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

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The discussion centers on understanding the optimality proof of the earliest finish time algorithm, particularly regarding the definitions of feasibility and optimality. The proof suggests that the greedy algorithm aligns with the optimal solution for job finish times up to a certain point, denoted as r, but fails at r+1. This indicates that if the greedy approach is not optimal, there must be a specific job count, r, where the discrepancy occurs. A contradiction arises when the finish time for the r+1th job is the same in both the greedy and optimal scenarios, challenging the initial definition of r. Clarifying these concepts is essential for grasping the algorithm's proof of optimality.
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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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