To see that the Petersen graph has no Hamiltonian cycle "C", consider the edges in the cut disconnecting the inner 5-cycle from the outer one. If there is a Hamiltonian cycle, an even number of these edges must be chosen. If only two of them are chosen, their end-vertices must be adjacent in the two 5-cycles, which is not possible. Hence 4 of them are chosen. Assume that the top edge of the cut is not chosen (all the other cases are the same by symmetry). Of the 5 edges in the outer cycle, the two top edges must be chosen, the two side edges must not be chosen, and hence the bottom edge must be chosen. The top two edges in the inner cycle must be chosen, but this completes a non-spanning cycle, which cannot be part of a Hamiltonian cycle. Alternatively, we can also describe the ten-vertex 3-regular graphs that do have a Hamiltonian cycle and show that none of them is the Petersen graph, by finding a cycle in each of them that is shorter than any cycle in the Petersen graph. Any ten-vertex Hamiltonian 3-regular graph consists of a ten-vertex cycle "C" plus five chords. If any chord connects two vertices at distance two or three along "C" from each other, the graph has a 3-cycle or 4-cycle, and therefore cannot be the Petersen graph. If two chords connect opposite vertices of "C" to vertices at distance four along "C", there is again a 4-cycle. The only remaining case is a Möbius ladder formed by connecting each pair of opposite vertices by a chord, which again has a 4-cycle. Since the Petersen graph has girth five, it cannot be formed in this way and has no Hamiltonian cycle.