Counting Spanning Trees
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If as you say, your code can be broken if vertices are strings, then you did not really solve your problem by taking this into account, as vertices can actually be of any immutable type. See for example patch where vertices are defined as Words which is a totally independent Sage object. This does not fit in the integer case, nor in the String case.
If I make no mistake remembering what is written in the book you mentioned, they also talk of a different way to compute the number of out-trees : you do not add this special vertex, but just consider the kirchhoff matrix of the first graph, then add 1 to the vertex you want to take as root.
It is I think an easier way to define your matrix in this case, without having to consider these types.. You just have to deal with the matrix! I'm sorry I can not write this patch myself now, I do not have the correct tools on the computer I use and have some urgent work to get done until tomorrow Hello Nathann, thanks for pointing me in the right way, I have made a new patch that just increases the diagonal value of the root vertex by 1, which indeed is a simpler solution.
As a bonus there is no longer any need to copy the graph, since we no longer do any changes to the original graph in our computations.
Nathann, I was wrong! I've ONLY dealt with Kirchhoff matrices in the context of electrical networks, in which loops are utterly insignificant. For me, the definition of the Kirchhoff matrix is that the diagonal is the row-sum of the weighted adjacency matrix. But, the definition you are using appears to be standard, and as noted here, more useful.
This patch should satisfy all of us, I hope. I thought of other ways to write it to preserve the code, but these ways could have at some point impaired the performances, as checks for the values of the variable indegree could have happened much more often. I write it this way as the function is still very short and this should not be a problem for its maintenance.
Looks fine to me. Besides, I wanted to do something about.
Counting Spanning Trees in Grid Graphs - Technische Informationsbibliothek (TIB)
With these lines, you are evaluating all the vertices at each look, just to return its jth element. As the vertices do not change, you could have stored the list of vertices in a variable, each time trying to find the jth element of this list without listing allt he vertices again. But with this new patch, you are just getting the index of the vertex you are interested in, and updating the matrix And with some luck, this patch is the last one Well, I'll take your "Looks really nice.
I'm fully satisfied with the patch" as as an answer Powered by Trac 1.
Counting degree sequences of spanning trees in bipartite graphs: A graph‐theoretic proof
Opened 10 years ago Closed 10 years ago Last modified 10 years ago. Description last modified by AJonsson This patch allows us to count the number of spanning trees in a simple graph, as well as the spanning out-trees from a user-defined root node in a digraph. Oldest first Newest first Threaded. Show property changes.
Changed 10 years ago by AJonsson. Here is a new patch based on yours, plus some modifications : I was not able at first to apply your patch, as it was based on an old version of Sage. This new version is now based on 4. Many graphs have vertices not among integers but among strings, or tuples, etc.
kereheemsbilec.cf The matrix tree theorem does not give the final answer to all problems concerning enumeration of trees because finding the determinant of a matrix is a complicated task for complicated graphs. So, this paper is useful to find the number of spanning trees of the complicated graph simply.
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