Consider the graph G given below to answer the following questions. Find WV. Find d g(el). Find N({d,eh}). Are ad neighbors? How do you know? Is G a multigraph? How do you know? Find path from n to m. Is G bipartite?
The Correct Answer and Explanation is:
Thanks for uploading the image of graph GG. Let’s go through each question based on the visual information provided.
Answers:
(a) Find EE:
The edge set EE includes the following undirected edges: E={(a,b),(a,e),(a,f),(a,g),(b,c),(b,e),(b,g),(c,d),(c,h),(d,e),(d,h),(e,f),(f,g)}E = \{(a,b), (a,e), (a,f), (a,g), (b,c), (b,e), (b,g), (c,d), (c,h), (d,e), (d,h), (e,f), (f,g)\}
Total: 13 edges
(b) Find ∣V∣|V|:
The vertices are: V={a,b,c,d,e,f,g,h}V = \{a, b, c, d, e, f, g, h\}
Total: 8 vertices
(c) Find deg(e)\deg(e):
Edges connected to ee: (a,e), (b,e), (d,e), (e,f)
So, deg(e)=4\deg(e) = 4
(d) Find N({d,e,h})N(\{d, e, h\}):
Neighbors of dd: {c, e, h}
Neighbors of ee: {a, b, d, f}
Neighbors of hh: {c, d}
Union (excluding d, e, h): N({d,e,h})={a,b,c,f}N(\{d, e, h\}) = \{a, b, c, f\}
(e) Are ee and hh neighbors?
No. There is no edge directly connecting ee and hh.
(f) Is GG a multigraph?
No. A multigraph has multiple edges between the same two vertices. Here, each edge is unique → G is not a multigraph.
(g) Path from aa to ee of length 5:
One such path: a→g→f→e→d→ea \to g \to f \to e \to d \to e
This is of length 5, but revisits ee. If simple path required (no repetition), then: a→b→c→h→d→ea \to b \to c \to h \to d \to e
✓ Valid, length = 5
(h) Is G bipartite?
To test bipartiteness, try to 2-color the graph such that no adjacent vertices share a color.
We cannot 2-color because there exists an odd cycle: a→e→f→aa \to e \to f \to a
Cycle length = 3 ⇒ G is not bipartite
Explanation:
Graph theory provides a structured approach to analyze relationships through vertices and edges. In this case, the graph GG contains 8 vertices and 13 undirected edges. Each vertex is connected according to the structure shown in the diagram. For instance, the vertex ee has four edges leading to a,b,d,a, b, d, and ff, giving it a degree of 4.
When considering the neighborhood of a set of vertices like {d,e,h}\{d, e, h\}, we find all vertices directly connected to any in that set, excluding the set itself. Here, the combined neighborhood is {a,b,c,f}\{a, b, c, f\}. Regarding adjacency, ee and hh are not directly connected by any edge, so they are not neighbors.
We also determined that GG is not a multigraph, as no vertex pair is connected by more than one edge. A multigraph would allow multiple parallel edges between the same vertices, which is not observed here.
In terms of pathfinding, a path of length 5 from aa to ee is possible. One such simple path is a→b→c→h→d→ea \to b \to c \to h \to d \to e, involving 5 distinct edges and no repeated vertices.
Finally, for bipartiteness, we assess whether the graph can be split into two sets where no two vertices in the same set are adjacent. A key property of bipartite graphs is the absence of odd cycles. However, GG contains the cycle a→e→f→aa \to e \to f \to a of length 3, which is odd. Thus, G is not bipartite.
