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I Graph Theory: (proof)conditions for Euler Circuit in Digraph

  1. May 29, 2016 #1
    I have read in many places that one necessary condition for the existence of a Euler circuit in a directed graph is as follows.

    Theorem: "A directed graph has an eulerian circuit if and only if it is connected and each vertex has the same in-degree as out-degree."

    However, I don't understand why the state of being connected is a necessary condition. I thought that a Euler circuit is a closed walk where all of the edges are distinct and uses every edge in the graph exactly once. Therefore, the disconnected graph shown below should satisfy the condition of being a Euler circuit. (All the edges in the graph are used exactly once).
    4zFle.png
    Can anyone explain what I did wrong here? Also it would be greatly appreciated if someone could give a mathematical proof to both parts of the theorem.

    Thanks!
     
  2. jcsd
  3. May 29, 2016 #2
    Your example graph contains a zero degreed vertex, one of whose properties your quoted theorem is still missing.
    Eulerian path and cycle or circuit.
     
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