How Many Spanning Trees Can a Graph Have?

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The number of spanning trees in a graph cannot be determined solely from the number of nodes and branches. However, if the graph's vertices and edges are labeled, the "deletion-contraction" theorem can be applied to recursively calculate the number of spanning trees. This theorem states that the total number of spanning trees is the sum of the spanning trees obtained by deleting or contracting an edge. Additionally, calculating a cofactor of the graph's Laplacian matrix is another method to find the number of spanning trees, but it also requires labeling the graph. Understanding these methods is essential for accurately determining the number of spanning trees in a given graph.
Marvin94
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How many trees are in a graph if the number of nodes and the number of branches (or meshes) is given? I refer to the trees which has to "walk" through the nodes along the given branches (so no tree between 2 nodes without a connecting branch between them)
 
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I don't think there is a way to count the number of spanning trees of a graph if you only have the number of nodes and branches.

However, If you are able to label the vertices and edges of a graph then there is the "deletion-contraction" theorem which says that the number of spanning trees of a graph, G, is equal to the number of spanning trees of G by deleting an edge plus the number of spanning trees of the G by contracting an edge. From there, it's clear to see that you can recursively apply this theorem until you end up with a bunch of small graphs with trees that you can count easily. Another method is to calculate any cofactor of the graph's laplacian matrix (https://en.wikipedia.org/wiki/Laplacian_matrix), but again to do so you must be able to label the graph.

I hope this helps a bit.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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