You can't use the method of "removing Hamiltonian cycle". You can think of it this way: even if a 4-regular graph works, then a 2-regular graph can't. Give a counterexample of the simplest 2- regular graph: 9 points, of which 4 points form a circle and the other 5 points form a circle. This is a 2- regular graph, but there is no Hamilton cycle, because these two cycles are independent and completely unconnected. The proof method is as follows. The proof method is four well-designed Hamiltonian cycles, and the simplest method is to draw four Hamiltonian cycles. In the classical method, the four Hamiltonian cycles are as follows: the above figure is 1 Hamiltonian cycle. 9 points, left 8, right 1. The eight points on the left are connected by red lines, and then the ninth point is connected by green lines from head to tail. The other three Hamiltonian cycles are to turn the subgraph of eight points on the left by 45 degrees, 90 degrees, 135 degrees respectively, and then connect them with the points on the right. BTW: With this construction method, it can be proved that there are m "edge-insensitive" Hamiltonian cycles for any complete graph with 2m+ 1 points.