# Are Euler trails and tours of a graph the same as Hamilton paths and cycles of the corresponding “edge graph”?

Is knowing information about Euler paths and Euler tours about a graph $$G$$ the same as knowing information about Hamilton paths and cycles of the graph $$H$$ obtained from $$G$$ such that vertices of $$H$$ are edges of $$G$$ and two vertices of $$H$$ are joined by an edge if the corresponding edges are adjacent?

An Eulerian walk in $$G$$ gives us a Hamiltonian path in $$H$$ (the line graph of $$G$$) and an Eulerian tour in $$G$$ gives us a Hamiltonian cycle in $$H$$. This is because consecutive edges $$uv, vw$$ in the Eulerian walk in $$G$$ correspond to adjacent vertices in $$H$$.
However, other Hamiltonian paths/cycles in $$H$$ may not correspond to Eulerian walks/tours in $$G$$. For example, if $$G = K_4$$ (with vertices $$1,2,3,4$$ and edges $$12, 13, 24, 23, 24, 34$$) then in $$H$$, we have a Hamiltonian cycle $$12, 13, 14, 34, 24, 23$$ which does not correspond to an Eulerian tour of $$G$$ (and in fact such an Eulerian tour does not exist).
The difference is that a path in $$H$$ is a sequence of edges of $$G$$ in which any two consecutive edges share a vertex. However, a walk in $$G$$ is more structured: the second endpoint of one edge must be the first endpoint of the next. Going from $$12$$ to $$13$$ to $$14$$ in $$H$$ (a valid part of a Hamiltonian cycle) cannot be made to follow this rule in $$G$$: if $$12$$ and $$13$$ share the vertex $$1$$, the next edge should include vertex $$3$$, not vertex $$1$$ again.