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
Answer is (C)
For the given condition we can simply design a K-MAP and mark an edge between every two adjacent cells in K-Map.(adjacency has to seen just as we do for minimization )
That will give us a Bipartite graph. chromatic number for this =2
Also from the same we can conclude that we need ,for a 'n' bit string, to traverse NO MORE than (n-1) edges or 'n' vertices to get a path b/w two arbitrary points.
So ratio is 2/n.
The given graph is actually hypercubegraph.
See problem 4 here:
for traversing from 000. . .0 ⟶ 111. . .1 , we require to traverse atleast 'n' edges.