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day1proof - PI is optimal solution to P(where PI is PI...

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We have a problem instance P, and our greedy choice was p_hat. We've already proven the greedy choice property, there is some OPTIMAL solution PI* that uses p_hat. Also, we've proven (by inspection) that after making greedy first choice p_hat for P, we are left with smaller instance P' of scheduling problem. What is left is to prove: If PI' is an opt solution to P', then PI' union {p_hat} is an optimal solution to P. Proof. (One way to prove A --> B is to "assume A, then prove B".) Assume PI' is an optimal solution to P'. Under this assumption, prove:
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Unformatted text preview: PI is optimal solution to P. (where PI is PI' union {p_hat}) Proof by contradiction: Assume PI is not optimal solution to P. Recall, from the greedy choice property (which we already proved), there is some OPTIMAL solution PI* that uses {p_hat}. |PI*| > |PI|, because PI is not optimal. |PI* - {p_hat}| > |PI - {p_hat}| |PI* - {p_hat}| > |PI'| but since PI* uses p_hat, PI* - {p_hat} is a solution to P'. So PI* - {p_hat} is a better solution to P' than PI' which is a contradiction, because PI' is assume to be optimal solution to P'...
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