Generating Uniformly Random Trees

Description of the Problem

Today, I would like to write an algorithm that would generate a random tree on $n$ vertices so that every such randomly created tree shows up equally likely among all possible trees on $n$ vertices.

The Algorithm

I will construct the tree as a topologically ordered directed graph and then remove the direction information.

I will call an undirected graph topologically ordered if there is an orientation of the edges such that the resulting directed graph is topologically ordered. In other words, if $G$ is a topologically ordered undirected graph if there is a topologically ordered directed graph $H$ such that when we forget about the directions of the edges of $H$ we get $G$ .

Assume we already have a random tree $T$ on $n - 1$ vertices. Pick a vertex $i$ uniformly random, and connect an edge $(i, n)$ to form a new graph $S$ . Since $T$ was a tree, it does not contain any cycles. Since $n$ is a new vertex added to $T$ , the new graph $S$ does not contain a cycle either.

As for the likelihood of our trees: There is only one tree on 2 vertices. So, picking a topologically ordered tree on 2 vertices by our algorithm is uniformly random with probability 1. Now, assume by induction that picking a topologically ordered tree on $n 1$ vertices with the algorithm we described above is uniformly random. Assume on the contrary that there is a topologically ordered tree $S$ on $n + 1$ vertices which is more likely to appear from our algorithm. We know that $S$ is obtained from a topologically ordered tree $T$ on $n$ vertices by adding a single vertex $n + 1$ and a single edge $(i, n + 1)$ where $i = 1, \dots, n$ . We also know that picking an edge $(i, n + 1)$ is uniformly random for $i = 1, \dots, n$ with probability $\frac{1}{n}$ . This means that $P (S) = \frac{P (T)}{n}$ and $T$ would have to appear more likely among all topologically ordered trees on $n$ vertices. This is a contradiction.

The Lisp Code

The lisp code is pretty straightforward.

(defun random-tree (n)
   (loop for i from 1 below n 
         collect (list (random i) i)))

RANDOM-TREE

(random-tree 7)

((0 1) (0 2) (1 3) (0 4) (1 5) (2 6))

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