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Proving a graph not simple and planar after adding edges

  1. Nov 3, 2015 #1
    1. The problem statement, all variables and given/known data
    YWeKXRJ.png

    Prove that no more than 3 edges can be added to G while keeping it planar and simple.
    (The graph in the image is graph G)

    2. Relevant equations
    For a connected, planar, simple graph G, with E edges and V vertices:
    E ≤ 3V - 6

    3. The attempt at a solution
    At first, we have 9 edges, and 6 vertices:

    9 ≤ 3(6) - 6 → 9 ≤ 12

    However, if I just add 4 vertices off to the side, say to the right, and draw four edges connecting them all to vertex E, we'd have 13 edges and 10 vertices.

    13 ≤ 3(10) - 6 → 13 ≤ 24 which is still true.

    Am I misunderstanding one of the definitions in this problem? I'm not quite getting how what I'm doing isn't keeping the graph planar and simple.

    Edit: Clarified equation
     
  2. jcsd
  3. Nov 3, 2015 #2

    haruspex

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    It does not say you can add any vertices.
     
  4. Nov 3, 2015 #3
    Ah, ok, not sure where my assumption that I could came from. So, does it suffice as mathematical proof to just show that:

    9 ≤ 12
    9 + 3 ≤ 12 - is true
    9 + 4 ≤ 12 - is not true

    Thanks!
     
  5. Nov 3, 2015 #4

    haruspex

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    No. The inequality is necessary but not sufficient. Consider K3,3. It has 9 edges, 6 vertices, but is not planar.
     
  6. Nov 7, 2015 #5

    epenguin

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    So, any ideas or plan?
    I think the 6 vertices are inviting you to try and show the result of adding more than 3 edges would contain K3,3 which is non-planar.
     
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