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9:03 AM
@JonasTeuwen It's a trefoil knot, with a few... decorations.
 
Hmm. Do you have your house fill of these images? 8-).
I bet it would get Stephen approval. A new kind of science.
 
@JonasTeuwen I think all the artwork I've done so far occupy ~ 4 GB of space on my hard drive. :)
 
@JM that's a lotta space for artwork!
 
@J.M. The... generating files?
 
@JonasTeuwen Yes, code and pictures...
I have been doing this for years, you see. :)
 
9:09 AM
@J.M. Oh yes :-). 4GB is not so much it is, but I would just store the text files.
 
Also, not all of my artwork has shown up as avatars.
@JonasTeuwen Well, yes, but I like looking at my gallery from time to time. :)
 
8-).
Print it! :P. Paste it to the wall. Might be a bit to... uh stimulating perhaps.
 
@JonasTeuwen Ink is expensive here, so unless I get funding, it'd be a while before I make posters out of my stuff.
 
So, in the good old wall and ceiling staring sessions you will see plenty of them.
Ink is expensive everywhere, but not to the extend that it would be too costly to do such a thing. But... yes.
Well, I think: "more expensive than gold" qualifies as "expensive".
 
@Nimza: hi, never
well, maybe one
but then I thought it's not that good idea
depends on your personal relationships
I need to talk to you, by the way - but seems like I lost your e-mail
could you send me some e-mail just to remind of your address?
 
9:14 AM
@Ilya wait, I send
 
@jay the organization we discussed earlier won't work, either. :(
Cuz of the nature of the equation, grouping them by mass will cause extensive messaging
... and now it's time for me to sleep on this problem.
 
@Ilya received?
 
@Nimza: yes, added you in gtalk
will you join?
 
@Ilya hah, this year is so strange. Monday lessons finish at 21-15
 
woahuet'
ppz
 
9:22 AM
I wrote you in gtalk, no reaction
 
mmm
 
ah... sorry
 
I can't see you there
?
 
I have a different mail for gtalk. чтобы случайно научнику или другим офиц лицам не написаь
 
@Nimza :)
 
9:25 AM
sent)
 
added
 
Crazy signs again.
 
От жеж ептить
 
жеж < whoah!
It must be drunk Russian with broken nose-smiley.
 
9:43 AM
@Jeff Do not worry too much about communication. It is dirt-cheap. Ok not dirt cheap, but it is very cheap. We did a basic gaussian quadrature benchmarking, where you have basically n weighted partitions which you have to sum up. And what we did was divided 1million partitions over 100 processors and compared with a million partitions over single processor and the result was consistently > 99.98 times faster. So less than 0.02% percent slowdown, so do not worry too much about it.
Finite element methods on the other hand do slow down and saturate over ~32 network nodes due to the large amount of data that is transferred there.
 
Data... evil. Serpens nequissimus.
 
Is that latin?
What the hell does it mean?
 
The most wicked serpent. Yea. Hell indeed! Terror! Torment!
 
Haha. Why do you say so?
 
I don't know.
 
9:48 AM
Google translate says "nequissimus " translates to worthless.
@JonasTeuwen To tell the truth HPC is tormenting.
 
@JayeshBadwaik Hmm. Is my Latin so rusty. That was one of the old names for our beloved Beelzebub.
HPC? Remind me.
 
High Performance Computing
 
Oh.
Then do something else, unless you like being tormented - I don't. I go out. Bye.
 
Its rewarding too. Many many insights.
Anyway, Bye. :-)
Its not ripping off methinks
 
@JonasTeuwen Hey
 
10:06 AM
@Matt Ohh, now I see what you mean.
 
@JonasTeuwen
 
10:42 AM
@JonasTeuwen I have a quick question
Suppose I know that $$\lim_{N \rightarrow \infty} \int_0^{Nx} \frac{\sin\theta}{\theta} d\theta$$ is $\pi/2$. Does this mean that my sequence of functions $f_N(\theta) = \int_0^{Nx} \frac{\sin\theta}{\theta} d\theta$ converges uniformly to $\pi/2$?
 
user19161
@Matt Not sure why that guy wants to do this, it is just wasting space on the internet.
 
StackExchange is a CC-BY-SA so it is not exactly plagiarizing. But it can be pulled up on the the grounds that it does not license its own site CC-BY-SA. (Unless it does, though I did not see a specific license page anywhere.)

Apart from that, it is pretty harmless in the sense that it is neither a better system nor is it advertising in a better manner.
 
@J.M. Thanks, yes, exactly that.
As for the rip-off site: it is a piece of shite and it's needless garbage lying around on the internet. Sites like the one I linked to have no right to exist.
Now I need JM's help since we don't want SE to link to garbage like that. Going to flag my previous comment for deletion.
@WillHunting Yes, I fully agree.
See you all later.
 
11:34 AM
Hi folks
 
hi
@OldJohn How did the party go?
 
@skullpatrol Great, thanks - just very tired after it all!
but today is a public holiday in the UK - and it is raining, so an easy day :)
 
@OldJohn I'm glad to hear it went well, enjoy your well deserved rest :)
 
@skullpatrol thanks! - more time for maths now :)
 
12:02 PM
@BenjaLim: you've deleted your main account?
 
Hello everyone!!
Hi @robjohn could you explain a bit about your answer on this question
 
@MonkeyD.Luffy what part do you find confusing?
 
this part $$ \frac14\int_{\gamma^+}\frac{1-e^{iz}}{z^2(z^2+1)}\mathrm{d}z
+\frac14\int_{\gamma^-}\frac{1-e^{-iz}}{z^2(z^2+1)}\mathrm{d}z\tag{2} $$
we have removable singularity at z=0
how did you add residue at z=i and z=0 along the positive curve
and subtract along negative curve?
 
The right side of $(1)$ is the whole line. Then we break up $1-\cos(z)=\frac12\left(1-e^{iz}+1-e^{-iz}\right)$
Then I closed the contours using curves along which the integral are $0$
The idea is to use $e^{iz}$ in the upper half plane, where it vanishes, and $e^{-iz}$ in the lower half plane, where it vanishes.
Yet retaining the decay at $z=0$
 
12:17 PM
this latex is not working ... let me log in through appropirate browser
 
@MonkeyD.Luffy have you installed the bookmark?
 
don't know how to do that on safari
 
@MonkeyD.Luffy Is there a bookmark bar?
where you can drag bookmarks?
 
logged in for ffx ... can see latex now :D
 
okay
Before we broke the line integral up, since $\frac{1-\cos(z)}{z^2}$ has a removable singularity at $z=0$, I moved the contour so that it didn't contain $z=0$
That is okay since the integrand vanishes at $z=\infty$ and the movement was a finite amount and the region enclosed contains no singularities.
 
so ... you moved down the line??
0,0 seems to be a bit above the line ...
 
@MonkeyD.Luffy Yes. The difference between the two integrals is the two pieces connecting the ends near $z=\infty$ (since there are no singularities contained in that long rectangle)
 
it seems that ... the residue of z=0 at both contour is same
 
@MonkeyD.Luffy only one contour circles $z=0$
 
no matter which curves contain z=0 ... would it yield up the same answer?
 
12:31 PM
The upper contour is counter-clockwise (so we add the residues it contains) while the lower contour is clockwise (so we subtract the residues it contains)
@MonkeyD.Luffy yes, whether I moved the line integral up or down, the answer would come out the same
@MonkeyD.Luffy I think the residue of the one function is the negative of the residue of the other function at $z=0$
 
so we are actually integrating along $y=1/2 i$ line instead of $y=0$ right??
 
@robjohn No
 
@MonkeyD.Luffy yeah, something like that. $\frac{i}{2}$ is as good as any
 
@BenjaLim Ah, so that still has not been resolved? Ick!
 
12:38 PM
Yeah I am now BenjaLim III
 
won't that be different than integrating along the real line?
 
@BenjaLim but not in chat.
 
hahahahahahah
 
@MonkeyD.Luffy No. That is what I was trying to point out earlier...
 
Where is tb
 
12:42 PM
@MonkeyD.Luffy consider the long thin rectangle whose top contour is left to right along the real line, whose bottom contour is right to left along the line $y=-1/2$ and whose left and right ends are arbitrarily far out near $+\infty$ and $-\infty$
@BenjaLim I haven't seen him in a while.
 
@WillHunting Jasp, is that you?
 
hmm ... amazing concept. @robjohn thank you for your time. I think i get it partially ...
will try to do it myself again ...
 
@MonkeyD.Luffy Let me know if you need more...
 
sure ... :D
 
@robjohn How was the zoo?
 
12:49 PM
@skullpatrol We had a very nice trip. We stayed a lot longer than I had anticipated.
@skullpatrol There are four new major exhibits (or renovations) since we were there last.
The gorillas and chimps each have nicer, new enclosures, and there are completely new reptile and Asian elephant exhibits.
 
@robjohn Glad to hear you had an interesting time :)
 
@robjohn
Can I ask you?
I want to evaluate $\int_0^\infty \sin x/ x dx$
I know it's $\pi/2$
the hint given in stein and shakarchi is to use the riemann lebesgue lemma
Now I an see that $1/\sin(\theta/2) - 2/\theta$ is continuous on $[-\pi,\pi$
If I multiply through the numerator by $\sin((N+1/2) x)$
I get the dirichlet kernel appearing
 
@BenjaLim let me think a bit...
 
Hey @rob!
 
@robjohn I think I got it
 
1:04 PM
@J.M. Hey there! It looks as if you used a lot of Mercedes Benzes to make that gravatar...
 
@robjohn By Riemann Lebesgue,
 
@BenjaLim okay... :-)
 
We know that $$\int_{-\pi}^\pi \sin( (N+1/2)\theta) ( 1/\sin(\theta/2) - 2/\theta) d\theta \rightarrow 0$$
as $N \rightarrow \infty$
so this would mean that as $N \rightarrow \infty$
$$\int_{-\pi}^\pi 2 \sin ((N+1/2) \theta)/\theta d \theta \rightarrow 2\pi$$
 
@robjohn Yeah, they look quite nice on the trefoil... :)
 
since we know that the integral of the dirichlet kernel from $-\pi$ to $\pi$ is $2\pi$ @robjohn
And so making a change of variables @robjohn
@robjohn you only need integrability
and the function is continuous
so we now have that by making a change of variables
 
1:07 PM
@BenjaLim Sorry, I forgot you're only on $[0,2\pi]$
 
well $[-\pi,\pi]$
but now we have
$$\int_{- (N+1/2)\pi}^{(N+ 1/2)\pi} \sin \theta/ \theta d\theta \rightarrow \pi$$
 
I have to go for a while
 
@robjohn but the function in the integrand is symmetric about the origin
 
@BenjaLim Yes, that does it I think :-)
 
@robjohn and so we have that $$\int_{0}^{ (N+1/2)\pi} \sin \theta/ \theta d \theta \rightarrow \pi/2$$
as $N \rightarrow \infty$
@robjohn what da ya think?
@robjohn I was going to answer this on math.se
 
1:20 PM
@robjohn Hi, congratulations! I see you're a blue.
 
@BenjaLim sorry, I had to leave for a bit..
@Gigili Thanks. It's been a while since I've seen you here.
 
Yes, that's why I didn't notice you being a mod.
 
@Gigili You're back!
 
@GustavoBandeira Heh, seems so!
 
@BenjaLim You are using the integral of the Dirichlet kernel there?
 
1:22 PM
Hello.
 
yes
@robjohn the fact that $$\int_{[-\pi,\pi]} D_N(x) dx = 2\pi$$
 
@BenjaLim yes. Since that is fairly well known, that should fly as an answer.
 
@robjohn yeah
I just discovered it tonight
@robjohn you know
how can I format latex so that the dx is not so bunched up with therest?
 
@BenjaLim I usually use something like \,\mathrm{d}z $${\huge\int} f(z)\,\mathrm{d}z$$
 
that is nice
how can I make the integral sign bigger?
 
1:28 PM
@BenjaLim like that?
 
like?
yeah
wow
@robjohn how?
 
It was awfully slow sometimes not able to connect to stack exchange site and chat for the past3-4 hours for me
Has anyone had this problem
 
1:51 PM
I am on a dialup and I had no problem. So, the sites seems okay.
 
hhh
2:10 PM
Does a linear-programming -problem always have a solution?
err I meant to ask: "Does LP -problem always have a solution or no solution?"
I think the answer is "No" because you can have degerate situation with cycling so it does no terminate but not sure. What do you think?
 
2:47 PM
@hhh It is possible that an LP problem has no solution. consider the two equations
$x+y>1$ $x+y<-1$ and then there is a degenerate solution which you have already described.
 
hhh
@JayeshBadwaik Ofc, check the next statement -- err I meant to ask "..."
 
@hhh What exactly is a "solution"? In my mind a solution is an object that is found to satisfy a given hypothetical property. By the law of the excluded middle, a problem ("find an object with property P") either has a solution or it does not have a solution.
2
 
@BenjaLim Sorry, I had to go there. If you right click on the rendered LaTeX and select "Show Math As > TeX Commands" you should see {\huge\int} f(z)\,\mathrm{d}z
@BenjaLim you could also use {\Large\int} and {\large\int} to get less extreme sizing.
 
@JayeshBadwaik Received. Thanks.
 
3:08 PM
${\huge\int} f(z)\,\mathrm{d}z$
the huge integral chases tiny differential $f(z){\rm d}z$
$f$ seems to run as fast as it can
 
hhh
3:46 PM
@anon good Q. I think you are right in that. Degecrazy is only a computational or more-implementation issue so the problem must still have a solution or no solution.
so the answer is "Yes".
The LP problem always has a solution or no solution according to the law of excluded middle.
@anon Yes, I would also like to know the definition of the "solution" of LP. Can a LP problem have two solutions?
(Is a set a solution? Or many solutions? Investigating my book on this...perhaps def -issue.)
 
A solution is a particular object that satisfies the property. A solution set is the set of all solutions. A general solution is a symbolic expression used to generate or explicitly construct the entirety of the solution set. Sometimes a text will drop "general" and simply use "solution" to mean the general solution, though this should be clear from context.
 
hhh
4:07 PM
But the Q is "Does there exist optimization problems with two optimal solutions?"

I would answer "Yes", for example, $\max |x| s.t. |x|\leq 1$ namely points $(1,1)$ and $(-1,1)$.
But if you define the term "solution" here as a set, then there does not exist optimal solutions but optimal solution. In which case, the answer is "No".

Which one is the correct answer?And why?
 
Solution is a singular noun and does not refer to a collection of things typically, unless one uses it to refer to a general solution (in which case technically we're still only referencing a single thing, an expression).
Hmm, I was going to edit that, not delete it.
In your situation solution should mean exactly one solution and no more. Hence multiple optimal solutions are possible.
 
4:23 PM
Hi @anon
 
hello
 
@MarianoSuárez-Alvarez Can I drop by before the exposition today??
 
4:41 PM
@jayesh are you here? (or someone who knows C who can help?)
 
@Jeff Yes, I am here.
Thought not for much long. Will go to sleep in around half an hour.
 
@jay (or anyone) C question: declaration inside of a function "realtype *z;". Then an assignment inside of the function is "z = data->z;" Is z an array with elements of type realtype? If so, how does the function know the size of the array?
 
the function does not know the size of the array from this operation that you have specified.
the source of such information must be something different.
 
Some detail. In main(), structure data is declared as "UserData data;" and type UserData is defined as typedef struct {
// realtype dx, hdcoef, hacoef;
int npes, my_pe, my_base, local_N;
MPI_Comm comm;
realtype z[NEQ+4];
} *UserData;
 
also data is struct
 
4:45 PM
Do you mean you think I typed the source wrong?
 
the size of z is NEQ+4
 
hold on a sec (otp)
otp=on the phone
 
hhh
Does someone know some Simplex Visualizer?

I feel it hard-learning to read just theoretical book about things such as Naive-Simplex, Revised-SImplex, Full-tableau-Simplex, Dual Simplex, etc-simplex -- about pages 100 in the Bertsimas.
 
@hhh it is hard learning simplex from theoretical textbooks. especially if you haven't read straight through from the beginning (like a novel). But, alas, I cannot help you (I just wanted to sympathize).
@jay but that's the size of data.z. what is the size of the array z in the function?
 
z = data->z;
this operation makes them equivalent
 
hhh
4:51 PM
(Precisely, my book uses notations such as $\bar{c}_j = c_j -\bf{c}^{'}_{B}\bf{B}^{-1}\bf{A}_{j}$ -- but I think I have somehow skipped the definition (p.84).

The vector -sign means here, not vector, but apparently optimal solution. While bolding means a vector and the sign $'$ means something different -- I wish someone else was reading the book

Introduction to linear Optimization, Dimitris Bertsimas, John N. Tsitsiklis)
 
basically, an array is acutally a syntactic pointer
for
int a[10];
basically a is a pointer which is allocated 10 memory locations of the size of int.
 
hhh
(latex bolding -cmd?)
 
@hhh \mathbf{stuff here}
 
hhh
@JayeshBadwaik Thank you.
 
@Jeff followed?
 
4:59 PM
@jay i'm confused. what are you asking me about "followed"
 
@Jeff I am asking if you got the reason why the size of z is equal to the size of data.z?
 
no. i don't. My understanding is that "realtype *z" is just a pointer to a variable of type realtype (which is just a long int). and that pointer is just to a single slot.
unless... i think i just figured it out (after I sent that last message)....
unless it's because you're setting the pointer z to point to the same place that data.z is pointing to! Is that it?
 
Yes, it is that.
And one more thing is this:
http://chat.stackexchange.com/transcript/message/5935745#5935745
http://chat.stackexchange.com/transcript/message/5935753#5935753
 
ahhhh! got it. so the function then is acting on it's data.z in memory - not on its own copy of z. i'm good to go now. TY
 
5:03 PM
@JayeshBadwaik i missed those four lines while i was otp. :D
 
:D I thought you responded to hhh and hence were off the phone :P
Anyway, good.
 
@jay i was still otp when i responded to him. that response didn't take any brainpower (not like understanding C, which takes a lotta brainpower).
 
@Jeff hmm. Anyway good. I will be going to sleep in some time so will be going off now. If you have any difficulties, you have my mail.
Also, grab a book "The C programming language" by Dennis Ritchie and Kerningham
it is a good book and you can use it for reference too.
 
yeah. thanks Jay. you've been helpful
 
Welcome. Good night / good day.
 
hhh
5:06 PM
Moved the reference-request about simplex -methods here.
 
@jay i have it. i'll dig it outta my closet. but sometimes it's helpful to talk to someone even after reading a book. i mean, i knew that pointers get passed around means you're accessing the same memory space. but not having the experience meant i didn't make the connection.
@jay good night!
 
@Jeff Yup, that is true.
 
 
2 hours later…
hhh
6:56 PM
How can vizualise "Reduced cost" in a LP -problem?
 
7:11 PM
At cinema watching batman
2
Was hoping to get shot but no luck so far
3
 
7:24 PM
@Jeff My apologies for not replying to your e-mail yet.
@Matt Good luck with that.
 
@Jonas it's cool. the email was entirely for you because you might like to read it. There's no obligation. :D
 
@Jeff Thanks :-). But I want to comment on it but I am not able to because of other things :-).
 
@jonas priorities! I totally understand :D
 
Good :-).
 
@anon yeah - covered in most undergrad introductions :)
 
Mental retardation so profound that even a simple intellectual task is beyond reach.
 
@anon "...Galois action on the etale cohomology..." - kids stuff :P
 
hhh
What does word "nonbasic indices" in LP -problem mean?
(p.83 in the book, perhaps something to do with the matrix $A$: $\min c'x$ so that $Ax\leq b$, $x\gel 0$)
err not \gel but \geq
<--- I am stuck to this word nonbasic.
Nonbasic variable means apparently a variable that is not in the basis but what does the basis mean in the simplex -problem? Does it mean the matrix $A$?
I ask "why" for the "Recall that... statement", I cannot recall what the author is speaking there.
 
user19161
8:31 PM
Yo @rob!
 
user19161
@JonasTeuwen Are you talking about yourself?
 
Yep.
 
hhh
It is trying to deduce the "reduced cost" in LP -problem. Shadow cost is the dual cost. Must read some other book to this...
 
@WillHunting Hmm - do I know you by another name? :)
 
user19161
@OldJohn Yes. Ah, I just feel like changing now and then for fun.
 
8:33 PM
@WillHunting Ah well - keeps us on our toes, I guess :)
 
hhh
"Reduced cost. In linear programming, it is the sum of the direct cost (c_j) of a nonbasic variable (j) and the indirect costs from induced changes in the levels of the basic variables (to satisfy the equations). For a dual price vector, p, the reduced cost vector is is c - pA."

http://glossary.computing.society.informs.org/index.php?page=R.html
<--- now I found a glossary, have to dig this :P
 
user19161
@hhh Have you checked the index?
 
hhh
@WillHunting Not yet "nonbasic indices in LP -problem", it has something to do with the "reduced cost" $$\bar{c}_j = c_j - \bf{c}'_j \bf{B}^{-1}\bf{A}_j$$ where $A_j$ is a shadow price (looked up from the MIT -material):
where $y_i$ is the shadow price that I can understand...
(Standard form usually has the X in text-books i.e. $\min c'x$ so that $Ax\leq y$. In its dual, X and Y get swapped where you switch the primal problem from considering amounts into dual problem that considers valuations)
Now the shadow prices are the $y_i$ terms
Now $A_j$ (Bertsimas notation) corresponds to $y_i$ (MIT -notation), I think.
$c$ is a cost. $\bar{c}$ is an optimal cost -- but what the non-basic index or non-basic cost is, I don't know yet.
1. DUNNO: "direct cost (c_j) of a nonbasic variable (j)"

<--- I don't understand the terms. What are they in the standard form LP -problem?

http://glossary.computing.society.informs.org/index.php?page=R.html
Hey!

Does it relate to the Simplex method -base? If you have N -amount of restrictions, you have N -basic variables. The variables in base gets changed in the procedure

... I cannot still see it totally but it may mean the base that is in the simplex method when executing the algorithm.
 
user19161
8:56 PM
@hhh You might be better off asking this on main, since it seems nobody in chat can give you a good answer now.
 
hhh
"The variables corresponding to the columns of the identity matrix are called basic variables while the remaining variables are called nonbasic or free variables. If the nonbasic variables are assumed to be 0, then the values of the basic variables are easily obtained as entries in b and this solution is a basic feasible solution."
Source here.
Now I understood the terms basic and nonbasic! BUT now I need to understand what it has to do with the "nonbasic indices"...have to remember a lot to dig this phrase.
$d_j=1$ apparently infers an $unit$ -matrix (ones in diagonal, otherwise zero).

But

2.DUNNO: why $d_i=0$ (p.83, Bertsimas)?
The unit-matrix means that you move every term in the element $x$ of the polyhedron by the same amount, we are finding the optimization "feasible direction" $\bf{x}+\theta \bf{d}$
**1. DEF**
"Let $\bf{x}$ be an element of a polyhedron P. A vector $d\in \mathbb R^n$ is said to be a feasible direction at $\bf{x}$, if there exist a positive saclar $\theta$ for which $\bf{x}+\theta \bf{d}\in P$." (p.83, Bertsimas)
Now this is the first time discussion about $d$ in this book and after that it states that odd statement "Recall...$d_i=0$ for all other nonbasic indices $i$. Then

$$\bf{0}=\bf{A}\bf{d}=\sum_{i=1}^n \bf{A}_i d_i=\sum_{i=1}^m \bf{A}_{B(i)}d_{B(i)}+A_j=\bf{B}\bf{d}_B +\bf{A}_j$$


...I drop out here because I cannot understand the premises about the direction $\bf{d}$.
I try to formulate my idea fully here before asking in the main, the next thing book explains is the "reduced cost" but I cannot yet understand this foreshadowing with the "feasible direction".

**Perhaps I could ask**

1. "Explain the 'feasible direction' in the deduction of the 'reduced cost' in the Simplex -method"

2. Explain the nonbasic 'feasible directions' $d_i$ where $i$ is nonbasic in Simplex -method, why $d_i = 0$?

3. Help me to understand the optimality conditions in Simplex method
 
hhh
9:37 PM
**Moved the clarification -question to the main, question

http://math.stackexchange.com/questions/187643/optimality-conditions-in-simplex-method-reduced-cost-and-nonbasic-feasible-inde

It is a bad confusing quesiton -- would be much easier to show with my thumb to someone that I cannot understand that point, help me with it.

Anyway hopefully it is clear.
 
user19161
9:53 PM
It's quiet in here...
 
10:07 PM
SO QUIET... OH SO STILLLL.
 
user19161
10:36 PM
@JonasTeuwen Are you not sleeping yet bro?
 
Sleep cannot.
 
10:48 PM
@JonasTeuwen Good morning. Just got up.
@WillHunting Morning. (Good is implied)
 
@JonasTeuwen Oh, Stille Nacht...
@WillHunting: Your avatar is certainly unblemished by any trace of variation...
 
Take something before you get banned for intolerance and discrimination against odd ones.
 
@JonasTeuwen to me?
 
@JonasTeuwen I pinged you last night so many times
@JonasTeuwen Can I ask you something
 
@robjohn No. Will
@BenjaLim Oh.
Man, I am exhausted, but try.
 
10:58 PM
Ok let me put on my clothes
 
@JonasTeuwen Sleep by over-exhaustion. Must have been a tough day.
 
Let me put off my clothes, so things stay in balance. We don't want to divide by 0.
 
@JonasTeuwen endurance training...
 
@robjohn 8-).
 

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