Let G(V,E) be a simple, undirected graph. A vertex cover of G is a subset V′ ⊆ V such that for every (u,v) ∈ E, u ∈ V′ or v ∈ V′. Let the size of the smallest vertex cover in G be k. Let S be any vertex cover of size k. For a vertex v ∈ V, which of the following constraints will always ensure that v ∈ S?
Correct Answer :
The degree of v is at least k +1
Solution :
The correct option is: The degree of v is at least k + 1.
Let us understand why this constraint always ensures that the vertex must be in any vertex cover of the minimum size .
By definition, a vertex cover of a graph is a subset of vertices such that every edge in the graph has at least one of its endpoints in .
Here, we are given that the size of the smallest vertex cover is , and is any vertex cover of size , so .
Let the degree of a vertex be denoted by . The degree of is the number of edges incident to .
Suppose that . This means there are at least distinct edges incident to , say where and are distinct neighbors of .
We want to prove that must belong to . Let us prove this by contradiction.
Assume for contradiction that .
Since is a valid vertex cover, it must cover all the edges incident to . That is, for every edge for , at least one endpoint of the edge must be in .
Because we assumed , the other endpoint must be in for every to cover each of these incident edges.
Therefore, we must have:
Since all are distinct, the size of the vertex cover must satisfy:
Substituting , we get:
But this contradicts our initial fact that the size of is ().
Since the assumption leads to a contradiction, the vertex must be in .
Thus, if the degree of is at least , is guaranteed to be in any vertex cover of size .
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