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El'endia Starman
El'endia Starman
4:00 PM
Didn't we already have a PPCG challenge on that puzzle?
 
mbomb007
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
Leaky Nun
@CᴏɴᴏʀO'Bʀɪᴇɴ What's the original question?
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
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
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
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
mbomb007
4:10 PM
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
trichoplax
@PhiNotPi I now understand that I understand less than I thought I did
 
PhiNotPi
PhiNotPi
That ping should of actually been directed to @zyabin101, since he was the person who originally posted those prices.
 
El'endia Starman
El'endia Starman
Do you mean toggling the lone square in the center of a 3x3?
 
mbomb007
mbomb007
@El'endiaStarman Yeah.
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Nice finds!
 
El'endia Starman
El'endia Starman
4:11 PM
You can do it in 9. Click all the squares. :P
 
Eᴀsᴛᴇʀʟʏ Iʀᴋ
Eᴀsᴛᴇʀʟʏ Iʀᴋ
HAI
 
trichoplax
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
Eᴀsᴛᴇʀʟʏ Iʀᴋ
relink the thingy?
the jsfiddle
 
zyabin101
zyabin101
@PhiNotPi $1.99 is the registration price for the domain.
 
El'endia Starman
El'endia Starman
@EᴀsᴛᴇʀʟʏIʀᴋ thingy
 
Eᴀsᴛᴇʀʟʏ Iʀᴋ
Eᴀsᴛᴇʀʟʏ Iʀᴋ
4:16 PM
thankee
took 4 moves.
Can it be solved in less than that?
 
mbomb007
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
zyabin101
@mbomb007 So I can finally solve the board in 25 moves! \o/
 
mbomb007
mbomb007
@trichoplax They're all similar, but my lights out game also toggles the diagonally adjacent tiles, whereas none of those do.
 
Eᴀsᴛᴇʀʟʏ Iʀᴋ
Eᴀsᴛᴇʀʟʏ Iʀᴋ
you can also solve it in less.
 
mbomb007
mbomb007
@EᴀsᴛᴇʀʟʏIʀᴋ Not necessarily.
 
zyabin101
zyabin101
4:21 PM
How much did it took for you?
 
PhiNotPi
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
zyabin101
O_o
 
mınxomaτ
mınxomaτ
Use internet.bs already :P
 
mbomb007
mbomb007
@EᴀsᴛᴇʀʟʏIʀᴋ If the starting configuration on a 3x3 is only the center tile, then solving it takes 3*3 = 9 moves.
 
Eᴀsᴛᴇʀʟʏ Iʀᴋ
Eᴀsᴛᴇʀʟʏ Iʀᴋ
@mbomb007 for this specific configuration (blank to full), it takes 4 moves for a 5*5 board.
 
zyabin101
zyabin101
4:23 PM
@PhiNotPi Well, actually, the $1.95 price was the price on the Add button. D:
 
mbomb007
mbomb007
@EᴀsᴛᴇʀʟʏIʀᴋ That's not the puzzle, though. That's trivial on any board size.
 
TimmyD
TimmyD
@mınxomaτ The Internet is all BS
 
Eᴀsᴛᴇʀʟʏ Iʀᴋ
Eᴀsᴛᴇʀʟʏ Iʀᴋ
@mbomb007 okay
 
mbomb007
mbomb007
The challenge is to get to blank from any starting configuration.
 
El'endia Starman
El'endia Starman
I modded the JS fiddle to cycle between three colors. It's weirdly kinda more interesting and kinda less interesting.
 
zyabin101
zyabin101
4:27 PM
@El'endiaStarman Share the fiddle?
I could try to make four colors! It will be very hard.
(To solve)
 
El'endia Starman
El'endia Starman
Four colors is probably fairly uninteresting.
 
zyabin101
zyabin101
D:
 
El'endia Starman
El'endia Starman
 
mbomb007
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
El'endia Starman
Ooh, okay.
 
mbomb007
mbomb007
4:30 PM
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
Bald Bantha
How to play the game?
 
mbomb007
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
Bald Bantha
what I figured when I saw the symetrical patterns and stuff. There should be a bigger board
 
mbomb007
mbomb007
@BaldBantha Make it bigger.
Change the number and click "Update"
 
El'endia Starman
El'endia Starman
Okay, I made the edit. Hmm. I should've forked it, hmm?
 
mbomb007
mbomb007
4:33 PM
Max is 20.
@El'endiaStarman Yeah, fork it.
 
El'endia Starman
El'endia Starman
 
Eᴀsᴛᴇʀʟʏ Iʀᴋ
Eᴀsᴛᴇʀʟʏ Iʀᴋ
You forgot the blank one.
 
El'endia Starman
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
mbomb007
@El'endiaStarman Yeah, I know.
Every configuration is still possible on a 3x3, I believe.
 
El'endia Starman
El'endia Starman
Yeah. You just have to repeat a sequence to toggle one square between the three colors.
 
mbomb007
mbomb007
4:39 PM
@El'endiaStarman Try to get only the center.
 
New Main Posts
New Main Posts
0
Q: Algorithm to generate mountain ranges with upstrokes and down-strokes (java)

salamanka44I 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
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
PhiNotPi
What's the pros/cons of whois privacy?
 
ArtOfCode
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
zyabin101
@PhiNotPi Information about you is hidden in the site's WHOIS response.
 
El'endia Starman
El'endia Starman
4:48 PM
@El'endiaStarman ...why can't I paste this into Python? I'm getting SyntaxError: invalid character in identifier. Weird.
 
PhiNotPi
PhiNotPi
does internet.bs have whois privacy?
 
zyabin101
zyabin101
idk
 
mbomb007
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
orlp
@MartinBüttner sorry for late reply
 
zyabin101
zyabin101
But 1&1 includes one for free.
 
orlp
orlp
4:50 PM
but I don't feel that embedding in a higher dimensional space adds power
 
TimmyD
TimmyD
@ArtOfCode If someone does that, file charges with the FBI (or equivalent) for threats and harrassment
 
zyabin101
zyabin101
ohai ~rys
 
orlp
orlp
since you can map Z^3 <-> Z
 
TimmyD
TimmyD
As an email administrator, having privacy WHOIS really hurts troubleshooting.
 
ArtOfCode
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
zyabin101
4:51 PM
> 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
orlp
I have an interesting chat challenge
 
zyabin101
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
orlp
maybe for @MartinBüttner @Dennis @Sp3000 @trichoplax
 
zyabin101
zyabin101
Or not.
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
@mbomb007 wow, interesting
 
orlp
orlp
4:52 PM
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
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
orlp
there are n! ways to shuffle them, that are all equal
 
El'endia Starman
El'endia Starman
But what are those?!
 
PhiNotPi
PhiNotPi
@PhiNotPi the answer is yes
 
orlp
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
TimmyD
4:55 PM
@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
ArtOfCode
@TimmyD people are terrible
 
TimmyD
TimmyD
People need more zen.
 
El'endia Starman
El'endia Starman
And more Jesus.
> Y'all need more Jesus!
 
orlp
orlp
hoooooooooooooly crap
I just realized something
log2(n!) ~ n log n
 
Martin Büttner
Martin Büttner
@orlp it does, because the projection at the end can create cycles even if g doesn't have any.
 
mbomb007
mbomb007
4:56 PM
@El'endiaStarman No idea.
Why are you trying to do that anyway?
 
PhiNotPi
PhiNotPi
@ArtOfCode I wonder if it's worth digging up some of the mod-related emails I've gotten.
 
orlp
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'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
PhiNotPi
Actually, I'm not going to do that.
 
orlp
orlp
since n log n is asymptotically bigger than n
@MartinBüttner you mean projection I hope?
 
El'endia Starman
El'endia Starman
4:57 PM
@mbomb007 Was gonna use Python to toss out duplicates.
 
orlp
orlp
@MartinBüttner either way, f can already do that
 
ArtOfCode
ArtOfCode
@PhiNotPi what for?
 
Martin Büttner
Martin Büttner
@orlp ummm...
 
orlp
orlp
consider f(x) = g(x) = f^-1(x) = x XOR 1
 
PhiNotPi
PhiNotPi
1 min ago, by PhiNotPi
Actually, I'm not going to do that.
 
Martin Büttner
Martin Büttner
4:58 PM
@orlp how is that?
 
orlp
orlp
f-1(g(f(1))) = 0
 
Martin Büttner
Martin Büttner
@orlp then g itself already has a 2-cycle
 
orlp
orlp
f-1(g(f(0))) = 1
 
Martin Büttner
Martin Büttner
If g doesn't then f-1 g f won't either
 
El'endia Starman
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
PhiNotPi
5:01 PM
@mınxomaτ I'm looking into that.
Is it a bad idea to log in with (one of) my gmail accounts?
 
El'endia Starman
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
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
El'endia Starman
34 mins ago, by mbomb007
@El'endiaStarman Use this one instead: https://jsfiddle.net/4o7vekpe/1/
 
zyabin101
zyabin101
@El'endiaStarman Danke sch'o'n.
 
orlp
orlp
@MartinBüttner I'm not convinced yet but I have to do more thinking
 
zyabin101
zyabin101
5:07 PM
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
El'endia Starman
It actually doesn't matter what order you tap them.
 
mınxomaτ
mınxomaτ
Adobe actually called their new software "Xd"...
 
Geobits
Geobits
Does that mean it's dead on arrival?
 
zyabin101
zyabin101
Welp, actually, [[2,2]] makes the same square... :P
I golfed 24 moves out of this moveset.
 
orlp
orlp
@MartinBüttner I managed to convince myself
 
Martin Büttner
Martin Büttner
5:12 PM
@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
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
Martin Büttner
yeah, pretty much, although I can't choose it freely. let's say it's just p(x,y,z) = x
 
orlp
orlp
what is this for btw?
 
Martin Büttner
Martin Büttner
it may or may not be for some future esolang :P
 
mınxomaτ
mınxomaτ
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
5:19 PM
   (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
orlp
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
Martin Büttner
does anyone have opinions on meta.codegolf.stackexchange.com/a/9219/8478 ?
 
RenderSettings
RenderSettings
Uh, isn't that just 64 bits?
 
orlp
orlp
@RenderSettings no, randomly generated sets of 64-bit integers
 
RenderSettings
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
orlp
5:28 PM
@RenderSettings the crux is the set
the numbers are unordered
 
RenderSettings
RenderSettings
:/
 
orlp
orlp
if you store 1000 64-bit numbers the usual way it's 64000 bits
 
RenderSettings
RenderSettings
Hardcode the list of numbers you are being tested on, reduce them into the index into the list.
 
orlp
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
RenderSettings
Ah, ok. I get it.
 
Geobits
Geobits
5:30 PM
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
orlp
@Geobits the crux is that any storage we humans do has order
but the set is unordered
 
Sp3000
Sp3000
To give an idea of how one might approach it
 
Geobits
Geobits
@orlp As well as any storage in a computer...
 
mbomb007
mbomb007
@El'endiaStarman I got duplicate removal to work in Python: repl.it/CUK5
Taking a break. Be back in an hour maybe.
 
Geobits
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
orlp
5:34 PM
@Geobits that's not true
let's say you store 3 and 5
that could also have been 5 and 3
 
Geobits
Geobits
Yes. And both take up the same space...
 
orlp
orlp
there's two possible pairs that encode the same value
that's one wasted bit
(not quite as 5 5 is unique)
 
Geobits
Geobits
Where exactly is this wasted bit going then? How could you possibly store them to "save" it?
 
orlp
orlp
well, that's the challenge
 
RenderSettings
RenderSettings
Just take out all the 0s from the binary! They're just wasted space!
2
 
Geobits
Geobits
5:36 PM
@orlp My premise is that it isn't really possible, though.
 
orlp
orlp
@Geobits you are definitely wrong
 
Geobits
Geobits
I'm willing to change my mind given evidence. I just haven't seen any.
 
orlp
orlp
well, you can always just enumerate all combinations with replacement
 
El'endia Starman
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
wizzwizz4
@orlp Challenge complete!
 
zyabin101
zyabin101
5:38 PM
@wizzwizz4 Which challenge?
 
El'endia Starman
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
wizzwizz4
2 mins ago, by orlp
well, that's the challenge
Ask @orlp.
 
zyabin101
zyabin101
@RenderSettings heh
 
orlp
orlp
@wizzwizz4 well
this isn't a practical algorithm
 
wizzwizz4
wizzwizz4
@orlp Whoever said anything about practical? This is PPCG, after all.
 
orlp
orlp
5:40 PM
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
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
Cᴏɴᴏʀ O'Bʀɪᴇɴ
@J users: is there a better way to write (y f n) g y tacitly than ] g~ n f~ ]?
 
wizzwizz4
wizzwizz4
@CᴏɴᴏʀO'Bʀɪᴇɴ Try naming specific J users. It might work -
Hmm...
 
Sp3000
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
wizzwizz4
Maybe we should add groups to chat!
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
5:49 PM
@wizzwizz4 I was thinking that
I could make a user script
 
wizzwizz4
wizzwizz4
I could make a meta post, then link your user script.
Ooh, I know! Add it to ZyabinVI! @zyabin101
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
How would a group ping work? {name}?
 
wizzwizz4
wizzwizz4
@CᴏɴᴏʀO'Bʀɪᴇɴ @@name?
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
@@wizzwizz4
Did that ping you?
 
wizzwizz4
wizzwizz4
... Ok, we'll go with yours.
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
5:52 PM
Maybe @{name}
a combination of both worlds
 
orlp
orlp
@Sp3000 now n numbers :)
or at least at 3 you'll save your first full bit
 
New Sandboxed Posts
New Sandboxed Posts
0
A: Sandbox for Proposed Challenges

weatherman115Primary 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
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
orlp
oh
then you're only halfway there
 
zyabin101
zyabin101
@wizzwizz4 Nope. zyabinVI is not a hosting for projects, it's an experimenting organization.
 
orlp
orlp
5:53 PM
the challenge is to write an algorithm that goes back and forth :P
 
Sp3000
Sp3000
I never said I was going to do your challenge :P
 
wizzwizz4
wizzwizz4
@zyabin101 Good point. :blush:
 
orlp
orlp
@Sp3000 well, you can't stop now
 
wizzwizz4
wizzwizz4
Oops, too much GitHub.
 
orlp
orlp
@zyabin101 why zyabin 6 if you're zyabin 5?
 
zyabin101
zyabin101
5:55 PM
@orlp I'm zyabin101.
 
mınxomaτ
mınxomaτ
...
 
zyabin101
zyabin101
I randomly chose zyabinVI, from some roman numbers I know. :P
 
wizzwizz4
wizzwizz4
Read the starboard, top to bottom.
 
orlp
orlp
for anyone that plays hearthstone
the biggest salt there is
 
Sp3000
Sp3000
f=lambda n,m:n*(n+3)//2-m sign error :P
 
Geobits
Geobits
6:01 PM
@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
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
Geobits
>_>
18
A: 4, 8, 15, 16, 23, 42

GeobitsJava 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
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
trichoplax
6:17 PM
@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
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
orlp
@MartinBüttner what is your PhD about btw?
 
flawr
flawr
@orlp Memory efficient source code.
 
Eᴀsᴛᴇʀʟʏ Iʀᴋ
Eᴀsᴛᴇʀʟʏ Iʀᴋ
6:33 PM
lol
 
Martin Büttner
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
Cᴏɴᴏʀ O'Bʀɪᴇɴ
anyone know how to play the ping noise for stack exchange?
 
trichoplax
trichoplax
@CᴏɴᴏʀO'Bʀɪᴇɴ Like this?
 
Bálint
Bálint
@CᴏɴᴏʀO'Bʀɪᴇɴ With what?
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
yes, but programmatically
 
orlp
orlp
6:38 PM
@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
Bálint
I don't even know that sound, can someoneping me?
 
Martin Büttner
Martin Büttner
@orlp more in the direction of IEEE VIS if you want to compare it to SIGGRAPH :P
 
trichoplax
trichoplax
@Bálint It's this
 
ArtOfCode
ArtOfCode
@Bálint ping
 
orlp
orlp
@Bálint ping
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
6:38 PM
@Bálint ping
 
Martin Büttner
Martin Büttner
some people in the group are also working on more SIGGRAPH'y stuff though ;)
 
Bálint
Bálint
Ok, enough
@CᴏɴᴏʀO'Bʀɪᴇɴ With wich language do you want to achieve it?
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Javascript
 
Sp3000
Sp3000
@orlp ideone.com/ZW8s4l I think I'll leave the decoder as an exercise, gotta sleep
 
Bálint
Bálint
The easiest would be to record it then
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
6:40 PM
I'm writing a userscript/javascript
 
trichoplax
trichoplax
@CᴏɴᴏʀO'Bʀɪᴇɴ Is it in the source of this webpage?
 
Bálint
Bálint
Then insert an audio tag in the HTML file, and play it
 
aditsu
aditsu
↰@​Bálint ping
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
@trichoplax idk, the source for se is large
 
Bálint
Bálint
@aditsu Fail, I already muted down my sound
 
aditsu
aditsu
6:41 PM
@Bálint fail, I didn't ping you
 
trichoplax
trichoplax
@CᴏɴᴏʀO'Bʀɪᴇɴ Maybe outsource the tedious searching to SO? Or use a text search?
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
@trichoplax that's a good idea
 
trichoplax
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
Cᴏɴᴏʀ O'Bʀɪᴇɴ
I think SE obfuscated their source code >_>
they aliased playSound twice
 
trichoplax
trichoplax
That's still only 2 to search for...
 
Bálint
Bálint
6:45 PM
Someone please ping me
 
trichoplax
trichoplax
@Bálint sure
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
@trichoplax it's a one-letter alias: a.
 
Bálint
Bálint
What event does occur, if we set the playing tag on an audio element to "play"?
 
trichoplax
trichoplax
@CᴏɴᴏʀO'Bʀɪᴇɴ Do you know what characters should come immediately before and after it when used?
 
Bálint
Bálint
Any ideas?
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
6:46 PM
@trichoplax maybe ( after, but that hardly helps
 
trichoplax
trichoplax
@Bálint I don't know JS sound...
@CᴏɴᴏʀO'Bʀɪᴇɴ Will it always be a space before?
 
Bálint
Bálint
@CᴏɴᴏʀO'Bʀɪᴇɴ Found it for you: cdn-chat.sstatic.net/chat/se.mp3
 
trichoplax
trichoplax
Nice :)
 
drobilc
drobilc
Hey!
 
Bálint
Bálint
Ok, I'm not going to lie, I opened the Chrome elements menu, then I searched for "audio"
 
trichoplax
trichoplax
6:49 PM
It worked...
 
Cᴏɴᴏʀ O'Bʀɪᴇɴ
Cᴏɴᴏʀ O'Bʀɪᴇɴ
@drobilc hi!
 
orlp
orlp
@MartinBüttner well, since that's your area of expertise, what do you think of this graph?
 
drobilc
drobilc
@CᴏɴᴏʀO'Bʀɪᴇɴ document.getElementsByTagName("audio")[0].play();
is anyone here familiar with german counting system?
 
orlp
orlp
@drobilc (n)ein
 
El'endia Starman
El'endia Starman
@drobilc We have a couple German speakers in here. :P @MartinBüttner, for one.
 
mınxomaτ
mınxomaτ
6:57 PM
And flawr, me, Dennis (sort of) ...
 
Bálint
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
orlp
quick! the mods are gone! post pictures of cute fluffy kittens
 
Bálint
Bálint
@drobilc In wich aspect do you mean? The "say the last numbers in reverse"?
 
drobilc
drobilc
@Bálint "say the last numbers in reverse" aspect :)
 
ArtOfCode
ArtOfCode
@orlp sorry what
:)
 
Bálint
Bálint
6:59 PM
@drobilc What's with it?
 
flawr
flawr
 
drobilc
drobilc
@Bálint Well, i was wondering if there was a challenge about that :)
 
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