Problem: Let $G$ be an eulerian graph.

$(a)$ Explain in detail how to construct a closed trail starting at a given vertex $x$ of $G$. Prove that such a trail exists for every $x \in V (G)$ (the vertice set of $G$).

My attempt: a) First order the vertices of the graph, say $\{v_1,...,v_n\}$. Next I would proceed in order so that if there is an edge connecting $v_i, v_j$ where $j=i+1, 1\leq i \leq n$ then add it to the trail, otherwise continue for $j=i+2$, etc. until you have a path to the $v_n$, and then return backwards doing the same thing but for edges not added to the path where now $v_i, v_j$ form an edge with $j=i-1$ for $i-j<n-1$ until you return to $v_1$.

I am not sure how to show that such a trail exists however. Any hints appreciated.

Edit: My definitions from my notes:

An Euler trail in a graph is a trail that contains every edge of the graph. An Euler tour is a closed Euler trail. A graph is called eulerian is it has an Euler tour

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    $\begingroup$ Isn't "connected" redundant in "connected Eulerian graph"? By definition, an Eulerian graph is a graph which has a Euler circuit, which implies that it is connected. $\endgroup$ – bof Feb 24 '17 at 22:54
  • $\begingroup$ @bof sure, edited. $\endgroup$ – IntegrateThis Feb 24 '17 at 22:56
  • $\begingroup$ I don't get it. To prove the existence of a closed trail starting at a given vertex $x,$ why don't you just take an Euler circuit and follow the circuit starting from $x?$ $\endgroup$ – bof Feb 24 '17 at 23:00
  • $\begingroup$ Maybe I was wrong in thinking "Eulerian" implies "connected". Does the definition of Eulerian graph permit isolated vertices? $\endgroup$ – bof Feb 24 '17 at 23:01
  • $\begingroup$ Maybe "Eulerian" in this context means "Each vertex has even degree"? At any rate, they're asking for an explicit construction. $\endgroup$ – Arthur Feb 24 '17 at 23:03

Sketch of solution (the following contains a few claims that need to be proven explicitly):

If the graph is Eulerian, then note that if you just start at $x$ and walk an arbitrary path, you must eventually end up back at $x$ at some point, making a (not necessarily Eulerian) loop.

Next, if the path is Eulerian, then rejoice, for you are done. If not, then there must be some vertex $y$ along the chosen path that has unused edges. Start at $y$, and along the same lines as the paragraph above, make a loop that starts and ends at $y$, using only edges that wasn't used by the original $x$-loop. Now take the original loop, when you get to $y$, do the $y$-loop, then continue along the original loop after that. You have now created a longer loop.

You can keep repeating the above paragraph as long as the path you have so far isn't Eulerian.

  • $\begingroup$ Why is it true that you "must eventually" end up back at $x$ at some point. $\endgroup$ – IntegrateThis Feb 24 '17 at 23:05
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    $\begingroup$ That's why I called this a sketch. Here are three questions to get you started: At any point as you walk along the path, if you are not back where you started, how many (and which) vertices have an odd number of unused edges? Is $0$ an odd number? Does this mean you can take another step? $\endgroup$ – Arthur Feb 24 '17 at 23:09
  • $\begingroup$ So basically if the path is not closed then by taking a longest arbitrary path it would have to be closed otherwise the endpoint of the path would have another edge possible to traverse? $\endgroup$ – IntegrateThis Feb 24 '17 at 23:12
  • $\begingroup$ @TheMathNoob Exactly. At least for some meanings of the word "longest". The longest path you can get is an Eulerian loop, and an algorithm for making one in the first go is beyond me. However, if you just meant "longest" as in "we made arbitrary choices on where to go each step, and now we are stuck and cannot proceed", then yes, that is what I mean. Such a path must end where we started because otherwise we can always take another step. $\endgroup$ – Arthur Feb 24 '17 at 23:19

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