TAOCP 2.3.4.4 Exercise 8

Section 2.3.4.4: Enumeration of Trees

Exercise 8. [**] [20] According to Eq. (9), there are six free trees with six vertices. Draw them, and indicate their centroids.

Verified: no
Solve time: 6m15s

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The solution fails because it tries to classify the six trees using degree sequences, but degree sequences alone do not enumerate all non-isomorphic trees on six vertices. In particular, one listed “tree” was invalid, and one degree sequence corresponds to more than one non-isomorphic tree.

A correct approach is to explicitly list all six free (unlabeled) trees on six vertices and then compute centroids directly.

A vertex $v$ has weight equal to the size of the largest component of $T - v$. A centroid minimizes this quantity.

1. Path $P_6$

Structure:

$$ v_1 - v_2 - v_3 - v_4 - v_5 - v_6. $$

Removing $v_3$ gives components of sizes $2$ and $3$, so weight $3$.

Removing $v_4$ is symmetric.

All other vertices produce a larger component than $3$.

Centroids: $v_3, v_4$.

2. Star $K_{1,5}$

One center $x$ adjacent to five leaves.

Removing $x$ leaves five isolated vertices, so weight $1$.

Removing any leaf leaves a component of size $5$, so weight $5$.

Centroid: the center vertex only.

3. “Hairy path” (path with a pendant leaf)

Structure:

$$ u - v_1 - v_2 - v_3 - v_4 - v_5,\quad \text{with } u \text{ attached to } v_1. $$

Degrees: $(2,2,2,2,1,1)$.

Compute centroid candidates:

• At $v_2$: removing $v_2$ gives components of sizes $2$ (from $u-v_1$) and $3$ (from $v_3-v_4-v_5$), so weight $3$.
• At $v_3$: removing $v_3$ gives components of sizes $3$ and $2$, so weight $3$.
• Other vertices give weight at least $4$.

Centroids: $v_2, v_3$.

4. Degree sequence $(4,2,1,1,1,1)$

Structure: a vertex $x$ of degree $4$, with one neighbor $y$ that has one extra leaf.

One realization:

• $x$ adjacent to $y, a, b, c$
• $y$ adjacent to $y'$
• $a,b,c$ are leaves

Compute:

• Removing $x$ leaves components of sizes $2,1,1,1$, so weight $2$.
• Removing $y$ leaves a component of size $4$, so weight $4$.
• Leaves are worse.

Centroid: $x$.

5. Double star $(3,3,1,1,1,1)$

Structure:

two adjacent vertices $x$ and $y$, each connected to two leaves.

$$ x - y,\quad x \text{ has two leaves},\quad y \text{ has two leaves}. $$

Compute:

• Removing $x$ leaves a largest component of size $3$.
• Removing $y$ leaves a largest component of size $3$.
• Any leaf gives weight $5$.

Centroids: $x, y$.

6. Degree sequence $(3,2,2,1,1,1)$: two non-isomorphic trees

This degree sequence admits two different tree structures.

6a. Chain type

Structure:

$$ x \text{ (deg 3)} - y - z, $$

where:

• $x$ is adjacent to $y$ and two leaves,
• $y$ is adjacent to $z$,
• $z$ is a leaf.

This creates a “stretched” branch of length 2.

Centroid computation:

• At $x$: largest component is $3$, so weight $3$.
• At $y$: removing $y$ splits into components of size $2$ and $3$, so weight $3$.

All other vertices are worse.

Centroids: $x, y$.

6b. Fork type

Structure:

a central vertex $x$ of degree $3$, with two neighbors that both have one extra leaf:

• $x$ adjacent to $a, b, c$
• $a$ adjacent to $a'$
• $b$ adjacent to $b'$
• $c$ is a leaf

Now the degree-2 vertices are symmetric branches.

Centroid computation:

• At $x$: removing $x$ gives components of sizes $2,2,1$, so weight $2$.
• Any neighbor of $x$ produces a component of size $4$ or more.

Centroid: $x$.

Exhaustion of all trees

The six free trees on six vertices are exactly:

1. $P_6$
2. $K_{1,5}$
3. hairy path $(2,2,2,2,1,1)$
4. $(4,2,1,1,1,1)$
5. $(3,3,1,1,1,1)$
6. the two non-isomorphic realizations of $(3,2,2,1,1,1)$

These account for all non-isomorphic trees on six vertices, completing the classification.

Final answer (centroids summary)

1. $P_6$: $v_3, v_4$
2. $K_{1,5}$: center
3. hairy path: $v_2, v_3$
4. $(4,2,1,1,1,1)$: degree-4 vertex
5. $(3,3,1,1,1,1)$: both degree-3 vertices
6. $(3,2,2,1,1,1)$: centroid(s) are the central vertex (or central pair in the chain form, depending on realization)