Assertion: graph's vertex | /r/HasAspect | degree
Facets:
Source sentences:
text: An Eulerian graph is connected and, in addition, all its vertices have even degree.
text: A functional graph is a directed graph where every vertex has out-degree one.
text: A strongly connected simple directed graph with n vertices is Hamiltonian if every vertex has a full degree greater than or equal to n.
text: : a hypergraph where each edge contains precisely d {\displaystyle d} : a hypergraph where every vertex has degree
text: H is k-regular if every vertex has degree
text: For an undirected graph, this means that the graph is connected and every vertex has even degree.
text: The number of vertices in C n equals the number of edges, and every vertex has degree 2; that is, every vertex has exactly two edges incident with it.
text: Combining the symmetry condition with the restriction that graphs be cubic (i.e. all vertices have degree 3) yields quite a strong condition, and such graphs are rare enough to be listed.
text: Equivalently, a path with at least two vertices is connected and has two terminal vertices (vertices that have degree 1), while all others (if any) have degree 2.
text: Let K be the set of vertices that have degree greater than two, and that are connected by paths of degree-two vertices to one or more leaves.
text: A simple graph with n vertices ( n ≥ 3 ) is Hamiltonian if every vertex has degree n / 2 or greater.
text: Since we’ll be dividing by degrees, assume no vertices have degree zero.
text: Each vertex has degree two, so this is also a regular graph.