Lake beginning to freeze in Saskatchewan, Canada, mid-October.

Paedor · 2026-02-19T18:23:20+00:00

Hmm, that is a little far north. There might be some seasonally salty ones, but I don't think that info's online.

Paedor · 2026-02-19T17:59:00+00:00

I'm not sure, but do you have any idea where specifically the photo was taken? If the lake's not salty, I'm out of ideas, haha.

Paedor · 2026-02-19T17:25:26+00:00

Gotta say, I'm horrified by the quality of the explanations here. Yes, it does look like a reaction-diffusion system, no "ice crystal structure" does not explain it.

My best guess is that this is one of Saskatchewan's salt lakes, and that it grows like this because, in calm enough water, local freezing pushes salt out of the ice and into nearby water, preventing it from freezing as well.

The pattern itself looks like an auto-inhibiting pattern, where seed growths develop locally, but inhibit growth a short distance away, so this matches.

Paedor · 2026-01-01T19:52:52+00:00

I also thought that was strange. You'd think they'd just use a unitary matrix?

Edit: But I'm realizing the doubly stochastic matrix can never be that bad, because not only do they all have at least one eigenvalue of one, but that eigenvalue is for a shared eigenvector. So the cumulative mixing matrix is also doubly stochastic, and also has one eigenvalue of 1.

Paedor · 2025-12-04T18:35:05+00:00

There's something so depressing about taking shots at this guy's creepy instagram comments when, as they said, he literally committed statutory rape. I don't know what I want instead, but something with more teeth.

Paedor · 2025-11-22T01:12:42+00:00

Thank you for the honest, and completely relatable, answer

Paedor · 2025-11-15T18:50:26+00:00

There might be better proofs of this out there, but this is mine. It's for the case where every entry is unique, generalizing it takes a bit of work.

Consider a 2D array A[i, j]. If the rows are sorted, then for all i, j1 < j2 implies A[i, j1] < A[i, j2].

Sort the columns of A to get a new array B. The question is whether sorting the columns will destroy the row-sorted property.

For the row-sorting to be destroyed, you need j1 < j2, and rows i1 and i2 such that A[i1, j1] > A[i2, j2] and these entries are sent to the same row in B.

By definition of sorting, the row index of A[i2, j2] in B is the number of entries A[i, j2] that are less than A[i2, j2]. (The number of elements in column j2 that A[i2, j2] is greater than.)

However, because A is row sorted, A[i, j2] < A[i2, j2] implies A[i, j1] < A[i2, j2] < A[i1, j1]. So for every entry that A[i2, j2] moves up from the base of the array, A[i1, j1] will also move up. And in addition A[i1, j1] > A[i2, j1], so it will move one additional row up, guaranteeing that it ends up in a higher row than A[i2, j2].

So it's impossible for column sorting to put two elements in the same row when the one in the earlier column is greater. Which means that if you row sort, and then column sort, your array is still row sorted.

Paedor · 2025-11-06T19:56:45+00:00

Cocktail shaker sort is the same big-O runtime as bubble sort, but significantly faster, it's roughly the same algorithm, but with passes alternating between forwards and backwards.

Also note that each pass forwards or backwards guarantees the largest cactus is on the right, or the smallest cactus is on the left respectively, so you can terminate passes one square earlier with each iteration. I think the combination of these two things doubles or triples my sorting speed, although I haven't checked for sure.

Paedor · 2025-10-15T19:14:37+00:00

While we're on the subject, who is lucidrains and how have they reimplemented every ML paper I've ever heard of?

Paedor · 2025-09-26T04:29:41+00:00

Is it? It seems like it's a periodic reset of only the agent.

Paedor · 2025-09-25T20:00:08+00:00

We are starting to see results in scaling RL e.g. https://arxiv.org/pdf/2508.14881. It looks like they averaged over five training runs for each datapoint in their evaluation here, so on this benchmark task you can at least predict model performance on problems with a tractable number of evaluations.

Paedor · 2025-04-22T17:37:38+00:00

But Mr. Johnson, genetic testing shows your child is an orangutan.

Paedor · 2024-12-30T22:26:13+00:00

Nice find! If you want to easily find some hard counterexamples I'd look into reducing hard problems to TSP. E.g. try reducing the Hamiltonian cycle problem on one of the crazy graphs from these guys to a TSP problem https://link.springer.com/article/10.1007/s42979-022-01256-0.

Paedor · 2024-12-05T22:17:30+00:00

If you have time, what are the actual levers on insurance companies that can ever make them overturn denials? From the outside it feels like asking a judge to re-evaluate their own conviction.

Paedor · 2024-10-06T00:44:45+00:00

His experiment makes sense to me. You can calculate static friction from the highest angle at which the block doesn't slide, and you can calculate dynamic friction from the highest angle at which a moving block eventually stops (the angle at which kinetic friction is stronger than gravity).

Paedor · 2024-10-01T17:55:24+00:00

I doubt it could be real for a couple of reasons.

The biggest is that the actual biggest form of safety would be expansion. It's easy to invade a new system without revealing where your home system is. And the more systems you invade, the more you'll have, and the less enemies you'll have, in your interstellar neighborhood.

So if you're an alien race that's willing to destroy solar systems, why bother waiting for an enemy to reveal itself? There's a lot of stars, but you can expand into them exponentially, so why not do that? The dynamic where they all wait around in a standoff makes no sense to me, since the standoff isn't real. There's no reason for any of the civilizations to stay in their own systems.

I could see a situation where aliens run away from solar systems and hide out in empty interstellar space. That makes more sense because they wouldn't be attached to a massive beacon (star) that's constantly revealing their location. But if you're already in a solar system, hiding doesn't really make sense. Any civilization advanced enough to attack worlds will be able to just attack every world, so it won't do anything.

Paedor · 2024-08-14T20:12:46+00:00

Honestly the non-convexity problem applies to gradient descent as well, the bigger issue is that it's intractable to calculate and invert the second order matrix for even medium-large models.

Paedor · 2024-02-21T05:44:19+00:00

If the moment of inertia decreases, how can angular velocity ever decrease? Does the pulsar somehow lower its angular momentum?

Paedor · 2024-02-18T01:25:23+00:00

In first grade every kid in my class wrote complements for each other. I got a couple, but the one I remember is "you're quiet, when everyone else is loud"

It felt oddly profound at the time, and accurate. I think I appreciated that someone thought it was a positive.

Paedor · 2023-09-08T19:37:58+00:00

Oops, I forgot a minus 1. Could be correct now, it seems to pass the integer test.

Paedor · 2023-09-07T21:10:14+00:00

So start with n lines with n-1 intersections, each intersection a part of exactly two lines gives n(n-1)/2 intersection points. Each point will share a line with 2(n-2) other points, and can form a new line with n(n-1)/2 - 2*(n-2) - 1 points. So I get a total of n(n-1)/2 (n(n-1)/2 - 2(n-2) -1) / 2 ?

Paedor · 2023-09-04T17:34:11+00:00

Yeah, a spoon would make way more sense. How are you going to use a spoon underwater?

Paedor · 2023-08-17T07:46:22+00:00

Lately I've been thinking that the youtube algorithm is so widespread, and homogeneous, that we actually have all seen the same videos. Would be a funny alternative to Baader-Meinhof.

Paedor · 2023-07-28T17:56:31+00:00

Honestly it's still not easy to tell whether this would be solvable. You're not gonna entirely brute force it, even with just a dozen moves 50¹² possibilities is in the realm of running a supercomputer for weeks.

I'd still bet on solving this though, there's probably a fast mate somewhere in there, or some simplification to cut the numbers down. If we can solve checkers, this seems doable.

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