The Nineteenth Byte

El'endia Starman

16:00

Didn't we already have a PPCG challenge on that puzzle?

mbomb007

@CᴏɴᴏʀO'Bʀɪᴇɴ I didn't write it out. But, for the 3x3 at least, I know a sequence to toggle each individual light, so combining those for the result you want gets you there.

Leaky Nun

@CᴏɴᴏʀO'Bʀɪᴇɴ What's the original question?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

I'm on mobile rn and can't link ya

@mbomb007 Oh, interesting! I'll work on a generic sequence for lights. It shouldn't be that hard, for say if two patterns toggle lights A0...AN and B0...Bm, then applying both patterns gives a pattern that toggles the states of the symmetric difference, leaving the intersection untouched

mbomb007

@CᴏɴᴏʀO'Bʀɪᴇɴ It might not work on 4x4, bc that's different. I'll post the sequences for 3x3 though. Rotational symmetry makes it quick.

PhiNotPi

@trichoplax I'm looking at Netfirms and Namesilo right now.

I'm kinda skeptical of Netfirms (like, what does the $1.99 price mean?)

mbomb007

16:10

Corner: [[0,2],[2,0],[0,0],[1,1]]
Edge: [[1,2],[1,0],[0,1],[0,2],[0,0],[1,1]]
Center: [[0,0],[1,0],[2,0],[0,1],[1,1],[2,1],[0,2],[1,2],[2,2]]

trichoplax

@PhiNotPi I now understand that I understand less than I thought I did

PhiNotPi

That ping should of actually been directed to @zyabin101, since he was the person who originally posted those prices.

El'endia Starman

Do you mean toggling the lone square in the center of a 3x3?

mbomb007

@El'endiaStarman Yeah.

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Nice finds!

El'endia Starman

16:11

You can do it in 9. Click all the squares. :P

Eᴀsᴛᴇʀʟʏ Iʀᴋ

HAI

trichoplax

Is the puzzle being discussed the lights out puzzle? We have various challenges inspired by it here, here, here and here (not necessarily making this a duplicate)

Eᴀsᴛᴇʀʟʏ Iʀᴋ

relink the thingy?

the jsfiddle

zyabin101

@PhiNotPi $1.99 is the registration price for the domain.

El'endia Starman

@EᴀsᴛᴇʀʟʏIʀᴋ thingy

Eᴀsᴛᴇʀʟʏ Iʀᴋ

16:16

thankee

took 4 moves.

Can it be solved in less than that?

mbomb007

Any position can be reached in n*m moves where n by m is the board dimensions, assuming that the given configuration is possible.

zyabin101

@mbomb007 So I can finally solve the board in 25 moves! \o/

mbomb007

@trichoplax They're all similar, but my lights out game also toggles the diagonally adjacent tiles, whereas none of those do.

Eᴀsᴛᴇʀʟʏ Iʀᴋ

you can also solve it in less.

mbomb007

@EᴀsᴛᴇʀʟʏIʀᴋ Not necessarily.

zyabin101

16:21

How much did it took for you?

PhiNotPi

@zyabin101 I found this on their site:

Domain .com, .net and .org sale pricing is $9.95/year for the first year. Domain will renew at $12.49 each year thereafter.

zyabin101

O_o

mınxomaτ

Use internet.bs already :P

mbomb007

@EᴀsᴛᴇʀʟʏIʀᴋ If the starting configuration on a 3x3 is only the center tile, then solving it takes 3*3 = 9 moves.

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@mbomb007 for this specific configuration (blank to full), it takes 4 moves for a 5*5 board.

zyabin101

16:23

@PhiNotPi Well, actually, the $1.95 price was the price on the Add button. D:

mbomb007

@EᴀsᴛᴇʀʟʏIʀᴋ That's not the puzzle, though. That's trivial on any board size.

TimmyD

@mınxomaτ The Internet is all BS

Eᴀsᴛᴇʀʟʏ Iʀᴋ

@mbomb007 okay

mbomb007

The challenge is to get to blank from any starting configuration.

El'endia Starman

I modded the JS fiddle to cycle between three colors. It's weirdly kinda more interesting and kinda less interesting.

zyabin101

16:27

@El'endiaStarman Share the fiddle?

I could try to make four colors! It will be very hard.

(To solve)

El'endia Starman

Four colors is probably fairly uninteresting.

zyabin101

D:

El'endia Starman

jsfiddle.net/xt3be066

mbomb007

@El'endiaStarman Use this one instead: jsfiddle.net/4o7vekpe/1

It has the added functionality by @CᴏɴᴏʀO'Bʀɪᴇɴ for automated moves given an input list.

El'endia Starman

Ooh, okay.

mbomb007

16:30

And it tracks your moves.

Then, if you want the simplified move set to get from an empty board to where you are, simply sort the moves, then remove pairs of duplicates.

Bald Bantha

How to play the game?

mbomb007

@BaldBantha It's less a game and more just for fun.

Normally, there'd be a random starting position, and you'd need to get to a blank board.

Bald Bantha

what I figured when I saw the symetrical patterns and stuff. There should be a bigger board

mbomb007

@BaldBantha Make it bigger.

Change the number and click "Update"

El'endia Starman

Okay, I made the edit. Hmm. I should've forked it, hmm?

mbomb007

16:33

Max is 20.

@El'endiaStarman Yeah, fork it.

El'endia Starman

3-color Lights Out

Eᴀsᴛᴇʀʟʏ Iʀᴋ

You forgot the blank one.

El'endia Starman

@mbomb007 Incidentally, this doesn't apply when you have 3 colors. You'd remove groups of three of the same. So basically, mod n, where n is the number of colors.

mbomb007

@El'endiaStarman Yeah, I know.

Every configuration is still possible on a 3x3, I believe.

El'endia Starman

Yeah. You just have to repeat a sequence to toggle one square between the three colors.

mbomb007

16:39

@El'endiaStarman Try to get only the center.

New Main Posts

0

Q: Algorithm to generate mountain ranges with upstrokes and down-strokes (java)

I tried to do the classical problem to implement an algorithm to print all valid combinations of n pairs of parentheses. And I found this program (which works perfectly) : public static void addParen(ArrayList<String> list, int leftRem, int rightRem, char[] str, int count) { if (leftRem < 0 ...

java

El'endia Starman

...oh, hrm.

I got center magenta with

[[2,2],[1,1],[0,0],[1,1],[0,2],[1,1],[2,0],[1,1],[0,0],[1,1],[2,2],[1,1],[0,2],‌[1,1],[2,0],[1,1],[1,0],[1,1],[1,2],[1,1],[0,1],[1,1],[2,1]]

.

Need to remove groups of 3.

PhiNotPi

What's the pros/cons of whois privacy?

ArtOfCode

@PhiNotPi If you're ever a mod, get it.

cuz people will look you up, find your phone number, and call you to yell at you for suspending them

zyabin101

@PhiNotPi Information about you is hidden in the site's WHOIS response.

El'endia Starman

16:48

@El'endiaStarman ...why can't I paste this into Python? I'm getting SyntaxError: invalid character in identifier. Weird.

PhiNotPi

does internet.bs have whois privacy?

zyabin101

idk

mbomb007

@CᴏɴᴏʀO'Bʀɪᴇɴ @El'endiaStarman On a 4x4, every configuration is possible:
Corner: [[0,0], [0,2], [0,3], [2,0], [2,2], [2,3], [3,0], [3,2], [3,3]]
Edge: [[2,0], [3,0], [2,2], [3,2], [2,3], [3,3]]
Inside: [[2,2],[2,3],[3,2],[3,3]]

orlp

@MartinBüttner sorry for late reply

zyabin101

But 1&1 includes one for free.

orlp

16:50

but I don't feel that embedding in a higher dimensional space adds power

TimmyD

@ArtOfCode If someone does that, file charges with the FBI (or equivalent) for threats and harrassment

zyabin101

ohai ~rys

orlp

since you can map Z^3 <-> Z

TimmyD

As an email administrator, having privacy WHOIS really hurts troubleshooting.

ArtOfCode

@TimmyD in theory that would be great. The reality is that they don't care.

It's happened to enough mods around here.

zyabin101

16:51

> A user buys privacy from the company, who in turn replaces the user's info in the WHOIS with the info of a forwarding service (for email and sometimes postal mail, done by a proxy server).

orlp

I have an interesting chat challenge

zyabin101

Domain privacy

Domain privacy is a service offered by a number of domain name registrars. A user buys privacy from the company, who in turn replaces the user's info in the WHOIS with the info of a forwarding service (for email and sometimes postal mail, done by a proxy server). == Level of anonymity == Personal information is typically collected by these registrars to provide the service. Some registrars take little persuasion to release so-called 'private' information to the world, requiring only a phone request or cease and desist letter. Others, however, treat privacy more seriously, and host domain names...

@orlp I have an interesting chat solution

orlp

maybe for @MartinBüttner @Dennis @Sp3000 @trichoplax

zyabin101

Or not.

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@mbomb007 wow, interesting

orlp

16:52

let's say I have a n 32-bit numbers with no discernible compression pattern

if I want to store them, normally I have to spend 32*n bits

but what if I don't care about their order?

El'endia Starman

>>> s = "[[2,2],[1,1],[0,0],[1,1],[0,2],[1,1],[2,0],[1,1],[0,0],[1,1],[2,2],[1,1],[0,2],‌[1,1],[2,0],[1,1],[1,0],[1,1],[1,2],[1,1],[0,1],[1,1],[2,1]]"
>>> len(s)
141
>>> sum(s.count(c) for c in '[],012')
139
>>> list(map(ord,s))
[91, 91, 50, 44, 50, 93, 44, 91, 49, 44, 49, 93, 44, 91, 48, 44, 48, 93, 44, 91, 49, 44, 49, 93, 44, 91, 48, 44, 50, 93, 44, 91, 49, 44, 49, 93, 44, 91, 50, 44, 48, 93, 44, 91, 49, 44, 49, 93, 44, 91, 48, 44, 48, 93, 44, 91, 49, 44, 49, 93, 44, 91, 50, 44, 50, 93, 44, 91, 49, 44, 49, 93, 44, 91, 48, 44, 50, 93, 44, 8204, 8203, 91, 49, 44, 49, 93, 44, 91, 50, 44, 48…

Well, there's my problem: those 8204 and 8203 in the middle of the string.

orlp

there are n! ways to shuffle them, that are all equal

El'endia Starman

But what are those?!

PhiNotPi

@PhiNotPi the answer is yes

orlp

so for 100 elements, there's 100! different representations of a single element, wasting log2(100!) = 524 / 3200 bits to encode that

that's almost 16%!

TimmyD

16:55

@ArtOfCode Wait, really? I'd always figured that was some sort of urban legend ... I didn't realize that people actually took time to do that.

ArtOfCode

@TimmyD people are terrible

TimmyD

People need more zen.

El'endia Starman

And more Jesus.

> Y'all need more Jesus!

orlp

hoooooooooooooly crap

I just realized something

log2(n!) ~ n log n

Martin Büttner

@orlp it does, because the projection at the end can create cycles even if g doesn't have any.

mbomb007

16:56

@El'endiaStarman No idea.

Why are you trying to do that anyway?

PhiNotPi

@ArtOfCode I wonder if it's worth digging up some of the mod-related emails I've gotten.

orlp

so if you store n copies of some fixed amount of bits without something to ignore ordering, you will use 100% of your storage to store only order, for big enough n

El'endia Starman

@El'endiaStarman In Unicode, 8203 is a zero width space and 8204 is a zero width non-joiner. Has to do with the line break in the JSfiddle.

PhiNotPi

Actually, I'm not going to do that.

orlp

since n log n is asymptotically bigger than n

@MartinBüttner you mean projection I hope?

El'endia Starman

16:57

@mbomb007 Was gonna use Python to toss out duplicates.

orlp

@MartinBüttner either way, f can already do that

ArtOfCode

@PhiNotPi what for?

Martin Büttner

@orlp ummm...

orlp

consider f(x) = g(x) = f^-1(x) = x XOR 1

PhiNotPi

1 min ago, by PhiNotPi

Actually, I'm not going to do that.

Martin Büttner

16:58

@orlp how is that?

orlp

f-1(g(f(1))) = 0

Martin Büttner

@orlp then g itself already has a 2-cycle

orlp

f-1(g(f(0))) = 1

Martin Büttner

If g doesn't then f-1 g f won't either

El'endia Starman

Okay, @mbomb007, with three colors, center magenta is [[0, 0], [0, 0], [0, 1], [0, 2], [0, 2], [1, 0], [1, 1], [1, 1], [1, 2], [2, 0], [2, 0], [2, 1], [2, 2], [2, 2]].

PhiNotPi

17:01

@mınxomaτ I'm looking into that.

Is it a bad idea to log in with (one of) my gmail accounts?

El'endia Starman

Accordingly, yellow center is [[0, 0], [0, 1], [0, 1], [0, 2], [1, 0], [1, 0], [1, 1], [1, 2], [1, 2], [2, 0], [2, 1], [2, 1], [2, 2]].

zyabin101

I kinda lost the two color fiddle... Link me again?

[[0,0],[2,2],[3,3],[1,1],[4,4]] makes a beautiful pattern.

El'endia Starman

34 mins ago, by mbomb007

@El'endiaStarman Use this one instead: https://jsfiddle.net/4o7vekpe/1/

zyabin101

@El'endiaStarman Danke sch'o'n.

orlp

@MartinBüttner I'm not convinced yet but I have to do more thinking

zyabin101

17:07

Tapping all squares in this order:

[[0,0],[1,0],[2,0],[3,0],[4,0],[0,1],[1,1],[2,1],[3,1],[4,1],[0,2],[1,2],[2,2],[‌3,2],[4,2],[0,3],[1,3],[2,3],[3,3],[4,3],[0,4],[1,4],[2,4],[3,4],[4,4]]

Makes a square in the middle.

El'endia Starman

It actually doesn't matter what order you tap them.

mınxomaτ

Adobe actually called their new software "Xd"...

Geobits

Does that mean it's dead on arrival?

zyabin101

Welp, actually, [[2,2]] makes the same square... :P

I golfed 24 moves out of this moveset.

orlp

@MartinBüttner I managed to convince myself

Martin Büttner

17:12

@orlp this is what we proved the other day. if h has a 2-cycle then for some x, h(h(x)) = x --> f^-1 g f f^-1 g f (x) = x --> f^-1 g g f (x) = x --> g(g(f(x)) = f(x), so g has a 2-cycle. the deduction also works the other way round which proves that h and g have exactly the same types and numbers of cycles

(sorry, I had already typed it up :P)

orlp

haha

yeah basically the same logic

the f and f^-1 cancel out in the middle

@MartinBüttner what exactly do you consider a projection?

an arbitrary function Z^3 -> Z?

Martin Büttner

yeah, pretty much, although I can't choose it freely. let's say it's just p(x,y,z) = x

orlp

what is this for btw?

Martin Büttner

it may or may not be for some future esolang :P

mınxomaτ

Cᴏɴᴏʀ O'Bʀɪᴇɴ

17:19

   (3&,$(],log2f,nlogn)@i.)11
0 1       2       3       4       5       6       7       8       9      10
0 0       1 2.58496 4.58496 6.90689 9.49185 12.2992 15.2992 18.4691 21.7911
0 0 1.38629 3.29584 5.54518 8.04719 10.7506 13.6214 16.6355  19.775 23.0259

(table of n, log_2 n!, and n log n)

orlp

@CᴏɴᴏʀO'Bʀɪᴇɴ en.wikipedia.org/wiki/Stirling%27s_approximation

maybe this is an interesting challenge?

write a compressor/decompressor that uses the least amount of bits on average to store randomly generated sets of 64-bit integers

whoever uses the least bits on average wins

Martin Büttner

does anyone have opinions on meta.codegolf.stackexchange.com/a/9219/8478 ?

RenderSettings

Uh, isn't that just 64 bits?

orlp

@RenderSettings no, randomly generated sets of 64-bit integers

RenderSettings

Can't reduce n bits of entropy below n bits. If you took the average for all 64 bit numbers "compressed" then it's either at or above 64.

orlp

17:28

@RenderSettings the crux is the set

the numbers are unordered

RenderSettings

:/

orlp

if you store 1000 64-bit numbers the usual way it's 64000 bits

RenderSettings

Hardcode the list of numbers you are being tested on, reduce them into the index into the list.

orlp

randomly generated

the test program will use random generation

and besides, that's a standard loophole

but log2(1000!) / 64000 bits are wasted

which is 13%

RenderSettings

Ah, ok. I get it.

Geobits

17:30

Wouldn't any savings have to be targeted at the set generated? What general-use compression can you really use for randomly generated integers?

orlp

@Geobits the crux is that any storage we humans do has order

but the set is unordered

Sp3000

Related?: stackoverflow.com/questions/12157578/…

To give an idea of how one might approach it

Geobits

@orlp As well as any storage in a computer...

mbomb007

@El'endiaStarman I got duplicate removal to work in Python: repl.it/CUK5

Taking a break. Be back in an hour maybe.

Geobits

For two 64-bit integers, you can store them back to back in a 128-bit array. The order doesn't "use up" any bits at all.

orlp

17:34

@Geobits that's not true

let's say you store 3 and 5

that could also have been 5 and 3

Geobits

Yes. And both take up the same space...

orlp

there's two possible pairs that encode the same value

that's one wasted bit

(not quite as 5 5 is unique)

Geobits

Where exactly is this wasted bit going then? How could you possibly store them to "save" it?

orlp

well, that's the challenge

RenderSettings

Just take out all the 0s from the binary! They're just wasted space!

2

Geobits

17:36

@orlp My premise is that it isn't really possible, though.

orlp

@Geobits you are definitely wrong

Geobits

I'm willing to change my mind given evidence. I just haven't seen any.

orlp

well, you can always just enumerate all combinations with replacement

El'endia Starman

3 requires two bits and 5 requires 3 bits. You could store them as 3 2, so long as you know that the numbers after the first are successive differences, and this requires only 4 bits, as opposed to 5.

wizzwizz4

@orlp Challenge complete!

zyabin101

17:38

@wizzwizz4 Which challenge?

El'endia Starman

The point is that maybe doing something with 5 3 will be even more efficient. (Not necessarily in this case since its so small.)

wizzwizz4

2 mins ago, by orlp

well, that's the challenge

Ask @orlp.

zyabin101

@RenderSettings heh

orlp

@wizzwizz4 well

this isn't a practical algorithm

wizzwizz4

@orlp Whoever said anything about practical? This is PPCG, after all.

orlp

17:40

if I give you 1000 64-bit elements, good luck enumerating those combinations

@wizzwizz4 the fact that it's a code-challenge

if your solution isn't practical it doesn't count

wizzwizz4

@orlp I've been waiting for 3 days and 2 nights to post that message here seamlessly into conversation. And it didn't work. :-(

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@J users: is there a better way to write (y f n) g y tacitly than ] g~ n f~ ]?

wizzwizz4

@CᴏɴᴏʀO'Bʀɪᴇɴ Try naming specific J users. It might work -

Hmm...

Sp3000

@Geobits Easiest way for 2 numbers is to enumerate (0,0) -> 0, (0,1) -> 1, (1,1) -> 2, (0,2) -> 3, etc., i.e. f=lambda n,m:n*(n+3)/2+m where n >= m. Then you can store the two numbers in a smidgen over 127 bits (recovery's a bit tricky, but for sufficiently large n you can do g=lambda k:int((2*k)**.5-1))

wizzwizz4

Maybe we should add groups to chat!

Cᴏɴᴏʀ O'Bʀɪᴇɴ

17:49

@wizzwizz4 I was thinking that

I could make a user script

wizzwizz4

I could make a meta post, then link your user script.

Ooh, I know! Add it to ZyabinVI! @zyabin101

Cᴏɴᴏʀ O'Bʀɪᴇɴ

How would a group ping work? {name}?

wizzwizz4

@CᴏɴᴏʀO'Bʀɪᴇɴ @@name?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@@wizzwizz4

Did that ping you?

wizzwizz4

... Ok, we'll go with yours.

Cᴏɴᴏʀ O'Bʀɪᴇɴ

17:52

Maybe @{name}

a combination of both worlds

orlp

@Sp3000 now n numbers :)

or at least at 3 you'll save your first full bit

New Sandboxed Posts

0

A: Sandbox for Proposed Challenges

Primary Chances Write a program that lists all possible outcomes for a general election with any number of candidates. Rules: Your answer must output how many states each candidate won in any format. Candidates can have any name. This is code-golf, so standard loopholes are forbidden, and s...

Sp3000

Hmm I think I got my formula a tad wrong. Close enough though, point still stands.

Eh, ceebs enumerating :P

I'm sure there's a usable formula for the encoding part, but backwards is a bit trickier

orlp

oh

then you're only halfway there

zyabin101

@wizzwizz4 Nope. zyabinVI is not a hosting for projects, it's an experimenting organization.

orlp

17:53

the challenge is to write an algorithm that goes back and forth :P

Sp3000

I never said I was going to do your challenge :P

wizzwizz4

@zyabin101 Good point. :blush:

orlp

@Sp3000 well, you can't stop now

wizzwizz4

Oops, too much GitHub.

orlp

@zyabin101 why zyabin 6 if you're zyabin 5?

zyabin101

17:55

@orlp I'm zyabin101.

mınxomaτ

...

zyabin101

I randomly chose zyabinVI, from some roman numbers I know. :P

wizzwizz4

Read the starboard, top to bottom.

orlp

HearthStone - "Leeroy Jenkins + Cold Blood" combo fail

►

for anyone that plays hearthstone

the biggest salt there is

Sp3000

f=lambda n,m:n*(n+3)//2-m sign error :P

Geobits

18:01

@Sp3000 Oh god, that's gotta be the least practical idea ever for a set of 1000 integers :P

I guess it does work, though, for not very much savings.

wizzwizz4

I've got a method:

Work out which PRNG is being used, then output only the numbers you needed to work that out and calculate the others.

i.e., hope it's the Mersenne Twister :-)

Geobits

>_>

18

A: 4, 8, 15, 16, 23, 42

Java I can't find a pattern in that sequence. If there's no recognizable pattern, we might as well just throw a bunch of small primes together, cram them into Java's built-in RNG, and call it a day. I don't see how that could possibly go wrong, but then again, I'm an optimist :) import java.uti...

Sp3000

@Geobits Yeah well, ~2! doesn't really help much :P

Hmm decoding might not be too bad with a bit of binary searching

trichoplax

18:17

@orlp Woo! I got put in a category with @MartinBüttner, @Dennis and @Sp3000!

@orlp That reminds me of this solution to tweet image compression - it used the permutation of the regions it stored in order to encode extra information

Sp3000

@orlp Yeah actually I think going both ways isn't too bad. The number of sets of k numbers where the largest number is n is just binomial(n,k) and you can use that recursively for encoding similarly to the two-number case. And I think you can binary search the reverse way

orlp

@MartinBüttner what is your PhD about btw?

flawr

@orlp Memory efficient source code.

Eᴀsᴛᴇʀʟʏ Iʀᴋ

18:33

lol

Martin Büttner

@flawr I wish...

@orlp I'm in a group for computer graphics and visualisation. Most of the research here is about uncertainty/ensemble visualisation.

Cᴏɴᴏʀ O'Bʀɪᴇɴ

anyone know how to play the ping noise for stack exchange?

trichoplax

@CᴏɴᴏʀO'Bʀɪᴇɴ Like this?

Bálint

@CᴏɴᴏʀO'Bʀɪᴇɴ With what?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

yes, but programmatically

orlp

18:38

@MartinBüttner if I understand you correctly, this is less in the direction of SIGGRAPH, but more in the direction of human-computer interaction and psychology?

Bálint

I don't even know that sound, can someoneping me?

Martin Büttner

@orlp more in the direction of IEEE VIS if you want to compare it to SIGGRAPH :P

trichoplax

@Bálint It's this

ArtOfCode

@Bálint ping

orlp

@Bálint ping

Cᴏɴᴏʀ O'Bʀɪᴇɴ

18:38

@Bálint ping

Martin Büttner

some people in the group are also working on more SIGGRAPH'y stuff though ;)

Bálint

Ok, enough

@CᴏɴᴏʀO'Bʀɪᴇɴ With wich language do you want to achieve it?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

Javascript

Sp3000

@orlp ideone.com/ZW8s4l I think I'll leave the decoder as an exercise, gotta sleep

Bálint

The easiest would be to record it then

Cᴏɴᴏʀ O'Bʀɪᴇɴ

18:40

I'm writing a userscript/javascript

trichoplax

@CᴏɴᴏʀO'Bʀɪᴇɴ Is it in the source of this webpage?

Bálint

Then insert an audio tag in the HTML file, and play it

aditsu

↰@Bálint ping

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@trichoplax idk, the source for se is large

Bálint

@aditsu Fail, I already muted down my sound

aditsu

18:41

@Bálint fail, I didn't ping you

trichoplax

@CᴏɴᴏʀO'Bʀɪᴇɴ Maybe outsource the tedious searching to SO? Or use a text search?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@trichoplax that's a good idea

trichoplax

@CᴏɴᴏʀO'Bʀɪᴇɴ Maybe look up "notification" in a thesaurus and search for all the variations

Or just search for whatever JS uses for sound

Cᴏɴᴏʀ O'Bʀɪᴇɴ

I think SE obfuscated their source code >_>

they aliased playSound twice

trichoplax

That's still only 2 to search for...

Bálint

18:45

Someone please ping me

trichoplax

@Bálint sure

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@trichoplax it's a one-letter alias: a.

Bálint

What event does occur, if we set the playing tag on an audio element to "play"?

trichoplax

@CᴏɴᴏʀO'Bʀɪᴇɴ Do you know what characters should come immediately before and after it when used?

Bálint

Any ideas?

Cᴏɴᴏʀ O'Bʀɪᴇɴ

18:46

@trichoplax maybe ( after, but that hardly helps

trichoplax

@Bálint I don't know JS sound...

@CᴏɴᴏʀO'Bʀɪᴇɴ Will it always be a space before?

Bálint

@CᴏɴᴏʀO'Bʀɪᴇɴ Found it for you: cdn-chat.sstatic.net/chat/se.mp3

trichoplax

Nice :)

drobilc

Hey!

Bálint

Ok, I'm not going to lie, I opened the Chrome elements menu, then I searched for "audio"

trichoplax

18:49

It worked...

Cᴏɴᴏʀ O'Bʀɪᴇɴ

@drobilc hi!

orlp

@MartinBüttner well, since that's your area of expertise, what do you think of this graph?

drobilc

@CᴏɴᴏʀO'Bʀɪᴇɴ document.getElementsByTagName("audio")[0].play();

is anyone here familiar with german counting system?

orlp

@drobilc (n)ein

El'endia Starman

@drobilc We have a couple German speakers in here. :P @MartinBüttner, for one.

mınxomaτ

18:57

And flawr, me, Dennis (sort of) ...

Bálint

@drobilc ein, zwei, drei, vier, fünf, sechs, sieben, acht, neun, zehn, elf, zwölf, dreizehn, vierzehn, fünfzehn, sechszehn, siebzehn, achtzehn, neunzehn, zwanzig, einundzwanzig, etc. (it becomes long from here)

orlp

quick! the mods are gone! post pictures of cute fluffy kittens

Bálint

@drobilc In wich aspect do you mean? The "say the last numbers in reverse"?

drobilc

@Bálint "say the last numbers in reverse" aspect :)

ArtOfCode

@orlp sorry what

:)

Bálint

18:59

@drobilc What's with it?

flawr

drobilc

@Bálint Well, i was wondering if there was a challenge about that :)

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