CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-03

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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∎

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