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Well my argument is sufficient to show that a solution where two adjacent filters don't follow this order can't be optimal, because otherwise swapping those adjacent filters f[i] and f[i+1] would improve the result.
So any optimal solution must follow this strict order criteria. ¹ qed! =)
Cheers Rolf PS: I'm glad I didn't start implementing the B&B algorithm :-) UPDATE: ¹) and it's easy to see that all ordered solutions (plural b/c adjacent filters can have the same weight) imply the same total cost. In reply to Re^5: Evolving a faster filter? (optimal!)
by LanX
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