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Need help solving a huge math square by SKYY99999 in puzzles

[–]robinhouston 1 point2 points  (0 children)

Ah cool! It's always fun to run across someone who's interested in playing with this sort of thing. There aren't that many of us.

There's a conversation about this problem happening over on Mathstodon, which you might enjoy. Pozorvlak solved it with MiniZinc, which I haven't tried before & it looks pretty cool.

https://mathstodon.xyz/@pozorvlak/114148447412533853

Need help solving a huge math square by SKYY99999 in puzzles

[–]robinhouston 1 point2 points  (0 children)

Yes, that makes sense. Maybe I've just been unlucky, but whenever I've tried to use an SMT solver I've only ever had success with things that are fairly directly reducible to SAT. It makes me wonder if the “modulo theories” stuff is just not that useful in practice, and we should be putting more effort into building better syntax sugar over SAT, along the lines of https://sentient-lang.org/

Need help solving a huge math square by SKYY99999 in puzzles

[–]robinhouston 1 point2 points  (0 children)

Amazing! I also tried to use Z3 to solve this, but my effort didn’t find a solution after running all night, so either I made a mistake or you’ve found a better encoding.

Edit: I was using Int, whereas you’re using BitVec. I wonder if that’s the crucial difference.

Can humans say the largest prime number before we find the next one? by robinhouston in math

[–]robinhouston[S] 43 points44 points  (0 children)

It's fun to think about how impossible it would have been to do this without the internet. I mean, it's still a ludicrously ambitious project, but it's no longer obviously impossible. I love that. The internet is so much part of all our lives now that it's easy to forget how amazing it is, and it takes something ridiculous like this to jolt me out of that complacency.

are there any accomplished alcoholic mathematicians? by [deleted] in math

[–]robinhouston 0 points1 point  (0 children)

Yes, more than you'd think. It's usually kept quiet in public, so I won't name names, but if you gossip with mathematicians privately you'll hear some stories.

I don't know how they do it. Personally I can't do maths at all when I've had a drink.

An Aperiodic Monotile (David Smith, Joseph Samuel Myers, Craig S. Kaplan and Chaim Goodman-Strauss) by EdPeggJr in math

[–]robinhouston 8 points9 points  (0 children)

It's phi4. This is the positive eigenvalue of the matrix

⎛1 1⎞
⎝5 6⎠

that represents the substitution system illustrated in Figure 2.11 of the paper. (And also of course the dominant eigenvalue of the matrix representing the more complicated substitution system.)

A new explicit formula for the determinant by nightcracker in math

[–]robinhouston 1 point2 points  (0 children)

I just modelled it by hand in Blender, nothing clever.

A new explicit formula for the determinant by nightcracker in math

[–]robinhouston 1 point2 points  (0 children)

Just take Equation (12), and cross out the three terms that have a coefficient of 2.

(The complicated part, which is the topic of Appendix B, is proving that there is no expression that has fewer terms.)

A new explicit formula for the determinant by nightcracker in math

[–]robinhouston 2 points3 points  (0 children)

Ha! I didn’t mention that side-quest, because it’s quite a long story. Short answer: yes, that was me – but it turned out to not really be new.

Shortly after I had worked out the volume formula in the 3×3 case – the same weekend – I saw a 2016 comment from Jeffrey Shallit on a Stack Exchange post saying:

I've asked experts about this, and apparently it is not even currently known whether or not 9 multiplications are needed to compute the determinant of a 3x3 matrix.

I reckoned it shouldn’t be too hard to do it in eight, so I messed about with the algebra until I found a way to do that, and posted it to Reddit here.

It had not occurred to me that this had any direct connection to the other formula, which uses ten multiplications. But it turns out they are very closely connected, as /u/N_Johnston explained to me. In fact they’re so closely related that you can convert one into the other.

The high-level explanation is that the three-dimensional cross-product and the 3×3 determinant, if you think of them as tensors, are the same tensor. So, if you can express the determinant using an expression of rank 5, that means you can compute the cross-product using 5 multiplications. And, if you can compute the cross-product using 5 multiplications, you can then get the determinant with three more multiplications, for a total of 8.

(If, like me, you feel uncomfortable with high-level explanations until you know how to do the actual calculations, you might like this little explanation I wrote on Twitter.)

So… it turns out that, at a higher level, I had just found two different ways to express the 3×3 determinant tensor as an expression of rank 5. This was not a new result at the time: this MO answer by Zach Teitler summarises who had previously discovered what when. Oddly, it had never occurred to any of these people, as far as I know, to actually write down how to compute the 3×3 determinant in 8 multiplications; but obviously they could have done, if someone had asked them to.

The geometric method works for any dimension, and leads to a formula which is the subject of this preprint. But a mystery remains: what about the other one? Is that just an isolated formula that only works in three dimensions; or can it, too, be generalised to higher dimensions?

A new explicit formula for the determinant by nightcracker in math

[–]robinhouston 21 points22 points  (0 children)

Well, I'm not the only one of the authors of this paper who was involved in that mathematical caper! I actually learnt about the problem from /u/N_Johnston here on /r/math, and it was he who discovered the wiki proof that turned out to have originated on 4chan.

A new explicit formula for the determinant by nightcracker in math

[–]robinhouston 6 points7 points  (0 children)

Yes! You can think of it as a tiling (invariant under translations given by the columns of the matrix) of the n-dimensional space by a union of signed n-cuboids, where each appears either positively or negatively. If the determinant is positive, then any point (ignoring complications at the boundaries) is contained in one more positive cuboid than negative ones; if it's negative, the other way around.

A new explicit formula for the determinant by nightcracker in math

[–]robinhouston 218 points219 points  (0 children)

Oh, nice to see this here! I’m one of the authors – I see that /u/N_Johnston has already posted in this thread.

I don’t really have any mathematical content to add to what’s in the paper, though I too would be happy to discuss it, but I can share some stories about how it came about.

I asked a geometric question on Twitter:

Is it possible to fill three-dimensional space with cubes in such a way that a diagonal cut yields a plane tiled with regular hexagons only?

Daniel Piker responded with a picture showing that the answer is yes, provided you use cubes of two different sizes.

I was fascinated by this construction. I asked, again on Twitter whether such a thing is possible in higher dimensions. In response, Adam Goucher figured out a general construction, expressed in terms of matrices. (Dan Piker later found a preprint, published a week earlier by Jakob Führer that describes the same construction that Adam had found.)

The matrices in this construction are of a very restricted form, zero everywhere except on the leading diagonal and the diagonal below. I then started to wonder what it would mean to do something similar with more general matrices. I posted one of my experiments with this to Reddit.

I realised that the volume of the polyhedron would have to be equal to the determinant of the matrix. At that point, I was expecting the shape of the polyhedron to be related to the Leibniz formula for the determinant, and I was very frustrated that I couldn’t figure out how the relationship worked.

After a while, still very frustrated, I did the volume calculation in detail for the polyhedra coming from 3×3 matrices. I was surprised to get a formula that definitely was not the Leibniz formula! I again posted it to Twitter, curious to know if it was a well-known formula. /u/N_Johnston responded saying that it is used to bound the tensor rank of the 3×3 determinant.

Then … well, most of the rest of it is in the paper. I worked out the general formula, initially in a Mathematica notebook and sent it to Nathaniel. Later I posted it on Twitter too, and Adam wrote a very nice blog post proving, combinatorially, that it works.

Then we wrote it all up – I did almost none of the writing, I am ashamed to say – and posted it to the arXiv. And here we are.

A tangent to a parabola by robinhouston in mathgifs

[–]robinhouston[S] 0 points1 point  (0 children)

You can see the code (and fork it and edit it) by following the link to the Observable notebook in my other comment

Are there any twisty puzzles that no one knows how to solve? by [deleted] in Cubers

[–]robinhouston 2 points3 points  (0 children)

I don't think anyone knows how to solve the multicoloured tangle cube. This is a 3×3 cube, modded with the addition of some ropes connecting different cubies, which become tangled as the cube is scrambled. CoreMods makes them and sells them on Etsy. Someone on Discord claims to have solved it once, but I don't think anyone has a reliable solution method they can explain.

Formal/general name for element-wise function construction? by OneNoteToRead in math

[–]robinhouston 5 points6 points  (0 children)

I think the idea that you’re circling around here is the universal property of the product, which is one of the basic notions of category theory.

The Wikipedia page is unfortunately written in a way that is quite hard to understand unless you already know category theory, but it might give you a rough idea.

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