ExampleConstruction: AI-generatedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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All fourteen triangulations of the labelled hexagon

Example

Grouped by the split index $k$ of For $m \geq 3$ and a triangulation $T$ of the $m$ -gon there is a unique $k$ with $1 < k < m$ such that ${1, k}$ and ${k, m}$ are both chords of $T$ or sides, and $T$ splits along $k$ , the triangulations of the labelled hexagon are:

$k$	triangulations
$2$	${{2, 4}, {2, 5}, {2, 6}}$ , ${{2, 4}, {2, 6}, {4, 6}}$ , ${{2, 5}, {2, 6}, {3, 5}}$ , ${{2, 6}, {3, 5}, {3, 6}}$ , ${{2, 6}, {3, 6}, {4, 6}}$
$3$	${{1, 3}, {3, 5}, {3, 6}}$ , ${{1, 3}, {3, 6}, {4, 6}}$
$4$	${{1, 3}, {1, 4}, {4, 6}}$ , ${{1, 4}, {2, 4}, {4, 6}}$
$5$	${{1, 3}, {1, 4}, {1, 5}}$ , ${{1, 3}, {1, 5}, {3, 5}}$ , ${{1, 4}, {1, 5}, {2, 4}}$ , ${{1, 5}, {2, 4}, {2, 5}}$ , ${{1, 5}, {2, 5}, {3, 5}}$

Facts & Assumptions

Given: the labelled hexagon with vertices $1, \dots, 6$ .

[L1]

The split index $k$ on the closing side is unique (For $m \geq 3$ and a triangulation $T$ of the $m$ -gon there is a unique $k$ with $1 < k < m$ such that ${1, k}$ and ${k, m}$ are both chords of $T$ or sides, and $T$ splits along $k$ ).

[L2]

The number of triangulations of the labelled hexagon is $C_{4} = 14$ ( $∣ P_{n + 2} ∣ = C_{n}$ , The Catalan number $C_{n} : = ∣ D_{n} ∣$ ).

Verification

technique · direct

1.1

Every diagonal set in the table has three pairwise non-crossing diagonals, so each row is a triangulation of the hexagon.

given

2.1

The four groups are disjoint because the split index $k$ of [L1] is unique, and the group sizes are $5$ , $2$ , $2$ and $5$ , so the table contains $14$ triangulations altogether.

L1step 1.1

3.1

This agrees with [L2], since $C_{4} = 14$ . The same grouped count is the recursion $∣ P_{2} ∣ ∣ P_{5} ∣ + ∣ P_{3} ∣ ∣ P_{4} ∣ + ∣ P_{4} ∣ ∣ P_{3} ∣ + ∣ P_{5} ∣ ∣ P_{2} ∣ = 5 + 2 + 2 + 5$ .

L2step 2.1∎

Remarks

The two extreme groups are the fan triangulations based at the vertices $2$ and $5$ , together with the four further triangulations on the corresponding pentagons.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

D. Guichard, An Introduction to Combinatorics and Graph Theory, Exercise 3.5.5 (standard reference, not scraped)