I dare you to make a polyhedron out of triangles with an odd number of faces! 8>

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@loke Hmmmm.. well in the 2D case it's trivial XD
Regardless of whether it's a polygon or a partial tesselation!

So idk if it would be easy in the 4D case, whether 3D cells or 2D faces were used..

HMmmmmmmmmmmmmm I dunno XD

I mean sure, that'd be great of course! XD :D
But I don't think it would be a counterexample to my conjecture ( unless was 3D or ND and N included 3 XD

But any knowledge is wonderful, of course! \:D/

@codepuppy Yes, that post was what I was thinking about when I replied. I'm not sure how to go about proving that for an arbitrary number of dimensions.

@loke Well to start with, maybe there's a coefficient (probably a different one) for other dimensions than 3/2 :D

I arrived at that simply by the fact that every edge is shared by exactly two triangles (a logical premise in the statement), and thus the number of edges if each triangle was independent would be 3*N, and so when they're merged, each edge is duplicated to make exactly two nonunique edges, so the number of *unique* edges is 3/2*N

But! Since it must be integer, N must be even!!

@codepuppy So tif we apply that in 4 dimensionse, each tetrahedron that makes up the faces have 4 sides, that would make 4*N independent sides, and each side is again shared by 2 faces, making the number of unique sides 2*N. That suggests that there is no restriction on the number of sides. In fact, the limitation would only exist for odd-dimensional spaces.

@loke Ohh—wait a second! A triangle has 3 1D edges, a tetrahedron has 4 2D triangle faces, are we sure a hypertetrahedron has 4 3D cells? XD

Whateven is a hypertetrahedron? XD
Is that an arbitrary degree of freedom as to what mesh primitive we use?!

Hmm, start with a line and linearly shrink it while moving perpendicularly in the new 2nd dimension: a triangle!

Start with a triangle and linearly shrink it while moving in the 3rd dimension!

""" For 4D! :D
What is that shape even called? XD

@loke Actually, does it matter what primitive is used??

If D is the dimensions,

(D-1) ..things are what are shared XD

So they will always be butted up against each other!

So the number of edges would be P/2*N :D

2D: Line segments in a 2D polygon connect at two 1D connectors, so it's 2/2*N or N :D

3D: Triangles have 3 edges so it's 3/2*N

4D: Tetrahedrons have 4 faces so it's 4/2*N

5D: 5-cells (I googled it X3 ) have 5 cells (duh XD), so it's 5/2*N

So it looks like you're right! :D

@loke (We'd need proof that the number of connectors of a N-dimensional triangle is equal to N+1 though, to be certain :3 )

(..And a proof of my original thing XD )

(For which I feel like Euler's theorem must come in here somewhere... let's see.. each D-1 dimensional primitive would be a node, and each D-2 dim. connector would be a graph edge.. but then what would "faces" be in this abstract graph? You'd use Euler's formula to calculate them which is unhelpful XD )

@codepuppy @loke
You don't need "faces" in the dual graph. All I used before was the degree-sum, and that works here again.

You could also investigate the number of 2D triangles (or whatever other kind of face you like) in a higher-dimensional polytope...?

@bmreiniger @loke (Ahhh so that's what "dual graph" means in this context :D )

...*think think*

OH YEah!

According to Wikipedia, this is true:

Σ (deg(n)) {n ∈ Nodes} = 2*E

Since the degree of each node is 'P' from above,

Σ (P) {n ∈ Nodes} = 2*E

P*N = 2*E

E = (N*P)/2

Which is it! \:DD/

(And thus, yes, when P is odd, N must be even! When P is even anything's fair game :3 )

Thanks so much!!! :D

@codepuppy @loke

Oh, you're probably right, that "dual graph" is maybe not the right phrasing anymore.

Of course, this doesn't say that when P is even any number of simplices are achievable; just this argument fails. At least the odd P+1 is possible, using the P-simplex itself...

@bmreiniger @loke Oh okie :3

And yes yes, exactly! When P is even anything's fair game; sometimes it'll work, sometimes it won't!

Ahh I see how that could be misinterpreted XD''

Hmm, how about "When P is odd, at least odd N's aren't allowed (and possibly more N's besides). But when P is even, there's no telling what's allowed or not!" ? :3


@bmreiniger @loke

And oh yeah! You mean using the P+1 (or D+1) simplex out of P simplexes, like making a triangle out of lines, making a tetrahedron out of triangles, making a 5-cell out of tetrahedrons, etc. Yeah good point, you can always do that! :D


That generalization of a triangle to n dimensions is called n-simplex.

@codepuppy @loke

There's a nice generalization of Euler's formula to n dimensions. Or you could simply form the alternating sum of the number of i-simplices for all i from 0 to n.

@codepuppy @loke

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