Can an Eulerian graph be of edge connectivity 1?
And can an Eulerian graph have vertex connectivity 1?
Please show step how you done it.
2026-03-25 20:40:03.1774471203
Eulerian graph confusing
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You'd have to go back to your source's definition, but I wouldn't think of a graph with only an Euler trail as being Eulerian. Any graph with an Euler circuit be such that every edge would be part of a cycle and therefore has edge connectivity greater than 1.
But an Eulerian graph can have a cut vertex. For instance, a "bow tie graph" of two triangles that share a vertex would meet that criterion.