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Help me find the shape

  1. Jan 4, 2012 #1
    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.
     
  2. jcsd
  3. Jan 5, 2012 #2
    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.
     
  4. Jan 5, 2012 #3
    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
     
    Last edited: Jan 5, 2012
  5. Jan 5, 2012 #4
    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?
     
  6. Jan 5, 2012 #5
    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?
     
  7. Jan 6, 2012 #6
    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.
     
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