Consider the undirected graph G defined as follows. The vertices of G are bit strings of length n. We have an edge between vertex u and vertex v if and only if u and v differ in exactly one bit position (in other words, v can be obtained from u by flipping a single bit). The ratio of the chromatic number of G to the diameter of G is
A. 1/(n-1)
B. 1/n
C. 2/n
D. 2/(n+1)
GATE 2006 · Discrete Mathematics · Graph Coloring · medium
Answer: C. 2/n
Show Q_n is bipartite => chi = 2: Partition vertices by parity of number of 1-bits: part A = even parity, part B = odd parity. Flipping one bit changes parity, so every edge crosses the partition. Thus Q_n is bipartite and chi(Q_n) = 2.
Compute diameter(Q_n) = n: The vertices 00...0 and 11...1 differ in all n bits. To get from one to the other we must flip each bit individually, requiring exactly n steps. No pair requires more than n steps. So diameter(Q_n) = n.
Form the ratio: chi(Q_n) = 2, diameter(Q_n) = n, so the ratio = 2/n.