The Nineteenth Byte

Kenny Lau

00:48

@Sherlock9 What does this mean?

in which language?

Mama Fun Roll

Everyone, I learned today that URLs can use 1 slash. For example, http:/google.com seems to work fine.

This could truly revolutionize golfing.

Kenny Lau

whaaaaaat

I never knew that

Quill

most golfing answers already have that

Quill$ open http:ppcg.lol

mınxomaτ

01:17

Slashes are optional per the RFC standard. Slash-less URLs work in almost every browser, including the parsed HTML.

Downgoat

@mınxomaτ are slashes then just good practice?

mınxomaτ

@Downgoat They are only optional for remote transfer. Local file paths etc. must have a certain amount of slashes. So it's good to use them to maintain compatibility and sanity ;)

Downgoat

> CheddarPrimitiveInterfaceDefaultScopeAliases

Do I get the award for longest variable name now

mınxomaτ

Maybe the most cheesy.

Downgoat

@mınxomaτ that joke has been made 14 too many times...

mınxomaτ

01:22

I guess we are on a roll ;)

Personally, I think cheese puns are just grate.

You need to choose your language name caerphilly to prevent any further jokes.

It's emmental, my dear Downgoat.

(That's it. I'm out of cheese puns)

LegionMammal978

01:52

I would be on slither.io

But my father is streaming videos on Netflix :(

Kenny Lau

1. 1 ∈ X
2. (n ∈ X) ⇒ (4n+1 ∈ X)
3. (6n+1 ∈ X) ⇒ (4n-1 ∈ X)
4. (6n-1 ∈ X) ⇒ (8n+1 ∈ X)

Prove that X contains every positive odd number

LegionMammal978

@KennyLau Homework?

Kenny Lau

Nope

LegionMammal978

What for then?

Kenny Lau

for fun

Have you ever seen anyone doing maths for fun lol

I thought every programmer here does

LegionMammal978

01:57

@KennyLau What is this idea that you call "fun"?

Kenny Lau

What you would call fun

@LegionMammal978 I could go to brilliant.com overnight

That's what I call fun

I meant brilliant.org

brilliant.org

Sherlock9

@KennyLau Bah, I was sure I wrote the word "Retina" last night

Kenny Lau

@Sherlock9 It's hard, but I'll try

Sherlock9

That's fair.

I thought it might be simple by virtue of the convert to base 5 algorithm you showed me yesterday, but with the 5 replaced with y. Then the stage after that would be ^10*$

Cᴏɴᴏʀ O'Bʀɪᴇɴ

+#->-[->-[<]>+>]<-@>+[------->++<]>++.----.-------.
----.--[--->+<]>--.++++[->+++<]>.--[--->+<]>-.+[->+
++<]>.+++++++++.---.------.++++++++.>-[>-<-----]>++
+&>+[------->++<]>--_+++_------------_--_--[--->+<]
>-_-----------_++++++_-_[------>+<]>_+[-->+<]>+++_[
->+++<]>++_++++++++++++__----_[-->+<]>++_----------
-__++[->++<]>_+++++++_+++++++++++_[++>---<]>_--[-->
+++++<]>--_-[--->+<]>_-[----->++<]>-_[--->+<]>-_+_+
_---_++++++++++++_-_--[++>---<]>_`:

This creates a button that rickrolls the user.

Kenny Lau

02:12

@CᴏɴᴏʀO'Bʀɪᴇɴ Which language is this?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

ITML

Kenny Lau

@Sherlock9 Which challenge are we talking about?

Sherlock9

@KennyLau Is x a power of y

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Kenny Lau

@Sherlock9 oh, ok

Sherlock9

02:15

@CᴏɴᴏʀO'Bʀɪᴇɴ Odd. Google can't find ITML

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@Sherlock9 I made it, but haven't published it yet ;)

Sherlock9

Ah ok

Very cool

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Thanks :3

ITML EXAMPLE PROGRAM
RICKROLLS THE USER
+#-					goes to the body element resets cell
>-[->-[<]>+>]<-		sets current cell to "144" (a button)
@					add said button
>					move to empty cell
+[------->++<]>++.
----.-------.----.
--[--->+<]>--.++++
[->+++<]>.--[--->+
<]>-.+[->+++<]>.++
+++++++.---.------
.++++++++.			"please click" text
>-[>-<-----]>+++&	onclick
>
+[------->++<]>--_+++_------------_--_--[--->+<]>-_-----------_++++++_-_[------>+<]>_+[-->+<]>+++_[->+++<]>++_++++++++++++__----_[-->+<]>++_-----------__++[->++<]>_+++++++_+++++++++++_[++>---<]>_--[-->+++++<]>--_-[--->+<]>_-[----->++<]…

Sherlock9

That is definitely going to be an interesting language \:)

Cᴏɴᴏʀ O'Bʀɪᴇɴ

merci. I make a gem every once in a while.

Sherlock9

02:20

Do you have every HTML tag covered?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Yes.

Sherlock9

So what's left before you publish?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Just the README.

github.com/ConorOBrien-Foxx/ITML

Kenny Lau

@Sherlock9 The problem is that the current base-conversion algorithm

the base needs to be hardcoded

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Readme updated

Sherlock9

02:23

@KennyLau Dang. That's a pain in the butt

@CᴏɴᴏʀO'Bʀɪᴇɴ Interesting how attr skips from 112 to 205

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@Sherlock9 Yeah. You'll notice how the other one ends at 204. >_> I was too lazy to redo my program

Ashwin Gupta

whats the total player health with bukkit?

like if I do Player.damage(x), I wanted to take 10% of the players health, do i just do 10?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Do 0.1

Ashwin Gupta

damage
void damage(double amount)

Deals the given amount of damage to this entity.
Parameters:amount - Amount of damage to deal

oh ok

Cᴏɴᴏʀ O'Bʀɪᴇɴ

idk tho

Ashwin Gupta

02:26

oh lol

well it may be out of 100 was what I was thinking

Cᴏɴᴏʀ O'Bʀɪᴇɴ

I thought you were making something

0.1 makes the most sense for a percentage.

Ashwin Gupta

I am, my friend asked me to make plugins for his server so I haven't worked on Rocket Blaster Squad this weekend.

Cᴏɴᴏʀ O'Bʀɪᴇɴ

isn't bukkit a minecraft thing?

Ashwin Gupta

yeah^

Sherlock9

@CᴏɴᴏʀO'Bʀɪᴇɴ Hm. That may be confusing for potential new users when this gets pushed, but stranger and more harmful quirks have been seen in "regular" languages. Good luck

Ashwin Gupta

02:27

I'm taking a break w/ my game tho to help my friend setup the plugins on his server.

Then back to rocket blaster squad.

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@Sherlock9 haha, thanks. which types of quirks?

I'll probably fix that now.

Sherlock9

JS and types, to start. Compared to that, a simple "the placement of some variables is odd" seems tiny

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Ah.

heyyy fixing my laziness saved a byte

quartata

@AshwinGupta Health is in half hearts.

Kenny Lau

@Sherlock9 with inputs swapped

quartata

02:32

so 20 health

meaning you want to do damage(2)

Sherlock9

@CᴏɴᴏʀO'Bʀɪᴇɴ Did you mean to repeat the following?

77: "ruby",
78: "rt",
79: "ruby",
80: "rtc",
81: "rb",
82: "ruby",
83: "rb",
84: "rt",
85: "rtc",
86: "ruby",

Cᴏɴᴏʀ O'Bʀɪᴇɴ

uh

darn

I scraped a webpage. Nice catch.

who forked it ? :o

Sherlock9

@KennyLau Very cool :D

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@Sherlock9 There's quite a lot of reps. I'll fix that.

Sherlock9

Is the group matching 1-indexed?

@CᴏɴᴏʀO'Bʀɪᴇɴ Godspeed o7

Cᴏɴᴏʀ O'Bʀɪᴇɴ

02:37

thanks

Kenny Lau

@Sherlock9 What does this mean?

Sherlock9

$#1 refers to the first group, $#2 to the second

Mama Fun Roll

I just found this

it's beautiful

Kenny Lau

@Sherlock9 yes they are 1-indexed

inputs not swapped

Sherlock9

Nice

What are $1 and $3 in this case?

Kenny Lau

02:41

The first capture group and the third capture group

$#2 is the number of matches of the second capture group

For example, if you apply (xy)+ to xyxyxy, you would only match xy

however, the match is actually an array

which looks like {"xy","xy","xy"}

$#1 => 3

$1 => xy

$.1 => 2 (length of match)

@Sherlock9 added

Sherlock9

Ah, I see. And what about \3?

Kenny Lau

backreference, inside the regex

Sherlock9

Ah. So it immediately uses divmod with the third reference (the base)

Kenny Lau

A tidier version: (x*)(\3)*;(x+)

note that the regex is applied from right to left

x is a power of y

this is the divmod stage

base conversion

Sherlock9

@KennyLau Yep, from the r

Kenny Lau

02:55

@Sherlock9 do you get it?

yes, but the converted form doesn't exactly look like that

Sherlock9

@KennyLau I think so. Maybe not enough to write some myself yet, but I get the principles

Kenny Lau

2 bytes saved by using a different delimiter

for example, eleven base three is now xx::x:;xxx instead of xx;;x;;xxx

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@Dennis Which (if any) parts of printable are not in Jelly's code page?

Sp3000

@KennyLau 1 ∈ X, (n ∈ X) ⇒ (2n ∈ X), (6n+4 ∈ X) ⇒ (2n+1 ∈ X), prove that X contains every positive integer?

Kenny Lau

@Sp3000 sure

@Sp3000 Can simplify 6n+4 to 3n+2

Sherlock9

03:01

@KennyLau And for the second line, $1 is the mod, $#2$*x is the div, $3 is the base and then the ` starts again

orlp

do you guys know photo shop?

Kenny Lau

@Sherlock9 bingo

orlp

what kind of shop sells photos?

Sherlock9

Why does : need to be between + and r?

Kenny Lau

1 byte saved

that was for debugging

forgot to remove it

to make 1 work

El'endia Starman

03:04

That's impressive.

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@KennyLau Almost done with the jelly keyboard

Kenny Lau

@CᴏɴᴏʀO'Bʀɪᴇɴ Thanks :D

Sherlock9

@KennyLau Magnificent. I think I've got it.

Kenny Lau

Nice

I just copied the divmod from the wiki, lol

Sherlock9

Unfortunately I need to be elsewhere in a half hour so I should pack, shower and go, in about that order

Kenny Lau

03:08

ok

bye

Mars Ultor

@CᴏɴᴏʀO'Bʀɪᴇɴ Keyboard?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

yeah

like a french keyboard, except, jelly

Sherlock9

@KennyLau Either way, the code is great. See ya o7

Cᴏɴᴏʀ O'Bʀɪᴇɴ

(keyboard layout)

Kenny Lau

@Sherlock9 bye

Mars Ultor

03:09

@CᴏɴᴏʀO'Bʀɪᴇɴ Like this?

Wait nvm

How easy will it be to use?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

it's basically a combination of ctrl+alt and sometimes shift

What I have so far

orlp

@Sp3000 stupid trick question mate

@KennyLau you can't prove that

it's the Collatz conjecture

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Kenny Lau

@orlp prove that I can't

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Alt+Gr = right alt, or ctrl+alt

orlp

03:17

@KennyLau 1. it's hard 2. QED

@Sp3000 why do you do this to kenny

Cᴏɴᴏʀ O'Bʀɪᴇɴ

I hear proof by intimidation is very useful here.

orlp

(and to me, I wasted some time on it before realizing it :()

Kenny Lau

@orlp I did this first

@orlp see what he was replying to

Cᴏɴᴏʀ O'Bʀɪᴇɴ

I found a counterexample to the collatz conjecture: 0.

:p

Sp3000

@orlp It was just the first thing I thought of when I saw Kenny's question :P

Kenny Lau

03:23

Is there any corollary of collatz conjecture lol

or we can make a induction based on collatz conjecture

orlp

@KennyLau I have a proof for your one

ohhh

Kenny Lau

@orlp But the margin is too small

orlp

odd number

nvm

I had a proof that 2 can not be generated :)

Kenny Lau

How do you prove that? :D

Cᴏɴᴏʀ O'Bʀɪᴇɴ

My proof is based on a proof that will come into existence in ten years.

orlp

03:26

@KennyLau you start with 1

you have 2 growing rules

6n + 1 applies, giving 3

n applies, giving 5

so all growing functions overshoot 2, and can not generate it

you also have one shrinking function

6n + 1 -> 4n - 1

but 4n - 1 can never generate 2

actually I just found a much simpler proof

(n ∈ X) ⇒ (4n+1 ∈ X)
(6n+1 ∈ X) ⇒ (4n-1 ∈ X)
(6n-1 ∈ X) ⇒ (8n+1 ∈ X)

every right-hand side is odd, thus 2 can not be generated :)

Kenny Lau

nice :D

Retina, 42 bytes

34 bytes

(testcase is the famous 27)

retina.tryitonline.net/…

Luis Mendo

@Dennis If I implement the base-95 to base-256 encoding in MATL, the byte count for n characters is either ceil(n*(log(95)/log(256))) or that minus 1 (depending on the number of leading 0 bits, which is determined by the first characters in the code). So the exact byte count is difficult to predict.

I'd like to keep posting the original code characters in the answers, rather than the hex dump (bytes) resulting from the encoding. But then there should be a way for people to check that the claimed byte count is correct.

One possible way would be to give a link to an online MATL program that does the encoding and decoding (for example this for the encoding and this for the decoding).

That's easy, but requires use of those "external" programs. Perhaps a better way would be to have the online compiler directly output the byte count, or the hex dump, when given a certain input argument. Or even take the hex dump with another optional input parameter, so it can be checked that the hex dump is really equivalent to the characters in the code.

Is it possible to pass the compiler a parameter (like currently there's matl -of ...) that can be selected from the page matl.tryitonline.net? How are the "arguments" in that page treated? Or do you suggest any other way? I don't really want to complicate things, but I'd like people to be sure that the encoding really works as claimed. Specially when people will see that the byte count is less than the character count.

(Sorry for long text)

Dennis

03:48

@CᴏɴᴏʀO'Bʀɪᴇɴ Jelly's code page contains (and is compatible with) printable ASCII.

@LuisMendo I could modify MATL's wrapper to pass the Arguments to matl, but I'd have to sanitize them somehow to prevent rhe execution of arbitrary Octave code. An easier way might be to print the information to STDERR, which can be made visible by toggling the Debug switch.

Doorknob

What's the best compression format if I want to be able to quickly extract a single file from the (potentially huge) archive?

Kenny Lau

compress each file separately?

Doorknob

No, I'm compressing my entire home directory as a backup of sorts (and for other purposes), and I want to be able to take out individual files quickly.

Kenny Lau

I think we should design a compact base-95 to base-256 encoding :)

Luis Mendo

@Dennis That's a better idea, thanks. I'll try with STDERR

Dr Green Eggs and Ham DJ

04:02

@Doorknob The solution is to switch to windows and setup automatic backups.

:P

Doorknob

@DrGreenEggsandHamDJ Then I'd end up with more problems than I started with. ;)

Luis Mendo

@KennyLau Well, just subtract 32 from the ASCII codes and convert from base 95 to base 256. See here for example

Dr Green Eggs and Ham DJ

jquery?

Doorknob

@DrGreenEggsandHamDJ brilliant

Mars Ultor

.tar.gz folders with <1k files/<1gb?

Dennis

04:03

@Doorknob Pretty much everything except compressed tarballs. Zip, RAR, 7z (without the solid option), etc. all compress individual files.

Kenny Lau

@LuisMendo And the decompresser?

Luis Mendo

@KennyLau Here. All input must be in a single line (no linefeeds)

Dr Green Eggs and Ham DJ

Also @Doorknob, your advice to setup a dotfiles repository? Probably the best advice I've ever taken.

Mars Ultor

LZMA (7z) offers the highest cmpression

Dennis

It's also slow as hell.

Doorknob

04:04

@Dennis Alright, thanks. All of those preserve file perms and such?

Mars Ultor

has ~6 dotfiles

Kenny Lau

@LuisMendo I'm thinking of something like decimal-64, how they store 3 decimal digits effectively as 10 bits

Mars Ultor

@Dennis Set it to a lower compression level then?

Doorknob

@DrGreenEggsandHamDJ :D Glad I could help!

Dennis

@Doorknob Um, excellent question. I'd have to try.

Dr Green Eggs and Ham DJ

04:05

@MarsUltor Is that too few or too many?

Mars Ultor

Way too few

Doorknob

@Dennis Ok, no worries, I can research/try that myself.

Luis Mendo

@KennyLau That's efficient because 1024 is only slightly larger than 1000. That doesn't happen with 128 and 95

Kenny Lau

@LuisMendo That's why I'm thinking

Luis Mendo

@KennyLau Why not direct conversion from base 95 to base 256? That way you don't waste any bit

Mars Ultor

04:06

It only works for 10n bits

Dr Green Eggs and Ham DJ

@MarsUltor Not really. It's more than I have (Unless you count vim-filetypes)

Kenny Lau

@LuisMendo sure..

Mars Ultor

Well, three actual dotfiles

Kenny Lau

95^9627 ~ 2^63248

Dennis

@Doorknob Both Info-ZIP (i.e., zip) and 7-Zip seem to preserve permissions.

Note that no format (not even tarballs) preserves ownership.

Kenny Lau

04:11

95^93 ~ 2^611

Doorknob

@Dennis It can if you tell it to, it seems.

Dennis

Oh, interesting.

Helka Homba

This is over a year old but I really couldn't believe how dumb a company could be - theverge.com/2015/2/5/7986327/…

Kenny Lau

@LuisMendo What if your program starts with a space?

Dr Green Eggs and Ham DJ

@HelkaHomba That's painfully stupid.

Sherlock9

04:18

2 hours ago, by Kenny Lau

1. 1 ∈ X
2. (n ∈ X) ⇒ (4n+1 ∈ X)
3. (6n+1 ∈ X) ⇒ (4n-1 ∈ X)
4. (6n-1 ∈ X) ⇒ (8n+1 ∈ X)

2 hours ago, by Kenny Lau

Prove that X contains every positive odd number

So, I'm not sure why you guys were looking at 2

Ah ok, reading back, I get it. I'll see if there are any counterexamples in X

Sp3000

@KennyLau Out of curiosity, have you already proven it or were you still trying to prove it?

Kenny Lau

@Sp3000 still trying to prove it

@LuisMendo compresser 37 bytes, decompresser 18 bytes

Dr Green Eggs and Ham DJ

@KennyLau What compression algorithm does it use?

Kenny Lau

The same.

Sherlock9

I can prove that 3 is not in X when we start from 1

Kenny Lau

04:27

go ahead

I think I made a typo

1. 1 ∈ X
2. (n ∈ X) ⇒ (4n+1 ∈ X)
3. (6n+1 ∈ X) ⇒ (8n+1 ∈ X)
4. (6n-1 ∈ X) ⇒ (4n-1 ∈ X)

Sherlock9

Wait, let me check my work again

Mars Ultor

@Sherlock9 ^^

Sherlock9

Well dang

Kenny Lau

mea culpa @Sherlock9

Sp3000

For the previous one, {3, 7, 13} is a cycle so that sounds about right

Sherlock9

04:31

In the corrected system, if you go from 1, Rule 2 to 5, Rule 4 to 3

@Sp3000 Yep, that's what I had

No longer sure about the corrected system

Not sure about the corrected system

Dr Green Eggs and Ham DJ

Coding-style question: Do you try to stick to columns of 80 or less characters?

Doorknob

Yep.

I have colorcolumn set in vim and usually try not to cross it.

Sp3000

100 here, just cos I find 80 restrictive

Sherlock9

@DrGreenEggsandHamDJ Not while golfing, but that's usually a given

Dr Green Eggs and Ham DJ

@Doorknob Haha, funny enough I was asking for that exact reason. I turned colorcolumn on, and I was wondering whether or not to keep it.

Kenny Lau

04:39

@Dennis Any chat mini-challenge?

Dennis

@KennyLau Take an even integer greater than 2 as input and output a pair of primes with that sum.

Dr Green Eggs and Ham DJ

@Dennis Is that guaranteed?

Kenny Lau

@DrGreenEggsandHamDJ That is Goldbach conjecture

Dr Green Eggs and Ham DJ

Goldbach's conjecture

Goldbach's conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states: Every even integer greater than 2 can be expressed as the sum of two primes. The conjecture has been shown to hold up through 4 × 1018, but remains unproven despite considerable effort. == Goldbach number == A Goldbach number is a positive integer that can be expressed as the sum of two odd primes. Since four is the only even number greater than two that requires the even prime 2 to be written as the sum of two primes, another form of the statement of Goldbach's...

Unsolved. haha, very funny dennis.

Dennis

@DrGreenEggsandHamDJ Why funny? Just assume the conjecture is true. If you find a counterexample, write a paper.

Kenny Lau

04:45

@Dennis Pyth, 14 bytes

Dr Green Eggs and Ham DJ

Challenge accepted.

jk, there's no way I could do that.

Kenny Lau

@Dennis I remember seeing you do it in Jelly

@Dennis Jelly, 11 bytes: ÆRṗ2S=¥Ðf³Ḣ

Pyth, 14 bytes: %RQ_Bf*P_TP-TQ

Dennis

8 bytes is possible.

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