What is the optimality proof of the earliest finish time algorithm?

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SUMMARY

The discussion focuses on the optimality proof of the earliest finish time algorithm, specifically addressing the conditions under which the greedy algorithm aligns with the optimal solution regarding job finish times. The proof indicates that the greedy algorithm can match the optimal solution for the first r jobs but fails at the r+1th job, where both algorithms yield the same finish time, contradicting the definition of r. This contradiction highlights the necessity for a clear understanding of the terms "feasible," "optimal," and "maximality of r" in the context of algorithmic proofs.

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Homework Statement


I am trying to understand the optimality proof of the earliest finish time algorithm. I have attached the pdf which I am reading. It's just 2 pages. I don't understand what they mean with solution still feasible and optimal (but contradicts maximality of r). An explanation from scratch would be nice...

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The proof supposes that the greedy algorithm matches the optimal solution, in regard to job finish times, up to and including the first r jobs, but not the first r+1. Clearly, if greedy is not optimal, there must be such an r, 0<=r<n. Note that optimal here just means that the finish time of the nth job is as early as it could be.
In the second diagram, the r+1th job finishes at the same time in both greedy and optimal. That contradicts the definition of r, since it was defined to be such that greedy was not as good as optimal at the r+1th job.
 

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