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OPG-401: circular chromatic number at most 20/7

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OPG401.root_problem

by hao jia · Sep 7, 2026 · Mathlib 0df444a (Lean v4.33.1)

circular-coloringcombinatoricsgraph-theoryopen-problem

Let GGG be any finite simple graph that is planar, triangle-free, and has maximum degree at most three. Then there exists a map φ:V(G)→Z20\varphi:V(G)\to\mathbb Z_{20}φ:V(G)→Z20​ such that every edge has cyclic color distance between 777 and 131313:

7≤d20(φ(u),φ(v))≤13for every uv∈E(G).7\le d_{20}(\varphi(u),\varphi(v))\le13 \qquad\text{for every }uv\in E(G).7≤d20​(φ(u),φ(v))≤13for every uv∈E(G).

Equivalently, every graph in the stated class has circular chromatic number at most 20/720/720/7. Disconnected and empty graphs are included.

Preamble
import Definitions.Def_opg401_circular_coloring
Formal statement
namespace OPG401

universe u

/-- OPG-401: every finite simple triangle-free planar subcubic graph has a
`(20,7)`-coloring. -/
theorem root_problem
    {V : Type u} [Fintype V] (G : SimpleGraph V)
    (hplanar : IsPlanar G) (htriangle : IsTriangleFree G)
    (hsubcubic : IsSubcubic G) :
    ∃ color : V → Fin 20, IsPQColoring G 20 7 color := by sorry

end OPG401
Source
Open Problem Garden / UnsolvedMath OPG-401, https://www.unsolvedmath.com/problems/OPG-401
Read-back

What the Lean code literally says, in plain math · gpt-5.6-luna

For a type V equipped with a Fintype instance, a simple graph G on V, and hypotheses that (i) there exists an injective map p : V → ℝ × ℝ such that no vertex distinct from both endpoints of an edge lies on the corresponding closed straight segment, and any two edges whose four endpoints are pairwise distinct have disjoint closed straight segments; (ii) for every a,b,c ∈ V, adjacency of a to b and b to c implies that c is not adjacent to a; and (iii) every vertex has at most three distinct neighbors, the declaration asserts that there exists a function color : V → Fin 20 such that for every edge u–v,

7≤min⁡((color(u).val+20−color(v).val)mod⁡20, (color(v).val+20−color(u).val)mod⁡20)≤20−7=13.7 ≤ \min\bigl((color(u).val + 20 − color(v).val) \mathbin{\operatorname{mod}} 20,\ (color(v).val + 20 − color(u).val) \mathbin{\operatorname{mod}} 20\bigr) ≤ 20 − 7 = 13.7≤min((color(u).val+20−color(v).val)mod20, (color(v).val+20−color(u).val)mod20)≤20−7=13.

Here the first hypothesis uses the closed segment consisting of all points ((1 − t)x₁ + t y₁, (1 − t)x₂ + t y₂) with t ∈ ℝ and 0 ≤ t ≤ 1; the second and third hypotheses quantify over all vertices, including the possibility that V is empty. No uniqueness of the coloring is asserted.

Human review
  • Endorsed by Shuze Chen · Sep 7, 2026

  • Endorsed by hao jia · Sep 7, 2026

    Confirmed by the mission captain (proposal self-audit).

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