The Nineteenth Byte

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

Dennis

@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

Dennis

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

TanMath

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

gist.github.com/Mego/5eba045109c0edf69f5c

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

TanMath

@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

TanMath

@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

o.O

Race? :P

I'd probably lose...

Anonymous

Btw I should use your algorithm in my solver :P

And people say nobody uses golfing languages for useful stuff

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

hehehe

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

Unwinding a spiral makes a spiral o.o

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

gist.github.com/Mego/5eba045109c0edf69f5c

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?

Side note: this is a thing that was published over here recently.

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

gist.github.com/Mego/5eba045109c0edf69f5c

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

Sparr

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.

Sparr

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?

Tensibai

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

A: Who is the sleepiest of them all?

9:13 AM

@Sp3000: I actually seriously (ha!) leveraged the 2D nature of Minkolang here! :D

0

Minkolang 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.

Q: Create the shortest BitShift

I can't compare stacks. :c I'll need to add that.

New Main Posts

0

Introduction 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...

code-challenge

randomra

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?

Tensibai

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.

randomra

11:27 AM

@ChristianIrwan we have a stack snippet which does something similar, it is often added to questions with many answers

e.g. Hello World question

Martin Büttner

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

Martin Büttner

see?

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

Martin Büttner

he deleted it

Optimizer

1:19 PM

Only one kind of Tabs are good - browser tabs.

Browser tabs are generally eight spaces wide, which is not good.

Optimizer

@Geobits generally they are more than 20 spaces wide

browser tabs != browser tabs

Optimizer

> false

Stefnotch

@MartinBüttner Ok, I will test it (and most likely find a few bugs in my interpreter) as soon as I have time!

PhiNotPi

1:37 PM

I have an idea for a BF derivative. Instead of a 1D array, memory will be an infinitely-dimensional array.

Stefnotch

With BigIntegers?

Q: How to change color in data-grid view cell base in condition

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?

New Main Posts

0

Hi 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...

code-golf code-challenge programming-puzzle c#