I'm confused about how the bead-tiling algorithm for the Dubins' TSP from this article works:     On  the  Dubins  Traveling Salesperson Problems:novel  approximation algorithmsKetan Savla, Emilio Frazzoli, Francesco Bullohttp://www-bcf.usc.edu/~ksavla/papers/KS-EF-FB:06h.pdf What does the algorithm return?  Does it return this path that has been going over all (meta-)beads almost log2(n) times? Or only the final path obtained from the last recursive phase? a  Dubins  tour  is  constructed  with  the  following  properties: ...  This  process  is  iterated log2n times, ...Also, does this result path go back to the top left corner since it started there?If yes, why waste some length of our path to go back to the top, instead of going to the first node encountered to obtain our Dubins tour?  ... it  visits  all  (meta-bead)  rows  in  sequence  top-to-down, alternating between left-to-right and right-to-left passes, ...A minor issue is also the complexity stated. it states the computational complexity is of order n. I personally thought it to be so much more. The  computational  complexity  of  the  RECURSIVE BEAD-TILING ALGORITHM is of order n.All in all, to express it in one simple question :What does the bead-tiling algorithm do?
Bead-tiling algorithm for Dubins' TSP
traveling salesman