I was trying to construct a graph $G_1$ and $G_2$, wherein both graphs $P_4$ is an induced graph. Graph $G_1$ is such that it contains exactly one vertex of eccentricity two and rest of the vertices with eccentricity three. Similarly, the graph $G_2$ is such that it contains exactly one vertex of eccentricity three and the rest of the vertices with eccentricity four. In both the cases, $P_4$ is induced in $G_1$ and $G_2$.
I tried in the following manner.
For $G_1$, I added $6$ vertices to $P_4$ and got the result, and for $G_2$, I added $10$ vertices to $P_4$ and got the result. However, later I got that $G_1$ can be obtained with the fewer number of vertices. Can $G_2$ be also obtained by adding less than $10$ vertices, if possible? Kindly help me to get the graph. Any hint or suggestion is helpful.
My attempt : (numbers are the eccentricity of the vertices)
Graph $G_1$ with less number of vertices:
For $G_2$, I also got an example. consider $C_6$ with vertices $1,2,3,4,5,6$. Add $5$ vertices $1′,2′,3′,4′,5′$ and make edges $1′2′,2′3′,3′4′,4′5′$, and $xx′$, where $x=1,2,3,4,5$. This also gives a graph with exactly one vertex with eccentricity $3$ and rest with $4$. The total number of vertices is $11$. Can we think of a smaller order? I am wondering that.