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:
:
- κ(G)\kappa(G): vertex connectivity (minimum number of vertices whose removal disconnects GG),
- λ(G)\lambda(G): edge connectivity (minimum number of edges whose removal disconnects GG),
- δ(G)\delta(G): minimum degree of GG.
You have two sets of conditions:
1. κ(G)=2,λ(G)=3,δ(G)=4\kappa(G) = 2, \lambda(G) = 3, \delta(G) = 4
Is there a graph GG with:
- Vertex connectivity κ(G)=2\kappa(G) = 2,
- Edge connectivity λ(G)=3\lambda(G) = 3,
- Minimum degree δ(G)=4\delta(G) = 4?
2. κ(G)=3,λ(G)=2,δ(G)=4\kappa(G) = 3, \lambda(G) = 2, \delta(G) = 4
Is there a graph GG with:
- Vertex connectivity κ(G)=3\kappa(G) = 3,
- Edge connectivity λ(G)=2\lambda(G) = 2,
- Minimum degree δ(G)=4\delta(G) = 4?
Background and Key Facts:
- Inequality relations:
For any graph GG, these inequalities always hold: κ(G)≤λ(G)≤δ(G).\kappa(G) \leq \lambda(G) \leq \delta(G).
- Vertex connectivity cannot exceed edge connectivity,
- Edge connectivity cannot exceed minimum degree.
- Interpretation:
- κ(G)≤λ(G)\kappa(G) \leq \lambda(G) means the number of vertices needed to disconnect the graph is never more than the number of edges needed,
- λ(G)≤δ(G)\lambda(G) \leq \delta(G) means that removing fewer edges than the minimum degree is impossible to disconnect the graph.
Analyze each case:
Case 1: κ(G)=2,λ(G)=3,δ(G)=4\kappa(G) = 2, \lambda(G) = 3, \delta(G) = 4
- Check inequalities: κ(G)=2≤λ(G)=3≤δ(G)=4,\kappa(G) = 2 \leq \lambda(G) = 3 \leq \delta(G) = 4, which is valid and respects the inequalities.
- Conclusion: Such a graph can exist.
Example:
- Consider a graph constructed by starting with a 4-regular graph (minimum degree 4) that is not 3-vertex-connected but is 3-edge-connected.
- For instance, take a graph composed of two 4-regular subgraphs connected by exactly two vertices (a 2-vertex cut), but with at least 3 edge-disjoint paths between the two parts (edge connectivity = 3).
- This graph would have κ=2\kappa = 2, λ=3\lambda = 3, and δ=4\delta = 4.
Case 2: κ(G)=3,λ(G)=2,δ(G)=4\kappa(G) = 3, \lambda(G) = 2, \delta(G) = 4
- Check inequalities: κ(G)=3≤λ(G)=2≤δ(G)=4,\kappa(G) = 3 \leq \lambda(G) = 2 \leq \delta(G) = 4, which fails since 3≤23 \leq 2 is false.
- Conclusion: No such graph GG can exist because vertex connectivity cannot exceed edge connectivity.
Summary
| Case | Can such GG exist? | Reason |
|---|---|---|
| κ=2,λ=3,δ=4\kappa=2, \lambda=3, \delta=4 | Yes | 2≤3≤42 \leq 3 \leq 4 satisfies inequalities |
| κ=3,λ=2,δ=4\kappa=3, \lambda=2, \delta=4 | No | 3≤23 \leq 2 is false, contradiction |
In graph theory, the vertex connectivity κ(G)\kappa(G), edge connectivity λ(G)\lambda(G), and minimum degree δ(G)\delta(G) satisfy the fundamental inequality κ(G)≤λ(G)≤δ(G)\kappa(G) \leq \lambda(G) \leq \delta(G). This reflects the intuition that disconnecting a graph by removing vertices is at least as hard as disconnecting it by removing edges, and both are limited by the minimum degree since no vertex can have fewer edges than the minimum degree.
For the first set of values (κ=2,λ=3,δ=4\kappa=2, \lambda=3, \delta=4), the inequalities hold as 2≤3≤42 \leq 3 \leq 4. Therefore, such a graph can exist. To construct one, imagine a graph where two vertices form a vertex cut (removing these two vertices disconnects the graph), but at least three edges must be removed to disconnect it, indicating edge connectivity is 3. The minimum degree of 4 implies every vertex is connected to at least 4 others, so the graph is relatively dense. This situation is plausible in graphs that are 4-regular but have small vertex cuts that do not translate into equally small edge cuts.
For the second set of values (κ=3,λ=2,δ=4\kappa=3, \lambda=2, \delta=4), the inequalities are violated because vertex connectivity cannot be greater than edge connectivity (3≤23 \leq 2 is false). Thus, no graph with these parameters can exist.
In conclusion, graph invariants must satisfy certain structural inequalities. Violations like κ(G)>λ(G)\kappa(G) > \lambda(G) cannot happen, ensuring consistency in connectivity measures. Hence, while the first example is possible, the second is impossible.
