The h Bar

SUBI R3,R0,1d
SUBI R6,R0,-1d
SUB R1,R3,R6
SUBI R4,R0,3d
B.EQ R4,R0,7d
SUB R2,R3,R6
SUB R3,R0,R6
SUB R6,R0,R2
SUBI R4,R4,-1d
ADD R1,R1,R2
B.NE R4,R0,-5d
SUB R2,R0,R1
B 0d

d denotes a decimal. RX are registers, basically memory cells that hold values

Cows

@0celouvsky do you think ACM got offended?

Bernardo Meurer

B.EQ means "Branch if Equal". And so it's comparing R4 and R0 and if they are equal it jumps seven intructions forward (to the end of the code)

Cows

@0celouvsky I was not trying to offend him

0celouvsky

offended is not the right word

but don't expect him to talk to you

Cows

12:43 AM

why is he like that?

0celouvsky

I agree with him 100%

So I'm not the best person to ask

Cows

whaaaaa

Bernardo Meurer

@Cows What did you do?

@0celouvsky So far so good?

Cows

@BernardoMeurer I think it is more about how I responded

I might have been a bit rude

0celouvsky

@BernardoMeurer I have no clue what I'm looking at.

Bernardo Meurer

12:45 AM

@0celouvsky Okay, this is how the syntax works:

Cows

I am second guessing myself because I don't quite know what his intentions were

Bernardo Meurer

OPERATOR RECEIVER,OPERANDS

So

ADD R1,R2, R3

Addds R2 to R3 and stores it in R1

Makes sense?

0celouvsky

SUBI?

B.EQ?

B.NE?

B?

Bernardo Meurer

SUBI means "Subtract Immediate", i.e. subtract a register and a fixed number

SUBI R1, R1,7d Stores in R1 the value of R1-7

the d is to show it's decimal

B means branch, i.e. go to a specific place in the code

Cows

@0celouvsky how do you recommend I respond in the future ? or should i not

Bernardo Meurer

12:47 AM

Just B is an unconditional branch, it just jumps to that place in code

B.EQ and B.NE are inverses of one another, Branch if Equal and Branch if not equal respectively

the syntax for a conditional branch is B.CONDITION OPERAND,OPERAND, JUMP

0celouvsky

@Cows I would suggest you go away and learn to converse in more than buzzwords, then return.

Bernardo Meurer

where JUMP is how many intructions forwards or backwards you'd like to go

0celouvsky

I tried to help you by suggesting a good book.

Cows

I am looking at it

I can't read the whole book in a day. It will take me a while

may be even a few months

0celouvsky

@BernardoMeurer Sorry, I am busy now. I'm reading a thesis that's saying something incorrect and I need to check it

Bernardo Meurer

12:50 AM

@0celouvsky Alrighty then

Cows

but I am making an effort to read it

0celouvsky

@BernardoMeurer I'm tracking down a Riemannian geometry book from 1949

Cows

it's not easy

0celouvsky

It's hard work

These guys don't use the right math words

Cows

i will take your advice

I think I will just go on a one month hiatus and see how far I get with the book

it's a big book though

so I will be away from this site for a bit

0celouvsky

12:54 AM

If you have specific questions I will be happy to answer them

so old

Cows

ok thank you. I will try to get better at this.

@0celouvsky send ACM my apologies hopefully in a month he will have forgotten my transgressions.

0celouvsky

ok

Cows

See you guys in a month or 2

0celouvsky

bye

Maxwell

@Ocelouvsky, hello

0celouvsky

12:57 AM

@Maxwell Hi?

Maxwell

@Ocelouvsky, Are Newton's three laws of motion independent from each other? And, we cannot derive one from the other?

0celouvsky

Don't know, don't care. Sorry.

Maxwell

@Ocelouvsky, Its ok

Hello @JohnRennie

Bernardo Meurer

@Maxwell JR is sleeping now

0celouvsky

@BernardoMeurer Ok it seems to be true but I'm a little worried I don't have a book that talks about it

the only book seems to be this 1949 book

ok, what's the issue?

Bernardo Meurer

1:06 AM

@0celouvsky Okay, let's look at the code again

SUBI R3,R0,1d
SUBI R6,R0,-1d
SUB R1,R3,R6
SUBI R4,R0,3d
B.EQ R4,R0,7d
SUB R2,R3,R6
SUB R3,R0,R6
SUB R6,R0,R2
SUBI R4,R4,-1d
ADD R1,R1,R2
B.NE R4,R0,-5d
SUB R2,R0,R1
B 0d

Do you understand the instructions here?

Maxwell

, hello @BernardoMeurer

Bernardo Meurer

@Maxwell Hello

Maxwell

Are Newton's three laws of motion independent from each other? And, we cannot derive one from the other?@BernardoMeurer

Bernardo Meurer

@Maxwell I'm not a physicist, can't help you. If you need help with computer architecture I'm your guy

0celouvsky

@Maxwell why do you care?

Maxwell

1:10 AM

@BernardoMeurer, oh. Its ok

0celouvsky

they're all true

@BernardoMeurer so is R0 a fixed number

Bernardo Meurer

@0celouvsky Oh,yeah, that's a special register that is always 0

dmckee

@Maxwell The first is a special case of the second, but the third is independent of the first two.

I've always assumed that Newton included the first as a explicit disclaimer to the effect that he was following Galileo and not Aristotle.

Maxwell

, @dmckee, physics.stackexchange.com/questions/66057/…

0celouvsky

@BernardoMeurer god dammit Reb wants me to read her homework

:(

sorry

Bernardo Meurer

1:14 AM

Dangit

It's alright :P

I'll be here for a while more

Maxwell

@dmckee, what about Timaeus answr?

E7_82_8E

can we ask physics questions here?

0celouvsky

no

E7_82_8E

ok, do you where I can to get quick answers? because the question I posted isn't getting a response but I need the question answered for tomorrow

ACuriousMind

@E7_82_8E Yes, you might get questions answered here when there's someone around willing to answer them, but "quick answers" isn't really what physics.SE is for. It might take a while until someone both willing and able to answer any given question happens upon in on the site.

0celouvsky

1:26 AM

ACM the madman

@ACuriousMind Cows sends his apologies. He has left for a month to train and seek your forgiveness.

Bernardo Meurer

@ACuriousMind Can you help me with this stuff?

I have new information

ACuriousMind

@BernardoMeurer What's the issue? (I remember how the instructions work)

Bernardo Meurer

@ACuriousMind Okay, here's what I have found out:

1. Branch operations finish on the EX step, meaning any two operations BEFORE a branch will always run.
2. Any arithmetic operation takes 3 cycles after it to complete. If I do ADD R1, R1, R2 I must have three command after before I can do SUB R3, R1, R1

So, I had the original ASM code:

SUBI R3,R0,1d
SUBI R6,R0,-1d
SUB R1,R3,R6
SUBI R4,R0,3d
B.EQ R4,R0,7d
SUB R2,R3,R6
SUB R3,R0,R6
SUB R6,R0,R2
SUBI R4,R4,-1d
ADD R1,R1,R2
B.NE R4,R0,-5d
SUB R2,R0,R1
B 0d

Which I rewrote to fit the issues with write and branch overhead

SUBI R3,R0,1d
SUBI R6,R0,-1d
NOP
NOP
NOP
SUB R1,R3,R6
SUBI R4,R0,3d
NOP
NOP
NOP
NOP
B.EQ R4,R0,7d
NOP
NOP
SUB R2,R3,R6
SUB R3,R0,R6
NOP
NOP
SUB R6,R0,R2
SUBI R4,R4,-1d
ADD R1,R1,R2
NOP
NOP
B.NE R4,R0,-5d
NOP
NOP
SUB R2,R0,R1
B 0d

NOP is the "do nothing" operation

@ACuriousMind Take a look at those, see if you understand why the NOPs were added

there may be some extra ones here and there

0celouvsky

@dmckee do you have an idea why theses are often in this horrible double space format?

to artificially increase the length?

ACuriousMind

@BernardoMeurer Shouldn't the numbers on the branches increase with the NOPs added? (I may not fully understand how branches work)

E7_82_8E

1:33 AM

Would the velocity of a non-viscous incompressible fluid right after exiting a pipe(into open air) be greater than the velocity inside the pipe if the static pressure inside the pipe was greater than atmospheric pressure?

Bernardo Meurer

@ACuriousMind Yes, my bad

Fuck I delivered it wrong

Well, yes, they should

@ACuriousMind Should be 16 and -9 respectively

ACuriousMind

Okay, so what's the issue?

Bernardo Meurer

@ACuriousMind So, I want to get rid of all the NOPs

I managed to get it down to 2, as follows

SUBI  R3,R0,#1
SUBI  R6,R0,#-1
SUBI  R4,R0,#3
SUBI  R1,R0,#2
NOP
SUB   R2,R3,R6
SUBI  R4,R4,#-1
B.EQ  R4,R0,#7
SUB   R3,R0,R6
ADD   R1,R1,R2
SUB   R6,R0,R2
B.NE  R4,R0,#-6
NOP
SUB   R2,R0,R1
B     #0

Here # is denoting decimal base

See if you understand what I did here

Keep in mind what I said about the branch overhead, every two ops before a branch are always executed

ACuriousMind

Jump numbers are wrong again, for starters :P

Bernardo Meurer

Fixed

The first branch never happens so I don't fix it usually :P

The binary code was write, just the commented assembly was screwed up

0celouvsky

1:42 AM

god dammit

I don't know how my professor let this guy get a degree with this shit

"by the maximum principle"

WHICH ONE

it's not even a linear PDE for christ's sake

Bernardo Meurer

*right

ACuriousMind

@BernardoMeurer I'm not sure what this code is actually supposed to do, but yes, I see what you did there

Bernardo Meurer

@ACuriousMind Oh, it does some pointless math. Here's the explicit steps it goes through:

r1 = -2
r2 = 0
r3=-1
r4=-3
r6=1

r1 = -4
r2 = -2
r3=-1
r4=-2
r6=2

r1 = -7
r2 = -3
r3=-2
r4=-1
r6=3

r1 = -12
r2 = -5
r3=-3
r4=0
r6=5


r1 = -12
r2 = 12
r3=-3
r4=0
r6=5

You can walk through the code (easier to do with the very first one, with no NOPs or anything) and see it follow these values

So, the challenge here is: How to get rid of those two NOPs?

I can't crack it

I only got those two left and I cannot get rid of it

ACuriousMind

Are you sure that it is possible to get rid of them?

I mean, it might be that this is the shortest version of the code, right?

dmckee

@Maxwell Timaeus' case for 1st law failing relies on a infinite force. He's pulled a fast one on us.

Bernardo Meurer

1:50 AM

@ACuriousMind The prof threw it in my face that he could do it without ANY nops

I don't see how though

dmckee

Classical mechanics does not actually deal well with cases where we postulate a system residing at an unstable equilibrium.

Bernardo Meurer

To be fair I think I'm already making wine from water here with these crappy overheads

@ACuriousMind For example, removing the second NOP

0celouvsky

How does string theory deal with Norton's dome?

Bernardo Meurer

That NOP is there to protect our write to R2 which depends on R1

But it's impossible, as far as I can see, to move R1 any further up

dmckee

@0celouvsky I suppose for the convenience of the Margin Lady when it came to hand-editing submitted copy.

Mine had to be that way. I played some LaTeX magic so I could print it that way for the binders and reasonably for everyone else.

ACuriousMind

1:54 AM

@BernardoMeurer Wait, wasn't the process of exectuing an instruction five steps, where in the third it read the registers and in the fifth wrote to them? Why do we need a three-operation gap between writing to R1 and accessing it?

Bernardo Meurer

@ACuriousMind Because we do an OP in R1 that means IF->ID->EX->MEM->WB, so from the moment we run the instruction, to the moment it "returns" there are 3 operations in between. For dependent instructions to be successful the second's EX stage must only come AFTER the first's WB stage

And they cannot happen at the same time, that doesn't work either

dmckee

@Maxwell Personally I don't worry about the difficulty CM has with unstable equilibrium states, because QM gives you a reason to understand them as subject to spontaneous loss of equilibrium (and the world is all quantum all the time under the hood). But it was a big philosophical puzzle for a long time.

ACuriousMind

@BernardoMeurer Okay, so the first must enter WB when the second enters ID. But as it's written now, the first enters WB when the second enters IF, unless I'm misunderstanding something.

Bernardo Meurer

@ACuriousMind Hmm, that seems correct actually

Let me try and simulate this again

ACuriousMind

So you can just cut the NOP without any effect.

Bernardo Meurer

1:59 AM

Okay, simulated with the NOP on instruction 12, runs fine, now let's remove it

0celouvsky

Jesus Christ

$V\ge 0$ and there's a $V^{-1}$

ayy

Bernardo Meurer

@ACuriousMind Oh hey, it works now!

Oh, no, wait

I was being silly

It just goofs

Maybe it's branch overhead

Let me try something...

ACuriousMind

You said:

35 mins ago, by Bernardo Meurer

1. Branch operations finish on the EX step, meaning any two operations BEFORE a branch will always run.
2. Any arithmetic operation takes 3 cycles after it to complete. If I do ADD R1, R1, R2 I must have three command after before I can do SUB R3, R1, R1

If you figured out 2 by testing, then something is wrong with my understanding of IF-ID-EX-MEM-WB

Bernardo Meurer

@ACuriousMind I'm being a goof in 2. Ignore that, it should be 4

ACuriousMind

Huh? No, I think it should be two, so that when the ADD is in WB, the first after it is in MEM, the second in EX, and the SUB in ID. Then when SUB enters EX, the WB is finished and the result should be available, no?

Bernardo Meurer

2:06 AM

Yeah, I have no idea how to get rid of those :P

ACuriousMind

Anyway, you'll figure it out, I'm off for now

Bernardo Meurer

@ACuriousMind Hmm, I guess so yeah

I'll keep trying

Thanks!

@dmckee Ideas? :P

0celouvsky

so this is all BS

the gravitational potential of the star can't be zero anywhere

but it has to be zero in the center

Maxwell

@dmckee, so whose explanation do you find reasonable and suitable?

dmckee

@Maxwell The first law says zero net force implies zero acceleration. The second law says acceleration is proportional to net force, and if you put zero in for either of those you get zero for the other and thus recover the first law.

As I said before the first law is a special case of the second.

Maxwell

2:22 AM

@dmckee, Explanatiom of Ondřej Černotík there is quite different. As i guess

dmckee

2:36 AM

@Maxwell I begin to suspect you are hoping that there is one single authoritative answer to that question. But something that you will learn about physics over time is that many parts of the conceptual structure we've built up can be read front to back or back to front and have just as much validity either way.

People routinely say "Hey, let's see what happens if we assume [this thing that is a conclusion in the traditional development]." and get around to showing that it gives rise to the same conceptual structure with one or more of the old postulates now appearing as derived results.

That gives people the freedom to lay different philosophical takes onto the structure, disagree with one another and none the less all be right at once.

Because agreement with observation is the final discriminator and if my structure and your give equivalent predictions then they pass or fail the same test.

Beyond that people bring ideas of simplicity, parsimony and beauty with them to chose their favorites, and a kind of pedagogical democracy decides how the discipline is taught, but alternate ideas lurk in old books and papers waiting for a fresh chance.

For instance Eric Mazur has re-introduced Mach's old momentum-first teaching order and since then I've seen at least one other new book taking the same approach.

0celouvsky

2:53 AM

@BalarkaSen hey are you around?

I need someone to talk to about some Morse theory

Sir Cumference

4:51 AM

Damn, I just found out that a friend of mine in high school is taking grad school classes for credit. And she's doing ok

Anonymous

@SirCumference Grad class in what subject?

John Rennie

@0celouvsky Giggs? Ryan Giggs the footballer?

@Maxwell I'm on UK time so I was fast asleep when you posted. I'm generally here from about six a.m. UK time.

Sir Cumference

5:07 AM

@blue She said something about QM

Secret

2

Q: Are eigenvectors always orthogonal each other?

When an observable $\hat{A}$ has only discrete eigenvalues, the eigenvectors are orthogonal each other. Similarly, when an observable $\hat{A}$ has only continuous eigenvalues, the eigenvectors are orthogonal each other. But what if $\hat{A}$ has both of discrete eigenvalues and continuous ones?...

The hamitonian has to be quite strange I guess

Phase

g'morning

Slereah

5:35 AM

@0celouvsky wot

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