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For a finite simple graph, choose an order of its vertices and process them one at a time. Give each vertex the smallest positive colour label not already used by its coloured neighbours, and keep all earlier assignments. This is greedy coloring: each step makes a locally available choice. The resulting graph coloring is proper because, for each edge, its later endpoint avoids the colour of its earlier endpoint.^[ref]

If the maximum degree of a graph is $\Delta (G)$, a vertex has at most $\Delta (G)$ previously coloured neighbours, forbidding at most that many distinct colours. At least one label among $1,\dots ,\Delta (G)+1$ remains available. Consequently every order givesThis is an upper bound, not a promise that greedy coloring attains the chromatic number.^[ref]

Consider the path graph $P_4$ with edges $12,23,34$. In the order $1,2,3,4$, greedy coloring assigns the colours $1,2,1,2$ respectively. In the order $1,4,2,3$, it assigns colours $1,1,2,3$ in processing order: vertices $1,4$ both get colour $1$, then vertex $2$ gets colour $2$, and finally vertex $3$ sees colours $2$ and $1$ at its two neighbours and must take colour $3$. Written in vertex order $1,2,3,4$, this second colouring is $1,2,3,1$.

The path graph has chromatic number two: the first assignment achieves two colours, and any edge rules out one. Its maximum degree of a graph is two, so three colours in the second assignment still satisfy the bound. The extra colour comes entirely from the processing order. The crown graph gives a larger family where a poor order can make greedy coloring use arbitrarily many colours although the chromatic number remains two.