On the size of P4-transversals of graphs A subset $T$ of vertices of a graph $G$ is called a $P_4$-transversal if $T$ intersects every $P_4$ of $G$. In the context of this question, we consider $P_4$ as an induced path on 4 vertices. 

Conjecture :
Given a graph $G$ of $n$ vertices, the minimal $P_4$-transversal of G is of at most $\frac{n}{2}$ vertices. 

We consider gneral simple graphs here. However, any results of subclasses of simple graphs are welcome.

Thoughts:
$P_4$-transversal can be seen as a special case of hitting set of sets of size 4. However, for general hitting set of 4-sets (i.e., 4-uniform hypergraph) on $n$ vertices, the minimum hitting set can be as large as $n-3$.

The original version of this question guessed the upperbound of the size of $P_4$-transversal as $\frac{n}{3}$. Thanks to the comments from R B and Jim Nastos, we have an example on 16 vertices and whose minimal transverval has 7 vertices.



Introduction and why I am interested in $P_4$-transversal of graphs.

Study of $P_4$-transversal is a part of study of $P_4$-structures of graphs, which plays an important role in the "Strong Perfect Graph Theorem". $P_4$-transversal of perfect graphs and chordal graphs were studied in perfect graph and chordal graph.

$P_4$ is a path of 4 vertices. A $P4$-free graphs is called a cograph, which is the minimal family of graphs generated from $K_1$ and complementation and disjoint union. Therefore, $P_4$ structures seems to be an interesting and important property of graphs.
Let H be any graph on N vertices and requires T vertices in a P4-transversal (so its ratio is T/N).
Construct G like in the previous examples: G is a P4 but substitute an H in for each vertex. Then the total number of vertices of G is n = 4N and the transversal size is N + 3T.

This gives a ratio of (N+3T)/4N = (1/4) + (3/4)T/N

So, the last example I gave in the comment of Peng Zhang's answer used H = $P_4$ with T/N = 1/4 and the ratio was 1/4 + 3/16 = 7/16.

Now use that 7/16 graph in for H and construct G. This uses 64 vertices, and the transversal size is 16 + 3(7) = 37. We get a ratio of (1/4) + (3/4)(7/16) = 16/64 + 21/64 = 37/64 which is already over 50%.

This seems correct to me (in that I don't see what is wrong with it) but it is troublesome, because repeating this process will get to a fixed point x = (1/4) + (3/4)x , which is x=1.0 (?)

Thoughts...?