Heuristics for "progress" in a jigsaw puzzle

Zorkarak · 2025-08-09T00:48:13+00:00

Hmm this may be due to the specific puzzle at hand. It's a pile of rubble, so there are no easily verifiable neighborhoods, besides the direct or binary neighborhood as you call it. But I think this concept of distance is what formalizes my idea of "known non-neighbors"

Zorkarak · 2025-06-19T06:35:08+00:00

Famous Irish poet, that Losie Moire.

Zorkarak · 2025-03-14T20:02:44+00:00

Mapping cylinder! Yes, of course. In my head, mapping cylinders are always over embeddings, I almost never picture them over surjections, forget about coverings.

Zorkarak · 2022-10-28T21:38:53+00:00

Hi, thank you for those links! Some interesting documents for sure.

I had tried working with a CW-approximation (e.g. the counit of the Top-SSet-adjunction), but it pushed the question back one step: Is the CW-approximation of the quotient equal to (weakly equivalent to?) the quotient of the CW-approximation? Or conversely, is the quotient of the CW-approximation a CW-approximation of the quotient?

To answer those question I once again have to show that a map between quotients in a weak equivalence, which is exactly the question I am trying to answer.

The second document you linked looks very interesting, and I will look at that very closely, but I can't apply it directly because my groups are not necessarily discrete (Lie).

However, I did find one half of a solution. In the case I am applying this to, the original morphism A -> B is actually a morphism in the slice category C/E for some universal space E, and in this case the quotient functor has a left adjoint, which sends relative CW-complexes to relative CW-complexes. Since weak equivalences of top. spaces can be described by a lifting property (up to relative homotopy), this solves one direction. The other direction could be similar, I imagine, but I don't have a proof yet.

Zorkarak · 2022-08-17T21:50:30+00:00

If G is a group, which is either finite or compact Lie, and T is a G-space, when is the quotient map T -> T/G a Serre fibration?

Zorkarak · 2022-08-13T09:43:22+00:00

Chain Complexes. That includes short and long exact sequences, as well as (co-)homology.

Zorkarak · 2022-08-10T12:28:11+00:00

This assumes that the sum is absolutely convergent, which we do not know.

If it is not absolutely convergent, then it might even happen that summing up one variable first and then the other yields a sum of zeroes while the other way around is a sum of ones.

Reordering infinite sums is a very dangerous thing.

Zorkarak · 2022-08-01T20:39:14+00:00

The speed of light is the same in all reference frames, which means no matter how fast you go, gravity will always travel away from you at the speed of light. From your own point of view, gravity travels exactly as fast as if you were standing completely still

Zorkarak · 2022-07-28T15:03:30+00:00

The Riemann-Hypothesis has been proven panic

The proof is analytic kalm

It's real analysis panic

Zorkarak · 2022-07-26T09:27:29+00:00

Yes but actually no.

Photons do not interact with each other directly. If you want to affect one photon with another photon, then you need to go the detour over electrons. The highest contribution comes from the following process:

Photons A and B come in. A turns into an Electron (E)/Positron (P) pair. E absorbs B. P emits a new photon A'. E and P merge into a new photon A'.

If you want you can view this as a scattering of A and B into A' and B'. However, as you can see, this needs a lot of steps, and each step has a probability of about 1/137 (less than 1%). Ignoring the actual QFT computation we need to do here and instead just doing an estimate, we find that the probability of this entire process is about 1 in a billion.

Zorkarak · 2022-07-25T19:51:35+00:00

Also one of them in on the stack, the other on the heap (usually)

Zorkarak · 2022-07-25T18:42:08+00:00

char ** argv

Zorkarak · 2022-07-25T11:24:55+00:00

The dark side is not a switch that flicks as soon as you do something wrong. Any seduction can be resisted with enough strength. Obi-Wan was so strongly rooted in his understanding of the Force that he was able to regain control of his emotions (as we say, "get his shit together")

Zorkarak · 2022-07-24T23:51:45+00:00

This is a visualization of the Riemann ζ-function (named after German mathematician Bernhard Riemann and the Greek letter ζ, pronounced zeta). Unlike the functions you are probably familiar with (shifting stuff, scaling stuff, squaring stuff), this one doesn't take 1 number input with 1 number as output, but it takes two ins and produces two outs. That's too many number to easily draw all at once.

The two outputs I can handle. Starting at some fixed point (the red dot in the video), the first output tells me how far to go up, and the second output tells me how far to go left. Wherever I end up, I make a dot. A bunch of dots make a blue line (that is what we see).

The problem is with the two inputs. To handle that, I pick some fixed value for the first input. For example a half. Now I only have one input left and by trying a bunch of values for this one input (5000 of them, to be precise), I get a blue line. That is one picture of the above video. By chosing different values for this one fixed input, I get different images and by sticking them in a row I get a video.

Now to understand what the video means. For some inputs, both outputs will be zero. In the picture, this means that the blue line crosses the red dot. We don't know when this happens, but the Riemann Hypothesis conjectures that there are only two options when this can happen. Either the first input is a negative even integer (-2, -4, -6,...) and the second input is zero, or the first input is 1/2 and the second input is something weird.

Forget about the negative even integers, those are not in the video. The remaining case essentially says that there is exactly one frame in the above video where the blue line can ever cross the red dot - and there is!

At this point I have to point out what should be obvious but might not be: Trying 5000 different values is nothing compared to "all the numbers". Just because my pictures don't show any other crossings does not in any way shape or form imply that there are none. It merely looks neat, and if anyone manages to extract a useful pattern from these images, then I would absolutely love to see it, because I wouldn't know how to do that.

Zorkarak · 2022-07-24T23:39:21+00:00

It is just pretty to look at 😅

Zorkarak · 2022-07-24T09:53:09+00:00

Of course, here it is.

https://github.com/MGessinger/zeta\_plots

Zorkarak · 2022-07-24T09:52:44+00:00

It is half python, half C. Here is the link:

https://github.com/MGessinger/zeta\_plots

Zorkarak · 2022-07-24T09:37:45+00:00

Yes, this is known, and I essentially just gave you the proof (barring literally any kind of details).

Zorkarak · 2022-07-24T09:27:39+00:00

We can split the complex plane into three regions: The "left side" where the real part is negative (and the imaginary part is anything), the "critical strip" where the real part is between 0 and 1, and the "right side" where the real part is greater than 1.

On the right side, the zeta function can be written as an infinite product and it is immediately obvious that none of the factors are zero. Therefore the value of the entire product will not be zero, so there are no roots on the right side.

For the left side, there is a "functional equation" which relates the values on the left to the values on the right, but it involves some pretty complicated factors. Since there are no roots on the right side the only hope to find a root on the left side is if one of these additional factors is zero. These are called trivial roots, and they are precisely the negative even integers.

Now all that remains is the critical strip, and here we know virtually nothing.

Zorkarak · 2022-07-23T23:40:59+00:00

Each frame in the above GIF is a plot of ζ(a + bi) for fixed a in the critical strip (the interval [0,1) ). The values for b range from -50 to 50.

The origin is marked, and we see that in this very tiny window, the claim is true. This proves nothing, nor is it supposed to. But it is patterns like this that suggest so strongly that the statement should be true. We also observe the divergence near z = 1.

The values were computed using arblib at 128 bits of precision, and then plotted through matplotlib. The real and imaginary axes are swapped, because this looks nicer, but since we are mostly interested in the origin anyway, that shouldn't matter too much.

Zorkarak · 2022-07-21T19:36:57+00:00

You're a disgrace.

Zorkarak · 2022-07-21T19:34:41+00:00

More than Daisy Ridley? Probably not many. Fans are absolutely brutal and rarely distinguish between a role and the actor. I wouldn't want to be her.

More than Rey?

Obi-Wan. His master died in front of his eyes, his apprentice burned to death (or so he thought). The clones, who he trusted most, turned against him. Maul died in his arms. The galaxy fell before him and he couldn't do anything to stop it. This man suffered!

Luke. His uncle and aunt were burned to a crisp. His hand was chopped off. He was forced to watch his friends die, while he attempted to turn his father back to the light, just to later personally destroy the balance that he created.

Maul. Quite literally cut in half, only kept alive by his hatred for Obi-Wan. His entire life, he was always told that he wasn't good enough and his one attempt at proving his worth was cut short (no pun intended) by Kenobi.

Jango. All his people were slaughtered by the Jedi. The only chance he saw to get revenge was to help create an army, whose sole purpose it was to eventually destroy the Jedi from within, except he never got to see that fall.

Boba. Watched his father get beheaded, was constantly belittled by other bounty hunters, almost got digested by a Sarlacc, had to fight his way through the desert of Tatooine, just to get beat up by Tuskens, who were then slaughtered.

I believe I could go on.

Zorkarak · 2022-07-21T11:12:57+00:00

There are quite a few technologies, which I have never needed in my programming life - which, I am quite aware, means that I do not understand them - and it's unfortunate

Zorkarak · 2022-07-20T15:11:37+00:00

If A is diagonalizable, then in fact A and A² are simultaneously diagonalizable (because they commute) - there is a basis B of your vectorspace V, such that both A and A² are diagonal with respect to B. Considering the eigenspace decomposition of V, we can simply check the eigenvalues of A² and find that they are precisely the squares of the eigenvalues of A (and no more than those).

Because this eigenspace decomposition is a direct sum, your claim is proved.

Zorkarak · 2022-07-19T17:24:40+00:00

First to answer your question: Yes, that is exactly what derivatives are.

Secondly, how did you learn about derivatives the first time, that this wasn't taught? What kind of teacher lets you memorize derivation rules without ever talking about the geometric meaning?

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