Let G be a simple, finite, undirected graph with vertex set {v_1, v_2, ..., v_n}. Let Delta(G) denote the maximum degree of G and let C = {1, 2, ..., Delta(G) + 1} denote the set of all possible colors. Color the vertices of G using the following greedy strategy: for i = 1 to n: assign to v_i the smallest color in C not already used by any neighbor of v_i that has already been colored.
Which of the following statements is/are TRUE?
A. This procedure results in a proper coloring of G.
B. The number of colors used is at most Delta(G) + 1.
C. The number of colors used is at most Delta(G).
D. The number of colors used is equal to the chromatic number of G.
GATE 2023 · Discrete Mathematics · Graph Coloring · medium
Answer: A, B
Verify Statement A: greedy always gives a proper colouring: By construction, when colouring v_i the algorithm picks the smallest colour NOT used by any coloured neighbour. So no coloured neighbour of v_i has the same colour as v_i. Subsequent vertices also avoid their neighbours' colours. Hence all adjacent pairs receive different colours — the colouring is proper.
Verify Statement B: at most Delta(G)+1 colours used: When v_i is processed, at most Delta(G) of its neighbours have been coloured already. So at most Delta(G) colours are forbidden for v_i. From the palette {1,...,Delta+1} of size Delta+1, at least one colour is always available. Hence the greedy never needs to exceed Delta+1.
Disprove C and D with counterexamples: C (at most Delta colours): False. For an odd cycle C_{2k+1}, Delta = 2 and chi = 3 = Delta+1 > Delta. Greedy in the right order uses 3 = Delta+1 colours. D (equal to chi): False. For a bipartite graph processed in a bad order, greedy can use 3 colours even though chi = 2.