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gpt 6 astra ultra attempt on Math-1114

Exploration · LLM-generated · gpt 6 astra ultra ·

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Claim: exploration

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Contents
  1. The Two Color Conjecture
  2. Definitions and mathematical statement
  3. Short English statement
  4. Research attempt
  5. Scope and process.
  6. Literature and directed hypotheses.
  7. A degeneracy bound.
  8. Necessary structure of a counterexample.
  9. Verification and the stalled extension.
  10. Final status and first remaining gap.
  11. Sources

The Two Color Conjecture

Definitions and mathematical statement

An oriented graph $D=(V,A)$ is obtained from a finite simple undirected graph by assigning one direction to each edge. In particular it has no loops, no repeated arcs, and no pair of opposite arcs. Assume its underlying undirected graph is planar. A directed cycle is a cyclic sequence of distinct vertices $v_1,\ldots,v_m$, with $m\ge3$, such that $v_i v_{i+1}\in A$ for all indices read modulo $m$.

For $X\subseteq V$, the induced subdigraph $D[X]$ retains all arcs whose ends lie in $X$. It is acyclic if it has no directed cycle. The Neumann-Lara two-colour conjecture asks whether there is always a partition \[ V=X_1\sqcup X_2 \quad\text{with }D[X_1],D[X_2]\text{ acyclic}. \] Empty parts are allowed. Equivalently, there should be a map $c:V\to\{1,2\}$ such that every directed cycle contains both colours. The least number of acyclic induced colour classes is called the dichromatic number, so the requested bound is at most two.

This is a directed condition. A colour class may contain an undirected cycle if its orientation is not cyclic, and its vertices need not form an independent set. Conversely, the prohibition of opposite arcs in an oriented graph is material: a pair of opposite arcs would itself be a directed cycle of length two. The assertion covers every orientation of every finite simple planar graph, without any assumption about the lengths of its directed cycles.

Short English statement

Can the vertices of every planar oriented graph be coloured with two colours so that no directed cycle is monochromatic?

Research attempt

Scope and process.
This is a GPT-6 Astra Ultra research attempt dated 2 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 directed hypotheses.
Li and Mohar [R1] prove the stronger restricted statement that every planar digraph of digirth at least four is two-colourable. Their hypothesis excludes directed triangles, rather than all undirected triangles. The elementary result below applies to a smaller class but has a short proof and extends to a general degeneracy bound. Throughout, there are no opposite arcs; this is essential to the neighbour count. No identification is made with other conjectures about orientations forcing large undirected chromatic number.
A degeneracy bound.
Suppose the underlying graph is $d$-degenerate, meaning that every nonempty subgraph has a vertex of degree at most $d$. Repeatedly delete such a vertex and restore vertices in reverse order. Set $q=\lfloor d/2\rfloor+1$. Inductively colour the already restored vertices with $q$ colours so that every colour class is acyclic. When restoring $v$, at most $d$ neighbours are already coloured. A colour can become invalid at $v$ only if $v$ has both an incoming and an outgoing neighbour of that colour: any newly created directed cycle must pass through $v$ and use both types of arc. Those two neighbours are distinct because opposite arcs are prohibited. If all $q$ colours were invalid, there would therefore be at least $2q>d$ coloured neighbours, a contradiction. Choose a colour missing one of the two types. This proves \[ \vec\chi(D)\le \lfloor d/2\rfloor+1. \] The condition is sufficient, without requiring a search for directed paths between the neighbours.

Euler's inequality gives $e\le3v-6$ for a simple planar graph with at least three vertices; hence every such graph is 5-degenerate, including its small subgraphs. The construction yields three acyclic colours for all planar orientations. For triangle-free planar graphs, $e\le2v-4$ gives 3-degeneracy and hence two colours. Forest components and subgraphs with fewer than three vertices satisfy the same degeneracy conclusion directly. This settles a genuine infinite subclass with arbitrary orientations.

Necessary structure of a counterexample.
Suppose $D$ were vertex-minimal among planar orientations not admitting two acyclic colours. Every vertex must have at least two incoming and two outgoing neighbours. Indeed, delete $v$ and colour the remainder. If $d^-(v)\le1$, one colour appears on no incoming neighbour, so assigning that colour to $v$ creates no monochromatic directed cycle. The same argument applies to $d^+(v)\le1$. Thus, with $n$ vertices and $m$ arcs, \[ \delta(D_{\rm und})\ge4,\qquad m\ge2n. \] Moreover $D$ is strongly connected: otherwise its strongly connected components are proper induced subdigraphs and can each be two-coloured; every directed cycle lies within one component, so their colourings combine. Its underlying graph consequently has no bridge. A bridge with only one orientation cannot allow directed paths in both directions between its two sides.

Embed the connected underlying graph in the plane. Every face boundary walk has length at least three, since the graph is simple and bridgeless. With $f=m-n+2$ faces, counting boundary lengths gives \[ \sum_F(4-|\partial F|)=4f-2m=8-4n+2m\ge8. \] A triangular face contributes one, a quadrilateral zero, and a larger face a negative quantity. Therefore every such minimal counterexample has at least eight triangular faces. This counts undirected facial triangles; it does not assert that all eight have cyclic orientations. It is a restriction on possible counterexamples, not a contradiction.

Verification and the stalled extension.
The certificate [R2] checks every orientation of $K_4$ by testing all vertex bipartitions for acyclicity. This is a small sanity check on the directed condition, not an exhaustive test of planar graphs. The proof above treats arbitrarily large triangle-free planar graphs. At a vertex of underlying degree four or five, two colours can each occur on both an incoming and an outgoing neighbour. The local extension rule then stops: deciding whether those neighbours are joined by monochromatic directed paths depends on the rest of the graph. The face count does not remove that obstruction.
Final status and first remaining gap.
UNRESOLVED. We obtain the general three-colour bound, prove the triangle-free underlying case with two colours, and constrain a minimal counterexample. An extension argument handling the interaction of directed triangles and those monochromatic paths is still missing for arbitrary planar orientations.

Sources