Crossings in an Eulerian Trail

Question

Exercise 11.2 in Graph Theory by Harary says

Every plane eulerian graph contains an eulerian trial that never crosses itself.

What does it mean for a trail to not cross itself? The book does not give a formal definition of this notion.

Answer 1

I don't know the "formal" definition, but informally it means just what you would think. If you regard the Eulerian trail as a curve in the plane, the curve does not cross itself, in the sense that the graphs of $y=0$ and $y=x$ cross at the origin, but the graphs of $y=0$ and $y=x^2$ touch without crossing.

For instance, consider the plane Eulerian graph with vertices $v=(0,0)$, $w=(1,0)$, $x=(1,1)$, $y=(-1,-1)$, $z=(-1,0)$, and straight-line edges $vw,wx,xv,vy,yz,zv$. The Eulerian trail $z,v,w,x,v,y,z$ crosses itself at $v$, but the Eulerian trail $z,v,x,w,v,y,z$ does not cross itself.

P.S. Following a suggestion by the OP, here's an attempt at defining a self-crossing for an Eulerian trail in a plane graph $G$. The trail crosses itself at a vertex $v$ if, among the edges of $G$ that are incident with $v$, there are four distinct edges $a,b,c,d$ which occur in the cyclic order $(a,b,c,d)$ in the clockwise ordering of the edges incident with $v$, and such that the edges $a$ and $c$ are traversed consecutively (though not necessarily in that order) in the trail, and the same goes for the edges $b$ and $d$.

Answer 2

A planar trail "crosses itself" if you cannot draw it without edges crossing. Connecting at a vertex does not count as crossing.