0

You would have to convert an instance of the seating problem to an instance of Hamiltonian circuit (cycle). Does this mean in terms of complexity if one takes a certain complexity it cannot be guaranteed that the respective other could be completed in the same complexity?

Nathan
  • 1
  • 1
  • Converting X to Y is not a symmetrical operation, consequently it doesn't make much sense to talk about "one" and "the other" without specifying which is which. – n. m. could be an AI Nov 07 '22 at 20:10
  • It could be considered in a few ways. One: If HAM has complexity X, does a similar seating problem have complexity X? Two: the opposite (vice versa). – Nathan Nov 07 '22 at 20:17
  • If X can be converted to Y, then Y has the same or bigger complexity than X. That's about the only thing you can say. – n. m. could be an AI Nov 07 '22 at 20:45
  • Not same and/or bigger? As in it is not guaranteed that it could be same..? – Nathan Nov 07 '22 at 20:48
  • I don't quite understand what "and/or" means in this situation. How could something be the same *and* bigger than something else? No, it is not guaranteed that they are the same, why would it be? – n. m. could be an AI Nov 07 '22 at 20:50
  • Usually you start with a problem that has a known complexity. That would be the Hamiltonian circuit problem. Then you consider an algorithm that has an unknown complexity. That would be the seating problem. If you can use the algorithm that solves the seating problem to solve the Hamiltonian circuit problem, then you can claim that the seating problem has the same or worse complexity than the Hamiltonian circuit problem. In short, you need to be able to convert the Hamiltonian circuit problem to the seating problem in order to say something about the complexity of the seating problem. – user3386109 Nov 07 '22 at 22:28

0 Answers0