TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-11

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Five colour theorem: every planar graph has chromatic number at most five

Statement

Every planar graph has chromatic number at most five in the sense of Proper vertex colourings and chromatic number. The induction uses vertex deletion from Vertex and edge deletion, edge contraction, graph minors, subdivisions and topological minors, Kempe chains from Kempe chains as connected components induced by two colour classes, and The principle of mathematical induction.

Facts & Assumptions

Given: A finite simple planar graph $G$ with a fixed plane embedding.

[L1]

Every nonnull simple planar graph has a vertex of degree at most five (Every nonnull simple planar graph has a vertex of degree at most five).

[L2]

Swapping the two colours on one Kempe component preserves a proper colouring (Swapping the two colours on one Kempe component preserves a proper colouring).

[L3]

Alternating Kempe paths between the first and third and between the second and fourth cyclic neighbours cannot both occur (For five cyclically ordered neighbours of a plane vertex, alternating Kempe paths between the first and third and between the second and fourth cannot both occur).

Proof

technique · induction

1.1

The null graph is five-colourable. For a nonnull graph choose by [L1] a vertex $v$ of degree at most five; by the induction hypothesis, $G - v$ has a proper colouring with colours $1, \dots, 5$ .

baseihL1

2.1

If fewer than five colours occur on the neighbours of $v$ , give $v$ a missing colour. Otherwise $v$ has degree exactly five, its five neighbours are distinct and use all five colours; list them $v_{1}, \dots, v_{5}$ in their cyclic plane order and relabel so $v_{i}$ has colour $i$ .

step 1.1

3.1

By [L3], either $v_{1}, v_{3}$ lie in different $1$ - $3$ Kempe components or $v_{2}, v_{4}$ lie in different $2$ - $4$ components.

step 2.1L3

4.1

In the first case, swap colours $1$ and $3$ on the component containing $v_{1}$ ; in the second, swap $2$ and $4$ on the component containing $v_{2}$ . By [L2] the colouring remains proper, and respectively colour $1$ or colour $2$ is now absent from the neighbours of $v$ .

step 3.1L2L3

5.1

Give $v$ the freed colour. Together with the immediate case in step 2.1 this extends a five-colouring at every induction stage, proving the theorem.

step 1.1step 4.1discharge-induction∎

Depends on

Used by

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

R. Diestel, Graph Theory, 6th ed., Proposition 5.1.2 (standard reference, not scraped)
R. Grassl and O. Levin, Exploring Combinatorial Mathematics, Activity 307 (standard reference, not scraped)