Section 7.2.1.6: Generating All Trees

Exercise 65. [38] [38] Generalize Rémy's algorithm (Algorithm R) to $t$-ary trees.

Verified: no
Solve time: 4m55s

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The earlier solution correctly described the growth process and the history-based bijection, but it made an unjustified leap from labeled histories to uniformity over unlabeled trees. The reviewer’s objection is decisive: the number of valid creation-time labelings depends on the tree shape, so one cannot cancel it as a constant factor. The correct resolution is to separate three distinct objects carefully: histories, increasing labelings, and unlabeled trees, and to compute probabilities at the level where the process is truly uniform.

1. The correct Rémy-type growth process for full $t$-ary trees

A full plane $t$-ary tree is a rooted ordered tree in which every internal node has exactly $t$ ordered child positions.

We work with the standard Rémy-style extended tree representation:

Each internal node has exactly $t$ ordered slots. Each slot is either an external node (available) or points to an internal node.

Let $E_k$ be the number of external nodes after $k$ steps. Initially, we start with a single external root, so

$$ E_0 = 1. $$

At each step $k \to k+1$:

Choose uniformly at random an external node $x$. Replace $x$ by an internal node $v$, and attach $t$ new external nodes as its ordered children.

Then

$$ E_{k+1} = E_k - 1 + t = E_k + (t-1), $$

so

$$ E_k = 1 + k(t-1). $$

This defines a well-posed Markov growth process on extended trees.

2. Histories and their exact probability

A history of length $n$ is a sequence

$$ (x_1, x_2, \dots, x_n), $$

where $x_k$ is the external node chosen at step $k$.

At step $k$, the choice is uniform among $E_{k-1}$ external nodes, hence every history has probability

$$ \mathbb{P}(\text{history}) = \prod_{k=0}^{n-1} \frac{1}{1+k(t-1)}. $$

This quantity depends only on $n$, not on any structural properties of intermediate trees.

So the process is uniform on histories.

3. What object is actually in bijection with histories

The key point is that histories do not correspond to plain unlabeled trees, and also do not correspond simply to arbitrary increasing labelings.

A history determines:

1. A final full $t$-ary tree.
2. For each internal node, the exact moment it was created.
3. The specific external slot that was expanded at each step.

This means a history is equivalent to a tree with a complete embedding of its growth history, not just a static labeled tree.

Conversely, given such a history-embedded tree, one recovers the unique sequence of expansions by repeatedly identifying the last created node via the unique external configuration it induces.

Thus:

$$ {\text{histories of length } n} ;\longleftrightarrow; {\text{growth-embedded full } t\text{-ary trees with } n \text{ internal nodes}}. $$

This is a true bijection.

4. Where the previous argument failed

The incorrect step was the identification:

$$ {\text{histories}} \leftrightarrow {\text{unlabeled trees}} \times n! $$

This is false because “label by creation time” does not freely permute. It encodes a partial order constraint, so different shapes admit different numbers of valid increasing labelings. The reviewer’s hook-length correction applies to increasing labelings, but even that is not the correct combinatorial object for histories, because histories contain additional slot-selection information.

Thus neither of the following is valid:

• “each tree has $n!$ labelings”
• “each tree has a shape-independent number of increasing labelings”

Both ignore structural constraints.

5. Correct probability of a fixed unlabeled tree

Fix a specific unlabeled full $t$-ary tree $T$ with $n$ internal nodes.

Each history that produces $T$ corresponds to:

• a choice of an order in which internal nodes are created,
• and a choice of which external slot is expanded at each step.

At step $k$, when the partial tree has $E_{k-1} = 1 + (k-1)(t-1)$ external nodes, the process selects one of them.

Therefore, every valid history contributes the same factor

$$ \prod_{k=0}^{n-1} \frac{1}{1+k(t-1)}. $$

So:

$$ \mathbb{P}(T)

H(T)\cdot \prod_{k=0}^{n-1} \frac{1}{1+k(t-1)}, $$

where $H(T)$ is the number of histories generating $T$.

6. The key combinatorial fact

The correct structural statement is:

$H(T)$ is not constant over all trees $T$.

Indeed, it depends on how many ways one can interleave expansions across the tree structure (equivalently, how many linear extensions exist for the induced growth partial order combined with slot choices). This dependence is exactly what invalidates uniformity.

So the process defines a non-uniform distribution on unlabeled full $t$-ary trees.

7. Final corrected conclusion

Rémy’s algorithm generalizes cleanly to full $t$-ary trees as a growth process on extended trees with external nodes. The correct invariant object is the set of growth histories, not unlabeled trees with naive labeling.

The correct statements are:

• The algorithm is well-defined and produces a random full $t$-ary tree.
• The probability of any particular history depends only on its length.
• The probability of a fixed unlabeled tree is proportional to the number of histories generating it.
• That multiplicity depends on the shape of the tree, so the induced distribution on unlabeled full $t$-ary trees is not uniform for $t \neq 2$ (and in general not uniform under this naive extension).

The earlier claim of uniformity fails precisely because it replaced a structured history count by an incorrect shape-independent labeling count.