# Prüfer Sequence from Labeled Tree

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## Algorithm

Given a finite labeled tree, it is possible to generate a Prüfer sequence corresponding to that tree.

Let $T$ be a labeled tree of order $n$, where the labels are assigned the values $1$ to $n$.

Step 1: If there are two (or less) nodes in $T$, then stop. Otherwise, continue on to step 2.
Step 2: Find all the nodes of $T$ of degree $1$. There are bound to be some, from Finite Tree has Leaf Nodes. Choose the one $v$ with the lowest label.
Step 3: Look at the node $v\,'$ adjacent to $v$, and assign the label of $v\,'$ to the first available element of the Prüfer sequence being generated.
Step 4: Remove the node $v$ and its incident edge. This leaves a smaller tree $T'$. Go back to step 1.

The above constitutes an algorithm, for the following reasons:

### Finiteness

For each iteration through the algorithm, step 4 is executed, which reduces the number of nodes by $1$.

Therefore, after $n-2$ iterations, at step 1 there will be $2$ nodes left, and the algorithm will stop.

### Definiteness

Step 1: There are either more than $2$ nodes in a tree or there are $2$ or less.
Step 2: There are bound to be some nodes of degree $1$, from Finite Tree has Leaf Nodes. As integers are totally ordered, it is always possible to find the lowest label.
Step 3: As the node $v$ is of degree $1$, there is a unique node $v\,'$ to which it is adjacent. (Note that this node will not also have degree $1$, for then $v v\,'$ would be a tree of order 2, and we have established from step 1 that this is not the case.)
Step 4: The node and edge to be removed are unique and specified precisely, as this is a tree we are talking about.

### Inputs

The input to this algorithm is the tree $T$.

### Outputs

The output to this algorithm is the Prüfer sequence $\left({\mathbf{a}_1, \mathbf{a}_2, \ldots, \mathbf{a}_{n-2}}\right)$.

### Effective

Each step of the algorithm is basic enough to be done exactly and in a finite length of time.

## Example

Let $T$ be the following labeled tree: This tree has $8$ nodes, so the corresponding Prüfer sequence will have $6$ elements.

### Iteration 1

Step 1: There are $8$ nodes, so continue to step 2.
Step 2: The nodes of degree $1$ are $8, 2, 6, 4, 3$. Of these, $2$ is the lowest.
Step 3: $2$ is adjacent to $1$, so add $\mathbf{1}$ to the Prüfer sequence.
Step 4: Removing node $2$ leaves the following tree: At this stage, the Prüfer sequence is $\left({\mathbf{1}}\right)$.

### Iteration 2

Step 1: There are $7$ nodes, so continue to step 2.
Step 2: The nodes of degree $1$ are $8, 6, 4, 3$. Of these, $3$ is the lowest.
Step 3: $3$ is adjacent to $7$, so add $\mathbf{7}$ to the Prüfer sequence.
Step 4: Removing node $3$ leaves the following tree: At this stage, the Prüfer sequence is $\left({\mathbf{1}, \mathbf{7}}\right)$.

### Iteration 3

Step 1: There are $6$ nodes, so continue to step 2.
Step 2: The nodes of degree $1$ are $8, 6, 4$. Of these, $4$ is the lowest.
Step 3: $4$ is adjacent to $5$, so add $\mathbf{5}$ to the Prüfer sequence.
Step 4: Removing node $4$ leaves the following tree: At this stage, the Prüfer sequence is $\left({\mathbf{1}, \mathbf{7}, \mathbf{5}}\right)$.

### Iteration 4

Step 1: There are $5$ nodes, so continue to step 2.
Step 2: The nodes of degree $1$ are $8, 6, 5$. Of these, $5$ is the lowest.
Step 3: $5$ is adjacent to $7$, so add $\mathbf{7}$ to the Prüfer sequence.
Step 4: Removing node $5$ leaves the following tree: At this stage, the Prüfer sequence is $\left({\mathbf{1}, \mathbf{7}, \mathbf{5}, \mathbf{7}}\right)$.

### Iteration 5

Step 1: There are $4$ nodes, so continue to step 2.
Step 2: The nodes of degree $1$ are $8, 6$. Of these, $6$ is the lowest.
Step 3: $6$ is adjacent to $7$, so add $\mathbf{7}$ to the Prüfer sequence.
Step 4: Removing node $6$ leaves the following tree: At this stage, the Prüfer sequence is $\left({\mathbf{1}, \mathbf{7}, \mathbf{5}, \mathbf{7}, \mathbf{7}}\right)$.

### Iteration 6

Step 1: There are $3$ nodes, so continue to step 2.
Step 2: The nodes of degree $1$ are $8, 7$. Of these, $7$ is the lowest.
Step 3: $7$ is adjacent to $1$, so add $\mathbf{1}$ to the Prüfer sequence.
Step 4: Removing node $7$ leaves the following tree: At this stage, the Prüfer sequence is $\left({\mathbf{1}, \mathbf{7}, \mathbf{5}, \mathbf{7}, \mathbf{7}, \mathbf{1}}\right)$.

### Iteration 7

Step 1: There are $2$ nodes, so stop.

The Prüfer sequence is $\left({\mathbf{1}, \mathbf{7}, \mathbf{5}, \mathbf{7}, \mathbf{7}, \mathbf{1}}\right)$.

## Also see

Compare with the example given in Labeled Tree from Prüfer Sequence.

These two results are pulled together in Bijection between Prüfer Sequences and Labeled Trees.