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How to construct a graph with girth = twice the diameter + 1

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  1. Apr 11, 2017 #1
    1. The problem statement, all variables and given/known data
    I am trying to verify this theorem for myself.
    Theorem: Every graph G containing a cycle satisfies g(G)≤2diam(G)+1

    2. Relevant equations
    N/A

    3. The attempt at a solution
    i have drawn graphs and i failed to verify the theorem. is it even possible, since increasing the girth directly increases the diameter of the graph...?

    However the theorem has a proof.

    Proof: Let C be a shortest cycle in G. If g(G)>= 2diam(G) + 2, then C has two vertices whose distance in C is at least diam(G)+1. In G, these vertices have a lesser distance; any shortest path P between them is therefore not a subgraph of C. Thus, P contains a C-path xPy. Together with the shorter of the two x-y paths in C, this path xPy forms a shorter cycle than C, a contradiction.
     
  2. jcsd
  3. Apr 12, 2017 #2
    This seems kind of obvious.

    Let the maximum eccentricity (diameter) be ## D## and the length of the shortest cycle be ## L##. Assume, for a contradiction, a finite graph ## G## is picked such that ##L>2D+1 ##. Then the graph ## G## contains two vertices with a distance of at least ##2D+1 ##, an immediate contradiction. Do you understand what the contradiction is?
     
  4. Apr 12, 2017 #3
    sorry for that, my book did not define diameter correctly which lead to my confusion.
     
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