For Travelling salesman problem, I've seen a couple of people claim that 3-opt yields a better solution than 2-opt (and 4-opt better than 3-opt, etc.) but I've never seen why and never seen any proof. It seems that 2-opt gives the same result as k-opt, k ≥ 2.
Do you have any counter-example? (preferably with images) I'd like ones that show 4-opt > 3-opt, 3-opt > 2-opt and 4-opt > 2-opt.