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12:00 AM
I frequent Imgur and I have seen that meme only a few times.
@ColdGolf I see you go to the school of Twisting Peoples' Words To Mean Something Else.
 
ignoring a user is sure a lot less effective when i still see people's responses to them
8
 
@ColdGolf only the stupid parts of reddit actually use that meme.
 
it just makes everyone look loony
 
@xnor Haha, I'll bet. There was someone who mentioned that yesterday...
 
Anonymous
12:01 AM
@El'endiaStarman Make it work, make it right, make it fast.
2
 
@EᴀsᴛᴇʀʟʏIʀᴋ Everytime I use a meme that SE doesn't know yet, I get flagged, and then probably kicked.
 
@ColdGolf no
it's not the new memes
 
@El'endiaStarman Aha, this.
 
it's you being obnoxious
 
12:02 AM
@EᴀsᴛᴇʀʟʏIʀᴋ Even the old memes, SE doesn't know them.
 
So I was bored and I found this: cafepress.co.uk/…
 
@EᴀsᴛᴇʀʟʏIʀᴋ Yep, this.
 
@ColdGolf no, we do
we don't use them though
 
You do , not we do.

Some people on SE just have not heard those memes.
 
12:03 AM
@Downgoat Found it, or made the design yourself? :P
 
everybody else doesn't use them because at the heart, SE is a learning site
 
@El'endiaStarman found it o_o
 
not r/circlejerk
@ColdGolf okay seirously stop editing that please
 
Anonymous
Also yes, no errors is clearly best, so Actually doesn't do anything if you tell it to do something dumb. It doesn't even tell you that you did something dumb.
 
@ColdGolf yes, and that's fine. Most of us have though, and jsut don't repeat it.
 
Anonymous
12:04 AM
(unless you pass the -d flag)
 
Most of you, from my experience, haven't.

You personally have, that's different.
 
@Mego actually "I'm"--dumb?
 
> from my experience
@xnor lel
 
12:04 AM
@EᴀsᴛᴇʀʟʏIʀᴋ A bad experience at that.
 
@ColdGolf i;long double a,b;main(){for(;;i++)for(;;b+=1e-18)for(;;a+=1e-18)if(pow(a,b)==i)printf("%LF ^ %LF == %d\n",a,b,i);} btw, it will take a while to finish.
 
@ColdGolf TL;DR: Those aren't welcome here. We are kindly asking you to stop.
 
@EᴀsᴛᴇʀʟʏIʀᴋ You mean aren't known, not just aren't welcome. I've stopped using them just because, while millions others know them, you guys don't.
@ReleasingHeliumNuclei How long is a while in this case?
 
@ReleasingHeliumNuclei -10/10 all of them are rational
 
12:08 AM
@ColdGolf print(sqrt(9)^(log(9,4)))
 
@ColdGolf Do we have to take an actual drill to your skull to get you to understand that we do know the memes? (Most of us anyway.) They aren't welcome when out of place and unconnected to any other conversation going on, or when posted repeatedly, or when in disregard of vulgarity, or when ignoring the requests of others to stop, etc...
 
@LeakyNun close enough :p
 
Julia, 27 bytes: print((2^.2)^(log(10,5)))
fixed
 
@ColdGolf up to 2147483647, increments 0.000000000000000001 at a time
 
@EᴀsᴛᴇʀʟʏIʀᴋ -10/10 sqrt(9) is 3
 
12:09 AM
...
 
@ReleasingHeliumNuclei Exceeds ideone's time limit.
 
i'm not fixing that one
@ColdGolf ... use gcc?
 
@EᴀsᴛᴇʀʟʏIʀᴋ you can use sqrt(2) to save 2 bytes
 
@ColdGolf How old are you?
 
12:11 AM
@LeakyNun also 2^.5
 
@EᴀsᴛᴇʀʟʏIʀᴋ I don't know Julia
 
@LeakyNun not without changing the exponent though
 
@Downgoat Seriously? So what?
 
12:12 AM
@EᴀsᴛᴇʀʟʏIʀᴋ which is why I said two bytes
 
Julia, 22 bytes: print((3^.5)^log(3,4))
 
is (√2)^((√2)^(√2)) rational?
 
@ColdGolf can I do a lambda with a dummy argument?
 
@EᴀsᴛᴇʀʟʏIʀᴋ Nope. Has to have a number.
 
12:14 AM
Not a dummy one.
 
@ReleasingHeliumNuclei remove last sqrt and yes
 
@ColdGolf i mean it takes a number as input, but doesn't use it
 
@Downgoat ;_;
 
takes anything as input really
 
@EᴀsᴛᴇʀʟʏIʀᴋ Doesn't use it? Not allowed.
 
12:15 AM
okay
 
This made me actually laugh.
 
@ColdGolf it just has to state whether irrational^irrational can equal rational, correct?
Julia, 8 bytes: print(1)
FOG, 1 byte: 1
 
@EᴀsᴛᴇʀʟʏIʀᴋ Nope.
 
then closed as unclear
 
@EᴀsᴛᴇʀʟʏIʀᴋ "Prove that there exists two positive irrational numbers x and y, such that x^y is rational, in as few bytes as possible."
 
12:17 AM
> prove
 
@EᴀsᴛᴇʀʟʏIʀᴋ cite any one of his CMCs that isn't unclear
 
closed as unclear
@LeakyNun closed as impossible
(can i do that?)
 
@EᴀsᴛᴇʀʟʏIʀᴋ Use the number 2, and display the output.
 
output: 1
which is truthy
 
Where's your code?
Seriously?
 
12:18 AM
1 min ago, by Eᴀsᴛᴇʀʟʏ Iʀᴋ
FOG, 1 byte: 1
 
@EᴀsᴛᴇʀʟʏIʀᴋ That displays the output but does no calculation.
So it doesn't count.
 
you don't need calculation
@ColdGolf then downvoted and closed as unclear
what is it supposed to calculate?
prove != calculate
 
It's not particularly funny anyways.

I'd downvote you if you did it.
 
that's fine
downvotes are fine
 
Hear hear
 
12:19 AM
@Geobits would agree
damnit
 
CMC: chess or go bot
 
VTCd as unclear.
 
Google, 1M bytes: message too long
 
There are full challenges for both of those on main.
 
12:20 AM
@xnor Is it even possible...
 
Mini-epiphany: x = sqrt(2)^x has two solutions: 2 and 4. This means that an infinite tower of sqrt(2)^sqrt(2)^... could be either 2 or 4...?
 
@El'endiaStarman Nope.
 
1+1-1+1-1+1...=0.5, so √2^√2^√2...=3?
 
@El'endiaStarman That limit is the fix-point of f(x) = sqrt(2)^x, starting as x=sqrt(2)
@ReleasingHeliumNuclei the first statement is not true so the second statement is useless
 
@ReleasingHeliumNuclei Somebody already did it, an entire chess program in 1KB of javascript:

for(B=i=y=u=b=i=5-5,x=10,I=[],l=[];B++<304;I[B-1]=B%x?B/x%x<2|B%x<2?7:B/x&4?0:l[i++]="ECDFBDCEAAAAAAAAIIIIIIIIMKLNJLKM@G@TSb~?A6J57IKJT576,+-48HLSUmgukgg OJNMLK IDHGFE".charCodeAt(y++)-64:7);function X(c,h,e,s){c^=8;for(var o,S,C,A,R,T,G,d=e&&X(c,0)>1e4,n,N=-1e8,O=20,K=78-h<<9;++O<99;)if((o=I[T=O])&&(G=o^c)<7){A=G--&2?8:4;C=o-9?l[61+G]:49;do if(!(R=I[T+=l[C]])&&!!G|A<3||(R+1^c)>9&&G|A>2){if(!(R-2&7))return K;n=G|(c?T>29:T<91)?o:6^c;S=(R&&l[R&7|32]*2-h-G)+(n-o?110:!G&&(A<2)+1);if(e>h||1<e&e==h&&S>
 
@ColdGolf -1 needs an AI
 
@ΛεγίωνΜάμμαλϠΨΠʹ It has a basic AI.
 
@LeakyNun grandi's series, 0.5 is true
 
> It is a divergent series, meaning that it lacks a sum in the usual sense. On the other hand, its Cesàro sum is 1/2.
 
12:24 AM
@ReleasingHeliumNuclei
 
The Cesaro sum is not the normal sum
 
In mathematics, the infinite series 1 − 1 + 1 − 1 + ⋯, also written ∑ n = 0 ∞ ( − 1 ) n {\displaystyle \sum _{n=0}^{\infty }(-1)^{n}} is sometimes called Grandi's series, after Italian mathematician, philosopher, and priest Guido Grandi, who gave a memorable treatment of the series in 1703. It is a divergent series, meaning that it lacks a sum in the usual sense...
 
> It should be noted that the term summation can be misleading, as some statements and proofs regarding Cesàro summation can be said to implicate the Eilenberg–Mazur swindle. For example, it is commonly applied to Grandi's series with the conclusion that the sum of that series is 1/2, a result that can readily be disproven.
 
12:25 AM
@ΛεγίωνΜάμμαλϠΨΠʹ Your purposely messed with it?
 
It works for me.
 
@ColdGolf No, that's what I get when running it in the FF console on about:blank
 
@ΛεγίωνΜάμμαλϠΨΠʹ Weird, works for me.
 
@ColdGolf You sure it wasn't copied wrong?
Where's the original source?
 
12:27 AM
@ΛεγίωνΜάμμαλϠΨΠʹ js1k.com/2010-first/details/750
It works for me on Firefox 48.
 
@ColdGolf Works when I copy from there
 
@ΛεγίωνΜάμμαλϠΨΠʹ Maybe SE messed up the formatting.
 
!RemindMe 8 hours
Oops this is not reddit
2
 
code blocks exist for a reason,
you \`know\\\`
 
@ΛεγίωνΜάμμαλϠΨΠʹ You know chat markdown sucks in general, though.
@ReleasingHeliumNuclei LOL
 
12:32 AM
@ColdGolf ;_;
 
@ReleasingHeliumNuclei Did you know reddit has a copycapy subreddit of ppcg?
 
@ColdGolf which one
 
r/tinycode existed before us, iirc
 
Serious mini-challenge: Does T(n) = T(m) - T(n) with 0 < n < m (integers) have any solutions where T(n) is the nth triangular number (1, 3, 6, 10....). e.g. n = 4, m = 6 is close: T(4) = 10, T(6) = 21, T(6) - T(4) = 11. 11 ~ 10
 
@EᴀsᴛᴇʀʟʏIʀᴋ So you're the copycats?
 
Also, the subreddit I was talking about was different.
 
they were made inpendently
@ColdGolf oh, which one?
 
@EᴀsᴛᴇʀʟʏIʀᴋ reddit.com/r/programmingchallenges
It's not this one either.
 
@HelkaHomba you mean T(m) = 2 * T(n)?
 
12:35 AM
Wait, lemme google it.
 
@ReleasingHeliumNuclei Yes. Same equation.
 
whoa, my caret pathefinder script still works through ignored messages
 
woah
wait relink me?
 
12:37 AM
@HelkaHomba n = 2, m = 3 works
 
Aha! Good. But are there other solutions?
 
@HelkaHomba n = 14, m = 20 works
 
@HelkaHomba You mean, are there any triangular numbers X and Y where X = 2*Y?
Aw shoot ninja'd.
 
you're a little late to the party :P
 
12:39 AM
2n(n+1) = m(m+1)
2n^2 + 2n - (m^2+m) = 0
determinant = 4+8m+8m^2, so 2m^2+2m+1 = m^2+(m+1)^2 must be a square
 
Well, at least I can contribute further! We're looking for n(n+1) = m(m+1)/2.
 
so it is equivalent to finding pythagorean triples where the shorter legs are adjacent
 
........really now.
I give up. Anything further I add is just going to be ninja'd. :P
 
@LeakyNun Are you talking about squared triangular numbers? (oeis.org/A001110)
 
@ColdGolf No
 
12:40 AM
@LeakyNun Then?
 
3,4,5 (first solution:2,3)
20,21,29 (second solution:14,20)
Just as @ΛεγίωνΜάμμαλϠΨΠʹ found
 
@LeakyNun dandy :)
 
n = 2, m = 3
n = 14, m = 20
n = 84, m = 119
n = 492, m = 696
n = 2870, m = 4059
n = 16730, m = 23660
 
there are 4 equation systems that give ns and ms
 
n = 97512, m = 137903
 
12:41 AM
(If I still had homework I swear I could get you guys to do it all :P)
 
brb finding a piece of paper
 
@HelkaHomba good idea. i will use some months later :p
 
@HelkaHomba Your teachers might get suspicious of the golfed code though...
 
A serious mini-challenge (not code): Prove that there are infinitely many triangular numbers that are the sum of two other triangular numbers.
 
CMC: solve 3x^2 - 4x + 5 = 0.2x^5 - x(x - 4)^4 for x
 
12:44 AM
@ConorO'Brien Use desmos
 
@LeakyNun (1) I don't think desmos solves (2) shhh it's math
 
@ΛεγίωνΜάμμαλϠΨΠʹ 2,3,14,20,84,119 not on oeis strangely
 
@HelkaHomba
2m^2+2m+1-a^2=0
det = 4-8(1-a^2) = 8a^2-4, so 2a^2-1 must be a square
@HelkaHomba try searching for just n or just m
 
@ConorO'Brien WA?
 
@HelkaHomba oeis.org/A053141 for n
 
12:45 AM
@Dennis can you update cheddar on TIO? (npm install -g cheddar-lang)
 
@ConorO'Brien with WA or paper or code
 
@ConorO'Brien missing the 3
 
> for n
 
@ReleasingHeliumNuclei his teacher probably intended paper when she gave it to him, so use that
 
@HelkaHomba It becomes Pell's equation:
Pell's equation (also called the Pell–Fermat equation) is any Diophantine equation of the form x 2 − n y 2 = 1 {\displaystyle x^{2}-ny^{2}=1\,} where n is a given positive nonsquare integer and integer solutions are sought for x and y. In Cartesian coordinates, the equation has the form of a hyperbola; solutions occur wherever the curve passes through a point whose x and y coordinates are both integers...
 
12:46 AM
@ReleasingHeliumNuclei idk :P bad joke
 
> Solution to b(b+1) = 2a(a+1) in natural numbers including 0
 
@ConorO'Brien oops sorry
 
> Prove that there are infinitely many triangular numbers that are the sum of two other triangular numbers.
 
@HelkaHomba and oeis.org/A001652 for m
 
12:46 AM
@ColdGolf A053141 doesn't end
 
so it's zip(A053141, A001652)
 
@ΛεγίωνΜάμμαλϠΨΠʹ A053141 isn't what my challenge talks about, is it?
 
@ΛεγίωνΜάμμαλϠΨΠʹ I don't understand how an OEIS link is proof.
 
@ColdGolf Why, it gives infinitely many closed-form solutions
 
@ColdGolf because the OEIS link contains a method to generate an infinite amount of solutions
 
@LeakyNun How is that the case?
 
> a(0)=0, a(1)=2 then a(n) = a(n-2) + 2*sqrt(8*a(n-1)^2 + 8*a(n-1) + 1).
 
Ohh.
How I'd prove it, though, in latex: $t_{n(n+3)/2 + 1}=t_{n+1}+t_{n(n+3)/2}$
 
@ColdGolf put $ around it
chatjax will work, i know @El'endiaStarman also has it
 
12:51 AM
@EᴀsᴛᴇʀʟʏIʀᴋ $ doesn't work.
 
I don't have chatjax, but then again, I don't see hsi stuff :P
 
yeah, you need an extension for it
@ConorO'Brien lel
 
@ColdGolf You have to install the Chrome extension "SE ChatJax".
 
Halp doesn't work on mobile
 
I'm on Firefox.
Is there a Firefox version of this extension?
 
12:52 AM
@HelkaHomba TL;DR: Yes.
 
@ColdGolf Perhaps. I don't know.
 
@El'endiaStarman Thanks anyways.
 
Where {x} is a non-integer part of x.

Prove the above image.
 
@SimonForsberg I actually returned and made a tiny levelset (under yet another alias) some time ago. ^^
 
12:54 AM
@ColdGolf WOAH dude calm down
 
@ColdGolf math? ;_; y u do dis
 
^^ warning, clickbait
 
@ColdGolf I'll just do k/n <= {x} < (k+1)/n:
 
> Ice.

So this one had to be cooked for various reasons. I boiled the rice and the ice, and I was left with rice.
10/10
 
12:55 AM
@EᴀsᴛᴇʀʟʏIʀᴋ most...2nd or 3rd upvoted?
 
@EᴀsᴛᴇʀʟʏIʀᴋ oatmeal >_>_>_>_>_>_
 
orange juice
 
@ColdGolf since we know floor(x)+k/n <= x < floor(x)+(k+1)/n
 
@ReleasingHeliumNuclei 3rd
the best part is that he tried them all
 
@ColdGolf So we have n floor(x) + k <= nx < n floor(x) + k + 1
 
12:56 AM
except beer + rice
 
which means floor(nx) = n floor(x) + k
 
@EᴀsᴛᴇʀʟʏIʀᴋ orange juice is not food
 
Neither is kimchi, but some people consider it to be.
 
@LeakyNun This is the proof.
Sorry for the bad latex formatting.
 
12:59 AM
@ReleasingHeliumNuclei does this have force awakening spoilers?
 
@Downgoat only the last sentence
 
@ColdGolf you should install chatjax (its latex for SE caht)
 

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