Give an example of a graph G with the following properties or explain why no such example exists. Here d(G) denotes the minimum degree of a graph G. i, ii, K(G) = 2, ?(G) – 3 and d(G) = 4 K(G) = 3, ?(G) = 2 and d(G) = 4
The Correct Answer and Explanation is:
We are asked whether there exists a graph GG that satisfies the following properties. Let’s examine each case:
i. κ(G)=2\kappa(G) = 2, λ(G)=3\lambda(G) = 3, and δ(G)=4\delta(G) = 4
Where:
- κ(G)\kappa(G): vertex-connectivity – the minimum number of vertices whose removal disconnects the graph.
- λ(G)\lambda(G): edge-connectivity – the minimum number of edges whose removal disconnects the graph.
- δ(G)\delta(G): minimum degree – the smallest degree among all vertices in the graph.
✅ Answer (i): Yes, such a graph exists.
Explanation:
By Whitney’s inequality: κ(G)≤λ(G)≤δ(G)\kappa(G) \leq \lambda(G) \leq \delta(G)
So if δ(G)=4\delta(G) = 4, then the edge-connectivity λ(G)\lambda(G) can be less than 4, and similarly, κ(G)\kappa(G) can be less than λ(G)\lambda(G). Therefore, having κ(G)=2\kappa(G) = 2, λ(G)=3\lambda(G) = 3, and δ(G)=4\delta(G) = 4 does not violate Whitney’s inequality.
Example Construction:
- Take two cliques of 5 vertices (each is K5K_5) with all nodes having degree 4.
- Connect the cliques with 3 edges between distinct pairs of nodes (one in each clique).
- Remove any two vertices — the graph remains connected (so κ(G)=2\kappa(G) = 2).
- But removing the three connecting edges disconnects the two K5K_5 parts (so λ(G)=3\lambda(G) = 3).
- Each node in K5K_5 has degree 4, satisfying δ(G)=4\delta(G) = 4.
So such a graph does exist.
ii. κ(G)=3\kappa(G) = 3, λ(G)=2\lambda(G) = 2, and δ(G)=4\delta(G) = 4
❌ Answer (ii): No, such a graph does not exist.
Explanation:
Again, by Whitney’s inequality: κ(G)≤λ(G)≤δ(G)\kappa(G) \leq \lambda(G) \leq \delta(G)
So if λ(G)=2\lambda(G) = 2, then κ(G)≤2\kappa(G) \leq 2. This contradicts the claim κ(G)=3\kappa(G) = 3. In other words, the vertex-connectivity of a graph cannot be greater than its edge-connectivity.
This is a strict constraint from graph theory and rules out the possibility of such a graph.
✅ Summary:
- (i): A graph with κ(G)=2\kappa(G) = 2, λ(G)=3\lambda(G) = 3, and δ(G)=4\delta(G) = 4 can exist.
- (ii): A graph with κ(G)=3\kappa(G) = 3, λ(G)=2\lambda(G) = 2, and δ(G)=4\delta(G) = 4 cannot exist due to violation of κ(G)≤λ(G)\kappa(G) \leq \lambda(G).
We are asked whether there exists a graph GG that satisfies the following properties. Let’s examine each case:
i. κ(G)=2\kappa(G) = 2, λ(G)=3\lambda(G) = 3, and δ(G)=4\delta(G) = 4
Where:
- κ(G)\kappa(G): vertex-connectivity – the minimum number of vertices whose removal disconnects the graph.
- λ(G)\lambda(G): edge-connectivity – the minimum number of edges whose removal disconnects the graph.
- δ(G)\delta(G): minimum degree – the smallest degree among all vertices in the graph.
✅ Answer (i): Yes, such a graph exists.
Explanation:
By Whitney’s inequality: κ(G)≤λ(G)≤δ(G)\kappa(G) \leq \lambda(G) \leq \delta(G)
So if δ(G)=4\delta(G) = 4, then the edge-connectivity λ(G)\lambda(G) can be less than 4, and similarly, κ(G)\kappa(G) can be less than λ(G)\lambda(G). Therefore, having κ(G)=2\kappa(G) = 2, λ(G)=3\lambda(G) = 3, and δ(G)=4\delta(G) = 4 does not violate Whitney’s inequality.
Example Construction:
- Take two cliques of 5 vertices (each is K5K_5) with all nodes having degree 4.
- Connect the cliques with 3 edges between distinct pairs of nodes (one in each clique).
- Remove any two vertices — the graph remains connected (so κ(G)=2\kappa(G) = 2).
- But removing the three connecting edges disconnects the two K5K_5 parts (so λ(G)=3\lambda(G) = 3).
- Each node in K5K_5 has degree 4, satisfying δ(G)=4\delta(G) = 4.
So such a graph does exist.
ii. κ(G)=3\kappa(G) = 3, λ(G)=2\lambda(G) = 2, and δ(G)=4\delta(G) = 4
❌ Answer (ii): No, such a graph does not exist.
Explanation:
Again, by Whitney’s inequality: κ(G)≤λ(G)≤δ(G)\kappa(G) \leq \lambda(G) \leq \delta(G)
So if λ(G)=2\lambda(G) = 2, then κ(G)≤2\kappa(G) \leq 2. This contradicts the claim κ(G)=3\kappa(G) = 3. In other words, the vertex-connectivity of a graph cannot be greater than its edge-connectivity.
This is a strict constraint from graph theory and rules out the possibility of such a graph.
✅ Summary:
- (i): A graph with κ(G)=2\kappa(G) = 2, λ(G)=3\lambda(G) = 3, and δ(G)=4\delta(G) = 4 can exist.
- (ii): A graph with κ(G)=3\kappa(G) = 3, λ(G)=2\lambda(G) = 2, and δ(G)=4\delta(G) = 4 cannot exist due to violation of κ(G)≤λ(G)\kappa(G) \leq \lambda(G).