AHL 3.16—Tree and cycle algorithms, Chinese postman, travelling salesman

Degrees are $4,3,2,3,2$. The odd-degree vertices are $B$ and $D$.

A connected graph with exactly two odd-degree vertices has an Eulerian trail but not a circuit.

The trail must start at one odd-degree vertex and end at the other, so it runs from $B$ to $D$ (or vice versa).

Since there are exactly two odd vertices, an Eulerian trail exists but no Eulerian circuit.

An Eulerian circuit requires all vertices to have even degree; here $B$ and $D$ are odd, so no circuit exists.

Exactly two odd vertices guarantees a trail exists, so claiming neither exists is incorrect.

$C$ has even degree, so a trail cannot both start and end there while covering every edge.

Only option B correctly identifies the trail and its endpoints.

A trail is a walk with no repeated edges, while a path is a walk with no repeated vertices.

Since repeating a vertex would force at least one edge to be repeated only in specific cases, but the fundamental restriction differs: trails restrict edges, paths restrict vertices.

Option B states trail = no repeated edges, path = no repeated vertices, matching the definitions exactly.

Therefore option B is correct.

This reverses the definitions - a trail actually restricts edges, not vertices.

Returning to the start describes a circuit or cycle, not the trail/path distinction.

Trails and paths are distinct concepts; every path is a trail but not every trail is a path.

Option B correctly captures the edge-vs-vertex distinction.