Let $G$ be a $2$-connected $3$-regular graph. Can $V(G)$ be partitioned into $V\_1$ and $V\_2$ where 
$G[V\_1]$(the induced subgraph on $V\_1$) is a cycle of $G$ and $G[V\_2]$ is a forest (Acyclic subgraph) of $G$?


**Edit:**


Since the counterexamples presented so far are $2$-connected, what happen if the connectivity of the graph is $3$ instead of two?



Can the Vertices of cubic graph be partitioned into and induced cycle and a forest?
I believe this is false.


**EDIT**


The previous counterexample was wrong, let me try again.


A program found counterexample on $10$ vertices and exhaustive search
confirmed it.


The edges are:



```
[(0, 3), (0, 5), (0, 7), (1, 4), (1, 6), (1, 9), (2, 6), (2, 7), (2, 8), (3, 5), (3, 7), (4, 8), (4, 9), (5, 8), (6, 9)]

```

It is $2$-connected.




---


According to a search the smallest $3$-connected counterexample
is on $12$ vertices with edges



```
[(0, 6), (0, 7), (0, 8), (1, 6), (1, 7), (1, 9), (2, 6), (2, 10), (2, 11), (3, 7), (3, 10), (3, 11), (4, 8), (4, 9), (4, 10), (5, 8), (5, 9), (5, 11)]

```

There are other counterexamples (modulo errors).


