Chapter 4

The motivation for interval scheduling is to schedule as many jobs as possible with no overlap between the intervals of time for each job. When designing a greedy algorithm for this problem, we base the algorithm on one simple rule; e.g. selecting jobs that start the earliest, or jobs with smallest interval times. In this problem, the simple rule that produces an optimal solution is selecting jobs that finish the earliest. The greedy algorithm looks like such:

   R = set of all jobs, A is an empty set
   while R != empty
        choose a job R(i) that has smallest finishing time
        add i to A
        delete all jobs from R that are not compatible with i
   return A
   

In this algorithm, we simply add the job i that ends as soon as possible, then remove all jobs that conflict with i. We then repeat this until there are no more jobs, resulting in a run time of O(n). We prove this algorithm returns a globally optimal solution using a Greedy Stays Ahead proof. Essentially, using a proof by contradiction, we declare an optimal set O that contains the most optimal scheduling of jobs compared to greedy's set A. If this were the case, O must have more jobs scheduled than A; however, if there was an additional optimal job, it would be in A, a contradiction. Thus, A must be as optimal as O.

A similar problem is the Interval Partitioning Problem, where we want to schedule all lectures in as few classrooms as possible. An important note about this problem is that the number of classrooms needed is at least the depth of the set of lecture intervals. Overall, the readability of this section was 7/10.

The motivation for this problem is to schedule jobs with corresponding deadlines in a way such that there is minimal lateness. Jobs are allowed to be late (scheduled so that they will end past their deadline), but this must happen as little as possible. Again, we must design a greedy algorithm based on a simple rule; e.g. schedule longest jobs first. The best rule in this problem is earliest job first. Algorithm:

   order jobs in order of their deadline
   f = s
   consider jobs i = 1, ..., n
        assign job i to the time interval from s(i) = f to f(i) = f + ti 
        let f = f + ti
   return set of scheduled intervals [s(i), f(i)] for i, ..., n
   

The schedule produced has no gaps in between jobs. On a different note, an inversion can occur in a schedule if there is a job i with deadline di that is scheduled before job j where dj < di. To prove that this greedy algorithm does not produce any inversions, we do an exchange proof, swapping inversions in an optimal schedule O until eventually we have a schedule that is identical to the one our greedy algorithm would have produced. Overall, the readability of this section is 6/10.