Theorem 6.4 1/4

1. Theorem 6.4 1/4 The Petersen graph is non-Hamiltonian. Proof：[proof by contradiction] Suppose that the Petersen graph, which we will denote by PG, is Hamiltonian. Then PG contains a Hamiltonian cycle C. This cycle contains ten edges. Two of the three edges incident with each vertex of PG necessarily belong to C. Certainly, C contains all five, some, or none of the edges uivi(1<= i <=5); so at least five edges of C belong either to C’ or to C”.

2. Theorem 6.4 2/4 Therefore, either C’ contains at least three edges of C, or C” contains at least three edges of C. Without loss of generality, assume that C’ contains at least three edges of C. First, observe that all five edges of C’ cannot belong to C since no cycle contains a smaller cycle as a subgraph. Suppose that C contains exactly four edges of C’, say the edges u4u5, u5u1, u1u2, u2u3. The cycle C must then contain the edges u4v4, u3v3 as well as v1v3, v1v4, implies that C contains an 8-cycle, which is impossible.

3. Theorem 6.4 3/4 One case remains then, namely that C contains exactly three edges of C’. There are two possibilities: (1) the three edges of C’ on C are consecutive on C’ and (2) these three edges are not consecutive on C’.

4. Theorem 6.4 4/4 It’s impossible in (1) as u1v1 is the only edge incident with u1 that could lie on C. It’s impossible in (2) since C would have to contain the smaller cycle u4, v4 , v1, v3, u3, u4. Therefore, as claimed, the Petersen graph is not Hamiltonian.