Construction of a graph from cycle $C_6$ and $C_7$ with specific properties (of specific eccentricities)

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This question is related to my previous problem asked:

Construction of a graph with specific properties (of specific eccentricities)

In the following given figures I tried to make a graph from $C_4$ and $C_5$ such that exactly two vertices having eccentricity five and rest of the vertices having eccentricity four. In my construction, I have added six vertices in each case. Bubbles filled with black color have eccentricity four, rest two vertices have eccentricity five.

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Similar construction is for $C_5$. I just need little help for the same if I consider graph $C_6$ and $C_7$ by adding six or lesser number of vertices. A little hint will help me a lot. Thanks for the help.

Note: Graphs where eccentricity of every vertex is $r$ except two vertices are known as Almost self-centered graphs. Rest two vertices have eccentricity $r+1$.

I tried for $C_6$ and got following. Not successful though.

enter image description here

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Start with a $C_8$ graph: vertices $u_0,u_1,u_2,u_3,u_4,u_5,u_6,u_7,$ edges $u_0u_1,u_1u_2,u_2u_3,u_3u_4,u_4u_5,u_5u_6,u_6u_7,u_7u_0.$ Add vertices $v_1,v_2,w$ and edges $u_0v_1,v_1v_2,v_2u_3$ and $u_4w.$ Now the subgraph induced by the vertices $u_0,u_1,u_2,u_3,v_2,v_1$ is a $C_6,$ the vertices $u_0$ and $w$ have eccentricity $5,$ and all other vertices have eccentricity $4.$

Oh, you wanted an induced $C_7$ too? Let me do it this way. Again, start with a $C_8$ graph: vertices $u_0,u_1,u_2,u_3,u_4,u_5,u_6,u_7,$ edges $u_0u_1,u_1u_2,u_2u_3,u_3u_4,u_4u_5,u_5u_6,u_6u_7,u_7u_0.$ Add a vertex $w$ and an edge $u_4w.$ The vertices $u_0$ and $w$ have eccentricity $5,$ all other vertices have eccentricity $4,$ but we don't have a $C_7$ yet.

Now delete the vertex $u_1$ (and of course the edges $u_0u_1$ and $u_1u_2$). Add $7$ new vertices $v_0,v_1,v_2,v_3,v_4,v_5,v_6$ and $21$ new edges: namely the edges $v_0v_1,v_1v_2,v_2v_3,v_3v_4,v_4v_5,v_5v_6,v_6v_0$ and the edges $u_0v_i$ and $v_iu_2$ for $i=0,1,2,3,4,5,6.$ Now everything is fine.

In the same way, given any graph $H,$ we can construct a graph $G$ having $H$ as an induced subgraph and having exactly two vertices of eccentricity $5,$ all other vertices having eccentricity $4.$