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Math-1095 · mathematics · Graph theory

Mapping planar graphs to odd cycles

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Statement

Definitions and mathematical statement

For an integer $k\geq1$, let $C_{2k+1}$ be the odd cycle on residues $\mathbb Z/(2k+1)\mathbb Z$, with adjacent residues differing by $1$ or $-1$. A homomorphism from a simple graph $G$ to this cycle is a function $f:V(G)\to\mathbb Z/(2k+1)\mathbb Z$ such that \[ f(v)-f(u)\equiv\pm1\pmod{2k+1} \qquad(uv\in E(G)). \] There is no injectivity or surjectivity requirement on $f$.

A finite graph is planar if it has a drawing in the plane without edge crossings. Its girth is the length of its shortest simple cycle, with girth $+\infty$ for a forest. The source conjecture is the assertion \[ \forall k\geq1\quad\forall G\text{ finite simple planar}: \operatorname{girth}(G)\geq4k \Longrightarrow G\to C_{2k+1}. \] Thus the cycle target becomes longer as the required girth increases.

At $k=1$ the target is a triangle and the desired map is exactly a proper three-colouring. For larger $k$, merely using $2k+1$ distinct labels is not enough: the allowed label pairs are only edges of the target cycle. The hypothesis here concerns ordinary girth, excluding short even cycles as well as odd ones. This preserves the supplied catalogue formulation; an odd-girth-only hypothesis is not silently substituted. The statement includes disconnected graphs and forests under the conventions above.

Short English statement

Must every planar graph with no cycle shorter than 4k map homomorphically to the odd cycle of length 2k+1?

Sources

Catalogue entry

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-1095-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-1095

Mapping planar graphs to odd cycles

Definitions and mathematical statement

For an integer $k\geq1$, let $C_{2k+1}$ be the odd cycle on residues $\mathbb Z/(2k+1)\mathbb Z$, with adjacent residues differing by $1$ or $-1$. A homomorphism from a simple graph $G$ to this cycle is a function $f:V(G)\to\mathbb Z/(2k+1)\mathbb Z$ such that \[ f(v)-f(u)\equiv\pm1\pmod{2k+1} \qquad(uv\in E(G)). \] There is no injectivity or surjectivity requirement on $f$.

A finite graph is planar if it has a drawing in the plane without edge crossings. Its girth is the length of its shortest simple cycle, with girth $+\infty$ for a forest. The source conjecture is the assertion \[ \forall k\geq1\quad\forall G\text{ finite simple planar}: \operatorname{girth}(G)\geq4k \Longrightarrow G\to C_{2k+1}. \] Thus the cycle target becomes longer as the required girth increases.

At $k=1$ the target is a triangle and the desired map is exactly a proper three-colouring. For larger $k$, merely using $2k+1$ distinct labels is not enough: the allowed label pairs are only edges of the target cycle. The hypothesis here concerns ordinary girth, excluding short even cycles as well as odd ones. This preserves the supplied catalogue formulation; an odd-girth-only hypothesis is not silently substituted. The statement includes disconnected graphs and forests under the conventions above.

Short English statement

Must every planar graph with no cycle shorter than 4k map homomorphically to the odd cycle of length 2k+1?

Research attempt

Scope and method.
This research attempt by GPT-6 Astra Ultra is dated 1 October 2026. It preserves the catalogue statement and distinguishes elementary proofs below from cited prior work. No novelty claim or formal proof-assistant verification is made. The final paragraph states the precise remaining scope.
Approach and source qualification.
The source [S2] attributes the conjecture to Jaeger and describes the subdivided-wheel obstruction below. We separate that sharpness example from an elementary positive extension method. Throughout, girth means the shortest cycle of either parity; replacing it by odd girth changes the problem. Put $q=2k+1$ and identify the target vertices with $\mathbb Z/q\mathbb Z$.
An exact path-extension lemma.
If a path has length $L$, assigning its edge steps to $\{1,-1\}$ gives endpoint differences $L-2j$ modulo $q$, where $0\le j\le L$. When $L\ge2k=q-1$, the first $q$ of these values represent all residues: if $L-2i\equiv L-2j\pmod q$, oddness of $q$ implies $i\equiv j\pmod q$. Therefore every assignment of endpoint colours extends over a path of length at least $2k$. This is a direct existence proof: choose the required number of negative steps and place them in any order. No injectivity is required of a homomorphism.

At length $L=2k-1=q-2$, exactly $q-1$ residues occur and the missing residue is zero. Indeed a zero would be an odd multiple of $q$ with absolute value at most $q-2$, which is impossible. Thus identifying the endpoint colours across such a path can prevent extension. The one-edge difference in this threshold matters in any reduction that suppresses degree-two vertices.

The lower-bound obstruction.
Take a rim $C_{4k-1}$ and a hub, replacing each spoke by an internally disjoint path of length $2k-1$. This is planar. A cycle avoiding the hub is the rim; a cycle using the hub contains two spokes and at least one rim edge. Its girth is consequently $4k-1$. In a hypothetical target mapping with hub colour $a$, no rim vertex could have colour $a$, by the preceding missing-residue calculation. The rim would then map to $C_{2k+1}-a$, a path and hence a bipartite graph. This cannot receive a homomorphism from an odd cycle. Thus girth $4k-1$ is insufficient. For $k=1$ the construction is simply $K_4$, which does not map to $C_3$. This argument verifies sharpness below the proposed threshold; it is not a counterexample at the requested girth $4k$.
A positive theorem for cactus graphs.
An even cycle maps onto any chosen edge of the target by alternating its endpoints. An odd cycle of length $m\ge q$ maps by traversing the target once and inserting $(m-q)/2$ backtracking pairs, giving a closed walk of length $m$. Rotation permits any desired image of one fixed vertex. In a cactus, every block is an edge or a cycle. Root the block-cutvertex tree and apply these constructions successively, agreeing with the already assigned image of the shared cutvertex. This proves that every cactus all of whose odd cycles have length at least $2k+1$ maps to $C_{2k+1}$. In particular the catalogue's conclusion holds for every cactus of girth at least $4k$, without needing a planar embedding.
Attempted induction and its failure point.
In a vertex-minimal counterexample one can remove a leaf and later extend its image. One can also remove the internal vertices of a sufficiently long degree-two path and extend any mapping of the remaining graph by the path lemma. But a general planar graph of the stated girth need not offer this precise removable path at each induction step. Suppressing a shorter thread can change girth and can force endpoint colours that fail the missing-residue test. The cactus argument avoids this issue because blocks meet in only one vertex. General planar cycles can overlap along paths or interact through several prescribed endpoints.
Verification and final status.
The root batch script computes all walk residues for $1\le k\le8$ and checks both path thresholds exactly. The symbolic arguments above cover all $k$, including the triangle target. UNRESOLVED. We verify the sharpness obstruction and a positive cactus theorem. A reduction preserving the $4k$ girth hypothesis and all necessary endpoint constraints is still required for arbitrary planar graphs.

Sources

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