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3:02 AM
@DrGreenEggsandIronMan I'm on the hive in main hub 1 (which you can select from the book in slot 9)(
 
OK, cool. Just give me a minute or two
@HelkaHomba How do I connect to the hive again?
 
From MC Multiplayer menu hit add server then paste in play.hivemc.com into the address spot (and name it w/e)
then save and play it
 
@StevenH. why do you need g and G?
it seems like wrapping a value in a function or forcing evaluation of a function isn't useful
 
Hello
 
Okay, I'm in the main hub. where are you?
 
3:13 AM
Are you sure you're in "Main Hub 1"? I should show up if you press tab
 
Oh, yeah I'm not (you don't show up)
 
It's mostly because some functions (ex. s) don't evaluate the function to retrieve arguments but rather use the function as the argument itself
In order to be able to pass a dynamic function as the sorting method to s, you'd need to use g. G may not be as useful, but I included it anyway as a just-in-case sort of thing.
 
@NathanMerrill If you can determine exactly how much greater, then you can bias the random choice "downwards" to balance it out.
 
@El'endiaStarman yes, you can calculate it. For integers, its simple, but for doubles its more of a range. That said, biasing seems easy for doubles (simply multiple), but I'm not sure how I'd bias for random integers
 
@NathanMerrill why didn't you tag me :(
I know a lot about random number stuff
 
3:20 AM
sorry :)
 
or algorithms in general
can you Each possible number has an equal chance of appearing?
 
explain it?
 
you didn't specify a range of numbers
there is no uniform distribution over the integers
because every integer will have an equal chance, zero, since there are an infinite number of numbers :)
 
I thought I said 1-10
 
oh you said that below the problem
 
3:21 AM
oh, it was in my example
 
so you have some finite range?
 
can we assume 0 <= x < MAX?
for simplicity
 
you still have to further clarify
does every number have an equivalent chance to be in the set at all
or does every number in the set need to be uniformly selected from all possible numbers at that position?
 
3:24 AM
not sure what you mean. I don't care about the probabilities of each possible set (except that each set has a non-zero probability)
 
@NathanMerrill consider a partial set {1, 10}
this still has room in the middle, assuming K = 3
 
its ok if 1,5,10 is less likely than 1,4,10
 
ah ok
that was basically my question
 
I believe its impossible to have a uniform probability over the sets AND over the integers
 
but the chance of a set containing 1 must be equal to that of one containing 2
 
3:27 AM
correct
 
then I believe the solution to be pretty simple
choose a number from 0 to MAX with uniform distribution
add it to the set
 
\o/ Numerical input works!
 
@Dennis I don't understand - the results are exactly as they should be.
 
proceed from here with any biased random method that results in a non-zero probability for the remaining members of the set
 
@El'endiaStarman Expected and "should" are not the same. :/
 
3:31 AM
@NathanMerrill does that solve your problem?
 
@orlp I'm not sure.
 
no wait it doesn't
 
Oh look, there's a featured poll I can answer! :D
"Star Wars or Star Trek?"
 
@NathanMerrill I believe a recursive solution can be quite elegant though
 
Ye guys are coders. Monte-Carlo it!
 
3:33 AM
@StevenH. Except for in this case. :P
 
@DrGreenEggsandIronMan :P
 
0       3,4,5,6,7,8,9
1       4,5,6,7,8,9
2       5,6,7,8,9
3       0,6,7,8,9
4       0,1,7,8,9
5       0,1,2,8,9
6       0,1,2,3,9
7       0,1,2,3,4
8       0,1,2,3,4,5
9       0,1,2,3,4,5,6
that chart shows all possible "second" numbers based on the first number
 
I voted!
 
6 only appears 5 times, while 0 appears 7 times
so, naively, you'd increase the probability of 6 appearing. However, if you do that, then you decrease the chance of its neighbors appearing
 
it's clear to me that only the edges matter
 
3:39 AM
I agree
 
to be more specific
x < k or x > MAX - k
 
correct
 
plus or minus off by one errors
I'm thinking of a recursive solution
 
Where does this problem come from?
 
assuming we chose a set with i members with uniform probability, can we use it to choose a set with i+1 members?
 
3:43 AM
@feersum my brain
 
You can nerd snipe yourself?
 
yep :)
if you're looking for a problem statement: chat.stackexchange.com/transcript/message/31286577#31286577
 
If N is small like 10, you could try converting it to a linear program
 
if you self nerd snipe is it caled nerd suicide? :P
 
Ask Randall Monroe and you might find out. Alternatively, you could be faced with an infinite field of 1-ohm resistors, but... It's worth the risk.
 
3:55 AM
@NathanMerrill I'll think more about your problem later
 
Anonymous
@StevenH. That one's easy. An infinite field of resistors has infinite equivalent resistance across it. A finite section of that field has an easily-computable, finite equivalent resistance.
 
first I gotta figure out repeating decimals
 
@Mego An infinite field of resistors has infinite equivalent resistance when measuring from one "end" to another, but the effective resistance over a finite distance within that infinite distance is extraordinarily difficult to compute because of the infinitely many paths that current can take along that field.
 
@orlp ?
@StevenH. it's not hard
 
@Mego See www.mathpages.com/home/kmath668/kmath668.htm
 
Anonymous
3:58 AM
Though, with a large enough section of resistors, if you get any current at the output, it's more likely that the current is arcing around the field of resistors
 
2 hours ago, by orlp
what is the repeating part of 987654321/123456789?
 
There is a finite resistance along a finite distance between two points, it's just hard to compute
 
@orlp didn't people already say how long it was or something?
@StevenH. how is it hard
 
3:59 AM
@MᴀʀsUʟᴛᴏʀ yes, but I'm looking for actual algorithms :)
 
Anonymous
@StevenH. That shows that it's actually quite easy to compute, but hard to prove.
 
@orlp algorithms?
 
@MᴀʀsUʟᴛᴏʀ to calculate the repeating part of a/b
@feersum the cycle length of the repeating part is the discrete log 10 of the divisor, if coprime to 10
so it's at least a believed to be hard problem
 
Anonymous
Essentially you can pretend that the grid is finite, because any resistors past a certain point will have such a negligible effect on the equivalent resistance that they may as well just be ignored.
 
@Mego that's not true
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... is a divergent series
 
4:02 AM
.....Randall's nerd-sniping ability is truly impressive. Nerd-sniping for years!
 
even though past a certain point any number's effect is negligible on the total sum
so your logic is fallacious :)
 
The repeat length is determined by finding the smallest number of form 10^m-10^n that is a multiple of the denominator
no idea if it can be simplified/if there's a way using smaller numbers
 
@MᴀʀsUʟᴛᴏʀ it's 10^k - 1 after removing 2/5 factors
which is also known as the discrete log of 10 :)
 
Anonymous
@orlp That's what I meant - you can estimate the total equivalent resistance fairly well by ignoring all resistors more than a certain distance from the source and destination nodes.
 
and the discrete log is believed to be a hard problem
@Mego but my point is that all that 'negligible' can add up
we're dealing with infinity here
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + ... = infinite
 
4:04 AM
@orlp oh, so you want a faster algorithm?
Would cycle detection work?
 
@MᴀʀsUʟᴛᴏʀ no
consider 0.1234512345 + 1/700000000000000000000000
or well
maybe cycle detection could work somehow
 
Would that be given as a fraction or a sum?
 
@MᴀʀsUʟᴛᴏʀ a fraction, I'm just writing it down here like this for easy explanation
instead of solving the discrete log maybe we can use an upper bound
and then cycle detect in that upper bound
 
Anonymous
@orlp Yes, which indicates an issue with the problem - you'd need infinite voltage to actually push a current through the infinite resistors. With finite voltage and real-world resistors, there's a point where current simply will not flow anymore, so the grid is in effect a finite one.
 
well, we'd first have to consider the implications of a universe with an infinite grid of resistors
that universe would contain infinite mass
what would the gravitational constant be?
is it even a relativistic universe?
if there is no relativism, will magnetism even exist?
how do electrons flow in a non-relativistic universe?
 
4:08 AM
@orlp How exactly do you convert d = k * (10^n - 1) to a discrete log?
 
@feersum 10^n = -1 (mod k)
 
But how do you know k?
 
1/k
maybe I swapped k and n
depending on what your definitions of k and n were
 
d is the denominator in my equation
of the fraction
 
what is k and what is n?
 
4:10 AM
n is the number of digits in the decimal expansion
 
10^n = 1 (mod d) I think in your notation
 
k is an arbitrary integer
 
sorry
1 instead of -1
for example for 1/7 we need 10^6 = 1 (mod 7)
which is correct, and n is the smallest integer that solves that
finding the smallest integer n in 10^n = 1 (mod d) is the discrete logarithm of 10 in field Z_d
 
Oh I wrote a stupiod equation
IT was supposed to be d * k = 10^n - 1
 
well then you're there
do that equation mod d
d * k = 10^n - 1 (mod d)
0 = 10^n - 1 (mod d) (left hand side is multiple of d)
10^n = 1 (mod d)
discrete log
 
4:14 AM
I see
 
but
maybe we don't need to solve the discrete log
actually
that's bullcrap
 
12 reps to 10k \o/
 
if we find the repeating part, we have effectively computed the discrete log
so the problem is at least as hard as the discrete log of 10 in field Z_p
 
Once we put it that way it seems too amenable to doing all the work with a library...
 
sequence identification time: 55 46 26 19 16 14 13 12 11 10 10 10 …
 
4:18 AM
oeis?
or is this a puzzle
or
 
Not in OEIS.
 
@feersum it's pretty interesting that the upper bound of the number of decimals of the repeating portion of 1/d is O(d)
 
And if you can efficiently compute discrete log, you can break a few encryption schemes
 
I would not have expected that
 
So if it's possible with a classical computer it would probably already have been done
 
4:23 AM
@CᴏɴᴏʀO'Bʀɪᴇɴ first difference negated: 9 20 7 3 2 1 1 1 1 0 ..
the 20 looks mysterious
 
@CᴏɴᴏʀO'Bʀɪᴇɴ what is the context?
 
@orlp because of fermat's little theorem
 
hint: its the first row in a table of values
 
@CᴏɴᴏʀO'Bʀɪᴇɴ does that even count as a hint
 
any of you guys know maple?
 
4:28 AM
nope. Im working on row two for you
 
Not me, but I can use google
 
cheddar> @symbol
< Instance of "Regex" >
wat
 
@Downgoat wat indeed
@Downgoat how does cheddar even break so often, you need to get a different cheese
 
@feersum I now understand why there is so little code snippets or w/e that solves this on the internet
I never figured it to be a pretty hard problem ^^
 
@MᴀʀsUʟᴛᴏʀ ;_; can't. I made dumb mistake and used language name in implementation
 
4:35 AM
@Downgoat just somehow do recursive replace
 
4:59 AM
in databases, we often have a context for relationships. For example, A could be the manager of B, but that's only relevant in the context of the company C. I'm planning on implementing relationships in my language, but I'm wondering whether all relationships should have a context (live in an object).
 
5:24 AM
@NewMainPosts Mod 11 please
 
@NewMainPosts come on
 
Maybe I should make a challenge out of that...
</joke>
 
I woke up at 6 AM here.
 
Good morning Zyabin!
113
Q: We need to improve the quality of our spam!

UndoRecently, I've noticed a downhill effect in the quality of spam posted on Stack Exchange websites. Take this as an example (found on Space.SE): There are a great many things wrong with this artifact: There is not one capitalized letter in the entire post. (-1 grammar point.) The only pun...

 
@DrGreenEggsandIronMan Good morning :3
 
5:43 AM
> [Chavacano: ]El hombre, con quien ya man encuentro tu, amo mi hermano.
Spanish: El hombre que encontraste, es mi hermano.
(The man [whom] you met is my brother.)
How did amo change from "I love" to "is" :o
 
5:54 AM
>>> 'a'[0][0][0][0][0]
'a'
TIL you can never get a character in python
 
@feersum The lowest complexity I've found so far for getting the recurring part length is factoring denominator d, then for every factor p of d, factor p-1
 
@orlp something similar to totient?
 
@LeakyNun yes
that's calculating the divisors of the totient
 
@orlp Don't you need to do prime powers?
 
@feersum what do you mean?
 
5:57 AM
Like if d = 3^5 * 7^6
 
from the wiki I've found that the length of the recurring part of 1/d must divide totient(d)
 
@orlp euler's totient theorem
 
once we factor d
we know that the factors of totient(d) are p^(k-1) * factors(p-1) for each factor p of totient(d) with multiplicity k
 
It's not really discrete log, is it?
 
@feersum it is :)
discrete log and factorization are closely related
if you break one you break the other
 
6:01 AM
Since you're solving 10^x = 1
not 10^x = an arbitrary value
 
yes, this is a particular instance
but there's other more general methods
 
This seems easier than discrete log
 
it's a particular instance of a discrete log
there are many, many examples under which the discrete log becomes easy
like if the modulus is a highly composite number
+1 or smt
it seems that no matter what you do in number theory, you'll end up at primes
one way or another
 
does knowing that p is prime make factoring p-1 easier?
 
6:07 AM
French with a Spanish accent :o
@orlp I don't think so, except that p-1 will be divisible by 2 (if p is not 2)
 
If p != 2 you know that 2 is a factor :P
 
@feersum ninja'd
 
exempt
 
@HelkaHomba depends on what you mean by pets
if you've captured them, they don't really count as pets, they just do what you tell them
 
6:13 AM
@MᴀʀsUʟᴛᴏʀ You have to feed them and shelter them and bathe them and all
 
on the other hand, if you didn't half of them would probably kil you
 
lol
> Yuzhong, Lanzhou, Gansu, China - From your search history
 
@HelkaHomba then the bigger ones will be such a pain to bathe
and feed and shelter
and then i don't think electric pokemon like electrode would react too well to water
 
@MᴀʀsUʟᴛᴏʀ Right but he'd be easy to buff with a damp cloth
Cleaning a Zapdos seems harder
 
Do you need two Electrodes in order to attack?
 
6:23 AM
Yay, I finally got the "proofreader" badge! I've been trying to get that for so long.
Nobody under 2K edits on this site.
 
Ah wait, on SO there are lots of < 2k edits...
 
No, I can't review edits on SO.
 
@HelkaHomba BTW "What was I replying to" bookmarklet: javascript:id=/^:([0-9]{8})/.exec(document.getElementById('input').value);if(id‌​)window.open('http://chat.stackexchange.com/transcript/message/'+id[1]+'%23'+id[1‌​],'_blank');
 
Although I did just get my first docs upvote on SO.
 
@Fatalize I just learnt today that words with aspirated h comes from Germanic :o
 
6:26 AM
@DrGreenEggsandIronMan I don't have doc upvote, I have been ninja'd each time I tried to write something down :p
 
Aww, that sucks. It probably helps that the only topic I feel like I have something worthwile to contribute to is a smaller tag.
 
@Dennis a question: in "mis versos", do you vocalize the first "s"?
 
@DrGreenEggsandIronMan I'm on Lua, but still :p
 
Hey, at least you're not on Java. :P
 
6:31 AM
wait
docs is open beta now?
 
@MᴀʀsUʟᴛᴏʀ Yup!
 
idk what to contribute to
 
Jul 21 at 10:12, by TùxCräftîñg
@zʏᴀʙiɴ101 topics at page 2 are mostly empty
 
Hi part of all!
 
6:37 AM
G'Night part of all!
 
I'm going to change the spec for Eseljik a bit more, then actually make the interpreter
I'm not entirely sure how to handle output and input at this point -~-
I think I'm going to rework that part entirely
 
So, the "Loop in the End" challenge is dead for three months. Three months, that's enough to show that everyone now has their attention at everything except me.
 
'~' not sure whether to remove binary i o
 
And that everyone now hates Nested Programs.
 
¯\_(ツ)_/¯
 
6:46 AM
@feersum ok
if a/b are coprime
and I know 1/b
(like, the recurring part and how many leading zeroes)
how do I get a/b
 
@orlp multiplication?
 
@LeakyNun it's not that simple
 
@orlp and then check for number of digits
and repeats
etc
 
The actual digits?
 
yes
 
6:48 AM
You just use a bigint division I guess
 
wait it might just be a multiplication
mod 10^k
where k is the length of the recurring portion
 
If you actually have to find the digits
 
@orlp you need to check if the repetend shifted
 
since the recurrence portion of a/b is equally long to 1/b if a and b are coprime
let me test that
 
You might as well generate d-1 digits and then find the period of the string :P
 
6:50 AM
@feersum nah, I want it proper :)
 
7:13 AM
@Dennis actually.tryitonline.net is borked:
Traceback (most recent call last):
  File "/opt/Actually/srs", line 3, in <module>
    from seriously import main
  File "/opt/Actually/seriously/__init__.py", line 1, in <module>
    from .seriously import *
  File "/opt/Actually/seriously/seriously.py", line 17, in <module>
    from Crypto.Cipher import AES
ImportError: No module named 'Crypto'
 
lol why would a golfing language use AES?
 
@feersum For an easter egg, of course.
 
Well, if you consider dying with an error an Easter egg.
 
Anonymous
Blame Dennis. He didn't do pip3 install -r requirements.txt or pip3 install seriously
 
Anonymous
Either of those would have properly loaded the dependencies :P
 
7:35 AM
I'm doing an Actually integer golf, and I have 14/1000 integers done.
However, I can't run Actually on my Windows computer, and Try it online's instance is borked, so...
 
@zyabin101 why not on your computer?
Windows supports Python fine
 
It uses pyshoco, which only runs on Linux. :(
 
TIL users just fix your bugs if you do nothing: github.com/The-Quill/PopcornSE/issues/5
 
2 reps till 10k :o
 
lol a flag voyeurism script?
 
7:44 AM
@LeakyNun Gotta upvote some post
 
@feersum gotta get your dose of drama somehow
 
Ideas for eseljik: `-halt with no output. & - output contents of variable b, encoded 7 bits per char ascii. question. Should I support utf 8 output?
 
@LeakyNun You're officially at 10k!
 
Thanks all of you
 
wait & is already being used :/
running out of good ascii chars to use :\
the eternal struggle of single char commands
I guess I could use numbers but ehhhh
 
8:08 AM
@feersum I misjudged the difficulty
my laptop computes that 987654321/123456789 = 8. (1*(10^6855006-1)/13717421 % 10^6855006)... in the blink of an eye
computing the decimal expansion of (1*(10^6855006-1)/13717421 % 10^6855006) takes a while
 
@TùxCräftîñg hello
 
why can't I do *FMsomeiterable in Pyth? :(
 
ven
@orlp because Pyth doesn't manage to parse *F as a pf1
maybe @isaacg's new lexer will/did change something
 
$ node src/neoscript.js
Neoscript REPL
neoscript> 10
Tokens:
[ { id: 'Number', value: '10', line: 1, column: 1 },
  { id: 'EOF', line: 1, column: 3 } ]
AST:
[ undefined ]
neoscript> 10 + 20
Unknow NUD for '+'
neoscript>
my parser is borked \o/
 
8:36 AM
What should the colour #FFA500 correspond to, yellow or orange?
 
orange
 
^ correct answer
 
orlow
yellange
 
ven
>Golisp has only a few grammatical constructs, like in Lisp, everything is a function.
what?
 
I agree that its hard to understand when its not quoted properly..
 
8:40 AM
> quote like this "> Quote"
 
$ node src/neoscript.js
D:\Neoscript\src\tokenizer.js:3
const REGEXS = new Map([
               ^

TypeError: Iterator value undefined is not an entry object
    at new Map (native)
    at Object.<anonymous> (D:\Neoscript\src\tokenizer.js:3:16)
    at Module._compile (module.js:541:32)
    at Object.Module._extensions..js (module.js:550:10)
    at Module.load (module.js:458:32)
    at tryModuleLoad (module.js:417:12)
    at Function.Module._load (module.js:409:3)
    at Module.require (module.js:468:17)
AND A BORK BORK BORK AND A BORK BORK BORK
tried to modify the tokenizer and get a wtf error
i love js
 
Someone should make a misleading add for stackylogic or bf and talk about lack of annoying syntax errors, and wtf types and wtf errors and hassle
that would be funny
 
ok i hate my compose key
$ node src/neoscript.js
D:\Neoscript\src\parser.js:3
        ["-"‚ 100],
            ^
SyntaxError: Unexpected token ILLEGAL
    at Object.exports.runInThisContext (vm.js:76:16)
    at Module._compile (module.js:513:28)
    at Object.Module._extensions..js (module.js:550:10)
    at Module.load (module.js:458:32)
    at tryModuleLoad (module.js:417:12)
    at Function.Module._load (module.js:409:3)
    at Module.require (module.js:468:17)
    at require (internal/module.js:20:19)
    at Object.<anonymous> (D:\Neoscript\src\neoscript.js:5:19)
UNICODE HOMOGLYPHS FTW \o/
 
I hate my compost key
it hardly opens the lock one in 10 tries.
 
@Optimizer compost key?
 
8:47 AM
you don't know what compost means?
 
@Optimizer But you have a key for your compost?
 
of course.. its locked up in a room
 
^^ and ^^^^ a new colour palette for country flags.
 
8:54 AM
LOL bullgard have quaraintened bash
 
@TùxCräftîñg Then prove to Bullgard that this app is not a virus, then obtain it back from him.
 
ik
but cmder have closed with vim with unsaved changes :/
 
@TùxCräftîñg Is the .*.swp file still there?
 
a swp file?
¯\_(ツ)_/¯ i have retyped my changes
 
Yup, a swp file!
 
9:02 AM
brb
 
9:28 AM
longest brb ever.. and counting
 
@LeakyNun Remember when we were discussing of metatables? Here's some documentation that explain some points better than I could
7
score
8
8 examples
3 contributors
 
wow... docs have the best onebox
 
@Optimizer I agree, I was amazed to see it ^^
 
Welcome to Programming Puzzles & Code Golf chat, @CaiusJard!
I'm interested what your QR code produces. :3
 
It looked like some white noise at first ^^
 
9:52 AM
@zyabin101 QR code?
 
@MᴀʀsUʟᴛᴏʀ Caius Jard had a QR code as his avatar.
 
@zyabin101 Link to profile?
 
i am back
 
@TùxCräftîñg You were away for long...
 
9:55 AM
¯\_(ツ)_/¯
 
Also, the RTRFTDs were released in chat.stackexchange.com/rooms/42936/….
 
@zyabin101 Text: I'm not here to give you a fish. I'm not even here to teach exactly how to fish. I'm giving you an old rod and good pointers. There are gaps;appreciate them because filling them in IS learning
 
@MᴀʀsUʟᴛᴏʀ Thanks. :3
 
tux crafting, I had great idea for new icon for you
 

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