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Math-1037 · mathematics · Geometry

Chromatic number of associahedron

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Statement

Definitions and mathematical statement

For an integer $n\ge3$, fix a convex polygon with $n$ cyclically labelled vertices. A triangulation is a maximal family of pairwise noncrossing internal diagonals, or equivalently a subdivision into triangles using only the polygon's vertices. It has $n-3$ internal diagonals. Each internal diagonal borders two triangles whose union is a convex quadrilateral. A flip removes that diagonal and inserts the other diagonal of the quadrilateral.

Let $\mathcal A_n$ be the finite simple graph with triangulations as its vertices and an edge between two triangulations exactly when one flip transforms one into the other. This is the one-skeleton of the associahedron associated with the $n$-gon; its polytope dimension is $n-3$. A proper colouring of $\mathcal A_n$ assigns colours to triangulations so that any two related by a flip get different colours. Its chromatic number $\chi(\mathcal A_n)$ is the least number of colours required.

The conjecture asks whether \[ \sup_{n\ge3}\chi(\mathcal A_n)=\infty. \] Explicitly, for every positive integer $C$, must there be a polygon size $n$ for which no colouring with $C$ colours is proper? The polygon's actual lengths and angles are irrelevant; only cyclic vertex order enters the graph. The problem concerns ordinary vertex colouring of the flip graph, not colouring the diagonals within one triangulation, and not colouring faces of the polytope. It does not specify a particular growth rate beyond unboundedness.

Short English statement

Is there no fixed number of colours that properly colours triangulations of every polygon when a single flip must change colour?

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Math-1037-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-1037

Chromatic number of associahedron

Definitions and mathematical statement

For an integer $n\ge3$, fix a convex polygon with $n$ cyclically labelled vertices. A triangulation is a maximal family of pairwise noncrossing internal diagonals, or equivalently a subdivision into triangles using only the polygon's vertices. It has $n-3$ internal diagonals. Each internal diagonal borders two triangles whose union is a convex quadrilateral. A flip removes that diagonal and inserts the other diagonal of the quadrilateral.

Let $\mathcal A_n$ be the finite simple graph with triangulations as its vertices and an edge between two triangulations exactly when one flip transforms one into the other. This is the one-skeleton of the associahedron associated with the $n$-gon; its polytope dimension is $n-3$. A proper colouring of $\mathcal A_n$ assigns colours to triangulations so that any two related by a flip get different colours. Its chromatic number $\chi(\mathcal A_n)$ is the least number of colours required.

The conjecture asks whether \[ \sup_{n\ge3}\chi(\mathcal A_n)=\infty. \] Explicitly, for every positive integer $C$, must there be a polygon size $n$ for which no colouring with $C$ colours is proper? The polygon's actual lengths and angles are irrelevant; only cyclic vertex order enters the graph. The problem concerns ordinary vertex colouring of the flip graph, not colouring the diagonals within one triangulation, and not colouring faces of the polytope. It does not specify a particular growth rate beyond unboundedness.

Short English statement

Is there no fixed number of colours that properly colours triangulations of every polygon when a single flip must change colour?

Research attempt

Literature check and attempted route.
Addario-Berry, Reed, Scott and Wood [A1] prove a logarithmic upper bound for associahedron colouring. Their upper bound does not imply the requested unbounded lower bound. The original problem [S2] discusses a lower bound of four for a particular polygon; this attempt does not improve it. Instead, it tests a natural amplification idea: insert many independent pentagon flip cycles and try to multiply their colouring requirements. The calculation below shows exactly why that amplification fails. Sources were checked 1 October 2026.
An induced product of pentagon cycles.
For $r\ge1$ a convex $(3r+2)$-gon can be dissected into $r$ pentagons by $r-1$ noncrossing diagonals: start with a pentagon and repeatedly glue another pentagon along a boundary edge, thereby adding three vertices. Fix the separating diagonals and consider the triangulations containing all of them. A triangulation of each pentagonal cell can be chosen independently, and each cell has five triangulations. Its flip graph is a $5$-cycle: there are two flippable diagonals per triangulation and the five triangulations are connected by successive flips. A flip between two triangulations that both retain every separating diagonal must occur inside one cell. Conversely every flip in one cell gives such an edge. Hence the induced subgraph is exactly \[ X_r=\underbrace{C_5\mathbin\Box\cdots\mathbin\Box C_5}_{r\text{ factors}}. \] The word "induced" matters: flipping a separator leaves this vertex set and cannot create an extra edge between two of its retained vertices.
An explicit three-colouring for every number of cells.
Label the vertices of each $C_5$ cyclically by $0,1,2,3,4$ and use colours $c=(0,1,0,1,2)$ in $\mathbb Z/3\mathbb Z$. Adjacent entries of this cyclic list differ, including the final pair $2,0$. Define \[ C(t_1,\ldots,t_r)=\sum_{j=1}^r c(t_j)\pmod3. \] An edge of $X_r$ changes exactly one coordinate along a cycle edge, so its colour difference is a nonzero element of $\mathbb Z/3\mathbb Z$. Thus $C$ is proper. Holding all but one coordinate fixed gives an odd cycle inside $X_r$, proving that two colours are impossible. We have proved \[ |V(X_r)|=5^r,\qquad \deg(X_r)=2r,\qquad \chi(X_r)=3. \] Exponential growth in the number of triangulations and unbounded degree in this family therefore produce no increasing chromatic obstruction.
The general limitation of independent-cell arguments.
Suppose fixed separators divide a polygon into cells of sizes $n_j$, and suppose each cell flip graph has a colouring into $k$ colours. Regard all these colours as residues modulo the same $k$. Their sum, exactly as above, properly colours the product graph. Each factor is also an induced subgraph, obtained by fixing every other cell. It follows that the chromatic number of this product is precisely the maximum of the factor chromatic numbers. Thus independent repetition cannot amplify any fixed cell obstruction, including one that requires four colours. Multiplying chromatic numbers would be the wrong graph-product rule for these single-cell flips.
Verification and first remaining gap.
A finite check enumerates all product states for $r=1,2,3,4$ and every coordinate flip, verifying the formula above on $5,25,125,625$ vertices. The proof covers all $r$. For $r=1$ it recovers the familiar pentagon, and for $r=2$ it checks that commuting flips form squares, not extra diagonal edges. To derive an unbounded lower bound, a construction must exploit flips that interact across cells or a different obstruction to bounded colouring. No such obstruction is established by this attempt; neither product size nor degree supplies it.
Reproducibility.
The finite calculations above are implemented in scripts/check_attempt_batch_2026_10_01.py; run it with Python 3 from the repository. It uses exact integer arithmetic and no external packages.
Final status.
UNRESOLVED. The independent-pentagon amplification strategy is ruled out by an explicit colouring, while the full associahedron conjecture is not settled. These product-colouring facts are elementary and no novelty claim is made. Completion estimate: 5%.

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