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Math-1060 · mathematics · Graph theory

Chromatic number of $\frac{3}{3}$-power of graph

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Statement

Definitions and mathematical statement

All graphs here are finite, simple, and undirected. For an integer $n\geq1$, the $n$-subdivision $G^{1/n}$ is obtained by replacing each edge $uv$ of $G$ by a path of exactly $n$ edges from $u$ to $v$. The $n-1$ internal vertices on each such path are new, and internal vertices belonging to different original edges are distinct. For an integer $m\geq1$, the $m$th power of a graph $H$ joins two distinct vertices when their distance in $H$ is at most $m$. Vertices in different components have infinite distance.

Define the fractional-power notation by \[ G^{m/n}=(G^{1/n})^m. \] In particular, $G^{3/3}$ means the third power of the graph obtained by subdividing every edge twice. The fraction in this notation is not reduced: $G^{3/3}$ need not be $G$.

Write $\Delta(G)$ for the maximum vertex degree of the original graph. A proper $q$-colouring assigns one of $q$ colours to each vertex, assigning different colours at adjacent vertices; $\chi$ is the least possible $q$. The conjecture is \[ \chi(G^{3/3})\leq2\Delta(G)+1 \qquad\text{for every }G\text{ with }\Delta(G)\geq2. \] Consequently the desired colours must separate every pair within three steps in the subdivided graph, including both original and new vertices. The degree in the bound is measured before either operation.

Short English statement

After replacing each edge by a three-edge path, can all vertices within distance three be distinguished using at most twice the original maximum degree plus one colours?

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Math-1060-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-1060

Chromatic number of $\frac{3}{3}$-power of graph

Definitions and mathematical statement

All graphs here are finite, simple, and undirected. For an integer $n\geq1$, the $n$-subdivision $G^{1/n}$ is obtained by replacing each edge $uv$ of $G$ by a path of exactly $n$ edges from $u$ to $v$. The $n-1$ internal vertices on each such path are new, and internal vertices belonging to different original edges are distinct. For an integer $m\geq1$, the $m$th power of a graph $H$ joins two distinct vertices when their distance in $H$ is at most $m$. Vertices in different components have infinite distance.

Define the fractional-power notation by \[ G^{m/n}=(G^{1/n})^m. \] In particular, $G^{3/3}$ means the third power of the graph obtained by subdividing every edge twice. The fraction in this notation is not reduced: $G^{3/3}$ need not be $G$.

Write $\Delta(G)$ for the maximum vertex degree of the original graph. A proper $q$-colouring assigns one of $q$ colours to each vertex, assigning different colours at adjacent vertices; $\chi$ is the least possible $q$. The conjecture is \[ \chi(G^{3/3})\leq2\Delta(G)+1 \qquad\text{for every }G\text{ with }\Delta(G)\geq2. \] Consequently the desired colours must separate every pair within three steps in the subdivided graph, including both original and new vertices. The degree in the bound is measured before either operation.

Short English statement

After replacing each edge by a three-edge path, can all vertices within distance three be distinguished using at most twice the original maximum degree plus one colours?

Research attempt

Scope and process.
This is a GPT-6 Astra Ultra research attempt dated 1 October 2026, using primary-source browsing and elementary mathematical checks. The canonical definition above is retained. Results below are restricted results or reductions, not a claim of a new solution to the entire catalogue question. No formal proof assistant or external mathematical referee checked this text.
Literature and scope.
Mozafari-Nia and Iradmusa [R1] prove the conjectured bound in the subquartic case and state the general $2\Delta+1$ question. We independently construct colourings for all graphs of maximum degree two and for all trees. The tree argument uses the actual three-edge subdivision; reducing the notation $3/3$ to $1$ would change the graph and invalidate these calculations.
Cycles: a five-colour construction.
Subdividing $C_m$ produces the cycle $C_{3m}$, with $m\ge3$. Let $N=3m$. Every such $N\ge9$ can be written $4a+5b$ with nonnegative integers $a,b$: $9=4+5$, and all integers at least $12$ have this form, since $12,13,14,15$ do and adding four preserves it. Arrange around the cycle a concatenation of $a$ blocks $(1,2,3,4)$ and $b$ blocks $(1,2,3,4,5)$, in any order.

No two equal colours have cyclic distance at most three. For colours $1,2,3,4$, consecutive occurrences are in consecutive blocks and separated by the preceding block length, four or five. For colour five, consecutive occurrences are separated by at least five; if it occurs just once there is no pair to check. The same reasoning applies across the closing boundary. Hence the colouring is proper for $C_N^3$ and uses at most five colours. For $m=3$, any independent set of $C_9^3$ has at most two vertices: three selected vertices would require three cyclic gaps each at least four, totaling at least twelve. Thus $C_9^3$ needs at least $\lceil9/2\rceil=5$ colours, so the bound is attained.

Paths and disconnected graphs of degree two.
A path with $m$ edges subdivides to a path with $3m$ edges. Colour its successive vertices periodically $1,2,3,4$. Vertices at distance one, two or three have different colours, giving a four-colouring of its third power. Isolated vertices need one colour. Every finite graph of maximum degree at most two is a disjoint union of these paths and cycles; separate components have no finite-distance adjacency, so palettes may be reused. This proves the target bound $2\Delta+1=5$ for $\Delta=2$.
A stronger bound for every tree.
Let $H$ be any tree, rooted at a vertex, and order its vertices by nonincreasing depth. The neighbours of a vertex $v$ in $H^3$ which occur later in this order form a clique. To check this, write $p$ for its parent. Such neighbours consist only of ancestors within three steps, siblings of $v$, and children of its grandparent of depth one less than $v$; some same-depth siblings may have been removed already. No later descendant of $v$ is possible. Every pair in this list is at distance at most three, as is seen by their paths through the parent and grandparent. The same description truncates harmlessly near the root. Thus colouring in reverse order needs at most the largest clique size: each vertex sees a clique of previously coloured neighbours.

A set of vertices pairwise within distance three in a tree lies within the closed neighbourhood of some edge, or within one closed vertex-neighbourhood. Indeed its minimal connecting subtree has diameter at most three, with its center an edge or vertex. For a central edge $uv$ the union $N_H[u]\cup N_H[v]$ has $d_H(u)+d_H(v)$ vertices. If now $H=G^{1/3}$ for a tree $G$ of maximum degree $\Delta\ge2$, every edge of $H$ either joins an original vertex to a degree-two subdivision vertex, or joins two degree-two vertices. The largest possible clique size is therefore at most $\Delta+2$. We obtain the stronger restricted bound \[ \chi(G^{3/3})\le\Delta+2\le2\Delta+1 \qquad(G\text{ a tree},\ \Delta\ge2). \]

Verification and remaining obstruction.
The script scripts/check_attempt_batch02_algebra_2026_10_01.py checks the cyclic block colourings for $9\le N\le300$ divisible by three. The proofs, rather than that cutoff, cover all sizes. The tree elimination argument uses unique paths. In a subdivided graph with cycles, vertices later in the order can acquire short routes around a cycle and need not form a clique. Consequently the tree bound does not extend by choosing a spanning tree: additional original edges introduce additional subdivision vertices and new distance-three conflicts.
Final status.
UNRESOLVED for arbitrary maximum degree and cyclic graphs. All degree-two instances and all trees are settled by explicit colourings or a proved elimination rule, with a sharp degree-two example and a precise failure of the spanning-tree shortcut.

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