The h Bar

It's a bug in the game code. I'd guess a type conversion error. I don't know what language the game has been written in so I don't know exactly where it's coming from.

@BernardoMeurer: that looks quite a nice idea for a simple game. Kind of arithmetical Tetris. Though it's going to take a loooooooooong time to get to 2048 :-)

Bernardo Meurer

@JohnRennie It doesn't take that long :)

The trick is, pick one direction not to move in

and then only do moves in the three remaining directions

Anyway, currently I hold my board on a 2-D array

Fawad

Keep big digit at one corner from biggning

Balarka Sen

11:12 AM

I got to 256

John Rennie

Isn't SDL an abstraction layer for multimedia devices? I have a vague recollection that the Linux build of Kodi used SDL to handle keyboards, joysticks, etc, though we've now moved to doing it natively.

Bernardo Meurer

but this is a problem because I can't write functions that return 2 values (i.e. return a position on the board)

So I wanted to make a struct

@JohnRennie SDL does a bunch of things, being a graphics library is one

Handling IO is another

Fawad

I first came to know from singing bananas video

Bernardo Meurer

@JohnRennie Would you be kind enough to take a look at my (commented) code and help me writing the struct?

I'm a struct noob

John Rennie

In C(++)?

Bernardo Meurer

11:14 AM

C99

I WISH we could use C++

John Rennie

Yes, of course. Though structs are pretty simple.

Bernardo Meurer

Whats you GH name again?

John Rennie

On GitHub or are you just going to post it here?

Bernardo Meurer

It's on GH, but private, so I have to add you as a collaborator

It'll be public after the delivery date

John Rennie

jhsrennie

Secret

11:16 AM

I do not terribly understood what happened here, but it seems that some guys have followed up the indefinite causal order experiment and directly demonstrate indefinite causal order is present via some kind of quantum eraser like experiment

Bernardo Meurer

@JohnRennie Added you, it's the project01 folder

John Rennie

In main.c ? I can't see any structs there ...

Bernardo Meurer

@JohnRennie There are none yet

In the main function do you see the board array

John Rennie

Ah, yes, all those function prototypes cluttering up the code. The joys of writing in C :-)

Bernardo Meurer

@JohnRennie :P

John Rennie

11:20 AM

It's usual to put all the prototypes in a header file so they don't clutter up your code. I'd stick them all in main.h.

Anyway, yes I see the board array.

Bernardo Meurer

So, suppose I want have a function that returns a position on the board

Say, the empty space I can move the block to when I move it up or down

Right now that's not very nice to do because it's a 2-D array and in C a function cannot return 2 values

How could I go about using a struct for board that would allow me to feed it coordinates like that? All at once

John Rennie

typedef struct _Position { int x; int y} Position;
Position p = GetPosition();

You don't have to use a typedef, but it makes for nice looking code.

Bernardo Meurer

And then Getposition is a function of type Position?

John Rennie

Position GetPosition(void)
{
Position p;
p.x = 0;
p.y = 0;
return p;
}

Bernardo Meurer

I see, hmm, sweet

John Rennie

11:25 AM

Hmm, I can't remember how to format code in chat

Bernardo Meurer

Once you paste there should be a fixed font button

But board cannot be of type position, can it?

John Rennie

Using typedef you can make the structs work very much like classes. It's a cool trick.

@BernardoMeurer Why would you want to make the board array an array of positions?

Bernardo Meurer

So that I can, somehow, specify a board square in one go

Well

I guess it can be a 2D array and I do board[foo.x][foo.y]

John Rennie

Ah I see, you mean use a Position as an index into the array.

No you can't do that natively. You'd need to use a function that accepted a position and gets or sets the value of the board at that position.

Bernardo Meurer

Here's my issue, maybe you can come up with a better way to do it than me

Look at the move function

I discretized a movement in the game as
1. Flush all pieces into one direction
2. Join neighbouring same-value pieces in that same direction
3. Flush again

John Rennie

11:31 AM

Isn't that function passing the whole board on the stack i.e. copying the entire board onto the stack before the call then reading it off again afterwards? That doesn't seem efficient. Can't you just pass a pointer to the board?

Oh no. Ignore me.

Bernardo Meurer

That's a pointer to the board I believe

Every array pased onto a function decays to a pointer

Alas I know I'm scrweing up that function declaration somewhere

But it works just fine :P

John Rennie

Anyway ... the move function?

Bernardo Meurer

So, All I'm trying to do is move these goddamn pieces the right way

without having to do two flushes

But I don't see how that's possible

I also what to replace board with some special type to hold it so it's nicer to deal with than a 2-D array somehow

John Rennie

Bearing in mind I'm not sure exactly how the game works ...

Suppose I've pressed UP. Assume we're looking at column 0 (0,0 is top left?) exactly what do I have to do with the numbers in column 0?

Bernardo Meurer

Kyle proposed something with a recursive function, but for that to work nicely I need to be able to do something like board{next_empty(foo,bar)}

@JohnRennie I'll give some instructive screenshots

What currently happens when UP is pressed

John Rennie

11:39 AM

Yes?

Bernardo Meurer

@JohnRennie If ti worked correctly the 4's would all be flushed up

they wouldn't join

but they would be together

i.e. 2 4's, on (0,0) and (0,1)

John Rennie

So if numbers have a space between them the space is removed leaving the numbers adjacent?

Bernardo Meurer

Yeah, i.e. a flush

John Rennie

And if two identical numbers are adjacent they get summed and the sum replaces the top number?

user228700

Eh, 2048? Time to brag!

user228700

11:42 AM

John Rennie

@Kaumudi.H Bernardo is trying to code it!

Bernardo Meurer

@JohnRennie In case of an UP movement yes, in a down movement it'd be the lower number, right the rightmost and so on

@Kaumudi.H only 4096? Sigh

user228700

@JohnRennie I see :-) Two of my friends did it in 12th (C++)

Bernardo Meurer

C++ is ez :P

user228700

@BernardoMeurer :-( I know. I remain a disappointment to friends, family and even acquaintances, as it turns out.

John Rennie

11:45 AM

So the code:
if (_board[i][j] == _board[i][j - 1] && _board[i][j] != 0) {
_board[i][j - 1]++;
_board[i][j] = 0;
}
finds adjacent identical numbers?

Bernardo Meurer

Yessir

Yashas Samaga

Yesterday, I wrote a program to compute the value of Euler's constant. It is fun to run the program for 24 hours to see how accurate it can get.

Few weeks ago, I did one for computing pi.

Bernardo Meurer

@JohnRennie And the if block above that flushes things up once

I have a conjecture that in a situation like the above the most the resulting sums will be apart from each other is once square

I think I will just use separate functions for flushing and joining

John Rennie

So if the column is (2,2, , , 1) the first for loop would give (4, , , , 1) and the second for loop would flush all the spaces to give (4, 1, , , ) ?

Bernardo Meurer

Using my current logic I would need one loop to go over columns

and then loop 3 times over each row

Let me show you

Hmm

I can't get the second flush to work

case UP: // Up
    for (int i = 0; i <= 4; ++i) {
        for (int j = 4; j > 0; --j) {
            if (!_board[i][j - 1]) {
                _board[i][j - 1] = _board[i][j];
                _board[i][j] = 0;
            }
        }
        for (int j = 0; j < 4; ++j){
            if (_board[i][j] == _board[i][j + 1] && _board[i][j]) {
                ++_board[i][j + 1];
                _board[i][j] = 0;
            }
        }
    }

then we need another flush after this

John Rennie

11:53 AM

How about you use a separate int[5] array

Start on the top row of your target column on the board. If this is not blank copy it into your buffer. If it is blank jump down to the next row on the board.

Then find the next non-blank number on the board. If it's the same as the number in your buffer add it and advance the buffer index. If it isn't the same advance the buffer index and copy the next number in.

You should be able to build up the buffer in one pass. Then copy it back to the board.

Bernardo Meurer

Wait, what's your idea?

Ah

I see

Interesting

John Rennie

You are trying to update the board in place, and that makes it unnecessarily complicated

Bernardo Meurer

Let me try

Okay, so one loop for the columns

0celou7

Wtf is this

Bernardo Meurer

Now go over the row and fill the array

John Rennie

12:03 PM

@0celou7 Since you're not keen on JEE discussions we thought we'd turn it into a programming chat instead :-)

Bernardo Meurer

    for (int i = 0; i <= 4; ++i) {
        for (int j = 0; j <= 4; ++j) {
            if (j > 0 && _board[i][j] == pos[index]) {
                ++pos[index];
            } else if (_board[i][j]) {
                pos[index] = _board[i][j];
                ++index;
            }
        }
    }

@JohnRennie Something like this I guess

John Rennie

Are you storing the power of two i.e. board[i][j] = n if that cell contains 2^n ?

0celou7

@JohnRennie that might be worse

John Rennie

@0celou7 I think we're pretty much done ...

Yashas Samaga

Can the breakdown voltage for a dielectric be calculated theoretically?

0

Q: Voltage of Arcing electricity

In Star Wars, Count Dooku electrocutes both Anakin and Yoda at a distance of about 5m. Obviously this is not meant to be scientifically accurate as it is a fictional movie, but how much voltage would be required for such an arc in real life, and how do I calculate this, and the charge expended ov...

Bernardo Meurer

12:09 PM

@JohnRennie Yep

where 0 stands for no value

        for (int i = 0; i <= 4; ++i) {
            for (int j = 0; j <= 4; ++j) {
                if (j > 0 && _board[i][j] == pos[index] && _board[i][j]) {
                    ++pos[index];
                } else if (_board[i][j]){
                    pos[index] = _board[i][j];
                    ++index;
                }
            }
            for (int k = 0; k <= 4; ++k) {
                _board[i][k] = pos[k];
            }
        }

@JohnRennie Lol, straight up doesn't work

John Rennie

I'm not sure I see what your code is doing ...

Bernardo Meurer

I created an array pos[5] and an index

I iterate over the columns

and then over the rows

Yashas Samaga

2

Q: Magnetic field in different medium

To solve this question we apply the same concept as that of light that the component magnetic field will be same which is perpendicular to the surface.But I could not understand the concept behind it.

magnetic-fields

could that be a homework question?

It is a conceptual doubt.

Bernardo Meurer

if we're over the first row, the value of this cell is the same as what we have in our current index and it's not 0 we increment the value of pos at the current index

otherwise, also assuming it's not zero, we

Ah

I see the problem I think

John Rennie

Let's take a concrete example. Suppose the user has pressed left and the top row of the board is (2, ,2, ,1). Is this OK as an example?

Bernardo Meurer

12:14 PM

Can we use UP? it's what I got set up for testing :P

Oh

Wait, no

John Rennie

Well I can't easily write column vectors in the chat :-) Does it matter? The method is the same in both cases.

Bernardo Meurer

@JohnRennie Fair :P

John Rennie

Ok do we proceed with my example or yours?

ACuriousMind

@YashasSamaga Honestly, I don't understand how that question got 3 upvotes. I can't even really tell what it's asking, let alone whether that question is homeworky or conceptual.

Bernardo Meurer

Let's try the one above because it captures two consecutive joins

Yashas Samaga

12:16 PM

@ACuriousMind I guess the OP is asking if magnetic fields refract just like light.

John Rennie

OK. Take int buf[5] and initialise it to zeros. Set the index bufnum to zero. Set colnum to zero, since we're looking at the first column, and rownum is the (duh) row number. So far so good?

ACuriousMind

@YashasSamaga I don't understand what that's supposed to mean.

Remember, if you have to guess what a question is asking, it's very likely unclear what you're asking!

Yashas Samaga

lol

Bernardo Meurer

@JohnRennie Uhum

Done

John Rennie

Start with rownum = 0 and use a loop that exits when rownum =5.

Bernardo Meurer

12:19 PM

int pos[5] = {0};
int index = 0;
switch (dir) {
case UP: // Up
for (int i = 0; i <= 4; ++i) {
for (int j = 0; j <= 4; ++j) {

}
break;

@JohnRennie 4 you mean?

John Rennie

Never mind the code for now, let's just get the algorithm right.

Bernardo Meurer

Alrighty

John Rennie

Is board[colnum][rownum] zero? If so rownum++ then if rownum > 4 exit the loop.

0celou7

@Slereah Let me put it this way: Lebesgue integration is not really useful for the new types of functions you can integrate in practice...no one cares if you can integrate the indicator function of $\Bbb Q$. It is useful for telling you certain operations on sequences are nicely behaved. For example, $f(x)=\lim_n f_n(x)$ is aways Lebesgue integrable if the $f_n$'s are (the integral might diverge of course), but in the Riemannian case this is not true. Furthermore, the Lebesgue integral [...]

Secret

Bernardo Meurer

12:21 PM

Okay, so if the cell is empty we just go look at the next one

John Rennie

If board[colnum][rownum] is non-zero copy it to buf[bufpos]. The array buf now looks like (2, , , , ) and bufpos = 0

Secret

Analysing the delayed choice quantum eraser experiment

0celou7

@Slereah [...] can tell you about the relationship between $\int f$ and $\lim \int f_n$ in certain cases. The Riemann integral can't say anything about it.

John Rennie

@BernardoMeurer we should probably take this to another room now that people have started talking physics ...

Bernardo Meurer

@JohnRennie Yessir

John Rennie

12:23 PM

Have you got a suitable room? Or would you like to create a room and invite me?

0celou7

@Slereah The important thing is that if you're using the Riemann integral, you have to check every time that your limits are Riemann integrable. And the condition for Riemann integrability is pretty horrible and you will not be able to do it most of the time. So the convergence properties of the Lebesgue integral are extremely important and it's why we use it. It's why $L^2$ is a Hilbert space, why you can use bra-ket notation.

ACuriousMind

@JohnRennie I'd be curious where you see any physics in the room currently ;)

Secret

If my interpretation is correct, then the reason why erasing the path information and you get interference patterns is simply because you have two inputs at those detectors. Since the photons are in some polarisation state, the photons that entered the detectors D1 and D2 are in the same hilbert space and thus they can superimpose onto each other, giving the interference pattern

John Rennie

@ACuriousMind OK maths then :-)

If no-one minds I'm happy to keep talking programming here :-)

0celou7

@Slereah Lebesgue measure/measure theory is a bitch to set up, but once you have it, it's a machine that keeps giving. The proofs get easier, and you can apply measure theory to far more abstract setting like locally compact groups, or with measures taking values in von Neumann algebras.

@Slereah For instance, proving that a continuous function is integrable in Lebesgue theory takes one line. For Riemann integrals it takes a page or two, just ask @BernardoMeurer

Secret

12:25 PM

whereas for D3 and D4, nothing is interfering with the photons and their state is already being projected due to the entangled pair being measured by D0 earlier, thus projecting away the entangled states into product states, thus the photons are no longer correlated after its partner hit D0. In particular, those photons have a well defined state randomly projected by the measurement at D0

Therefore no interference pattern

0celou7

I don't know why measure theory gets such a bad rap. I was taught to hate it before I knew anything about it. @ACuriousMind hates it and won't tell me why

Bernardo Meurer

@0celou7 ACM is a hateful german

yuggib

@0celou7 measure theory $\cong$ (commutative) probability theory

Secret

In conclusion, I suspect how far D0 is from the source relative to that of D1,D2,D3,D4 is irrelevant as long path length to D0 > D1,D2,D3,D4, because the key might be the measurement of D0 projected away the entangled pair, and thus its partner must end up with some well defined pure state after the measurement at D0

(assuming there are no noise that is)

0celou7

@BernardoMeurer I thought he was a special snowflake?

Sir Cumference

12:34 PM

Ugh, I hate waking up in crappy weather

Secret

Now, my suspicion is that if D0 < D1,D2,D3,D4, the interference patterns at D1 and D2 will end up correlated

I'll calculate later to see if it is indeed the case

0celou7

@yuggib in which category?

I heard the term noncommutative probability the other day and I'm not sure I want to know what it is

Secret

> Moreover, the apparent retroactive action vanishes if the effects of observations on the state of the entangled signal and idler photons are considered in the historic order. Specifically, in the case when detection/deletion of which-way information happens before the detection on D0, the standard simplistic explanation says "The detector Di, at which the idler photon is detected, determines the probability distribution at D0 for the signal photon". Similarly, in the case when D0 precedes detection of the idler photon, the following description is just as accurate: "The position at D0 of …

ACuriousMind

@0celou7 You don't - it's quantum mechanics :P

Secret

ok it seems I understood this correctly (if the wikipedia is correct)

10

A: 'Quantum' approach to classical probability

Yes. You can define a(n algebra dual to a) noncommutative probability space as a complex $^{\ast}$-algebra $A$ together with a positive linear $^{\ast}$-functional $\mathbb{E} : A \to \mathbb{C}$ (the expectation). The quantum examples occur when $A$ is a suitable $^{\ast}$-algebra of linear oper...

0celou7

12:43 PM

@ACuriousMind idk, the wiki article makes it seem like pure math

Secret

Free probability

Free probability is a mathematical theory that studies non-commutative random variables. The "freeness" or free independence property is the analogue of the classical notion of independence, and it is connected with free products. This theory was initiated by Dan Voiculescu around 1986 in order to attack the free group factors isomorphism problem, an important unsolved problem in the theory of operator algebras. Given a free group on some number of generators, we can consider the von Neumann algebra generated by the group algebra, which is a type II1 factor. The isomorphism problem asks if these...

this one?

ACuriousMind

@0celou7 See e.g. Valter Moretti's answer here

0celou7

@yuggib but f(x) = l.i.m. f_n(x) doesn't mean pointwise convergence...so wh you write it as an a.e. convergence?

TrY iS CheM

@YashasSamaga Yes ,I mean that , with reason

Yashas Samaga

You should edit your question and make it explict then.

TrY iS CheM

12:59 PM

@YashasSamaga is it now fine

Secret

Suppose I have a singlet state and Alice want to measure spin in the x direction while Bob does nothing, how to write the measurement operator for this operation?

I know the observable will be $\sigma_x^A \otimes I_B$ but how to convert it into a measurement operator so that when acting on a two particle wavefunction, it projects it out to some product state?

ACuriousMind

@Secret What is a "measurement operator"?

Secret

We knew that measurement will project the state to some eigenstate (or for degenerate cases, the eigenspace of the corresponding observable). How to write this mathematically as an operator if I had a state made of many tensor products?

By "measurement operator", I want an operation on the wavefunction so that once it act on the wavefunction, it effectively performs a measurment and project it to some eigenstate of the observable

ACuriousMind

I don't understand the question. If you just want to project to an eigenstate, then you should know enough linear algebra to write down a projector.

Secret

1:15 PM

Ok I will try

0celou7

@ACuriousMind linear algebra is hard

ACuriousMind

Alas, if you want a quantum mechanical operator that "performs a measurement", then that's different since you'd need to also consider the measurement apparatus as a quantum object, and write down a time evolution of hteir combined system that lets you end up with the system in a state that represents the result of the measurement. It can be done, but it is rarely necessary - pretending (strong) measurement just instantaneously projects to eigenstates is usually fine.

Secret

Ok

ACuriousMind

@Secret Well, what is there to "try"? If I hand you a state $\lvert \psi\rangle$, then what is the projector onto it?

I suspect your problem is not writing down the projector, but I can't tell what the actual issue is because I have no idea what you really trying to do

Slereah

Best notation to write components of the coordinate chart?

Secret

1:25 PM

I am trying to do a measurement of the following type: Alice measure the spin of a singlet state along the x direction while bob does nothing. I am trying to investigate how mathematically a measurement can project an entangled state to a product state when only one party is doing a local measurement (we know that by non communication theorem all is good about causality)

Since we knew an entangled state of two subsystems is of the form $\sum_i \lvert \psi_i \rangle \otimes \lvert \phi_i\rangle$, and any local operations on subsystem 1 can only affect the subsystem 1 portion of the composit…

0celou7

(x^i)

Slereah

I could use $x^\mu$ but it feels a bit weird

0celou7

Why?

Secret

Since by no communication theorem, a local operation can only affect a subsystem, therefore for a local measurement to end up projecting the entangled state into a product state (and thus removing the correlation), it means any measurement has to be inherently nonlocal (even if the other parties are doing nothing)

This sounds weird, thus I want to double check again how well the maths can justify this thought

Secret

In short, I want to understand how a local measurement end up destroying all the correlations in an entangled state, which is nonlocal

ACuriousMind

1:29 PM

@Secret Say Alice measures the spin up state $\lvert +\rangle$. Then the projector is $P_+ = \lvert +\rangle\langle +\rvert$ on Alice's subspace, and $P_+\otimes\mathrm{id}_B$ on the entire space, where $\mathrm{id}_B$ is the identity on Bob's space.

@Secret That's not what the no-communication theorem says. It says you cannot transport information, not that quantum mechanics is local.

Secret

Hmm, based on what you said above, does that means a local measurement is actually nonlocal if a system is part of an entangled state, because other parties can be considered to be applying an identity operation on their subsystems despite they are doing nothing?

Slereah

@0celou7 'cause it's not a tensor

ACuriousMind

@Secret The point of the no-communication theorem is that if I have a projector $P_\text{tot} = P_A\otimes\mathrm{id}_B$, then if $\rho$ is the density matrix of the state before measurement, we have that $\mathrm{tr}_A(\rho) = \mathrm{tr}_A(P_\text{tot} \rho P_\text{tot}^\dagger)$.

0celou7

@Slereah so?

Tensors are not defined by indices

ACuriousMind

That is, if you trace out Alice's space, which means looking at how the state looks from Bob's viewpoint, then the state before and after Alice's measurement looks exactly the same

0celou7

1:38 PM

You should object to writing christoffel symbols otherwise

ACuriousMind

@Secret How is "applying an identity operation" different from "doing nothing"? Don't imbue the mathematical representation with physical meaning it doesn't have.

Hey @Prahar, @AccidentalFourierTransform recently wanted to tell you something:

Mar 18 at 16:55, by AccidentalFourierTransform

I wanted to talk to Prahar, but he's never been here in the chat :-(

Mar 18 at 16:58, by AccidentalFourierTransform

I just read the lecture notes by Strominger from yesterday, and they're very nice

Mar 18 at 16:59, by AccidentalFourierTransform

I just wanted to tell him that, well, I liked them

0celou7

No homo?

Slereah

Also how to write the frame bundle

Secret

Suppose I have the singlet state $\frac{1}{\sqrt{2}}(\lvert + -\rangle - \lvert - +\rangle)$ and the projector $P_A \otimes \textrm{id}_B$. Then
$$P_A \otimes \textrm{id}_B(\frac{1}{\sqrt{2}}(\lvert + -\rangle - \lvert - +\rangle))=\frac{1}{\sqrt{2}}(\lvert +-\rangle-\lvert ++\rangle)=\frac{1}{\sqrt{2}}\lvert +\rangle \otimes (\lvert -\rangle-\lvert +\rangle)$$
Uh, let me check the maths again, I got a really strange product state...

Slereah

$FM$, $O(M)$ or $L(M)$

I've seen all of those

ACuriousMind

1:51 PM

@Slereah Just put $M$ inside a box. Then it's framed.

Slereah

Heyooo

Secret

$$P_A \otimes \textrm{id}_B(\frac{1}{\sqrt{2}}(\lvert + -\rangle - \lvert - +\rangle))=\frac{1}{\sqrt{2}}(\lvert +-\rangle-0\lvert ++\rangle)=\frac{1}{\sqrt{2}}\lvert +-\rangle$$
Much better

hmm... so mathematically I do end up with a product state as expected, but how does the correlation get broken, let me think about it...

actually no, the correlation is still there, except that the two subsystems are no longer correlated afterwards for future measurements

0celou7

@ACuriousMind god damn physicist

@Slereah $\mathfrak L\mathscr M$

@ACuriousMind How do I form quotient rings?

how do I know such a ring exists?

@BernardoMeurer ??

ACuriousMind

2:09 PM

@0celou7 By what do you want to quotient?

0celou7

@ACuriousMind I have a ring of functions where addition is pointwise addition and multiplication is kinda convolution

I want to be able to divide in this ring

ACuriousMind

Ah, you don't want to take a quotient in the sense of algebra, you want to localize the ring/form its field of fractions

0celou7

Oh, yeah

The field of fractions

ACuriousMind

The field of fractions exists iff there are no zero divisors in your ring.

0celou7

So how do I do this? I know the formal idea

ACuriousMind

2:12 PM

:36310099 No, you need $rs = 0\implies (r=0 \vee s=0)$.

0celou7

What exactly do I need? I forgot what a zero divisor is

Prahar

@ACuriousMind Hey ACuriousMind, i've been meaning to be on chat more but that doesn't happen as often. Thanks a lot! I will chat with him when he's here!

0celou7

Well, I have $(rs)(t)=\int_0^tr(t-u)s(u)\,du$

Yashas Samaga

0

Q: Gravitational lensing or lensing by refraction?

I was watching a vedio in which a guy claimed that gravitational lensing is not real rather the effect is caused by refraction.. He cited some Indian scientist, can someone debunk this?

general-relativity

is that on topic?

he is doubting the validity of gravitational lensing

0celou7

@ACuriousMind maybe I don't want fractions

I think I just want to divide by integral operators

@ACuriousMind Let $h$ denote the operation of integration in this ring $\mathscr C$.

This can be identified with the multiplicative action of the function $h(t)=1$

ACuriousMind

2:17 PM

@0celou7 Well, then you want to localize. Do these operators form a multiplicative subset, i.e. is hte product of two such operators again such an operator?

0celou7

Then Yosida shows that if $k\in H=\{h^1,h^2,\dotsc\}$ and $kf=0$, then $f=0$.

@ACuriousMind Yeah, multiplying two integral operators just means doing more integrals

ACuriousMind

I see. So you know it's a multiplicative subset and it contains no zero divisors. Yeah, that's enough to be able to localize and know the loalization is not trivial

0celou7

What does that mean?

ACuriousMind

(If you localize w.r.t. a set with a zero divisor, you get the trivial ring)

@0celou7 Basically, it means you can pass to the ring where you have fractions, but only your integral operators are allowed in the denominator

0celou7

Yeah, that's what Yosida says

But what does that ring mean?

How does one construct it

ACuriousMind

2:20 PM

the fractions are equivalence classes of numerator-denominator pairs, cf. Wiki

But you should just treat them as fractions, because that's what they're meant to be :P

0celou7

@ACuriousMind oh, equivalence classes of things that satisfy the cross multiplication property?

ACuriousMind

Yes, exactly

0celou7

aha

works for me

thanks

Slereah

@0celou7 a bit much

yuggib

@0celou7 $L^2$ convergence is an a.e. convergence

0celou7

2:24 PM

@Slereah $\mathcal FM$

yuggib

(it is a convergence of equivalence classes)

Slereah

that is what wikipedia uses

0celouvsky

@yuggib $L^2$ convergence gives a.e. convergence for a subsequence

Slereah

Hawking uses O(M)

although...

That might be the orthonormal frame bundle, actually

0celouvsky

Hawking's notation is terrible

Slereah

2:26 PM

$\frac{\partial }{\partial t}$ is probably the worst

ACuriousMind

"0celouvsky"? This is getting out of hand!

5

0celouvsky

I got rid of the 7 at the end

looked silly

ACuriousMind

I...kinda like it

0celouvsky

@ACuriousMind indeed, it has a nice ring to it

@ACuriousMind lol I should name my next pet Rellich-Kondrachov

what is going on with that $j\equiv 1$ and that $L^2$ lol

Slereah

the $j$ is equivalent to $1$, obviously

They have the same truth value

ACuriousMind

2:44 PM

> ______, Spin structures on manifolds, Enseignement Math. (2) 9 (1963), 198–203.

Apparently, "_______" is an author now

Slereah

is he related to et al?

ACuriousMind

Oh, I cannot believe this laziness! It's the same author as of the paper listed before it in the bibliography (Milnor), so it appears instead of writing the name again, they just drew a line

It happens at another point in the bibliography, too.

Why would you do this? Is that some odd feature of a bibTeX style that's on by default and no one bothered to turn it off?

Slereah

is it a recent paper?

ACuriousMind

@Slereah Yes, from 2012

Slereah

odd

0celouvsky

2:50 PM

@ACuriousMind hmm? That's the usual style.

My first thought was "who was the author right above"

In fact that's how I was taught to write a bibliography

ACuriousMind

...wat

0celouvsky

What's confusing?

ACuriousMind

Perhaps American vs. European style then, I've never heard this before

And I've noticed it for the first time now, so it's not that ubiquitious

Doesn't really explain why you would draw a line, but whatever

0celouvsky

Rest assured that it's standard, so not complete laziness on the part of the authors

Slereah

the proper definition of singularities isn't easy man

ACuriousMind

2:56 PM

lol, tex.SE has both how do I draw the line and how do I get rid of the line

alas, I find nothing about the origin of this style

Bernardo Meurer

@Slereah Do you know C?

Slereah

yes

Bernardo Meurer

Is there a way to add a case in a Switch statement that always executes

I'm making 2048 in C, and I have a switch that selects between the move operation (up, down, left, right) but independent of that the function "add_random" should be called

Slereah

Why can't you just add it after the switch

Bernardo Meurer

because then it will run even if the keypress wasn't one of the move keys

Guess I'll just add an if

0celouvsky

2:59 PM

@BernardoMeurer I was robbed, any idea who did it?

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