Activity Selection Problem

• Note that f_j always has the maximum finish time of any activity in A
• Greedy-AS takes $\Theta(n)$ time
• This algorithm is greedy because it always picks the activity with the earliest compatible finish time (leave as much time as possible)
• Optimal? Yes
• Proof
  • Note that if ordered by f_i, activity 1 has earliest finish
  • Show that there exists an optimal solution that begins with a greedy choice (activity 1).
  • Let A $\subseteq$ S be an optimal solution. If the first activity in A is (not greedy), then there exists another optimal solution B that begins with 1.
  • B = A - {k} $\cup$ {1}
  • Because , activity 1 is still compatible with A
  • $\mid$B$\mid$ = $\mid$A$\mid$, so B is also optimal. Thus there exists an optimal solution that begins with a greedy choice.
  • Now prove optimal substructure. Once we make the first greedy choice, now the problem is , with optimal solution A' = A - {1}.
  • This solution must be optimal. If B' solves S' with more activities than A', adding activity 1 to B' makes it bigger than A, which is a contradiction.
  • Therefore, greedy choice yields an optimal solution.