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1:04 AM
if you wanted to produce an image of a living human brain with enough clarity to discern each and every neuron and synapse without killing or damaging the brain tissue of the person involved, must the required technology be more than simply increasing the numbers in the specifications of an MRI or C-scan?
 
1:16 AM
@0xFFF1 So, the MRI spin flip (T2-limited resolution) is around 10 um. You/d have to do some super resolution witchcraft, maybe like STED to get higher resolution.
10 um is pretty terrible, compared to light microscopes, not to mention FIB-SEM, basically MRI doesn't have enough resolution and will never have enough resolution.
 
same story for Cscans?
 
xrays have really high resolution, although it depends on your belief in the reversibility of fields
 
high frequency EM waves bounce off bone though?
 
Not really
 
hm
 
1:21 AM
xrays go through, mostly
The ultraviolet catastrophe, also called the Rayleigh–Jeans catastrophe, was the prediction of late 19th century/early 20th century classical physics that an ideal black body at thermal equilibrium will emit radiation in all frequency ranges, emitting more energy as the frequency increases. By calculating the total amount of radiated energy (i.e., the sum of emissions in all frequency ranges), it can be shown that a blackbody would release an infinite amount of energy, contradicting the principles of conservation of energy and indicating that a new model for the behaviour of blackbodies was needed...
Problem is the signal is scrambled when it "bounces" around a lot
Also to get the coherence properties of an xray source that will let you do sub micron imaging, you gotta build a cyclotron (or simliar)
 
how many zones of lines-of-sight do you have to peer into a living human cranium while only piercing soft tissue? Through the eye holes and nose hole, and underneath the jaw?
 
You have zero because the brain is covered?
 
i mean there's ways in that dont have to go through bone
looking up cyclotron, looks like it's either so much energy, or too long an exposure to xrays to be safe
 
Its not about the intensity, its about the coherence requirements
 
yeah, but we dont want to kill the person
at least, I dont
 
1:29 AM
You probably don't understand what I'm talking about. Its not about the intensity, its about getting the perfect xray out.
Even then its not clear if you can reverse the media turbidity.
 
coherence is about being able to constrain the trajectory of a particle outgoing an accelerator, to get more information about the image that it produced from it bouncing back?
 
no
The illumination isn't perfect, its a bunch of waves with some spread. This spread reduces interferometric contrast (among other things). If you don't have the right coherence you see something like a low resolution shadow map.
So, when they do xray imaging of cells or semiconductors they hook up a microscope to end of a massive ring that makes the fuck sure the xray field has no spread in frequencies.
If they didn't do that, they wouldn't see anything.
 
tubidity seems like a simple enough concept, a bunch of free particles in a fluid cause em packets to bounce around. This would obviously ruin the image quality
 
So, an image is the interference of a scattered and transmitted field. Turbidity in a sample ruins the ability of these field to interfere while also producing a third, "non-interferometric" signal.
 
is that unwanted third part of the signal mostly random?
 
1:44 AM
That's a loaded question. Some people (perhaps me) believe that the mechanism that make it introduce true randomness and the signal is gone forever. Its also typically quite intense compared to the remaining interference portion (which vanishes all the way as you make your sample thicker).
 
i suppose I should use the word random here in its common usage (variance that is beyond our control to calculate or modify). I think actual randomness doesn't exist. (IE the copenhagen interpretation is bullshit and we should instead use pilot-wave)
well, an image of an entire brain would be pretty thick
so there isn't hope of being able to resolve such an image? or just not with xrays?
 
So, if you use an xray you can get more depth penetration and the right resolution, the question is if the turbidity can be resolved.
 
would taking the image multiple times under the same parameters help?
 
no
 
then comparing
 
1:53 AM
The loss of coherence (in this case) comes from the variable path the photon will take moving through the sample.
 
 
3 hours later…
5:08 AM
Physics all night long
Taking this stuff to the next level
 
 
2 hours later…
7:24 AM
@JohnRennie Hii
I have a question
 
I'm working at the moment I'm afraid. Back in an hour or so.
 
Why in C program we sometimes use return 0; or return 1; in the code ?
@JohnRennie np
reply me when you are free .
 
@Koolman So that a program calling this program knows that it ended correctly
 
Example is given in above link of C program
 
Your program may be called by another program. Generally, it's considered that if it returns 0, it has ended correctly
If it returns another value, that value is an error code
For which hopefully you have documentation
 
7:27 AM
@Slereah Where it return 0 ?
 
yes
 
@Slereah What do you mean by documentation ?
I am a beginner .
 
Well in the best case scenario, the program should have a document indicating what the return values mean
Although for simple programs, such as the ones you're doing as a beginner, there's not much need to worry and you can just return 0
 
@Slereah How could we know it will written 0 or any other integer
But in this code they have used both the integers
 
Anonymous
@Koolman You can obviously check in the console output...
 
7:29 AM
At the level of the OS, a C program works somewhat like a function
You can call it with arguments
and it returns a numeric value
which you can check if you know what you're doing
 
Anonymous
Or yeah, just call it from another function
 
@Blue Does on console output there it is written 0 or 1 ?
 
Anonymous
Should be written, yes
 
That's why the main program is written like a function
int main (int argc, char *argv[])
although technically it's not mandatory
The main function can just return nothing
It depends what the purpose of the program is
In what kind of environment it has to run
 
Anonymous
@Koolman Why are you learning C tho? Part of syllabus?
 
7:33 AM
@Blue Actually I have almost completed C but never thought of it .
Now I am starting with JAVA
 
Anonymous
What does "completing C" even mean? :P
 
Anonymous
Anyhow, I vouch for Python ;)
 
Sid
That ^
 
Python isn't gr8
No typing
it can be a bit of a mess
 
Sid
7:35 AM
@Slereah Python has an easy learning curve.
 
It comes at a grave price!
 
@Slereah Also we sometimes use void main() and int main()
 
yes
void is when you don't require your function to output anything
So that either your program doesn't get called by another program, or if it does, they can't know how the execution went
 
Anonymous
@Slereah It isn't strongly typed, true. But I really haven't faced many issues with it. In which type of cases does it cause a problem (other than it's difficult to keep track of which variable is of which type)?
 
Thanks
Now I understood it
 
Sid
7:39 AM
@Blue That's one of the primary issue with it. But, it's not a dealbreaker for me.
 
Anonymous
Additionally jupyter notebooks, numpy and pandas are the best thing ever :)
 
@Blue you just said it :p
Real men program in Ada!
that shit won't compile if there's any type error
 
Anonymous
Anyhow, I sort of got bored of picking up new programming languages now :P I'll stick to C and Python till they become the new COBOLs
 
Anonymous
@Slereah lol...nice
 
Sid
@Blue I don't have any liking to C. I will just do Python and R.
 
Anonymous
7:46 AM
The good thing about C and C++ is that almost every programming problem you'll ever face, is already coded in those two languages by someone, somewhere :P
 
@Blue C has been the new COBOL for 50 years
 
Sid
I was just seeing my subjects for next semester and only a few look particularly interesting.
 
C ain't going away
they're not gonna recode Linux in Java
 
Anonymous
lol
 
The three things that never die are COBOL, FORTRAN and C
and of course the Space Jam website
Although I guess technically Ada also never dies, but it's a much more narrow field of application
Ada is all for like plane hardware and that kind of stuff
 
Anonymous
@Sid Dude, I convinced my professor to learn C and forget FORTRAN :P Give me a medal (not that they have still forgotten their love for FORTRAN)
 
Anonymous
Well, tbh FORTRAN does have some advantages
 
Anonymous
But it has so many different versions that it gets difficult to keep track of
 
Anonymous
"You'll get the sample code and look at it, and it will be completely incomprehensible. They all are. All Fortran codes are incomprehensible."
 
Anonymous
This is soooo true, lol ^
 
8:01 AM
FORTRAN is basically to do math
It's not very friendly for any other application
 
Anonymous
Indeed. When you just need to churn out the numbers...
 
"Actual Fortran legacy code I received once.
All four of the comments are in French.
I dare you to figure out what it means."
I am shocked by this racism against my people
 
Anonymous
The French are (in)famous for many things :P
 
"The convention is (usually) that names beginning with letters A-H or O-Z are implicitly floating point, while names beginning with letters I-N are implicitly integers."
My sides
 
Sid
@Blue Awards you gold medal. :P
There's also this
 
8:11 AM
analogy and catalogy
open(unit=1234, iostat=stat, file=file, status='old')
if (stat == 0) close(1234, status='delete')
 
Anonymous
@ACuriousMind Is your new display picture for checking whether people will still address you as "Sir" on the main site? ;)
 
lol
 
 
1 hour later…
9:25 AM
"Perl is “some assembly required”. Python is “batteries included”. PHP is “kitchen sink, but it’s from Canada and both faucets are labeled C“."
3
"0x0+2 produces 4. The parser considers the 2 as both part of the hex literal and a separate decimal literal, treating this as 0x002 + 2. 0x0+0x2 displays the same problem. Strangely, 0x0 +2 is still 4, but 0x0+ 2 is correctly 2."
Jesus
 
10:32 AM
0
Q: Why do the laws of physics fail to predict the behavior of frustrators?

AllureThis is my attempt to make an earlier question less broad. This question takes the form of a thought experiment, and is based on this video. Suppose you are given: The positions, velocities, wavefunctions, masses, whatever you want to know of every single particle in the experiment Unlimited c...

seed of light should put a limit
 
11:05 AM
0
Q: Can I ask if my calculation is correct?

myradioLet's say I've been working in a calculation which is reasonable simple but I didn't found it done anywhere else. Can I post is in physics.SE and ask what the community think about it? I know in other SE communities that would not be allowed but in physics that might be, because check whether th...

 
rob
11:23 AM
@Slereah That's a funny way to spell "the parser is garbage"
 
 
1 hour later…
12:25 PM
R.I.N.G.
Resentment
Induced
Nonlocal
Genocide
That will make a good thriller
 
If suppose $z = iAB^*$ where z, A, B are all complex. Does $Im(z) = AB^*$ or $Im(AB^*)$ ?
 
The imaginary part function does not distribute
 
Im of z would be imaginary itself?
 
@MohammadAreebSiddiqui it is certainly not the first one
Since $AB^*$ may not even be real
While $\operatorname{Im}$ is always real
 
yeap thats what confusing
so it would be the second one?
 
12:32 PM
nope
Because Im does not distribute over multiplication
This is easy to check by noting Im (w) = (w + w*)/2i
 
i think its (w - w*)/2i
 
Ah right, sign error
 
also how is it distributing? The imaginary part of z is $AB^*$
but that itself could be imagianry
so I cant really say $Im(z)$ is imaginary
or the imaginary part of z is not well defined?
 
Because $Im(iAB*) = (iAB* - (iAB*)*)/2i = (iAB* + iA*B)/2i = 2iAB*/2i=AB*$ ok nvm I rest my case
uh...
 
how is $iAB^* + iA^*B = 2iAB^*$
arent $AB^*$ and $A^*B$ 2 different terms?
 
12:37 PM
ah f888 I keep making typos
$ = i (AB^* + A^*B)/2i = (AB^*+A^*B)/2 = (AB^*+ (AB^*)^* )/2 = Re (AB^*)$
 
so its 0?
wow thats weird
ohhhhh
right
$Im(z) = Re(AB^*)$
wait
 
If $A$ and $B$ are complex then $AB^* = \mathrm{Re}(AB^*) + i \mathrm{Im}(AB^*)$ so that $z = i AB^* \leftrightarrow \mathrm{Re}(z) + i \mathrm{Im}(z) = i [\mathrm{Re}(AB^*) + i \mathrm{Im}(AB^*)] = - \mathrm{Im}(AB^*) + i \mathrm{Re}(AB^*) $ i.e. $\mathrm{Re}(z) = - \mathrm{Im}(AB^*)$ and $\mathrm{Im}(z) = \mathrm{Re}(AB^*) $ no?
 
Just try a simple example
 
wow that's much cleaner
 
that makes sense XDD
 
12:43 PM
for $A = i$, $B = 1$, $Im(iAB*) = Im(-1) = 0$
and not $Im(AB^*) = 1$
 
@Slereah there is no getting around it, this amazon.com/… (along with these youtube.com/playlist?list=PLo6qq_MJTOZ2ZHdr-aMdZNcsjkbGD65uR for the beginning) is the only option to know Lie theory, and it's not easy :(
 
so now what if $z = kAB^*e^{2ikx}$ ? $\Im(z) = k\sin(2kx)\Re(AB^*)$ ?
 
gets around this
 
12:59 PM
@ACuriousMind hi, one question was about how to explain the fact that Faraday cage effect is only a macroscopic phenomenon in classical electrodynamics and that at microscopic level of electrons (where tunneling occurs among other QM phenomena) it makes no sense to speak of such effect? I'm trying to learn how to put this into a meaningful physical reasoning. Look forward to your insights on this :)
 
20 mins ago, by Mohammad Areeb Siddiqui
so now what if $z = kAB^*e^{2ikx}$ ? $\Im(z) = k\sin(2kx)\Re(AB^*)$ ?
why do I have a strong sense of deja vu...
2 days ago, by Mohammad Areeb Siddiqui
if $\Psi = Ae^{ikx} + Be^{-ikx}$ where $A, B \in \mathbb{C}$ and $k \in \mathbb{Z}$ and I have to find the probability current:
in Mathematics, 50 mins ago, by Mohammad Areeb Siddiqui
What would be the result of $Im(i\cdot AB^*ke^{2ikx})$ where A and B are complex constants.
I don't think there is a shortcut. Just use bolbteppa's method
 
1:26 PM
@Secret seeing the same question asked twice in two chat rooms doesn’t necessarily bother me. OTOH if they do it repeatedly then I lose my patience pretty quick
 
I really have to deal with my hypersensitivity issues... >_>
My answer is genuine though, because there is really no easier way
 
I wasn't looking for an easier way xD
im sorry for posting it simultaneousl
ly* on 2 groups
@Secret and yes im stuck at this :'(((
the worst part is i do not have the answers keys to these problems so i kinda have to verify the answers too hence all these stupid questions
 
Hmm... let's see...
 
1:48 PM
I don't think there is any nice form either: The kAB* being complex prevent any sin from bubbling out
 
Wassup. I just came to say that I will have to do my geography homework now. So see you later guys :)
(Btw, I hate geography, so please help)
Ok, I'm going, see you, I love this community.
 
@Secret so i just leave it as $k\Im(AB^*e^{2ikx})$ ?
 
k is real?
otherwise, yeah, you cannot do much simplification until maybe somewhere along the line another term get add to it so you can factor out something
 
this originated from $J(x) = (A^*e^{-ikx} + B^*e^{ikx})\cdot(iAke^{ikx}-iBke^{-ikx}) = ik(|A|^2 - |B|^2) + 2i\Im(iAB*ke^{2ikx})$. i think this is it
 
2:10 PM
sounds about right, the imaginary part is the interference term
don't know why there is no cos popping out though
perhaps it is because you are computing J not the probability
 
2:44 PM
@JohnRennie hi :)
 
@user929304 Hi
 
@JohnRennie I had a naive question about electromagnetism if you have time possibly
 
@user929304 yes, of course. What's the question?
 
@Secret yes maybe it would appear in the density function
 
@user929304 maybe ask in the problem solving room
 
2:47 PM
@JohnRennie It's simply about how to put it in physical terms that the Faraday cage phenomenon is restricted to classical EM theory. In other words, it is observed macroscopically in the limit of many photons (thus classical waves), that electrons redistribute to cancel out the internal electric field. How can I emphasize that? That it s not meaningful to talk about faraday cage at QM level.
 
Another question, if $\Psi$ is a superposition of 2 other normalizeable wavepackts. And the overlap integral $\gamma(t) = \int_{-\infty}^{\infty}\Psi_1(x,t)^* \Psi_2(x,t) dx$ is very small at t=0. It would remain the same for all t right? because time evolution does not hinder the normalization condition?
 
@user929304 I'm not sure how meaningful the question is. I'm not sure I agree that a Faraday cage is only meaningful in the classical limit.
 
@JohnRennie Maybe for more context, I mean for instance imagine two conductive nanoparticles (they could be metallic) dispersed in a polymer-environment. Then it is commonly known that if they are sufficiently close to one another, then electrons can tunnel from one to the other, so eventually the system can conduct, which is a purely QM situation. Now at this level and scale, I am trying to reason correctly as to why it doesn't make sense to talk about Faraday-cage type of events playing a role.
 
I'm afraid I still don't see what you are getting at.
 
3:05 PM
@JohnRennie well with this we have an electric field (or a voltage simply) across the medium (barrier) through which the electrons tunnel, which may lead one to argue that the electrons will redistribute as to cancel this field, like in a Faraday Cage. But that s not the case at such microscopic level but the problem is I don't know how to explain it. :/
@JohnRennie to put it differently, we associate the Faraday cage effect to Lenz law, right? and that in itself is a purely classical EM wave theory concept, is that a sound way of putting it? this is at the level I'm trying to go about explaining it
 
3:42 PM
Guys, I know that in definite integrals, where two points on the horizontal axis are specified, you can find the area under the curve, between those points. But what about indefinite integrals, where you get just a number, what does it represent? (Since the indefinite integral has no boundaries?)
 
represents an anti derivative maybe?
 
$\int f(x) dx$ is the antiderivative of $f(x)$
 
Ok, then what about definite integrals?
 
Wait, what do you mean by 'you get just a number' ?
 
3:48 PM
Well, in the definite integral, you get a number, don't you?
 
Well, that's what I meant.
 
It could mean different things. For example if $f(t)$ gives the acceleration of some object. Then $v(t) = \int f(t) dt$ would give its velocity. And in the same way $\int_a^b f(t) dt$ would mean the change in velocity between the time interval $a$ to $b$ because $v(b) - v(a) = \int_a^b f(t) dt$
The definite integral usually also represents the Riemann Integral.
 
Wait, which was the opposite of derivatives, the definite or indefinite integral?
 
the definite is the opposite :P
 
3:57 PM
And what about indefinite?
 
@MohammadAreebSiddiqui please stop misleading newcomers.
@NovaliumCompany Both, each in its own way.
It depends, partly, on how your reference text has defined the indefinite integral, as there are multiple possible (incompatible) approaches.
But basically the definite and indefinite integrals are pretty much the same thing
 
Well, one sets boundaries and the other doesn't?
 
i.e. $\int f(x) \mathrm dx = \int_{x_0}^x f(x')\mathrm d x'$, where $x_0$ takes the role of the additive constant $+C$
 
ξ what's that weird symbol?
 
definite integrals are indefinite integrals evaluated over a region. That's pretty much all there is to it...
 
4:00 PM
Yep, Makes sense ^
 
That's xi
 
@NovaliumCompany a dummy integration variable, the Greek letter xi
 
the most annoying to draw of greek letters
well maybe that or zeta
$\zeta$
probably $\xi$ is a bit worse...
 
@EmilioPisanty im a newcomer too :P sorry
although
 
@enumaris $\displaystyle\mathrm{difficulty}_\mathrm{draw}(\xi) > \mathrm{difficulty}_\mathrm{draw}(\zeta),$ definitely
 
4:02 PM
that's what i wrote no?
 
@EmilioPisanty I can't actually view TeX atm lol
I see your source code tho so I get your point :P
 
Well, I imagine indefinite integrals as they select the infinite structure under the curve and find their area, which makes no sense.
 
@NovaliumCompany No, that's an incorrect picture.
 
I know
but what's the correct one?
 
Indefinite integrals are just definite integrals, seen as a function of their upper integration limit.
But strictly speaking, indefinite integrals are not functions, because of the additive constant $+C$ at the end.
 
4:04 PM
Still don't understand, sorry ;\ I guess I'll watch some vids on indefinite and come back.
 
i.e. you can say $F(x) = \int \cos(x) \mathrm dx$ and then $F(x) = \sin(x) +C$, which is nice enough as a symbolic statement, but if you want to give a numerical value to $F(2.6)$ then you can't, because you don't know what $C$ is.
@NovaliumCompany That's just beating around the bush. Read an actual textbook, from the beginning, instead.
 
or take a class
 
Seriously: take your learning seriously. Stop relying on strangers on the internet to repeatedly fill in the many gaps that get left by a haphazard learning path.
 
oh i see what i wrote wrong
 
Please, let me choose my way of learning.
 
4:07 PM
my bad
 
@NovaliumCompany you can choose your way of learning. Just be aware that there are ways of learning which are ultimately rude to the people you're in contact with, and don't be surprised when that fact is pointed out to you.
 
blinks
 
yes. The kind of learning gap that you just asked others to explain is the kind of gap that comes from thinking that "just watching some vids" is a suitable replacement for the structured learning that comes from reading a textbook.
 
tbf reading a textbook can be painful depending on the textbook
 
4:11 PM
I don't understand how me watching videos is offending anyone. Maybe I don't want to dive deep into integrals or calculus, how do you know. Please, just let me mind my own business.
 
and "some vids" might be like an online course which could be quite good
 
We are here to help on pointwise questions. Asking for a five-minute explainer to make up for an entire textbook chapter that you skipped in favour of a few videos is a rather different proposition.
 
Ok, sorry, is there anything you want from me?
 
I'd like a copy of MTW shipped to my house kthx
 
@NovaliumCompany I'll just stop at this advice: learning calculus well requires some serious structure, because there are many interconnected concepts and each different course takes them in different order, so if you mix-and-match then you don't see the structure as intended by each author. It can be videos or a textbook or whatever, but take it seriously and take it in structure, or you're seriously jeopardizing the depth at which you can learn the material.
@enumaris gen lib rus ec =)
 
4:15 PM
well I can get an electronic copy of MTW, but since @NovaliumCompany was offering...I figured I'd get a physical copy :P
 
Is mitopencourseware channel down for everybody?
 
@enumaris ah, gotcha
 
@EmilioPisanty Ok, I will do that, the problem is that I didn't have time until now, school was taking over me and I was so interested in calculus that I just wanted to learn it, so I started, but as you can probably guess, I didn't have time to read textbooks. Now my summer holiday starts, so I'll much time to read so, thanks for the advice :)
 
good =)
in other news
0
Q: Suggestion for books about mathematics of quantum entaglement and computation

Shahabeddin MostafanazhadI am reading book called Mathematics of Quantum Computation by Ranee K. Brylinski and Goong Chen, and it's really what I need since I am working on quantum entanglement and this kind of stuff, but the problem is that I found the book really vague and I can't really understand it. I wonder if some...

how did this absolutely terrible question get upvotes?
 
It doesn't have upvotes?
 
4:18 PM
it deserves to sit at -2 where its downvotes put it
 
@EmilioPisanty What about watching all Khan's Academy videos on Calculus, that should do the job as well?
 
@NovaliumCompany I don't think videos are anywhere near a replacement to textbooks, but that's a personal choice
just do an actual course starting from the beginning
 
Okay :)
 
↑ and consider this and its relationship to any misconceptions that you're already carrying about calculus
reading a textbook and doing the problem forces you to confront and fix them
 
having learned deep learning through Coursera's online courses and reinforcement learning by reading a textbook through and through. I feel either method can be good, it just depends on the quality of the instruction.
 
4:21 PM
I think I learn better from videos, but nothing stops me from trying a textbook :)
 
watching videos often doesn't
 
but I have a pretty heavy mathematical background already so none of that was an issue for me so ymmv
 
@NovaliumCompany so do the people in that veritasium piece
 
I'm watching the vid now.
Got it, so the main idea of the video is that learning from videos is bad. Well, if I have to be honest, I'm more hyped to learn from a textbook right now, because the things said in the video are actually true, as I have experienced them.
 
make sure u get a good textbook tho
some textbooks are trash
 
4:30 PM
Ok, I'll see you later guys :) Thanks for the advice and help.
 
see ya
 
4:41 PM
figuring out the best way to structure a Reinforcement Learning Library seems very non-trivial...hmmm...
 
@EmilioPisanty Learning math is not about doing "exercises". Anyone can do that. The biggest problem at hand is the conceptual grasp, which videos can often help
Far too many textbooks overlook that imo
@NovaliumCompany To be honest, watching videos alongside a textbook gave me a much stronger grasp on calculus and linear algebra than learning from either source alone would
Videos are a very powerful supplement, though not a replacement
 
5:00 PM
To steal a Philip Anderson title: More is different.
 
@NovaliumCompany watch all the Khan videos, they are extremely useful
 
 
1 hour later…
6:26 PM
@Blue No, it's because with Pillars of Eternity 2 I finally got a new source for character portraits :P
@user929304 I don't think it's true that the Faraday cage "makes no sense" at a quantum level. If one carefully and correctly calculated the expectation value of the electromagnetic field inside sucha cage one should eventually find that it is not much dependent on the outside fields. But why would you insist on a complicated quantum computation when the classical approximation perfectly suffices?
 
6:43 PM
Cus Quantum > Classical
Like the word is just cooler
 
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