However, two Hamiltonian cycles are considered to be equivalent if they connect the same vertices in the same cyclic order regardless of the starting vertex, while in the ménage problem the starting position is considered significant: if, as in Alice's tea party, all the guests shift their positions by one seat, it is considered a different seating arrangement even though it is described by the same cycle.
Year 2000 problem | Waring's problem | The Final Problem | The Problem with Popplers | Hume and the Problem of Causation | Dirichlet problem | Boolean satisfiability problem | The Dog Problem | Tammes problem | Species problem | problem solving | Problem gambling | Packing problem | packing problem | chess problem | Znám's problem | Year 10,000 problem | Yamabe problem | Weber problem | Undecidable problem | Travelling salesman problem | travelling salesman problem | The Problem of Thor Bridge | The Problem of the Media: U.S. Communication Politics in the 21st Century | The Problem of Social Cost | Tarski's circle squaring problem | Steiner tree problem | Shortest path problem | RSA problem | Quadratic eigenvalue problem |