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6:00 AM
whysoparanoid :P
 
it's only 6 am UTC
 
Accepts don't count in the 200, which is the problem
 
Yeah, you can't cancel out the -15 with upvotes.
 
huh? why not?
 
Dennis would end on 200 - 15 = 185, basically
 
6:04 AM
Because you can only earn 200 rep from upvotes in a day, but accepts and bounties are not affected by the rep cap.
 
Anonymous
You could just get another accept
 
6 mins ago, by Sp3000
Lies Dennis, lies.
 
Already did, thanks to Sp3000. Without an accept, I would have been screwed though.
 
Are you planning to surpass Martin?
 
6:07 AM
Hoping would be a better word. :P For now, I just want to get Legendary.
I'm currently at 144 / 150 days.
I've only missed the rep cap four times since September 14, but it wasn't enough to beat him to Legendary...
 
I wonder what one could get for selling a high rep ppcg account, like people do with mmorpg game accounts
 
Banned
5
 
Well sure, just curious
 
tbh I think bounties would be the easier commodity to sell, because you don't have the long history of answers which aren't yours
... let's pretend I never said that
 
Anonymous
Purely hypothetical discussion here, because none of us are dumb enough to try this
 
Anonymous
6:18 AM
Post a really hard question that you have a really clever, very-golfed solution for
 
@Sp3000 But you'd just pretend to be the person when you're on their account
 
Anonymous
After 7 days, offer a large bounty, give the solution to your client
 
Anonymous
They post it, get the bounty
 
@Mego Hey! Maybe I am!
......
wait, no
 
Anonymous
No, I promise you aren't
 
Anonymous
6:19 AM
You have a diamond, so you really better not try it
 
Oh hey, that's something I can sell!
Hey @Mego, wanna be a mod for $1000?
 
@Calvin'sHobbies I guess you could do that, but given the size and state of the current community, said banning seems likely and you probably won't have much of a market to sell to :P Or hopefully, at least
 
Anonymous
ಠ_ಠ
 
How good is your buyer's acting/IP spoofing skills?
 
Oh wait, you're broke like I am.
 
Anonymous
6:20 AM
Even if I had $1000, I wouldn't buy a diamond
 
Anonymous
I'm on Calvin's side - having a diamond is too much work
 
But it's SO MUCH FUN!
You get to ban people!
 
Anonymous
Now if you paid me $1000 to have a diamond... That's another story
 
Anonymous
I've moderated enough communities to know that it's a soul-draining job
 
@El'endiaStarman For example, people that bought diamonds. :P
 
6:22 AM
TBH i brought it up because I was selfishly wondering if people would rather buy a 'mostly answers' account like Dennis/Sp's or a 'mostly questions' one like mine ;D
 
Anonymous
Or a 'mostly downvotes' one like Geobits's
 
Moderating PPCG is pretty calm. For every two questions that get posted per day, we have a mod.
 
They lined up perfectly!
 
Anonymous
Generating a bunch of test data for the syzygy challenge takes forever :/
 
Anonymous
6:25 AM
I have to generate it and solve it to make sure it's solvable
 
I wonder what the smallest non-trivial non-solvable set is.
As in, 2,[3,3,3] is trivial.
 
@Calvin'sHobbies Well I guess it'd depend on the motive for the purchase? e.g. pretending to be you is pretty damn hard
 
Anonymous
@El'endiaStarman They should all be theoretically solvable. Because of the condition that the hands start aligned, after a finite amount of time, they should become aligned again, even with the restriction of integer times.
 
Anonymous
Well
 
Well, I'm thinking about how the number that align must be exact, so maybe you specify that 3 hands align, and every alignment you get is either 2 or 4 hands...
 
Anonymous
6:29 AM
Unless you have a set of k>n hands who are not coprime but are coprime to all the other hands
 
@Mego Generating test cases isn't that bad is it? (I gave a bunch myself... o_O)
 
Anonymous
But even still, that could still be solvable
 
Wow. The sleepiness challenge is giving me so much trouble.
Up to 79 bytes with a solution that doesn't yet work. ಠ_ಠ
 
Anonymous
For the test cases, I'm imposing the restriction that the alignment must happen within prod(n largest elements of L) seconds (where n is the number of hands to align and L is the period list)
 
@El'endiaStarman waits for sleepless pun
 
Anonymous
6:30 AM
So that I can actually verify in finite time if they're solvable :P
 
Anonymous
@Calvin'sHobbies That would be Alex. He's the one in Seattle, after all.
2
 
@Calvin'sHobbies ...well now that you've said it, I can't exactly use it, can I?
 
I mean what are you doing with the test cases, randomly generating or something?
 
...brilliant, @Mego, brilliant...
 
Anonymous
bows
 
Anonymous
6:32 AM
My bad code:
 
Anonymous
#!/usr/bin/env python
from __future__ import division
import itertools, fractions, random, operator
total=0
while total<25:
    k=random.randint(2,7)
    l=random.sample(xrange(100),random.randint(k,7))
    n=1
    m=reduce(operator.mul,sorted(l,reverse=True)[:k],1)
    while 1:
        if n > m:
            break
        s=map(lambda y:fractions.Fraction(360*n,y)%360,l)
        if sum([s.count(y)>1 for y in s])==k:
            print "%d, %s -> %d"%(k,str(l),n)
            total+=1
            break
 
Er... what type of test case are you trying to generate, anyway?
As in, what are you trying to test - large inputs?
 
Anonymous
I'm just trying to generate a bunch of test cases to add to the post (separate from the examples)
 
I feel like you should focus on adding test cases where a program is likely to fail (e.g. one for groups, one for possible floating point errors, etc.)
 
Need a spongy animal challenge
 
6:39 AM
hey all!
 
Hello!
We do need a spongey animal challenge. o-o
@Mego Hmmm... What if we... ;D
 
Anonymous
@VoteToClose wut
 
Anonymous
How did that happen 0_0
 
Anonymous
ZeroDivisionError
 
Anonymous
Oh duh
 
Anonymous
6:47 AM
@Sp3000 This doesn't seem right... Can you double check it? 7, [1, 23, 19, 4, 7, 10, 16] -> 244720
 
Anonymous
@El'endiaStarman ^
 
You need all 7 and the primes are pretty spaced out, so I wouldn't be surprised
 
000000000 1
000000001 23
000000010 19
200000000 4
000100000 7
101000000 10
400000000 16

401100011 16*5*7*19*23 = 224720
 
Anonymous
Cool
 
Anonymous
6:50 AM
How's that for some test data?
 
15 mins ago, by Sp3000
I feel like you should focus on adding test cases where a program is likely to fail (e.g. one for groups, one for possible floating point errors, etc.)
Having more test cases isn't necessarily better if it doesn't test anything different, e.g. 2, [13, 18] -> 234 and 2, [13, 12] -> 156 might as well be one case
(or even scrapped, since it's no different from 2, [3,4])
Well... I guess it doesn't hurt to have more, but you should emphasise which ones were crafted for specific situations
 
@Mego i believe it is easier if we have a wiki on your github repo...
and why are we supposed to write in RST?
 
Anonymous
@TanMath Because that's what phase started, and I'm letting him take the lead on the docs
 
@Mego oh well... fine...
 
Anonymous
@Sp3000 This challenge would be way harder if there were stationary hands
 
6:59 AM
?
 
Anonymous
Well, I guess the 1-period hands sort of work like that
 
Anonymous
Nvm just rambling
 
Oh, you mean like hands fixed at a certain position around the circle?
 
Anonymous
Yeah
 
Wouldn't be too different
 
Anonymous
7:01 AM
It would invalidate a lot more cases
 
Well what I mean is my code (and probably your/El'endia's code) wouldn't have to change much
 
I have no code for this. :P
 
... algorithm
 
hehe ;)
 
(No code, seriously?)
 
7:03 AM
Seriously.
I just look at the prime factors.
 
Anonymous
@Sp3000 @El'endiaStarman ಠ_ಠ
 
I'd imagine you're looking at groups with prime factors in common as possible overlap groups, but calculating for a large set seems like a pain
 
Oh, that's for sure. But I haven't had to do that. :P
 
Anonymous
5, [128, 848, 780, 326, 627, 730, 543, 236]
 
Anonymous
7:05 AM
Have at it, my solver is still churning
 
Race? :P
 
I'd probably lose...
 
Anonymous
Btw I should use your algorithm in my solver :P
 
7:06 AM
sweeet :P
I just used Wolfram|Alpha. :P
 
... I do believe the solution's one large LCM of 5 numbers
So that'll be why
... oh wait no
 
Anonymous
@El'endiaStarman Your algorithm is just take the exponents of the prime factorization of each number, take the largest exponent for each prime, and then get the product of the first n exponents?
 
That sounds about right.
But you do get to pick-and-choose, so it's not quite that simple.
 
Anonymous
:/
 
I conjecture it's 16312433280.
Hands 1,2,3,5,8.
 
Anonymous
7:12 AM
That is a solution, not sure if it's the minimal solution
 
right, yes
 
Anonymous
Here's another particularly evil one: 9, [860, 680, 660, 669, 112, 323, 285, 259, 613, 793]
 
Anonymous
After seeing these I decided to shrink the test cases to x < 100
 
The problem isn't the size of the x, it's the prime factors in common
 
Anonymous
7:15 AM
And the larger the x, the more prime factors it is likely to have in common with the others
 
Yeah, but that's why I'm asking - why the randomness? :/
 
Anonymous
Just to generate a bunch
 
Anonymous
I want hand-crafted ones, but I also want a decent amount total
 
I still think it'd be better to have more useful test cases rather than just more test cases...
 
Anonymous
7:19 AM
Have any suggestions?
 
What sort of test case do you want?
 
Anonymous
I added in yours from earlier that breaks with floating points
 
Anonymous
I suppose one that tests for proper detection of groups would be good
 
Anonymous
But yours from earlier also does that well :P
 
16312433280  <- old
3785321056   <-new
New estimate!
This time it's hands 1,2,8, and hands 5,7.
Wait, I think I can do better.
 
Anonymous
7:24 AM
I'm running exhaustive search because that's how I roll
 
On checking...I realized that I already checked that (my idea). :P
 
Anonymous
def print_same_line(val):
    sys.stdout.write('\r'+val)
    sys.stdout.flush()
 
Anonymous
One of my favorite pieces of code
 
6, [2, 4, 7, 10, 20, 21, 52, 260] <-- case for groups that shouldn't give float problems
(add a 3 to the above list and you get a 3 group test case)
 
Anonymous
65 for that?
 
7:28 AM
Yup
And 35 for the alternative version
 
Anonymous
Yep
 
Anonymous
Cases so far:
 
Anonymous
I wanna add the massive one that @El'endiaStarman is working on, but I also want to verify the solution first :P
 
I'm actually trying to figure out Sp3000's 65.
 
7:31 AM
I'd try, but I don't want to have to pull out Chinese remainder theorem :/
 
Anonymous
Who needs Chinese remainder theorem when you have force brutes?
 
Ah-ha, figured out the 65.
 
Anonymous
(me. the answer is me.)
 
Anonymous
7, [33, 19, 25, 36, 21, 18, 37] is turning out to be a nasty one
 
Anonymous
2 perfect squares, 2 primes, 4 products of two primes, 4 multiples of 3...
 
7:35 AM
@Mego But you need all 7. That makes it easy.
 
Anonymous
Oh true
 
2^2*3^2*5^2*7*11*19*37=48717900
22 mins ago, by Mego
Here's another particularly evil one: 9, [860, 680, 660, 669, 112, 323, 285, 259, 613, 793]
One of those is a prime (613). LCM the other 9.
 
Easy one for floats, no groups: 3, [171, 1615, 3420]
 
Anonymous
Changing it to 6 out of the 7 makes it harder
 
@Mego That was a response to the "None of us are stupid enough to do that."
 
Anonymous
7:38 AM
@VoteToClose oh
 
@Sp3000 3060?
 
Anonymous
I also get 3060
 
@Mego what does "fastest pythonista in the west" refer to
 
Anonymous
@quintopia Fastest Gun in the West, Python style
 
7:42 AM
Yeah, with meeting at 18/19
 
@Mego 109725, I think.
 
@Mego Not what it means. What it refers to. What prompted your saying it to me.
(it was in the middle of a conversation that did not include me, so i wonder if it was a mistake)
 
Anonymous
@El'endiaStarman Nope
 
You found one lower?
 
Anonymous
@quintopia Link? It's really far back
 
7:44 AM
Can't wait for the day when all computer ports can use the same type of cord (usb c?)
 
I'd try but I just stepped out
 
Anonymous
@El'endiaStarman No, my checker didn't find that one
 
6 hours ago, by Mego
@quintopia Fastest pythonista in the west
 
Anonymous
@quintopia That was a reply to an earlier message you sent about me getting the first Python answer in the odd then even challenge
 
@Mego Huh. It'd be hands [1,5],[2,3],[4,6].
 
7:45 AM
@Mego Ah, got it
 
Anonymous
@El'endiaStarman I got to 506315 before I ctrl-c'd it
 
ahh
Y'know, I think this is actually a really interesting problem in its own right.
 
Anonymous
Hmm?
 
Anonymous
This is what I'm going with as test data, unless anyone objects
 
7:47 AM
How do you know that you have the optimum without doing an exhaustive search?
 
Anonymous
I am doing an exhaustive search
 
I know.
But, I'm saying, that's the interesting problem.
 
Factorise -> generate groups -> for each set of groups adding to n, calculate optimum using CRT
Take min of all those
 
Anonymous
This is a non-trivial extension of the subset-sum problem
 
Anonymous
So I suspect it's also NP-complete
 
Anonymous
7:48 AM
Or at least NP-hard
 
You can write a program that greedily picks the best prime-sharing groups, but there might be a better one because the products are lower if you split them up differently.
 
I was just going to brute force all groups :P
 
Anonymous
Wait this wouldn't even be in NP
 
Well, combinations of groups
 
Anonymous
Well, maybe
 
7:50 AM
@Mego You're thinking it might be in E?
 
Anonymous
Since you have a strict upper and lower bound, at worst, verifying a solution is O(n)
 
The main characteristic of NP is that verifying the solution is polynomial but finding it is exponential (or worse).
Do we even have polynomial verification?
Verifying that it is a solution, certainly. But minimum?
 
Anonymous
If n is a valid syzygy point, then the optimal solution is in [1,n]. At worst, you go through all n integers and verify - O(n)
 
Anonymous
@El'endiaStarman Not necessarily. There are things between exponential and polynomial.
 
Not much.
 
Anonymous
7:52 AM
(I think)
 
Anonymous
But yeah, generally exponential algorithms
 
@Mego But we don't have that n!
 
Anonymous
The worst in NP-hard have factorial algorithms :P
 
Anonymous
@El'endiaStarman If you're verifying a solution, you do have the n.
 
That still doesn't sit right with me.
 
Anonymous
7:53 AM
For example, 9, [860, 680, 660, 669, 112, 323, 285, 259, 613, 793]
 
Why not just use the same algorithm to find n?
 
Anonymous
You came up with a valid n: 3785321056
 
Anonymous
So that acts as an upper bound
 
Anonymous
@El'endiaStarman Hmm... Yeah, this isn't NP.
 
Anonymous
This is a very bad P.
 
7:54 AM
So, then, take all integers in the set and multiply them together. Ta-da, upper bound!
That's not right.
 
Anonymous
That is right. Either you find a solution by then, or the solution doesn't exist.
 
What we really have is k, the number of aligned hands, and m, the upper limit for each element in the set.
 
Anonymous
Really the more interesting thing is to look at the decision problem: Given n and L, decide if a solution exists.
 
No no, wait, think about this:
 
Anonymous
Though that's really not much different :/
 
7:56 AM
Let's say the maximum is 100.
 
Anonymous
For finding a solution or verifying a solution?
 
We start with 3 of them. So the search space is ~100^3.
Now add another one. Search space is now ~100^4. That's exponential. O(m^k).
 
Anonymous
Hmm
 
Anonymous
Yeah
 
Anonymous
You're right
 
Anonymous
7:58 AM
So this is beyond NP - it's in EXP
 
That's the search part.
If the verification is also in EXP (or NP, I think), then the whole thing is in EXP. If verification is in P, the whole problem is in NP.
Roughly. There's probably minor details I'm glossing over.
 
Anonymous
The verification can't be NP, I think, because you can just define the oracle to wrap the other one
 
Yeah, true.
But anyway, our only hope is to find a polynomial verification algorithm. :P
 
Anonymous
So really, it comes down to: can a solution be verified as optimal in polynomial time?
 
Yep, that.
 
Anonymous
8:01 AM
You're in charge of posting this to CS.SE
 
...d'oh.
 
Anonymous
There might be a simpler way to do that
 
Anonymous
So the verification can be done in exponential time
 
Anonymous
(i.e. it's in EXP at the worst)
 
I think brute force is in EXP.
 
Anonymous
8:04 AM
Yep
 
Anonymous
It's the same problem as finding a solution, just with a lower upper bound
 
interviewer asked me to set my own 1 hour coding challenge
I hate that sort of creative question :/
 
Anonymous
If verification is EXP-complete, it's not in P, therefore the problem isn't in NP, therefore it's EXP.
 
spent all day thinking, all i came up with is reimplementing logrotate
or hunt the wumpus, although I'd have to add features for it to take an hour
 
Also, I just realized that just because the search space is exponential doesn't require that the search algorithm be exponential. If you can pick all possible/likely solutions in polynomial time and verification is polynomial, the whole thing is in P. That's the deal with primality checking, I think. If verification is exponential, the whole thing is exponential.
@Sparr Black and White Rainbows, Euclidean algorithm? :D
I mean, that took me at least an hour to figure out.
 
Anonymous
8:09 AM
I dunno
 
Anonymous
This is making my head hurt
 
Anonymous
-.-
 
heh heh
 
Anonymous
I should really go to bed - I'll post the challenge to main in the morning if there are no more issues with it
 
The chat's always dying when I wake up. -.-
@Sparr Make gamma without gamma?
 
8:21 AM
Howdy
 
Howdah.
 
8:43 AM
I have a challenge proposal that I may write up in the sandbox tomorrow if you guys like it.
Basically: "write a Purple interpreter" as golf. esolangs.org/wiki/Purple
 
Hmm. Looks good, but keep in mind that "Write an interpreter for <insert esolang name here>" may be marked as duplicate.
 
why so?
 
It's been done many a time, and people may not consider it different enough.
 
you would think it would be different every time
golfing interpreters is always fun :P
so do you think it's different enough, or shall I just abandon the idea?
 
I think it's different enough. I always up vote stuff like this - it's a question of whether others will consider it different enough. Post it up - this is what the sandbox is for.
 
8:54 AM
ok. sleep now tho
 
Testing my new autocorrect features: Vitsy
Yuuus.
Automatically converts Vitsy to the markdown link Vitsy
 
9:13 AM
@Sp3000: I actually seriously (ha!) leveraged the 2D nature of Minkolang here! :D
0
A: Who is the sleepiest of them all?

El'endia StarmanMinkolang 0.12, 122 bytes At first, I tried doing this short and really golfy. I gave up and went for more "fun". >2@fv$oI2:[9[i$z3[iciz1+q=]++3=tt"^"3zpt]$x$x]IX3140w o.o1F o.=1$ =.o1+ =.=12 o.-1: -.o11 =.-1+ -.=13 -.-1[ /1id< \gqO]. Try it here! Explanation (tomorrow) But really, click on...

Okay, I'm off to bed! Time to become -.-...
 
I was about to say "yay!" but then I saw 'I gave up and went for more "fun".'
 
It's still relatively golfy...
3 hours ago, by El'endia Starman
Up to 79 bytes with a solution that doesn't yet work. ಠ_ಠ
The problems I had with shorter solutions: 1) which face within a group was unknown, 2) the ordering was wrong, and/or 3) the eyes weren't in the right order.
I'm pretty pleased with my current solution, honestly.
 
I can't compare stacks. :c I'll need to add that.
 
0
Q: Create the shortest BitShift

BasIntroduction BitShift is an esolang, created by me, to output strings. Sounds boring, is boring. However, the only instructions available to BitShift are 0 and 1. Therefor it can be challenging to golf in it. To make matters even more interesting, the language only supports 4 bitshifting instruc...

 
9:34 AM
stumbled upon a (security related) code golf usage:
Tracker Infection: Limitations
Max 17 bytes. Is that enough?
Yes:
Crash Pentium Trojan (2004): 4 bytes
Mini DOS virus (1991): 13 bytes
Not enough for an advanced botnet though ;)
 
9:58 AM
Do someone know when I can find documentation for "Seriously"?
 
I do believe there's a command list here: github.com/Mego/Seriously/blob/master/commands.txt
 
Thanks
 
10:17 AM
Why does we choose to use SE?
 
o-o Wut?
 
Updated my sandbox challenge
 
10:49 AM
I think SE is a bad choice for our community. We need many accepted answer for different language in a challenge. We need to sort answer based on length of code. etc. etc.
 
11:01 AM
I am thinking of a golfing language based on Haskell.
 
11:27 AM
@ChristianIrwan we have a stack snippet which does something similar, it is often added to questions with many answers
 
12:19 PM
I think I asked this before, but is there a non-vulgar term for "dick move"? context
 
Aug 29 at 16:48, by Martin Büttner
I've been wondering for a while... is there a non-vulgar but similarly catchy equivalent of "dick move"?
Hmm I can't remember what the result was tbh
 
You're probably not going to find something as catchy and non-vulgar... :P
 
And I do remember this not helping
Calling them a jerk is probably the closest thing I can think of, but that doesn't really refer to the move itself
 
he deleted it
 
1:19 PM
Only one kind of Tabs are good - browser tabs.
 
Browser tabs are generally eight spaces wide, which is not good.
 
@Geobits generally they are more than 20 spaces wide
 
browser tabs != browser tabs
 
> false
 
@MartinBüttner Ok, I will test it (and most likely find a few bugs in my interpreter) as soon as I have time!
 
1:37 PM
I have an idea for a BF derivative. Instead of a 1D array, memory will be an infinitely-dimensional array.
 
With BigIntegers?
 
Maybe I'm being naive here, but how do you access, say, the first entry in an infinitely-dimensional array? I imagine something like [0][0][0][0]....., but obviously that won't work.
Or are we just traversing, with no random access?
 
0
Q: How to change color in data-grid view cell base in condition

EllaHi Guys can i ask question ? i hope someone can answer me. How to change color in data-grid view cell base in condition ? Using C# this codes that im using its always error with this codes int val = Convert.ToInt16(dataGridView1.Rows[i].Cells[2].Value.ToString()); RowsColor(); } publ...

 
@NewMainPosts That doesn't answer my question at all.
 
1:52 PM
LOL!
 

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