Given a sequence A = {a_1, ..., a_n} of points on the real line. The task is to determine the smallest set of unit-length (closed) intervals that contains all of the input points. Consider the following two greedy approaches: (a) Let I be an interval that covers the most points in A. Add I to the solution, remove the points covered by I from A, and recurse/continue. (b) Add the interval I = [a_1, a_1 + 1] to the solution, remove the points covered by I from A, and recurse/continue. One of these approaches is correct, the other one is not. Show which of the approaches is not correct by finding a counter-example. (The counter example consists of an example input and two "solutions" - one is the actual optimal solution, the other is the solution computed by the greedy algorithm, which is not as good as the optimal solution.)