To construct a graph using path graph $P_4$

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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)



enter image description here



enter image description here



Graph $G_1$ with less number of vertices:



enter image description here






combinatorics discrete-mathematics graph-theory







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    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)



    enter image description here



    enter image description here



    Graph $G_1$ with less number of vertices:



    enter image description here






    combinatorics discrete-mathematics graph-theory







    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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)



      enter image description here



      enter image description here



      Graph $G_1$ with less number of vertices:



      enter image description here






      combinatorics discrete-mathematics graph-theory







      share|cite|improve this question













      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)



      enter image description here



      enter image description here



      Graph $G_1$ with less number of vertices:



      enter image description here






      combinatorics discrete-mathematics graph-theory




      combinatorics discrete-mathematics graph-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Sep 4 at 10:57









      monalisa

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