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]
would improve the result.
So any optimal solution must follow this strict order criteria. ¹
PS: I'm glad I didn't start implementing the B&B algorithm :-)
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.
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