151
submitted 4 days ago by chunes@lemmy.world to c/science@lemmy.world

Here's a nice explanation from /u/gameryamen on reddit:

Say you have a flat arrow pointing up. You spin it 3/4ths of a rotation clockwise, so it's pointing to the left. The simple way to undo that rotation (meaning, get back to the starting point) is to simple rotate it counter clockwise the same amount. But another way to do it is to rotate it 1/4 of a turn clockwise.

Another way to describe that last 1/4 turn is as two 1/8th turns, right? We're scaling the amount of rotation down, then doing it twice. The factor we need to scale down by is pretty easy to work out in this simple example, but it's much harder when you're working in 3D, and working with a sequence of rotations.

However, this paper shows that for almost all possible sets of rotations in 3D space, there is some factor by which you can scale all of those rotations, then repeat them twice, and you'll wind back up at the starting position. A key thing here is that we still have to find or calculate what that factor is, it's going to be a very specific number based on the set of rotations, not any kind of constant.

Why does that matter? Well, besides just being a neat thing, it might lead to improvements in systems that operate in 3D spaces. Doing the two 1/8th turns takes less work than doing a backwards 3/4ths turn. Even better, it allows us to keep rotating in the same direction and get back to the start. If calculating the right scaling factor is easy enough, this could save us a bunch of engineering work.

Edit: The most common question is "why do two 1/8th rotations instead of just one 1/4 rotation?" The reason is because the paper deals with a sequence of rotations in 3D, not a single rotation in 2D. But that's kinda hard to wrap your head around without visuals. This is going to be a little tortured, but stop thinking about rotations and imagine you're playing golf. You could get a hole in one, but that's really hard. A barely easier task would be aiming for a spot where you could get exactly halfway to the hole, because you could just repeat that shot to reach the hole. There's still only one place that first shot can land for that to work, it still takes a lot of precision.

But if you change your plan to "Take a first shot, then two equal but smaller shots", there's a lot more spots the first shot could land where that plan results in reaching the hole on your third shot. Having one more shot in your follow up acts as kind of a hinge, opening up more possibilities. This is what the "two rotations" is doing in the paper, it's the key insight that let the researchers find a pattern that always works.

top 15 comments
sorted by: hot top controversial new old
[-] PlutoniumAcid@lemmy.world 1 points 21 hours ago

TLDR: Two wrongs don't make a right, but three lefts will.

[-] Lumidaub@feddit.org 50 points 4 days ago

mind-bendingly consciousness-expanding unspeakably eldritch non-euclidian occult incantations

simple trick

[-] whaleross@lemmy.world 22 points 4 days ago

Everything is easy when you know how to bend spacetime.

[-] applebusch 6 points 3 days ago

This article is... difficult to critique. There's too many things to refute so I'll just undermine the title. We've known about an operation that can undo any rotation for a lot (hundreds?) of years. 3D rotations represented as quaternions or rotation matrices are trivially invertible, and their inverse literally undoes the rotation. For a rotation matrix the inverse is simply the transpose of the matrix, and for a quaternion it's the complex conjugate (3 of the 4 numbers have their sign flipped). These operations have been used in computer algorithms likely for as long as we have had computers. Honestly this whole thing feels like a big fat nothing so I'm gonna stop letting it steal my time.

[-] loppy@fedia.io 5 points 2 days ago

Read the actual actual article: https://arxiv.org/abs/2502.14367

The authors prove that given any sequence of rotations W "almost always" there is a sequence of rotations W' formed by scaling every rotation angle in W by the same positive factor, such that the sequence W'W' is the identity (that is, apply all the rotations in W' and then apply all of them again).

The issue isn't the result, it's popsci writing.

[-] loppy@fedia.io 11 points 4 days ago
[-] loppy@fedia.io 6 points 3 days ago

The example given in the OP is incorrect. /u/gameryamen is implying something like: given a sequence of rotations W there is a scale factor a>0 such that W(a)W(a)W = 1, with W(a) the same sequence of rotations as W but with all rotation angles scaled by a.

This is not what the paper does. The paper finds an a such that W(a)W(a) = 1.


His whole post seems bunk, honestly. Example:

Having one more shot in your follow up acts as kind of a hinge, opening up more possibilities.

This seems completely irrelevant. It seems that maybe they're referring to the probabilistic argument the authors give to justify why their theorem should be true (before giving a complete proof), but this argument involves repeating the same exact rotation two times, not two different rotations in sequence.

[-] Goretantath@lemmy.world 6 points 4 days ago

So like taking a clock and going from 12 to 10 then just continuing to 12 and saying you never went to 10 in the first place.

[-] loppy@fedia.io 2 points 3 days ago

No, the original paper is saying something like: given the sequence say 12 -> 3 -> 7, there is a sequence 12 -> 3x -> 7x -> 7x + 3x -> 7x + 3x + 4x where 7x+3x+4x = 12. Obviously here x=6/7, but doing something like this with arbitrary rotations in 3D isn't so simple.

[-] ieatpwns@lemmy.world 4 points 4 days ago

Does it work on a Rubik’s cube?

[-] floofloof@lemmy.ca 7 points 4 days ago

I imagine the answer would be mathematically yes, but mechanically no.

[-] loppy@fedia.io 1 points 3 days ago

If you're referring to doing moves on a Rubik's cube, no that's irrelevant to the theorem. If you're referring to applying a sequence of rotations to a whole solid cube, then yes.

[-] Imacat@lemmy.dbzer0.com 3 points 3 days ago

Did I miss it in the paper, or do they never really say how to find that scaling factor? Seems like there’s some magic number that theoretically exists for almost all rotations but no good way of finding it.

[-] loppy@fedia.io 3 points 3 days ago

They use continuity and the fact that F(0) = 1 to conclude that, since it also takes on a negative value (what they put effort into arguing), it must also attain 0. So no, they don't find it directly, but it is technically possible to express F explicitly and then numerically find a root.

[-] Machinist@lemmy.world 2 points 3 days ago

Woah, this could be really cool for certain kinds of machine motion if it is computationally easy/fast/cheap.

this post was submitted on 20 Oct 2025
151 points (100.0% liked)

science

22199 readers
537 users here now

A community to post scientific articles, news, and civil discussion.

rule #1: be kind

founded 2 years ago
MODERATORS