Originally Posted by
exwannabe
No. The "shortest path" problem is with some pre-assigned value to any leg. It does not depend on distance, or any assumptions of an additive metric.
Seth is still technically wrong though. The said logic does work to search for fare breaks, but not solve the entire married fare problem.
Example. A-B-C or A-C (only routes). Shortest path algs would be able to easily search A-B(stop)B-C vs A-C. But the A-C fare itself could route via B with different metrics I would guess that a simple solution of the married fare problem would actually be quadratic in routes.
This is still a useless conversation. In practice the problem is trivial. If DL starts flying 1000000000 routes a day, then let's come back here.
this doesn't address this
once I've found a way to get from AAA to BBB for $222, any route that's already over $222 doesn't need to be extended, because it already lost.
at all, but you are right that this thread is basically useless 1s and 0s