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You have been invited to an $n$-dimensional universe, where $n$ is an integer greater than $1$.

Here, matchsticks are idealised lines of unit length, and they can pass through each other.

They want you to construct a ruler for them. (Never mind that they are already rulers themselves of this universe.)

A ruler consists of a structure made out of matchsticks (of unit length). Matchsticks can be stuck together at their ends (with magical glue). Every pair of endpoints which are rigid with respect to each other can be used to make a measurement.

For example, when $n=2$, two equilateral triangles are affixed together, the $2$ furthest endpoints are rigid with respect to each other, and can measure a distance of $\sqrt{3}$.

Two equilateral triangles

You are not limited in the number of matchsticks you want to use in your ruler, the more matchsticks, the grander the ruler, of course.

Of course, a ruler would be useless if it could not measure lengths. What sort of lengths can your ruler measure, for each number of dimensions?

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  • $\begingroup$ Although this is a nice question and has a nice answer, I don't know of any way to resolve it that doesn't involve quite highbrow mathematics. Element118, do you? $\endgroup$ – Gareth McCaughan Aug 10 '16 at 16:21
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I think the answer is that the ruler can measure

any algebraic distance, in any number of dimensions.

Take a look at this Wikipedia page under the heading beginning "Representation of algebraic numbers"; it claims that

for any algebraic number A " for any two points p and q at distance A, there exists a finite rigid unit distance graph containing p and q such that any transformation of the plane that preserves the unit distances in this graph preserves the distance between p and q"; the corresponding configuration of matchsticks will serve as a ruler for distance A. So in the plane, at least, we can measure any algebraic distance with such a configuration of matchsticks.

On the other hand,

all distances measured by such rulers in any number of dimensions are algebraic, because if we fix one matchstick so as to rule out rigid motions of the whole structure (in the plane; in higher dimensions, we need to fix more but it is easy to ensure that all the coordinates of the points we fix are algebraic) then the coordinates of the ends of the others satisfy polynomial constraints given by the edge lengths, and the particular configuration we have is an isolated solution to these constraints (else the ruler wouldn't be rigid), and these conditions imply that the coordinates are algebraic numbers. This fact is pretty easy to prove and probably well known but the only actual reference I have is to my own PhD thesis, because when I happened to need that theorem a little bit of poking around didn't find it anywhere standard :-).

We still need to consider

whether we can measure arbitrary algebraic numbers in more dimensions. It feels as if the answer should be yes -- there should be a way to turn any rigid structure in dimension d into one in dimension d+1 by adding some sort of "bracing" -- but I'm not at all sure I should trust that intuition. The original Beckman-Quarles theorem, which says that preserving all unit distances implies isometry, applies in euclidean space of any dimension, but of course that's not at all the same thing. ... Aha! Although the original paper cited by the Wikipedia article doesn't appear to be publicly available, a followup by the same author is -- https://archive.org/details/arxiv-math0102023 -- and it indicates that the theorem does hold in any number of dimensions.

So, to recap:

Whatever the (finite) dimensionality of your space, you can measure any algebraic distance and no other distances with rulers of this kind.

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  • $\begingroup$ The "bracing" from 2 to 3 dimensions can be done with octaeders. An octaeder includes a rigid square. So you can extend a 2D structure to 3D by placing on each stick a rigid square that has two sides parallel to the additiona dimension. With rigid squares you can make rigid cubes and hypercubes. So you can probably extend this to any number of dimensions. $\endgroup$ – Florian F Aug 10 '16 at 16:00
  • $\begingroup$ Ah yes, that sounds like it would work. $\endgroup$ – Gareth McCaughan Aug 10 '16 at 16:20
  • $\begingroup$ @FlorianF: I don't think octahedrons are rigid in >3D. You might be able to use general cross polytopes for the job, although I don't actually know if those are rigid. $\endgroup$ – user2357112 supports Monica Aug 10 '16 at 16:47
  • $\begingroup$ you can't measure number like Pi, at least in 2D, since they're transcendant numbers because you can't create such a number without circular shape. But you can't, you only have lines of unit length $\endgroup$ – rmilville Aug 10 '16 at 16:48
  • $\begingroup$ @rmilville: No one is claiming to be able to measure transcendental numbers like $\pi$. (Though I think saying "you can't create such a number without circular shape" is going too far; it's not inconceivable that there might be some configuration that happens to produce a distance of $\pi$, though it turns out that there isn't.) $\endgroup$ – Gareth McCaughan Aug 10 '16 at 16:51

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