Maths WorkshopMaths problems with full solutions
Exercise 636Combinatorics › Graph theoryM1Added 5 Nov 2026

How many colours does a graph with m edges need?

#graph colouring#chromatic number#extremal

Statement

A proper colouring of a graph colours its vertices so that adjacent vertices get different colours. Let GG be a graph with mm edges and let k=χ(G)k=\chi(G) be the smallest number of colours of a proper colouring.

  1. Prove that (k2)≤m\dbinom k2\le m, hence k≤1+8m+12k\le\dfrac{1+\sqrt{8m+1}}2.
  2. What does this give for m=10m=10? Is the bound attained?
Hint

Take a proper colouring with the minimum number kk of colours. If two colour classes had no edge between them, what could you do?

Show solutionHide solution

Solution

1. Take a proper colouring with exactly kk colours and consider two colour classes Ci,CjC_i,C_j. If there were no edge between CiC_i and CjC_j, we could recolour CjC_j with the colour of CiC_i and still have a proper colouring with k−1k-1 colours, contradicting minimality. So for each of the (k2)\binom k2 pairs of classes there is an edge joining them, and these edges are distinct. Hence m≥(k2)m\ge\binom k2, which gives k≤1+1+8m2k\le\dfrac{1+\sqrt{1+8m}}2.

2. For m=10m=10: k≤1+92=5k\le\dfrac{1+9}2=5. The bound is attained by K5K_5, which has 1010 edges and needs 55 colours.

Remark. Equality m=(k2)m=\binom k2 forces exactly one edge between any two classes; KkK_k is the typical example.

Lesson. In a minimal colouring every pair of colour classes must be linked: use minimality to produce edges.