Supposing you have a simply connected planar graph of even number of vertices with no loops edges labeled like so:

I found one can set one edge leading to one vertex, say A arbitrarily to 1, then write equations from Kirtchoff's laws for the graph...

http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws

Kirchoff's laws say that the sum of the edges connected to each vertex is 0 and the sum around each face is 0.

It's kind of interesting that after you set one edge to 1, this will always be the right number of equations because of the graph equation V+F=E+2, and we have V+F equations to solve for E unknowns, The plus two in this formula is canceled by the fact that usually the formula counts the region outside of the graph as a face, but we don't use write an equation for that, and we've set one edge value to 1 so there is one less unknown edge.

Note there will be no solution if c,g,l, or m in this example is the edge chosen to be 1. But for any the other edges it will either be this solution or -1*all the values.

Then we can plot these solutions...

Now we can follow the flow of the graph around through every vertex exactly once by starting with A which is 1, then following the path from that vertex that is -1, and from there to the vertex following the path of 1, and so on alternating values. Since there are an odd number of vertices this works out evenly.

If you had an odd number of vertices to start out with, you could break the edge that you set as 1 into two edges one -1 and 1...

**Reasoning**

I don't know how to prove it, it just seems to work in every case. It makes sense to me electrically that if there is a nonzero flow around each face's edges, that some faces will combine and form one flow around the outside. I don't know how to prove that that will always go through every vertex though.

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