endiaStarman Is it possible to have a continuous function on a closed interval with infinite arc length?

I think not.

Martin Ender

@NathanMerrill I think at that point it becomes a bit too similar to codegolf.stackexchange.com/q/43018/8478 (at least I don't see any other approaches than those used on that challenge)

El'endia Starman

19:34

@flawr Koch curve?

flawr

@El'endiaStarman Right, or a hilbert curve, there we go=)

Maltysen

@flawr would the hilbert curve be a function?

Nathan Merrill

heh, true

El'endia Starman

Even if it has to be a well-defined x -> y function, I'm sure it can still be done.

Maltysen

@El'endiaStarman well without being a function, you could just have x=4

Martin Ender

19:37

@Lynn I just realised, I overlooked the possibility of just swapping the tops of the two adjacent stacks

flawr

@Maltysen A function from [0,1] to [0,1]^2

(Injective, but not surjective)

Maltysen

true

Nathan Merrill

is it possible to have a bijective function between ragged arrays and integers?

probably is, but I can't think of one

flawr

what do you mean by 'ragged'?

(existence and actually finding one are two totally different things)

Nathan Merrill

ragged meaning an array containing zero or more arrays (which can each contain zero or more, and so on)

Maltysen

19:42

so between the set of all possible ragged arrays and Z?

Nathan Merrill

yes

Martin Ender

@NathanMerrill multiply all integers by 2. then recursively use a bijection from flat lists to odd integers.

Nathan Merrill

er, does Z include negatives?

Maltysen

yeah

Martin Ender

(you'll end up with an odd integer, but that's easy to fix)

Nathan Merrill

19:42

oh, I don't want negatives

@MartinEnder not sure I understand. What would 8 map to?

Martin Ender

that depends on the specific bijection between lists and odd integers

Nathan Merrill

which odd integers are we talking about?

16-> 15 and 1?

flawr

@NathanMerrill it is basically a subset of NxN (N = natural numbers), which is countable.

Martin Ender

ummm... let's start from the beginning. do we agree that bijections between the positive integers and arbitrary finite lists of positive integers exist?

Nathan Merrill

I believe so, and according to flawr, it does

oh wait

Martin Ender

19:45

right. so we can easily turn that into a bijection between lists and odd integers

flawr

(as a visualization: consider the countability argument of the rationals)

Nathan Merrill

these are lists without numbers

actually, that doesn't make a difference

because you can replace each number N with a list of length N

Martin Ender

now I'm lost

Maltysen

can you give us an example of one of the ragged lists?

Nathan Merrill

ok, let me give you a couple examples of ragged arrays

Lynn

19:47

@NathanMerrill Sure; just go from ragged arrays ↔ string representations ↔ natural numbers.

Martin Ender

@Lynn that's a bit boring? :P

Nathan Merrill

@Lynn that goes one way, but not the other

not every integer maps to a ragged array

Maltysen

@NathanMerrill why not

Martin Ender

I think that depends on your string representation

Nathan Merrill

[]
[[],[]]
[[[[[]]]]
[[],[[]]]

ok, there are 4 possible ragged arrays

Lynn

19:49

Imagine iterating over all strings in {comma, left bracket, right bracket}*, and keeping only the valid ones.

Nathan Merrill

sure, that would work

Lynn

Now you have a bijection between ℕ and ragged arrays!

This trick also works nicely for trees and stuff.

Nathan Merrill

it's quite inefficient, though :/

Martin Ender

I liked mine because it doesn't require iterating through all valid lists ;)

flawr

Which leads us to my favourite topic of the week: Shortest path problems on Graph structures on uncountable sets.

Lynn

19:51

Absolutely. ^^

Nathan Merrill

@MartinEnder I don't understand yours. I'm not sure what odd integers have to do with anything

Lynn

So, if you remove the commas from Nathan’s examples, you get words in the Dyck language.

Martin Ender

@NathanMerrill okay so the basic idea is to start from the bottom up and recursively turn flat lists into integers.

but to go the other way round you need to distinguish the numbers that represent lists from the numbers that represent leaves in your ragged array

so you use even numbers represent the leaves in the ragged array and odd numbers to represent nested lists

Maltysen

@MartinEnder but lists are leaves?

Nathan Merrill

so, [] is 2?

Martin Ender

19:54

@Maltysen do you mean empty lists?

Maltysen

yeah

oic

Martin Ender

well it would be 1 (because odd numbers)

Maltysen

you're calling empty lists leaves

flawr

@Maltysen Think the string of the brackets as tree traversal path.

Martin Ender

if anyone can give me a simple bijection between flat lists and integers, I can work through an example later

flawr

19:55

What are flat lists?

Nathan Merrill

@MartinEnder isn't there only 1 flat list, []?

PhiNotPi

update on my #yolo domain investment: I've parked the domains and put them for sale for some overly inflated price (the auto-suggested sale price, which I will probably lower in a couple months to be more reasonable).

Lynn

Couldn’t you do, like, [5, 3, 0, 2] ↔ 1111101110011 ↔ 8051

Oh, leading zeros and stuff.

Oh, just add a 1 bit in front

Nathan Merrill

@PhiNotPi link?

Lynn

[0, 0, 5, 3, 0, 2] ↔ 1001111101110011 ↔ 40819.

That’s a bijection between positive integers and lists of non-negative integers.

Maltysen

19:59

@Lynn what about 13

1101

is that 1, 101 or 110, 1?

Lynn

Chop off the leading 1 bit, you get 101. Then split over 0s, and count the lengths of the groups of 1s. You get [1, 1]

Nathan Merrill

yeah that works

Dennis

11

Q: Map a list of indefinite size to a number!

It is well known, in the field of Mathematics studying infinity, that the Cartesian product of any finite amount of countable sets is also countable. Your task is to write two programs to implement this, one to map from list to integer, one to map from integer to list. Your function must be bij...

PhiNotPi

@NathanMerrill for example: qx3.xyz (warning: is literally ads)

wizzwizz4

HELLO I AM MOD FEED ME MUDDER

Nathan Merrill

20:04

@PhiNotPi lol, my adblocker blocked it

@PhiNotPi what's the origin of that domain? (aka, why qx3.xyz?)

Maltysen

@NathanMerrill 2¢ .xyz's

Nathan Merrill

ah ok

so, simply mass purchasing domains

wizzwizz4

@Zizouz212 What's wrong with being a bot?

@NathanMerrill Yes, do that.

PhiNotPi

@NathanMerrill yeah, there's no more 3-character .xyz domains left.

I got one of the last ones.

Lynn

Now you need to decide what qx3 stands for.

Nathan Merrill

20:08

clearly, a copter

Maltysen

@NathanMerrill he should sell it to them for a lots of moneys

Lynn

.xyz domains are so shady

Bálint

@Lynn They're pretty much free

Considering that registering a new TLD costs $185k

Lynn

http://qx3.xyz/jx824v5/not_a_virus.mp3.exe

Maltysen

@PhiNotPi if I get it this year, will I only have to pay 2¢/year forever, or only for this year?

mınxomaτ

20:10

First year

Lynn

I got a .moe, and I don’t regret it. ♥

Bálint

@Lynn That really depends on w you name the domain

mınxomaτ

Otherwise $ 1.99 / year.

Maltysen

@mınxomaτ that's still really good...

Lynn

.xyz is Alphabet’s, right?

Bálint

20:11

I'm not sure if google's adsense works or not. Whenever I watch a video, it plays a wix ad in front of it, and I feel myself a bit offended because of that

mınxomaτ

@Lynn They don't own it. Alphabet just has a domain ending with xyz.

Lynn

Oh, wow, it isn’t

Yeah, I was under the impression that they made .xyz just for the shameless abc.xyz meme

Maltysen

@Lynn nah IIRC, google TLD was before alphabet

Bálint

If you can afford to run a home server, then creating a website with a "proper" domain name is close to free

@Maltysen google even registered .youtube too

Maltysen

@Bálint why is a.youtube giving localhost?

and b.youtube

Lynn

20:13

If you can afford X, X is close to free.

mınxomaτ

Alphabet owns .android, .cal, .channel, .chrome, .dclk, .docs, .drive, .google, .goog, .gle, .gmail, .gbiz, .guge, .hangout, .moto, .page, .play, .prod, .nexus, .youtube

Maltysen

and also c.youtube :/

Bálint

@Maltysen They're all google's domains

Maltysen

@Bálint yeah but why are they sending me to localhost?

Lynn

They don’t do that for me :o

Maltysen

20:14

ALL OF THEM ARE LOCALHOST WTH

mınxomaτ

@Maltysen Misconfigured your DNS or hostfile.

Maltysen

@Lynn a.youtube

Lynn

Is your hosts file weird?

Bálint

@Maltysen There was a good reason for it somewhere...

Maltysen

@Lynn probably

Bálint

20:15

namecheap.com/promos/2016/crazy-88.aspx These are all domains for 88 cent

mınxomaτ

@Bálint plus ICANN fee of $0.18

Read the smallprint.

Lynn

> Every option on Namecheap's list of 88-cent TLDs is a world-class extension that you'll be proud to have on your website.

Bálint

@PhiNotPi There are still 1 letter long xyz, for example a.xyz....wich costs $65k a year...

Lynn

> .cricket

mınxomaτ

@Bálint That's not free (i.e. available for registration). That's aftermarket.

Lynn

20:17

(Who even gets this money??)

(DNS is messed up.)

Bálint

@Lynn ICANN

Maltysen

@Lynn sitters

mınxomaτ

@Bálint No.

@Lynn The company who squatted the domain in the first place.

Bálint

@mınxomaτ That's technically in every domain, so it doesn't really count

@mınxomaτ So the one, who started the registering process?

flawr

Why is this so damn cheap over there????

Lynn

20:18

Donuts (company)

Donuts or Donuts Inc. is a start-up company that was created to apply and run new generic top-level domains (gTLDs) as made possible by ICANN's gTLD expansion program; it was co-founded by Paul Stahura, Jonathon Nevett, Richard Tindal, and Daniel Schindler. In April 2011, the company was in stealth mode and raising capital; based on the company's filing with the Securities and Exchange Commission (SEC), Donuts raised US$1 million equity financing. Since then it has submitted 307 gTLD applications and secured an initial investment of $100 million in financing, and a subsequent round of an undisclosed...

mınxomaτ

@Bálint Yes it does, if it's an additional fee.

Lynn

These people must be rich as fuck.

Bálint

@mınxomaτ But whatever you buy, you always need to pay that 18 cents

mınxomaτ

@Bálint No.

Case in point is the 2ct xyz promo.

@Lynn Aftermarket has the most money. E.g.: mediaoptions.com/domains-for-sale

Bálint

You don't say "something on the internet costs $500 + $1", because the transaction has a fee.

PhiNotPi

20:20

@Maltysen Only for this year: $10/year after that.

Bálint

> ICANN (the Internet Corporation for Assigned Names and Numbers) charges a mandatory annual fee of $0.18 for each domain registration

For each domain, doesn't matter which you buy, ,.com or .cricket

PhiNotPi

This is a "flash sale" for the 2-year anniversary of .xyz

Maltysen

@PhiNotPi wait, minxomat said that it would be 1.99 after

which isnt bad

PhiNotPi

@Maltysen for .xyz?

Maltysen

$10 sucks tho

10 mins ago, by mınxomaτ

Otherwise $ 1.99 / year.

Bálint

20:21

If I would register my name as a domain, it would cost me 15 cents

Lynn

@mınxomaτ Jeez. °_°

I hope you’ve all heard the saga of sex.com.

Sex.com

Sex.com is an Internet domain name and web portal currently owned by Clover Holdings LTD. The domain name was the focus of one of the most publicized legal actions about ownership of domain names. Kieren McCarthy, a journalist who followed the case, wrote the book Sex.com, which was published in 2007. == History == On October 18, 1995, entrepreneur Gary Kremen (who also founded Match.com) registered sex.com with Network Solutions, but did not develop the site as he focused on growing Match.com. Later in 1995, Network Solutions transferred, without permission, the domain to Stephen M. Cohen, who...

> Sex.com was reportedly sold to Escom LLC in January 2006.[10] At a reported $14 million price, the domain name had widely been cited as the highest priced domain sale.[11]

(The site itself is obviously NSFW.)

Martin Ender

@NathanMerrill [1, 2, 3] is also a flat list. by flat list I just mean a list containing no other lists.

mınxomaτ

@Bálint No they don't. @PhiNotPi Really bought their domains for 0.02. That's the final price - no transaction fee or any additional payment. Namecheap charges you their promo price plus the ICANN fee, which is usually payed by the registrar and not you.

Martin Ender

since you only want ragged lists of lists (without any integers anywhere) that definitely simplifies things

then you really don't need to use a bijection to odd integers

Bálint

@mınxomaτ Where did he bought those domains?

Maltysen

20:24

@Bálint internetbs.net

PhiNotPi

^ there

I spent a total of $0.10 on .xyz domains that I intend to get rid of within the next year (before it costs me "real money" to renew).

Maltysen

the hell took brainfreeze.xyz?

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Zizouz212 yes. he did. and stopped to eat food

PhiNotPi

there's lists of "premium" .xyz domains that are still available here: gen.xyz/premiums#standard

flawr

www.abcdefghijklmnopqrstuvw.xyz

mınxomaτ

20:27

@PhiNotPi There's no central list for aftermarket domains. Many are sold by auctions on domain.com, namecheap, united domains, SEDO etc.

Bálint

@PhiNotPi $0.02 for a domain, at least it's unusable

Except for math websites

PhiNotPi

I also got independents.xyz (a one-word URL), which I can probably figure out how to sell for a couple bucks sometime.

Bálint

Just sell it to some anti-trump guy and boom, profit

Lynn

@Nathan arxiv.org/pdf/1002.2625 part 2 might be relevant

Wait, it isn’t very useful, I think.

Alex A.

@Lynn wat, new avatar

Zizouz212

20:40

@wizzwizz4 You can't taste delicious ice cream.

Mint chocolate chip ice cream is very delicious.

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Lynn liked the old one better. :/

in Mos Eisley, 7 mins ago, by Eᴀsᴛᴇʀʟʏ Iʀᴋ

PSA: Watermelon Pop-Tarts exist. They are life.

Lynn

Aw. I liked both, but changing avatars now and then is fun!

Eᴀsᴛᴇʀʟʏ Iʀᴋ

> every now and then

cough cough @Quill cough

Alex A.

@Lynn They're both nice. :)

Zizouz212

Guys.

Alex A.

20:44

Gals.

Zizouz212

There is a flag.

On OS. Capture the flag!

Eᴀsᴛᴇʀʟʏ Iʀᴋ

?

Alex A.

Zizouz212

Why is there an eagle there?

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Zizouz212 okay? There. But I CAN"T HANDLE TEH FLAGZ

Alex A.

20:45

America

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Zizouz212 american mascot thingy

Zizouz212

Why mix eagle and flag?

Alex A.

Double America

Zizouz212

Who mixes animals with their flags? My goodness americans are weird.

Fatalize

this is an irrespectuous use of the American flag

/s

Zizouz212

20:46

Ah, oui. Capture the flag.

Alex A.

Zizouz212

:D

The flag. It's. It's. It's gone!

Oh no. I already cleared it a few hours ago. I was looking at an old window -_-

flawr

Alex A.

Flawless

Zizouz212

@flawr o_O

wizzwizz4

20:50

@Zizouz212 Just simulate it. It's quite simple, you -

nothing.

Zizouz212

This is what I feel like:

Alex A.

Hey @Doorknob, you know Tim Pope's Commentary plugin? I found a far superior commenting plugin (though surely you've known about it for way longer than I have): NERDCommenter. It's really nice.

Doorknob

I don't use commenting plugins

Visual block mode is plenty

Alex A.

Yeah, I learned how to use that earier.

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@El'endiaStarman re: that rupert's drop: (another hydraulic press)

http://i.imgur.com/IBUPIhC.gif

Alex A.

20:53

You should use more plugins.

flawr

2

Alex A.

Make Vim great again

3

flawr

\m/

Eᴀsᴛᴇʀʟʏ Iʀᴋ

Also, Ender's game in Halo 5:

http://i.imgur.com/mcn6bhW.gif

big gifs

Maltysen

@EᴀsᴛᴇʀʟʏIʀᴋ o_O

that poor waltermelon

wizzwizz4

20:54

Too many gifs...

Zizouz212

@wizzwizz4 lol

Fatalize

@AlexA. Should Emacs pay for it?

3

Alex A.

hahahaha

Eᴀsᴛᴇʀʟʏ Iʀᴋ

^

flawr

Maltysen

20:55

\o/ DN no longer requires you to resave your decks for new ban lists!

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@Maltysen ikr very sad

@Maltysen nice?

@flawr ._. but but the birdie

Doorknob

@AlexA. I keep my plugin usage to a minimum

Alex A.

@Doorknob I know

3 mins ago, by Alex A.

You should use more plugins.

Doorknob

14 is already uncomfortably many :P

flawr

@Doorknob Still incredibly small number compared to most other numbers there are.

Transcript for