Question

Let \(G(V, E)\) be a simple, undirected graph. A vertex cover of \(G\) is a subset \(V' \subseteq V\) such that for every \((u, v) \in E\), \(u \in V'\) or \(v \in V'\). Let the size of the smallest vertex cover in \(G\) be \(k\). Let \(S\) be any vertex cover of size \(k\).
For a vertex \(v \in V\), which of the following constraints will always ensure that \(v \in S\)?

• A. The degree of \(v\) is at least \(k+1\)
• B. The vertex \(v\) is on a path of length \(k+1\)
• C. The vertex \(v\) is on a cycle of length \(k+1\)
• D. The vertex \(v\) is a part of a clique of size \(k\)

Hint:

Use a proof by contradiction on the degree condition, and construct small explicit graphs (a path, a triangle, a complete graph) to test whether the other conditions can fail.

Answer 1

The Correct Option is A

Step 1: Understand what "always ensure \(v \in S\)" means.
We need a condition on \(v\) such that, for EVERY simple undirected graph \(G\) and EVERY minimum vertex cover \(S\) of \(G\) (of size \(k\)), if the condition holds for \(v\), then \(v\) must belong to \(S\). It is not enough for the condition to hold in one example; it must be logically impossible for \(v\) to be excluded from any minimum vertex cover while the condition holds. Each option is tested with a rigorous proof (if true) or an explicit counterexample (if false).

Step 2: Analyze Option (A) -- degree of \(v \ge k+1\).
Suppose, for contradiction, that \(v \notin S\) even though \(\deg(v) \ge k+1\). Since \(S\) is a vertex cover, every edge incident to \(v\) must have its OTHER endpoint in \(S\) (because \(v\) itself is not in \(S\)). Vertex \(v\) has at least \(k+1\) distinct neighbours, and each of these neighbours must individually be in \(S\) to cover the corresponding edge. So \(S\) must contain at least \(k+1\) distinct vertices just from \(v\)'s neighbourhood alone. But \(|S| = k\), so \(S\) cannot contain \(k+1\) or more vertices -- a contradiction. Hence \(v \notin S\) is impossible, so \(v\) MUST be in \(S\). This argument uses no specific graph structure, so it holds for every graph and every minimum vertex cover: Option (A) is TRUE.

Step 3: Analyze Option (B) -- \(v\) is on a path of length \(k+1\), via a counterexample.
Take \(k=2\) and let \(G\) be exactly the path \(v_1-v_2-v_3-v_4\) (3 edges, a path of length \(k+1=3\)). The minimum vertex cover size of this path is \(2=k\) (e.g. \(\{v_2,v_4\}\) covers all 3 edges: \(v_1v_2\) by \(v_2\), \(v_2v_3\) by \(v_2\), \(v_3v_4\) by \(v_4\)). Take \(v=v_1\), which lies on this length-3 path. But the minimum vertex cover \(S=\{v_2,v_4\}\) does NOT contain \(v_1\). So the condition "on a path of length \(k+1\)" holds for \(v_1\) while \(v_1 \notin S\) for this minimum vertex cover: Option (B) is FALSE.

Step 4: Analyze Option (C) -- \(v\) is on a cycle of length \(k+1\), via a counterexample.
Take \(k=2\) and let \(G\) be exactly a triangle \(v_1v_2v_3\) (a cycle of length \(k+1=3\)). The minimum vertex cover size of a triangle is \(2=k\) (any 2 of the 3 vertices cover all edges). Take \(v=v_1\), which lies on this length-3 cycle. The minimum vertex cover \(S=\{v_2,v_3\}\) is valid (it covers \(v_1v_2\) by \(v_2\), \(v_2v_3\) by either, \(v_1v_3\) by \(v_3\)) and does NOT contain \(v_1\). So the condition holds for \(v_1\) while \(v_1 \notin S\): Option (C) is FALSE.

Step 5: Analyze Option (D) -- \(v\) is part of a clique of size \(k\), via a counterexample.
Let \(G\) be the complete graph \(K_{k+1}\) on \(k+1\) vertices. Its minimum vertex cover size is \(k\) (covering \(K_{k+1}\) requires all vertices except exactly one, since leaving out two vertices would leave the edge between those two uncovered). Every subset of \(k\) vertices of \(K_{k+1}\) is itself a clique of size \(k\) (all subsets of a complete graph are complete), so every vertex \(v\) of \(G\) is "part of a clique of size \(k\)". However, the minimum vertex covers of \(K_{k+1}\) are exactly "all vertices except one", so there is a minimum vertex cover that excludes any particular chosen \(v\), namely \(S = V \setminus \{v\}\). So the condition holds for \(v\) while \(v \notin S\) for that choice of \(S\): Option (D) is FALSE.

Step 6: Conclusion.
Only the degree-based condition in Option (A) is a logically airtight guarantee across all graphs and all minimum vertex covers. \[ \boxed{\text{Option (A) only}} \]