====== Section 4.2 Scheduling to Minimize Lateness: an Exchange Argument ======

This problem gives a set of requests which all may begin at time s and must use the resource for a specific interval of time. The difference from the previous problem is that instead of a start and end time, these requests only come with a deadline before which they must be completed. These criteria allow us a few possible rules by which we may define the greedy algorithm.

We may order the jobs by increasing length. This will not work because it is insensitive to deadlines. So one short job with a long deadline will be scheduled before a long job with a short deadline which may lead to the second job being scheduled after the first and the second job being late while the first job has excess slack. 

Maybe they should be scheduled in order of slack, or the difference between the deadline and the length of the interval. This seems better since it takes deadline into account, however it will still not work. The counter example to this solution is a set of intervals, one which takes one hour and is due after two hours and one which takes three hours and is due after three hours. The second interval mentioned would be scheduled first since it has no slack and the other interval would be two hours late. If the two intervals were switched, that would result in minimizing maximum lateness, since the second interval would only be one hour late. 

The rule we should use is scheduling the intervals with the earliest deadlines first. The algorithm will schedule one task as soon as the previous one is completed, and so there will be no idle time. Additionally, two different schedules with neither inversions nor idle time will only differ only in the order which they schedule jobs with identical deadlines and jobs with the same deadlines are scheduled consecutively. This does not affect maximum lateness.  

Since there is an optimal schedule that has no inversions and no idle time and all schedules with no inversions and no idle time have the same maximum lateness, the schedule produced by the greedy algorithm which will have no inversions and no idle time will be an optimal schedule. 

This algorithm will run in O(n log n) time, since the costliest step in the algorithm will be sorting the intervals by deadline. 

This was a dense section which took me a while to get through due to the level of technicality and specification of some of the proofs. It required a good deal of attention and focus at some points, including the proof for statement 4.9.