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✓ 2 results · all verified · 0 also independently AI-judged

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Graph Colouring — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-03Open item page →

The complete graph $K_{n}$ has chromatic number $n$

Example

For every $n \in N$ , including $n = 0$ , the complete graph $K_{n}$ satisfies

$χ (K_{n}) = n .$

Facts & Assumptions

Given: The complete graph $K_{n}$ on the vertex set $n$ .

[L1]

Every two distinct vertices of $K_{n}$ are adjacent (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

[L2]

A proper colouring gives different colours to adjacent vertices, and $χ$ is the least size of an available colour set (Proper vertex colourings and chromatic number).

Verification

technique · direct

1.1

In a proper colouring of $K_{n}$ , [L1] and [L2] force all vertices to receive distinct colours, so at least $n$ colours are required; this is also true at $n = 0$ , where the assertion is vacuous.

L1L2

1.2

The identity function $n \to n$ is a proper $n$ -colouring because distinct vertices receive distinct values.

L1L2

2.1

The lower bound in step 1.1 and the colouring in step 1.2 give $χ (K_{n}) = n$ .

step 1.1step 1.2L2∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-03Open item page →

A vertex order on $P_{4}$ makes greedy colouring use three colours although $χ (P_{4}) = 2$

Counterexample

Let $P_{4}$ have vertices $0, 1, 2, 3$ and consecutive edges. In the order

$0, 3, 1, 2,$

the least-available-colour greedy rule uses three colours, although $χ (P_{4}) = 2$ .

Facts & Assumptions

Given: The path graph $P_{4}$ with edges ${0, 1}$ , ${1, 2}$ and ${2, 3}$ .

[L1]

The path-graph convention gives exactly these vertices and edges (Empty and complete graphs, complete bipartite graphs, and the convention that $P_{n}$ and $C_{n}$ have $n$ vertices).

[L2]

The greedy rule assigns the least colour absent from the earlier neighbours (The greedy colouring bound $χ (G) \leq Δ (G) + 1$ for every nonnull finite graph).

[L3]

A proper colouring assigns different colours to adjacent vertices, and the chromatic number is the least number of colours in a proper colouring (Proper vertex colourings and chromatic number).

Verification

technique · direct

1.1

In the displayed order, vertices $0$ and $3$ are nonadjacent and both receive colour $0$ ; vertex $1$ is adjacent to $0$ and receives colour $1$ ; vertex $2$ is adjacent to $3$ of colour $0$ and to $1$ of colour $1$ , so it receives colour $2$ . Thus greedy uses three colours.

L1L2

1.2

The assignment $0, 2 \mapsto 0$ and $1, 3 \mapsto 1$ is a proper $2$ -colouring, while the edge ${0, 1}$ rules out a $1$ -colouring. Hence $χ (P_{4}) = 2$ .

L1L3

2.1

In the order $0, 1, 2, 3$ , greedy assigns colours $0, 1, 0, 1$ and uses only two colours. Together with step 1.1 and step 1.2, this shows that the number used by greedy colouring depends on the vertex order and can exceed the chromatic number.

step 1.1step 1.2L1L2∎

Sources

Springer, Graph Theory Fundamentals, Section 1.7.1

Graph Colouring — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The complete graph KnKn​ has chromatic number nn

Example

Facts & Assumptions

Verification

A vertex order on P4P4​ makes greedy colouring use three colours although χ(P4)=2χ(P4​)=2

Counterexample

Facts & Assumptions

Verification

Sources

The complete graph $K_{n}$ has chromatic number $n$

A vertex order on $P_{4}$ makes greedy colouring use three colours although $χ (P_{4}) = 2$