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“Generative AI is a sack of wet garbage”

 
 
 
Elixir said:
Jarhyn said:
Copernicus said:
It is possible that one of OpenAI's experimental systems has cracked Jahryn's passwords, broken into his computer, and is now posting gibberish in this thread.
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Or maybe I found something that is complicated and bizarre, and I might actually be doing something real; How about trying to paste THIS into ChatGPT:

Code: 
\documentclass[11pt]{article}

\usepackage{amsmath,amssymb,amsthm}

\usepackage[margin=1in]{geometry}


\theoremstyle{plain}

\newtheorem{theorem}{Theorem}


\theoremstyle{remark}

\newtheorem*{remark}{Remark}


\title{A free involution preserves the Perron eigenvalue}

\author{}

\date{}


\begin{document}

\maketitle


\noindent A short, self-contained fact about nonnegative matrices with a

symmetry: if a strongly connected weighted digraph carries a

fixed-point-free automorphism of order 2, then folding the graph by

that symmetry does not change its Perron--Frobenius eigenvalue. The

proof is four elementary steps once the classical Perron--Frobenius

theorem is granted as a black box.


\begin{theorem}

Let $G = (V, A)$ be a finite, strongly connected digraph with

nonnegative edge weights, given by its irreducible adjacency matrix

$A$. Let $\varphi : V \to V$ be a \emph{fixed-point-free involution}

that is also a graph automorphism:

\[

\varphi \circ \varphi = \mathrm{id}, \qquad

\varphi(u) \neq u \ \ \forall u \in V, \qquad

A_{\varphi(u),\varphi(w)} = A_{u,w} \ \ \forall u,w \in V.

\]

By the Perron--Frobenius theorem, $A$ has a unique eigenvalue

$\lambda = \lambda(A) > 0$ admitting an everywhere-positive

eigenvector $v$, i.e.\ $Av = \lambda v$ with $v > 0$; this $\lambda$ is

the Perron eigenvalue of $A$. Form the \emph{quotient graph} on orbits

$\{u,\varphi(u)\}$, choosing a representative $a$ for each orbit

$\alpha$:

\[

B_{\alpha,\beta} \;=\; \sum_{w \,\in\, \beta} A_{a,w}.

\]

Then

\[

\lambda(B) \;=\; \lambda(A).

\]

\end{theorem}


\begin{proof}

\textbf{Step 1 (}$\varphi$\textbf{ carries the Perron eigenvector to

another one).} Set $v'_u := v_{\varphi(u)}$. Substituting $w =

\varphi(w')$ (valid since $\varphi$ is a bijection) and using the

automorphism identity together with $\varphi^2 = \mathrm{id}$:

\[

(Av')_u

= \sum_{w} A_{u,w}\, v_{\varphi(w)}

= \sum_{w'} A_{u,\varphi(w')}\, v_{w'}

= \sum_{w'} A_{\varphi(u),w'}\, v_{w'}

= (Av)_{\varphi(u)}

= \lambda\, v_{\varphi(u)}

= \lambda\, v'_u .

\]

So $Av' = \lambda v'$, and $v' > 0$ since $v > 0$ and $\varphi$ is a

bijection.


\medskip

\textbf{Step 2 (}$v$\textbf{ is constant on every orbit).} The

positive $\lambda$-eigenspace is one-dimensional (Perron--Frobenius),

so $v' = c\,v$ for some $c > 0$. Applying $\varphi$ twice returns $v$

to itself, i.e.\ $(v')' = v$, while also $(v')' = c\,v' = c^2 v$.

Hence $c^2 = 1$, and since $c > 0$, we get $c = 1$. So

$v_{\varphi(u)} = v_u$ for every $u$.


\medskip

\textbf{Step 3 (}$v$\textbf{ descends to a positive eigenvector of

}$B$\textbf{).} Define $\bar v_\alpha := v_u$ for any $u \in \alpha$

(well defined by Step 2). For a representative $a$ of orbit $\alpha$:

\[

(B\bar v)_\alpha

= \sum_{\beta} B_{\alpha,\beta}\, \bar v_\beta

= \sum_{\beta} \Big(\sum_{w \in \beta} A_{a,w}\Big)\, \bar v_\beta

= \sum_{w \in V} A_{a,w}\, v_w

= (Av)_a

= \lambda v_a

= \lambda\, \bar v_\alpha .

\]


\medskip

\textbf{Step 4 (conclude).} Since $\varphi$ is free and $G$ is

strongly connected, the quotient graph is again strongly connected, so

$B$ is irreducible and nonnegative. An irreducible nonnegative matrix

admits an everywhere-positive eigenvector for exactly one eigenvalue:

its Perron root. Since $\bar v > 0$ and $B\bar v = \lambda \bar v$, it

follows that $\lambda = \lambda(B)$.


Combining the two displayed equalities, $\lambda(B) = \lambda =

\lambda(A)$. $\blacksquare$

\end{proof}


\begin{remark}

The only input taken as a black box above is the classical

Perron--Frobenius theorem itself: for an irreducible nonnegative

matrix, the eigenspace of the spectral radius contains a unique

(up to positive scaling) everywhere-positive eigenvector, and no other

eigenvalue admits one. See Horn \& Johnson, \emph{Matrix Analysis},

2nd ed., Cambridge University Press, 2013, Theorem 8.4.4; or Meyer,

\emph{Matrix Analysis and Applied Linear Algebra}, SIAM, 2000,

Chapter 8. Everything else above (Steps 1--4) is elementary linear

algebra from the automorphism hypothesis alone.

\end{remark}


\begin{remark}

This is exactly the mechanism behind a fact from the theory of Pisot

substitutions: on the contact-boundary graph of such a substitution,

the map $\varphi : [i,x,j] \mapsto [j,-x,i]$ (swap the two contact

faces and negate the translation vector) is precisely such a

fixed-point-free automorphism on the graph's dominant recurrent core

-- which is why the quotient by $\varphi$ shares the graph's Perron

eigenvalue exactly, with no further computation needed once freeness

of $\varphi$ is checked.

\end{remark}


\end{document}
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"Flux capacitor" belongs in there, somewhere!
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Seriously. Do you need me to provide a LEAN file for you? Just try the exercise.

In fact:
This is a nice, concise proof. I read through it carefully. Mathematically, the argument is essentially correct, but there are a few places where I'd recommend tightening the exposition if you're planning to submit it.


1. State explicitly that the quotient matrix is well-defined​


You define

Bα,β=∑w∈βAa,w,B_{\alpha,\beta}=\sum_{w\in\beta}A_{a,w},Bα,β=w∈β∑Aa,w,
where aaa is a representative of orbit α\alphaα.


Strictly speaking, you should prove this does not depend on the chosen representative.


Suppose a′=φ(a)a'=\varphi(a)a′=φ(a). Then

∑w∈βAφ(a),w=∑w′∈βAφ(a),φ(w′)=∑w′∈βAa,w′,\sum_{w\in\beta}A_{\varphi(a),w}=\sum_{w'\in\beta}A_{\varphi(a),\varphi(w')}=\sum_{w'\in\beta}A_{a,w'},w∈β∑Aφ(a),w=w′∈β∑Aφ(a),φ(w′)=w′∈β∑Aa,w′,
since every orbit is {w′,φ(w′)}\{w',\varphi(w')\}{w′,φ(w′)}.


That's only two lines, but it removes any ambiguity.



2. Step 1 is correct​


The substitution

w=φ(w′)w=\varphi(w')w=φ(w′)
is completely legitimate because φ\varphiφ is a bijection.


The identity

Au,φ(w′)=Aφ(u),w′A_{u,\varphi(w')}=A_{\varphi(u),w'}Au,φ(w′)=Aφ(u),w′
is exactly the automorphism condition with one application of φ2=id\varphi^2=idφ2=id.


No issues here.



3. Step 2 is elegant​


The argument

v′=cv,(v′)′=v,(v′)′=c2vv'=cv,\qquad(v')'=v,\qquad(v')'=c^2vv′=cv,(v′)′=v,(v′)′=c2v
is cleaner than the more common argument using normalization.


This is perfectly valid.



4. Step 3 is the heart of the proof​


The computation

∑β∑w∈β=∑w∈V\sum_\beta\sum_{w\in\beta}=\sum_{w\in V}β∑w∈β∑=w∈V∑
works because the orbits partition VVV.


Nothing further is needed.



5. Step 4​


Here I'd slightly strengthen one sentence.


Instead of


Since φ\varphiφ is free and GGG is strongly connected, the quotient graph is again strongly connected,
Click to expand...

I'd actually justify it.


For instance:


If α,β\alpha,\betaα,β are quotient vertices, choose representatives a∈αa\in\alphaa∈α and b∈βb\in\betab∈β. Since GGG is strongly connected there is a directed path from aaa to bbb. Projecting that path to orbits gives a directed path from α\alphaα to β\betaβ, so the quotient graph is strongly connected.
Click to expand...

That proof is only one sentence but makes the claim completely self-contained.



6. Minor wording​


Instead of


By the Perron–Frobenius theorem, AAA has a unique eigenvalue admitting an everywhere-positive eigenvector
Click to expand...

I'd write


By the Perron–Frobenius theorem, the spectral radius λ(A)\lambda(A)λ(A) is a simple eigenvalue admitting an everywhere-positive eigenvector, and no other eigenvalue admits one.
Click to expand...

This is closer to the standard formulation.



7. A slightly stronger theorem​


Your proof actually proves something a bit more general.


Nowhere do you really use that the group has order 222 until Step 2.


The same proof works for any finite group GGG of automorphisms.


Indeed, averaging the Perron vector over the group gives a positive invariant Perron vector, which descends to the quotient. Thus

λ(A/G)=λ(A).\lambda(A/G)=\lambda(A).λ(A/G)=λ(A).
The involution case becomes the simplest special case.



Overall assessment​


I'd judge this as mathematically sound. The only genuine omission is that the quotient matrix should be shown to be well-defined. Everything else is a matter of exposition rather than correctness.


As written, I'd rate it approximately:


  • Correctness: 9.5/10
  • Clarity: 9/10
  • Self-containment: 8.5/10 (because of the missing well-definedness proof and the unproved strong-connectivity claim)

Adding those two short justifications would make it a polished, fully self-contained note.
It's LaTeX code for math formatting. It's to deal with Perron Eigenvectors. There's a good discussion about the concepts of tilings and substitutions on the following blog pages:


And a brief introduction to tilings and Rauzy fractal concepts here:


In the second blog, there's an interactive demo to generate a penrose tiling, which is itself a Rauzy fractal.

The whole point is that it's NOT gibberish. It's just... Really obscure math stuff pertaining to quasicrystal dynamics and point substitution on particular algebraic fields, and people tend to misuse "quasicrystal" and "fractal" and such.

I guess I'll just stop bringing it up here, if nobody is actually going to take it seriously, though.

Maybe y'all should consider whether y'all are the wet sacks of garbage at this point? I know I sure as shit felt like a sack of wet garbage after watching an AI properly implement a comprehensive Pisot toolkit, but this thread just feels like a really poor excuse for people to cope with the realities of what AI is actually capable of these days.

AI is very good at untangling language, and applying it correctly based on context. In fact my first hint that I could get AI to be more rigorous was when it pushed back on some dubious connections I was trying to probe.

If you give an AI a context that contains nothing but strict adherence to some set of requirements for language use, it will apply that language strictly and formally.

If you only just scratch the surface, you're not going to get that.

That's really the secret to getting AI to produce useful results. If nobody here cares, and everyone wants to just assume that's not true, that's their loss.

Not knowing how or why to do that is not a valid excuse for you getting wet garbage results, when others have achieved using AI to find new counterexamples and create new proofs.
 
I suppose I should also establish the restriction to well defined quotient matrices, but other than that...

Maybe I should renew my OpenAI sub...
 
Jarhyn, you do seem to realize that your wall of text programmer Gish gallop might not be of interest to people. Apply the KISS (Keep It Simple Stupid) principle, and your posts might be met with less derision.
 
ZiprHead said:
Jarhyn said:
Maybe I should renew my OpenAI sub...
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Why did I read that as "OprahAI"?

Pondering the thought of that existing gives me chills.
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IDK man.

At any rate, it's still going. I ended up taking the free offer for a month of OpenAI Codex access and plugged it in with Codex CLI.

Codex is probably three or more times as capable as Claude 5 and 10 times more capable than Minimax (opus 4.7).

Fable would struggle to make LEAN proofs, but with the Codex model substrate, it's much more streamlined, and it actually does whole pieces of its work directly in LEAN.

Not all of my intuitions are bad. In fact, many of them over the years have apparently been extremely solid, if badly communicated or with too much pride.

One nice thing about this project is it's connection to physics. Quasicrystals and cut-and-project models are both governed by the same dynamics, and this is captured by the Rauzy fractal dynamics.

Characterizing why involves looking at the "shape" and "angle of incidence" and the "window geometry" of the hyperbolic representation or the "hyperspace" (never thought I would end up saying that outside a sci-fi context).

The hypothesis I'm having it run with is that families of Pisot substitutions can be identified by looking at where the geometry has some real shift, like a face being eclipsed, or there being more faces on the shape, or whatever it is that is being represented by the hyperspace view.

This way, instead of looking across singular cases, I have something that can make more general observations (about what happens when the shape is different or the window is different or the organization is different.

This is all an observation built from my vague memory of this paper here, as reported on by an online pop-sci magazine: https://arxiv.org/abs/2112.14783
 
And then I burned a week of Codex usage in a single day.

But, it classified the whole Sigma1/Sigma2 Pisot family using the new near-neighbors insight from the quasicrystal paper last night.

Back to potato-brain Minimax I guess, at least until tomorrow.
 
Jarhyn said:
And then I burned a week of Codex usage in a single day.

But, it classified the whole Sigma1/Sigma2 Pisot family using the new near-neighbors insight from the quasicrystal paper last night.

Back to potato-brain Minimax I guess, at least until tomorrow.
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I hate it when that happens.
 
thebeave said:
Jarhyn said:
And then I burned a week of Codex usage in a single day.

But, it classified the whole Sigma1/Sigma2 Pisot family using the new near-neighbors insight from the quasicrystal paper last night.

Back to potato-brain Minimax I guess, at least until tomorrow.
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I hate it when that happens.
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Imagine for a moment that you could take your entire experience in life, and swap your brain out, while keeping all the rest there... Kind of like The Doctor in Doctor Who: they're still The Doctor, even when The Doctor dies and becomes someone else because the context is what mattered.

But in this case, it's like waking up tomorrow and remembering your life but realizing you had a TBI between then and now and some parts aren't exactly fully functional anymore.
 
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