The h Bar

0celouvsky

8:00 PM

linear algebra killed me

@Slereah moving on to the "analytical theory of semi-groups"

sounds ominous

Good god, it's a generalization of stone's theorem

Emilio Pisanty

@Anonjohn a Born-Oppenheimer energy curve would come mighty close to a full answer to the question.

However, it's not completely obvious that there is an electronic ground state for all internuclear distances, particularly given the first paper in Wolpertinger's comment

Are asking about the full 3-body bound states or for the electronic bound states in the Born-Oppenheimer approximation? In the latter case the following may be related, although certainly not answer the question in full. sciencedirect.com/science/article/pii/0009261479800068 and possibly (I have no access to that one) researchgate.net/publication/… — Wolpertinger 3 hours ago

However, do note that just because the equations are formally symmetric it doesn't mean that there is full symmetry between them because, as Siddhart notes, the domains are different.

0celouvsky

lol, one-parameter semigroups in locally convex spaces

@EmilioPisanty help

Emilio Pisanty

@Anonjohn Regarding your answer, though if $R_j$ is the distance to the $j$th nucleus, then the swap $R_1\leftrightarrow R_2$ doesn't swap $\xi$ and $\eta$ - it's just the inversion $\eta\to-\eta$.

0celouvsky

I don't understand liner algebra

Anonjohn

Yes, the symmetry will not work. I think i will write some code and actually try to plot the Energy- Internuclear distance curve.

Emilio Pisanty

8:10 PM

@0celouvsky what about it?

0celouvsky

@EmilioPisanty the rank nullity theorem

Does it have any meaning in infinite dimensions?

I don't think it does

Emilio Pisanty

@0celouvsky I don't think it does, either

other than saying that at least one of the two needs to be infinite

0celouvsky

@EmilioPisanty I mean, the rank nullity theorem is just the first isomorphism theorem

and that makes sense for any (abelian?) group

Emilio Pisanty

damn mathematicians and their notation

lemme google that too

0celouvsky

So for $T:X\to X$, we have $R(T)\cong X/N(T)$, where $R$ is the range and $N$ is the kernel

Now assuming choice, we can actually invert that: $X\cong R(T)\oplus N(T)$

in finite dimensions you can take $\mathrm{dim}(\cdot)$ of both sides and get the familiar equation

Emilio Pisanty

8:14 PM

@0celouvsky by rank and kernel you mean spaces or dimensions of those spaces?

0celouvsky

range, not rank, sorry

so I mean the spaces

Emilio Pisanty

ok

0celouvsky

I think the issue is that in infinite dimensions, that $\cong$ is a pretty worthless result

One cannot always make it an $=$

Emilio Pisanty

@0celouvsky $\cong$ is about as strong a result as you can hope for

what does $=$ even mean?

0celouvsky

@EmilioPisanty Ah, but that's the thing

$R(T)$ and $N(T)$ are subspaces of $X$

So $X=R(T)\oplus N(T)$ does make perfect sense

Emilio Pisanty

8:16 PM

@0celouvsky ah, I see the point

@0celouvsky except it's not a natural decomposition

0celouvsky

@EmilioPisanty It's just that in functional analysis, one often sees results like $X=\overline{R(T)}\oplus N(T)$

Emilio Pisanty

case in point $$T=\begin{pmatrix}0&0\\1&0\end{pmatrix}$$

0celouvsky

And I think the big deal is that it's an honest to god $=$

Emilio Pisanty

@0celouvsky no, it's not

cf example above

0celouvsky

@EmilioPisanty for what?

Emilio Pisanty

8:18 PM

$R(T)=N(T)$ there

0celouvsky

I'm telling you that for certain operators, there is a legit $=$ decomposition

But in general it's just $\cong $

Emilio Pisanty

@0celouvsky those would have to be some pretty special operators, though

@0celouvsky $\mathbb R(0,1)$

0celouvsky

Yeah :P

Can't into matrix multiplication

Emilio Pisanty

@0celouvsky can't into grammar either, apparently

=P

0celouvsky

It's a meme, never mind

Emilio Pisanty

8:20 PM

ok

Anonjohn

@EmilioPisanty Yes, I was terribly mistaken with that. Hence the deletion!

0celouvsky

@EmilioPisanty For instance the Koopman operator associated with an euqimeasure transformation on $L^2$

Emilio Pisanty

@0celouvsky anyways, I think you're getting sidetracked with having the same space for domain and codomain

$R(T)$ and $N(T)$ live in fundamentally different spaces

0celouvsky

@EmilioPisanty not if the map is an endomorphism

Emilio Pisanty

$R(T)\oplus N(T)$ is never going to make natural sense

0celouvsky

8:21 PM

@EmilioPisanty but it does if $T:X\to X$

what's wrong with that

Emilio Pisanty

@0celouvsky that's the thing, formally it's the same, but you shouldn't make that identification

@0celouvsky you can

but it won't help you

0celouvsky

@EmilioPisanty I'm telling you that it's a very common thing in functional analysis

But I'm trying to figure out why it isn't trivial, and it's because the abstract algebra only gives a $\cong $

Emilio Pisanty

@0celouvsky sure, but here you need to put your geometer cap on

if you have one, that is

if not, go buy one

0celouvsky

Apparently I'm known as a geometer around these parts

Emilio Pisanty

@0celouvsky so put on your geometer cap and benchmark all your results against that $T$ above

0celouvsky

8:23 PM

I don't see what's verboten. $X=R(T)\oplus N(T)$ just means that for $x\in X$, we have $x=Ty+x_0$ uniquely, where $Tx_0=0$.

Emilio Pisanty

if they don't hold for a two-by-two over $\mathbb Z_2$, they're not going to hold in fancier spaces

@0celouvsky it's possible, sure

but it's very rare

0celouvsky

Ok, so we're agreeing

Emilio Pisanty

@0celouvsky that it's possible, yes

0celouvsky

yes

but in general not true

and you gave a good example

Emilio Pisanty

but it's not true in general, so if you're trying to prove it for general $T$ then it's not gonna work

0celouvsky

8:25 PM

yeah now that's clear, thanks

Emilio Pisanty

that said

0celouvsky

it's also a strange thing with infinite dimensional spaces

Emilio Pisanty

I don't see how you can be left with anything more helpful than $\cong$

0celouvsky

you can have a linear isomorphism $Y\to \bar Y$ even if $Y$ is not closed

($Y\subset X$ some TVS)

Emilio Pisanty

@0celouvsky $\bar Y$ being topological closure in $X$?

0celouvsky

8:26 PM

@EmilioPisanty that's where the functional analysis comes into play,

@EmilioPisanty yeah

Emilio Pisanty

@0celouvsky any clean examples?

0celouvsky

@EmilioPisanty Depends on how much you like measure theory

Emilio Pisanty

@0celouvsky ultimately though, doesn't that statement just say "when we say 'continuous operator' we add the 'continuous' for a reason y'know"?

@0celouvsky I like it, but if it involves measures then it's probably not gonna make the bar for clean

0celouvsky

@EmilioPisanty yeah, plain old linear isomorphism is pretty weak

Emilio Pisanty

OK, so, with all that settled, what's bothering you?

0celouvsky

8:29 PM

Although, any two infinite-dimensional separable Hilbert spaces are isometrically isomorphic

@EmilioPisanty Ah, I thought I needed to assume $X=R(T)\oplus N(T)$ for a proof, but that resolved itself

Emilio Pisanty

@0celouvsky ;-) it was never gonna be necessary because it's not true

0celouvsky

@EmilioPisanty Clearly.

Actually what I needed was $X=R(T)\oplus R(I-T)$

And for $T$ idempotent, $R(I-T)=N(T)$.

Emilio Pisanty

@0celouvsky sounds rather more reasonable

0celouvsky

legitimate equality there

Emilio Pisanty

though not necessarily true for arbitrary $T$

take $T=2I$ for instance

0celouvsky

8:32 PM

Uh, I'm pretty sure it is always true

Since $T+(I-T)=I$

and $R(I)=X$ trivially

Emilio Pisanty

@0celouvsky direct sum requires uniqueness in the representation

yuggib

@0celouvsky yes

0celouvsky

@EmilioPisanty $x=2x+(-x)$ is pretty unique in my book

@yuggib the issue resolved itself, thanks

@yuggib Oh, did you see my $C^\infty(\Bbb R)\otimes C^\infty(\Bbb R)$ question from the other day?

Emilio Pisanty

@0celouvsky $X=R(T)\oplus R(1-T)$ says that for all $x\in X$ there exist unique vectors $y,z$ in $X$ such that $x=T(y) + (I-T)(z)$. Now take $T=2I$, let $x\in X$, and set $y=3x$ and $z=5x$.

0celouvsky

@EmilioPisanty No

that's not what it says

Emilio Pisanty

8:37 PM

@0celouvsky what is it you think it says?

0celouvsky

it says that $x=x_1+x_2$, $x_1\in R(T), x_2\in R(I-T)$ such that $x_1$ and $x_2$ are unique.

Emilio Pisanty

@0celouvsky yes

now take $T=2I$, let $x\in X$, and take $x_1=x$ and $x_2=0$

or $x_1=0$ and $x_2=x$

Danu

If I think of a map $X\to X$ as a graph in the product $X\times X$, then that graph is closed. And therefore the locus where two maps agree is closed. Does that sound right?

0celouvsky

@EmilioPisanty then $Tx=2x\ne x$

assuming $x\ne 0$

Emilio Pisanty

@0celouvsky no. $x_1=T(\frac12 x)$ is still in $R(T)$.

0celouvsky

8:39 PM

@Danu Why should the graph be closed?

Danu

@0celouvsky I'm assuming the map to be continuous. Isn't it just homeomorphic to the diagonal?

0celouvsky

Ah, then you need $X$ to be Hausdorff as well

Danu

Yeah, always assume that, of course.

If it ain't Hausdorff.... It's either something algebraic geometric or I don't care :D

Emilio Pisanty

@0celouvsky ok, yeah, scratch unique on that comment above and put in unique modulo $N(T)$, but the point stands.

0celouvsky

what's the point, I'm confused

@Danu It sounds right.

Emilio Pisanty

8:42 PM

@0celouvsky the point is that when you say $X=R(T)\oplus R(I-T)$, you need to decouple the action of the two operators

you're saying $X=U\oplus V$ where $U,V\leq X$ are subspaces

0celouvsky

good god $\le $ for spaces???

Emilio Pisanty

which happen to be $U=R(T)$ and $V=R(I-T)$, but that doesn't matter much

@0celouvsky yeah. You had some other use planned for $U\leq X$?

Danu

Ah, there we go

Closed graph theorem

In mathematics, the closed graph theorem is a basic result which characterizes continuous functions in terms of their graphs. There are several versions of the theorem. == The closed graph theorem == In mathematics, there are several results known as the "closed graph theorem". For any function T : X → Y, we define the graph of T to be the set { ( x , y ) ∈ X × Y ∣ T x = y } . {\displaystyle \lbrace (x,y)\in X\times Y\mid Tx=y\rbrace...

theorem :D

Emilio Pisanty

@0celouvsky In any case, for $T=2I$, and arbitrary $x\in X$, both $x$ and $0$ are in $U=R(T)$ as well as in $V=R(I-T)$.

which makes the decomposition nonunique

0celouvsky

@Danu ah, I only know the functional analytic version

Danu

8:44 PM

I was trying to understand why an isometry is determined by its value at a point & the tangent map there.

0celouvsky

looks like you need one of your spaces to be compact for that to work

@Danu Ah

Danu

I was struggling with why the locus where they agree is closed :D

topology 101 how do I

0celouvsky

You can understand it without the graph

Danu

of course

It's a trivial thing

0celouvsky

um, ok

why were you wondering then?

Danu

8:46 PM

the non-trivial part is why the set is open

because "topology 101 how do I"

0celouvsky

@Danu exponential map or something

Danu

yeah something like that

just need that evry point has a nbh that you can cover by geodesics emanating from the center

loltospoon

If I'm given a step potential and I try to use Euler's method to generate an allowed energy and plot the corresponding wavefunction and it turns out like this: imgur.com/a/B4W14

0celouvsky

a normal neighborhood

loltospoon

Should I conclude that this was not an allowed energy?

Danu

8:48 PM

it has a name? ok

yuggib

@0celouvsky no

Anonjohn

@EmilioPisanty One more Comment: If you took the Anti proton to infinity, then, we know the existence of a bound state solution. If However, you made the internuclear distance 0, we know that we go down to the free particle case

0celouvsky

@EmilioPisanty You're right, one needs $T=T^2$ as well

good catch

luckily my operator satisfies that

Anonjohn

Further to first order in inter nuclear distance the hamiltonian is positive definite.

0celouvsky

@yuggib

yesterday, by 0celouvsky

@yuggib Is $C^\infty(\Bbb R)\otimes C^\infty(\Bbb R)$ dense in $C^\infty(\Bbb R^2)$ in the $\mathscr C^\infty$ topology? (This is the topology of uniform convergence for each multi-derivative on compact sets.) I can't find it online. There's a sketch of this on Math Stack for compactly supported functions, but I don't think the proof generalizes.

yuggib

9:12 PM

@0celouvsky don't know, probably yes

heather

@SirCumference same here - thanks to Khan Academy. now though, i know the way of Python

Emilio Pisanty

@Anonjohn $c=0$ is too strong; you don't actually care about that

You do care about the limit where $c\ll a_0$, though, and there you have a weak dipole but you do have a nonzero well

@0celouvsky toldja

you're proving that all idempotents are projectors?

that's possibly the most beautiful result in all of linear algebra, I think

0celouvsky

I'm not proving that, no

Emilio Pisanty

@0celouvsky shame

0celouvsky

@EmilioPisanty I already know that

@yuggib does Yosida like being cryptic

this book is infuriating at times

ACuriousMind

9:23 PM

@0celouvsky I was...otherwise occupied. :P

0celouvsky

it better have been a woman

ACuriousMind

What if it was a man?

Or an AI

0celouvsky

like 97% of the population is straight, so it's easier for me to say woman and you correct me I assumed wrong

@ACuriousMind pretty sure you can do that while doing other things

but a woman would demand your whole attention

yuggib

9:37 PM

@0celouvsky maybe, that's the japanese style probably

0celouvsky

@yuggib yeah I really don't understand this semigroup growth bound stuff

supposedly for a $C^0$ semigroup one has $||T_t||\le Me^{\beta t}$ for $M>0$ and $\beta<\infty$

no clue how he wants me to prove that from $\lim_{t\to\infty} \frac{1}{t}\log ||T_t||=\inf_{t>0}\frac{1}{t}\log ||T_t||$

Jaime Gallego

@heather That makes sense, most programmers' first contact is with HTML/CSS/JS or Python.

Do you know C?

0celouvsky

honestly what is with all of these people in high school learning how to program?

what the hell are you programming

Jaime Gallego

I have a code folder somewhere

Ah yes, found it

Mostly programs I created during CS50

ACuriousMind

My first contact with programming was Pascal, or rather Delphi.

0celouvsky

9:50 PM

@yuggib did you ever read chapter IX "semigroups" in Yosida?

heather

@JaimeGallego eh, sort of, not really. I've used the Arduino, which uses a version of C, but I couldn't tell you what a pointer is.

well, I mean, I guess I sort of have a vague idea what it is. But not really.

Jaime Gallego

If you have some time to spare

ACuriousMind

Heh, MO can be funny sometimes

loltospoon

Anyone know of a free program I can use to model wavefunction using arbitrary potentials?

dmckee

@loltospoon More or less any numeric PDE solver will do the job at one level.

The devil is always in the details.

Jaime Gallego

10:01 PM

@heather I recommend the course I mentioned, CS50x by Harvard University.

I have good memories of it. There was this problem

0celouvsky

@ACuriousMind I give up. Suppose $f(t)\ge 0$ satisfies $\lim_{t\to\infty}\frac{1}{t} \log f(t)=\inf_{t>0}\frac{1}{t}\log f(t)$, which might be negative infinity. Then why are there constants $M>0,\beta<\infty$ with $f(t)\le Me^{\beta t}$?

That limit gives me no information about the function as far as I can tell

Jaime Gallego

Where you had to recover JPEG images from a binary file. The task was to write a program to process the data and output each image in a separate file

dmckee

@heather At the lowest level a pointer is just a bit of memory that can be interpreted at identifying another bit of memory.

In all but the lowest level of representation that identification also come with an understanding of what kind of data is stored at the identified location in memory, but that's a frill.

You "dereference" a pointing by going to the memory it points to and getting the value there for some kind of use.

yuggib

@0celouvsky no, I read about semigroups in another book

Pazy

0celouvsky

@yuggib but do you know about this growth property?

@yuggib lol this book makes no jokes, straight into the analysis on page 1

You know it's a good book when it begins with "let X be a Banach space"

loltospoon

10:11 PM

@dmckee throughout my physics studies my professors have shown me those programs (like a Java applet or something) that you can use to plot wavefunctions for any type of potential. Sadly, I never paid much attention to where I can find them ::sigh:: . I specifically need one of those types of programs.

0celouvsky

@yuggib Nice, he has a much simpler proof. I still wonder what Yosida is going for in his "proof"

ACuriousMind

@0celouvsky Suppose $f > M\exp(\beta t)$ for all $M,\beta$. Then $\frac{1}{t}\log(f(t)) > \frac{\log(M)}{t} + \beta$, meaning as $t\to\infty$, $\frac{1}{t}\log(f(t)) > \beta$, meaning $\lim_{t\to\infty} \frac{1}{t}\log(f(t)) = \infty$. Contradiction.

0celouvsky

@ACuriousMind See I thought that, but the negation of that statement is not what you wrote...it's that for each $M,\beta$, $\exists t>0$ such that $f(t)>Me^{\beta t}$

Unless I'm missing something

negation of $\forall t$ is $\exists t$

ACuriousMind

@0celouvsky Ah, you're right. But if those $t$ for which that occurs don't occur towards infinity, then your function is unbounded in some $(0,T]$. I suppose you know that $f$ is continuous on $(0,\infty)$ or something?

0celouvsky

It's bounded on every compact interval

@ACuriousMind Maybe I should try to dominate by $ne^{nt}$ for $n=1,2,\dotsc$ and see if that contradicts

using what you just said

ACuriousMind

10:21 PM

@0celouvsky Right. So as $\beta\to\infty$ for some fixed $M$, you must have that $t_\beta$ for which that occurs shimmies off to infinity. So for the limit to exist at all the function must then diverge to $\infty$ as $t\to\infty$, which gives the same contradiction, no?

0celouvsky

@ACuriousMind yeah modulo details that's right

thanks

heather

@dmckee, ah, and is the stack like the stack of data, and the pointer points to a bit of data in the stack?

@JaimeGallego, thanks, I'll take a look at that course

0celouvsky

@ACuriousMind It works well with $ne^{n t}$, because then you get a sequence $t_n$ that is either bounded or goes to infinity.

So some $Ne^{Nt}$ must bound

Jaime Gallego

@heather The malloc function actually gives addresses to the heap, not the stack

dmckee

@heather Yes. The push operation adds another chunk of data to the "top" of the stack and then moves the pointer. The push operation returns the topmost data and moves the pointer the other way.

0celouvsky

10:24 PM

@JaimeGallego what is this, algebraic geometry?

ACuriousMind

No heaps in geometry. Stacks do occur, though, although I don't think I know what a stack is. :P

0celouvsky

@ACuriousMind you can totally have a heap of wheat

therefore it should be something in algebraic geometry

ACuriousMind

@EmilioPisanty 36 seconds from posting to closure :P

0celouvsky

and the fact that you refer to algebraic geometry as just geometry is a little gross

yuggib

@0celouvsky I know about the growth property, yes, but only a bit

and yeah, Pazy does not lose time in futile introductions

heather

10:29 PM

@ACuriousMind geesh ::round of applause::

ACuriousMind

@0celouvsky My recent adventures in it have made me marvel how funnily you can express e.g. curvature conditions on Kähler manifolds

0celouvsky

@ACuriousMind do I want to know?

heather

I was so angry at it I fixed the grammar and spelling. It hurt my eyes.

Jaime Gallego

"Galilayo gallayaie"

heather

::shivers::

ACuriousMind

10:34 PM

@0celouvsky I dunno: The Ricci class of a Kähler manifold is $c_1(M)$, which is by definition $c_1(TM)$, which is the same as $c_1(\wedge^n TM)$, and $\wedge^n TM$ is what one calls the "anticanonical bundle", and positivity of the Chern class (read, essentially the positivity of the curvature) is equivalent to ampleness of the bundle/sheaf from the algebro-geometric viewpoint.

0celouvsky

What's a Ricci class? I'm assuming $c_1$ is the first Chern class

ACuriousMind

The Ricci class is just the cohomology class of the Ricci form, and yes, $c_1$ is the Chern class

0celouvsky

I'll be honest, I don't know how $c_1(\Lambda^n TM)$ is defined

heather

> never know could be history here

::facepalms::

ACuriousMind

@0celouvsky From a differential viewpoint, you take a (holomorphic) connection on the line bundle, and it's the cohomology class of its curvature divided by $2\pi$

0celouvsky

10:41 PM

::looks up axiomatic definition of Chern class in Kob-No::

Jaime Gallego

@heather At first, pointers do not seem to have any practical use, but they are really neat. Suppose we are executing a program on a machine. Processors have what's called an instruction pointer, which is a pointer (just a number interpreted in a particular way) that tells which step of the program is being executed

0celouvsky

yeah..I remember why I forgot this

ACuriousMind

(Of course, in that approach you need to show that the choice of connection doesn't matter)

0celouvsky

@ACuriousMind that's just Chern-Weil theory

Jaime Gallego

So the program adds one when going to the next step

But

ACuriousMind

10:41 PM

@0celouvsky "just" ;P

0celouvsky

@ACuriousMind That's something I know! So yeah, just :P

@ACuriousMind So what connection are you putting on $\Lambda^n TM$?

i.e. what's the curvature

my first thought was $\wedge^n \mathcal R$, but that's not even a 2-form lol

ACuriousMind

Anyway, the marvel is not in the $c_1$, but that it can be phrased in an algebraic terms as some tensor power of this line bundle having "sufficiently" many sections.

0celouvsky

@ACuriousMind I'm working my way through that sentence, I haven't gotten to your ampleness yet

Jaime Gallego

@heather The pointer is just a bit of data anyway, so it can be altered as well! So, when dealing with things like conditional statements, like ifs or elses, the program itself overwrites the instruction pointer with new information that tells where to continue (depending on the condition being true or not), and the program continues its execution there like nothing has happened

ACuriousMind

@0celouvsky Well, it doesn't matter since $c_1$ is topological, does it? I read something where the connection was essentially $\bar{\partial}$, but I'm too lazy to look it up exactly now

Jaime Gallego

10:47 PM

Thus, one gets control over which code gets executed.

0celouvsky

@ACuriousMind yah but how do you know that $c_1(TM)=c_1(\Lambda^nTM)$?

Jaime Gallego

I shouldn't have said "a bit of data" there. I meant a small quantity of data

heather

huh, interesting

so the pointer itself can be manipulated to "slide along" the stack?

0celouvsky

@ACuriousMind presumably you put/lift a connection on $c_1(\Lambda^nTM)$ that has curvature form $\mathcal R$

or...something like that

oh wait, it's gonna be $\Omega^nM$ valued curvature

Jaime Gallego

Yes, that's what is going on under the hood when going from one statement to the next.

0celouvsky

10:51 PM

so $\wedge^n \mathcal R$ might be the curvature

who knows

ACuriousMind

@0celouvsky It's a general fact that the determinant bundle has the same $c_1$ as the original bundle

0celouvsky

@ACuriousMind I see

ACuriousMind

It follows using the identities for tensor products and direct sums for Chern classes

0celouvsky

@ACuriousMind Yeah I remember the direct sum thing but I don't really see what direct sums have to do with determinant bundles

ACuriousMind

Ah, in this case the direct sums are not relevant, you just need to know how it behaves with respect to $\otimes$, or more precisely, $\wedge$

0celouvsky

10:59 PM

@ACuriousMind Ugh. Well I remember how the Chern character responds to $\otimes$, but I'm not even sure what that thing (the character) is right now

ACuriousMind

The direct sums are relevant when your original bundle is not a line bundle, then you have to use the splitting principle to reduce it to the question of direct sums of line bundles.

I wasn't going for a discussion of the details here, actually :P

0celouvsky

@ACuriousMind Well you made me curious

ACuriousMind

@0celouvsky A formal sum of Chern classes? For what we need, it's $1+c_1+\text{stuff that doesn't matter}$

0celouvsky

@ACuriousMind You're in luck, I just realized I have circuits homework due tomorrow that I haven't started

Bernardo Meurer

Heya

@dmckee Are you around?

dmckee

11:06 PM

Maybe.

Depends when my better half sends me out.

Bernardo Meurer

@dmckee Would you mind checking is my rewriting of a code in MIPS assembly to C is correct? If you don't know MIPS don't worry, the commands are similar to all other assembly languages

It's like 30 lines, but I'm a noob at asm

6 hours ago, by Bernardo Meurer

I have this assembly code:

dmckee

I've never done MIPS assembly, but the first thing I notice is that your initial assignments all appear to rely on R0 having a value of 0. Do you know that this is true?

Bernardo Meurer

@dmckee I do, yeah, I have all my registers set to 0

I designed the CPU that way :P

dmckee

Second thing: use labels for your branches even in assembly.

Bernardo Meurer

@dmckee Oh, that Assembly is "fake"

I'm writing in machine code

dmckee

11:14 PM

Who wants to count every time (not to mention that counting is error prone).

Bernardo Meurer

my instruction decoder doesn't work lol

I write in binary

dmckee

@BernardoMeurer Uhg. Fix that soon. Build the tools to build the tools and all that.

Bernardo Meurer

@dmckee I know, I know it's just that I am currently debugging the instruction decoder itself

and there's something fishy going on

So I wanted to test my code on C to see the correct result

I think I'm screwing up the pipeline timing, because MIPS is a bitch with this stuff

IF->ID->EX->MEM->WB

0celouvsky

@dmckee for complex impedance I just use ohm's law like I usually would, right?

dmckee

I don't see anything obviously wrong, except that the forward jump ends one short---but that would be correct is the PC is incremented after the operation finishes.

Again, I don't know MIPS.

Bernardo Meurer

11:18 PM

It is, yeah, PC get's

Wait a second

Hmm

Stupid five stage pipeline

AH

signal storage: storage_type := (
     --    OPCODE  &   DR   &   SA   &   SB   &        KNS        -- ASSEMBLY CODE
     0 => "000011" & "0011" & "0000" & "0000" & "00000000000001", --        SUBI  R3,R0,#1
     1 => "000011" & "0110" & "0000" & "1111" & "11111111111111", --        SUBI  R6,R0,#-1
     2 => "000010" & "0001" & "0011" & "0110" & "00000000000001", --        SUB   R1,R3,R6
     3 => "000011" & "0100" & "0000" & "0000" & "00000000000011", --        SUBI  R4,R0,#3
     4 => "010111" & "0010" & "0100" & "0000" & "00000000000111", --        B.EQ  R4,R0,#7    ; --> if (R4 = R0) go…

How I code right now lol

fuck my life

@dmckee Debugging this is literally hell

dmckee

11:52 PM

Finding, characterizing and fixing bugs is the biggest part of real programming.

Large parts of the lore of programming and the many theories on what constitutes the right Process are about keeping control of that aspect.

For instance, test driven design is about knowing with certainty where bugs are not and never re-introducing them.

Bernardo Meurer

@dmckee Aha! I found it!

It was a bug on my decoding multiplexer for the branch condition

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