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[thematrixiam]

Hey guys.

I am trying to find a shape that suits my needs.

The amount of vertexes can be any, but the fewer the better.

The edges are special in that each end has an A or a B. like this A------B

I need the vertexes to have more A than B inputs from edges.

I need all vertexes to have the same amount of edges.

I have tried various 3dimensional shapes to no avail. Maybe someone here can help me. I have even tried shapes inside of shapes.

Thanks in advance.

 

[JHamm]

A-------A

If each edge needs to have an A and a B then unless you can connect two edges together like A-------BB-------A I don't see how you can take an equal number of As and Bs and have more As than Bs at your vertices.

 

[thematrixiam]

ya, that's the way it's looking.

Keep in mind this can be in any dimension. 2, 3, 4,5. just as long as I get more vertexes with more A than B.

A-----A could be possible, assuming the end result was still the same. Same with
B-----B.

each vertex should look like
Let n < m
Axn
Bxm

technically it could have
A----A
A----B
B----B
B----A
or even
/-1/2B
a--
\--1/2B
But the last one would make things a real mess.

what ever method is used, though, the same ratio and values of A:B has to be on each vertex.

edit: right now I am looking at 2-d shapes. before I was looking at 3d.

edit2: in 2d I have found the pattern of vertexes = A +1, where B = 2A. for example 4 vertexes with one a and 2 b. This works when the edges are allowed to change. I remember last night dreaming something about repeating decimals, so now I have to figure out what that meant. I think the lower the repeating decimal the better? but I could be wrong.

Edit 3: I just realized I wrote out that pattern wrong. # of vertexes = A+1+B

 

[thematrixiam]

Doh!... Wasn't thinking.

The issue comes up that one A is actually existing as two. So over all the vertex would still have A=B. which doesn't work.

Back to the drawing board.

Anyone know of geometry software that could try to calculate that?

 

[thematrixiam]

I have noticed that Cayley Graphs allow for Arrows, but only allow one going out and one going in. Does anyone here know of a theory that allows for set number of out arrows and set number of in arrows?

 

[Edgardo]

I don't have a solution but these definitions or terms may be helpful:

- If every edge connects exactly one vertex A and one vertex B this means your graph is 2-colorable.

- For an undirected graph, i.e. edges don't have arrows, the number of edges incident to a vertex is called degree.

- For a directed graph (edges have arrows):
The number of edges going out of a vertex is called outdegree.
The number of edges pointing to a vertex is called indegree.

Maybe you can look for a theorem using these terms.