Counting Spanning Trees of a Graph

Description of the Problem

Given a undirected graph $G = (V, E)$ , a spanning tree $T$ of $G$ is a connected subgraph of $G$ which uses the full set of vertices but uses a minimal set of edges such that $T$ has no cycles while still being connected. Today, we are going to count all spanning trees in a given graph.

Laplacian Matrix

In almost all of the graph algorithms I considered so far, I assumed my graphs were sparse: the number of edges were comparible to the number of vertices. Therefore, I almost invariably used a list of edges to represent a graph such as

(defvar G '((0 1) (1 2) (1 3) (2 4) (3 4)))

G

which represents the following graph

Of course there are other representations. One of them is the adjacency matrix which is a $n \times n$ matrix which consists of 0’s and 1’s and defined as $a_{i j} = {\begin{cases} 1 & if there is an edge between i and j \\ 0 & otherwise \end{cases}$ and yet another one is the Laplacian of a graph which is also an $n \times n$ matrix given by $a_{i j} = {\begin{cases} \deg (i) & if i = j \\ - 1 & if there is an edge between i \neq j \\ 0 & otherwise \end{cases}$

Kirchoff’s Theorem states that the determinant of any minor of the Laplacian will give us $n$ times the number of spanning trees of $G$ .

Code

The following function returns the Laplacian matrix of a given graph represented as a list of edges. We assume edges are represented by lists of vertices of size 2, and that the vertices are numbered starting from 0.

(defun laplacian (G)
   (let* ((n (1+ (apply #'max (loop for x in G append x))))
          (m (make-array 
               (list n n) 
               :initial-element 0)))
     (loop for edge in G do
        (destructuring-bind 
             (a b) edge
          (incf (aref m a a))
          (incf (aref m b b))
          (decf (aref m a b))
          (decf (aref m b a))))
     m))

LAPLACIAN

Let us test:

(laplacian G)

#2A((1 -1 0 0 0) (-1 3 -1 -1 0) (0 -1 2 0 -1) (0 -1 0 2 -1) (0 0 -1 -1 2))

For the number of spanning trees, we will need to compute the determinant of any minor. In the following, I will delete the first row and the column.

(defun number-of-spanning-trees (G)
   (let* ((m (laplacian G))
          (n (array-dimension m 0))
          (x (grid:subgrid m (list (1- n) (1- n)) '(0 0))))
      (round (lla:det x))))

NUMBER-OF-SPANNING-TREES

(number-of-spanning-trees G)

4

Now, let us test something larger

(defvar H
   (remove-duplicates 
      (loop repeat 100 collect 
          (let* ((a (random 12))
                 (b (+ 1 a (random 9))))
             (list a b)))
      :test #'equal))
H

and the number of spanning trees for this graph is

(number-of-spanning-trees H)

51292592574547

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The Kitchen Sink and Other Oddities

Counting Spanning Trees of a Graph

Description of the Problem

Laplacian Matrix

Code

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