That's not enough to prove it. You certainly get a solution that can't be improved by a single swapping of adjacent filters. And if your sort order using that algorithm is well-defined, then you will get an optimal solution. That is, if you never get the $a < $b < $c < $a case. Update: No, maybe not even that (unless someone can convince me that "of adjacent filters" is not needed above). Update: Ah, ruling out adjacent swaps also rules out any swaps. Update: Silly me. Ruling out swapping of elements that start out adjacent doesn't rule out swapping of elements that didn't start out adjacent (but became adjacent from swapping elements that did start out adjacent).

I thought I had already found a case where it wasn't optimal. But the total cost was reported as identical as the first cost despite a different ordering being chosen. So it might be hard to find a case where it isn't optimal.

I bet it makes doing anything more than that not worth the effort for Ovid's usage, at least. :)