Is There any Untraceable Generalized Petersen Graph?

Question

The Petersen graph is one of the example of graph which is not Hamiltonian. Can we find an example among the generalized Petersen graph which doesn't have Hamiltonian path (untraceable)?

Answer 1

The graph $G(n,k)$ is defined to have:

• Vertex set $\{u_0, u_1, \dots, u_{n-1}\} \cup \{v_0, v_1, \dots, v_{n-1}\}$.
• Edge set $\{u_i u_{i+1} : 0 \le i \le n-1\} \cup \{u_i v_i : 0 \le i \le n-1\} \cup \{v_i v_{i+k} : 0 \le i \le n-1\}$, where the addition in "$i+1$" and "$i+k$" is done modulo $n$.

Let's define a few paths in this graph first. Let path $P(i)$ be the path starting at $u_i$ which goes from there to $v_i$, then $v_i$ to $v_{i-k}$ to $v_{i-2k}$ to $v_{i-3k}$ and so on until getting to $v_{i+k}$, then from there to $u_{i+k}$. (All addition and subtraction is modulo $n$.) Let path $Q(i)$ be the reverse of path $P(i+k)$: starting at $u_i$, it goes to $v_i$, then to $v_{i+k}, v_{i+2k}, v_{i+3k}, \dots$ until getting to $v_{i-k}$, and then from there to $u_{i-k}$.

If $n$ and $k$ have no common factors, then paths $P(i)$ and $Q(i)$ all end up going through all $n$ of the $v$-vertices $v_0, \dots, v_{n-1}$. In this case, we can find a Hamiltonian path in $G(n,k)$: start at $u_1$, go through the $u$-vertices to $u_2, u_3, \dots, u_k$, follow $Q(i)$ back to $u_0$, then go through the $u$-vertices $u_{n-1}, u_{n-2}, \dots, u_{k+1}$.

More generally, if $\gcd(n,k)=d$, then each path $P(i)$ and $Q(i)$ goes through only $n/d$ of the $v$-vertices: the vertices $v_j$ with $i \equiv j \pmod d$. So we can visit all of the $v$-vertices by a more complicated route:

1. Start at $u_0$ and follow $P(0)$ to $u_k$.
2. Go to $u_{k+1}$ and follow $Q(k+1)$ to $u_1$.
3. Go to $u_2$ and follow $P(2)$ to $u_{k+2}$.
4. Go to $u_{k+3}$ and follow $Q(k+3)$ to $u_3$.
5. And so on, until we've followed $d$ of the $P$- or $Q$-paths and seen all the $v$-vertices.

If $d$ is even, then this path ends by following $Q(k+d-1)$ to $u_{d-1}$; we've visited vertices $v_0, \dots, v_{n-1}$, as well as $u_0, \dots, u_{d-1}$ and $u_k, \dots, u_{k+d-1}$. Now, we can turn the path into a Hamiltonian path by continuing from $u_{d-1}$ to $u_d, u_{d+1}, u_{d+2}, \dots, u_{k-1}$, and also by adding the segment $u_{k+d}, u_{k+d+1}, \dots, u_n, u_0$ to the start.

If $d$ is odd, then instead the path ends by folowing $P(d-1)$ to $u_{k+d-1}$. By continuing from here to $u_{k+d}, u_{k+d+1}, \dots, u_n, u_0$, we get a cycle! Call this cycle $C$. It's not necessarily a Hamiltonian cycle; if $k>d$, then vertices $u_d, u_{d+1}, \dots, u_{k-1}$ are missing from $C$. But now, we can get a Hamiltonian path by starting with $u_d, u_{d+1}, \dots, u_{k-1}, u_k$, then following $C$ all the way around starting at $u_k$. (It's a cycle, so we can follow it all the way around from anywhere we like.

In all cases, we've found at least a Hamiltonian path in $G(n,k)$, so it is always traceable.