Not necessarily, no - the "perfectly efficient" algorithms often have prohibitively high constants in front of the theoretically perfect big-O, making them only

Not necessarily, no - the "perfectly efficient" algorithms often have prohibitively high constants in front of the theoretically perfect big-O, making them only useful/faster for numbers with so many digits that you can't realistically represent a lot of them, even on modern super computers. On top of that, caching effects and different implementations can make algorithmically "worse" code perform better for longer than naively assumed when comparing their big-O.