Condensing incidence matrices

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Condensing incidence matrices

Postby ubersketch » Thu May 02, 2019 7:49 pm

The incidence matrix of the triangle is
Code: Select all
0   0   0   1   1   0
0   0   0   1   0   1
0   0   0   0   1   1
1   1   0   0   1   1
1   0   1   1   0   1
0   1   1   1   1   0

Condensing it (making it so every vertex is the same) gives us this:
Code: Select all
0   2
2   2


What happens to an incidence matrix when you condense it?
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Re: Condensing incidence matrices

Postby Klitzing » Fri May 03, 2019 7:03 am

Dear ubersketch,
why do your matrices assume that each element wouldn't be incident to itself?

I'd assume that you'd have to add all ones to the diagonal.
And then the "condensed form" will be exactly my Incidence Matrix of a triangle:
Code: Select all
3 | 2
--+--
2 | 3

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Re: Condensing incidence matrices

Postby ubersketch » Fri May 03, 2019 11:08 am

Alright, let’s say they are incident to themselves, but how are incidence matrices condensed?
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Re: Condensing incidence matrices

Postby Klitzing » Fri May 03, 2019 11:22 am

ubersketch wrote:... but how are incidence matrices condensed?


By symmetry equivalence. - A triangle e.g. might have 3 distinct vertices and 3 distinct sides. But for isoceles triangles you can count the 2 base vertices together, as well as the 2 skew sides. And for a regular triangle all 3 vertices are equivalent and all 3 sides are equivalent:
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general triangle
1 * * | 1 1 0
* 1 * | 1 0 1
* * 1 | 0 1 1
------+------
1 1 0 | 1 * *
1 0 1 | * 1 *
0 1 1 | * * 1
Code: Select all
isoceles triangle
1 * | 2 0
* 2 | 1 1
----+----
1 1 | 2 *
0 2 | * 1
Code: Select all
regular triangle
3 | 2
--+--
2 | 3

--- rk

PS: or even for regular triangles you might want to apply different / alike colorings etc.
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Re: Condensing incidence matrices

Postby ubersketch » Fri May 03, 2019 2:30 pm

Alright, fair enough, but how would you condense a strange incidence matrix like
Code: Select all
1 1 0
1 1 0
1 1 1
?
I'm interested in incidence matrices without the underlying polytopes since I think there exist incidence matrices that can't be represented by a polytope (even with edges with multiple vertices, multiple facets in one ridge etc.)
Also making so elements be not incident to themselves can allow for negative elements that connect a point to itself while not being an edge or any higher dimensional element. I only consider one element incident to another if there is a lower dimensional element connecting it with the other one, or contains the lower dimensional element (things can't contain themselves). Another thing is how would you represent polytopes on different surfaces. What is the difference in the incidence matrix of a square tiling on a torus, and a square tiling on an infinite euclidean plane?
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Re: Condensing incidence matrices

Postby Klitzing » Fri May 03, 2019 3:20 pm

In fact it happens that incidence matrices provide an equivalent description to the abstract polytopes.
And as such incidence matrices cannot encode the type of realization.

OTOH a square tiling on a torus and a square tiling on the euclidean plane surely differ in the absolute counts of elements:
finite versus infinite.

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Re: Condensing incidence matrices

Postby ubersketch » Fri May 03, 2019 8:59 pm

Alright. I supposed we could pair a incmat with a condensed incmat, and get maximum info. But how do you condense any arbitrary matrix?
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Re: Condensing incidence matrices

Postby Klitzing » Sat May 04, 2019 7:54 am

There is no THE condensation.
All depends on the interpretation on what your matrix shall represent, and what a condensed form therefrom shall represent.
That is, which elements can be considered to be equivalent. And thus might be combined into a single entry each.
It is the equivalence relation of the real objects which induces the according operation on the matrices.
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Re: Condensing incidence matrices

Postby ubersketch » Sun May 05, 2019 6:11 pm

Well I think we could make a graph like object and then figure out the symmetries from there.
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