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For a nonempty finite simple graph $G$, the chromatic number $\chi (G)$ is the least positive integer $k$ for which $G$ has a proper vertex graph coloring with at most $k$ colours. To establish an exact chromatic number, give a colouring using that many colours and explain why fewer cannot suffice. A colouring alone gives only an upper bound.^[ref]

These small graphs show why the chromatic number measures more than the presence of triangles in a graph:

For a cycle graph of even length, alternating two colours closes consistently; for an odd cycle, it does not, and a third colour on the final vertex suffices. More generally, the two-colourability criterion for bipartite graphs says that $\chi (G)\le 2$ exactly when $G$ is a bipartite graph, and the odd-cycle characterization of bipartite graphs identifies the obstruction as an odd cycle. A nonempty graph with no edges has chromatic number one; a bipartite graph with at least one edge has chromatic number exactly two.

In the pairwise-conflict scheduling model of graph coloring, $\chi (G)$ is the least number of interchangeable slots needed. This turns a drawing into an optimization question: how many groups of compatible activities are unavoidable? Greedy coloring constructs a valid grouping quickly, but its number of colours can exceed the chromatic number.

A crown graph is a complete bipartite graph with a perfect matching removed. For $n\ge 2$ it has chromatic number two, but greedy coloring in the order $a_1,b_1,a_2,b_2,\dots ,a_n,b_n$ uses $n$ colors: the matched pair gets the same new color, and every subsequent pair sees all earlier colors.

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.