Consider the following generalization of the maximum matching problem, which we call Strict-Matching. Recall that a matching in an undirected graph G = (V, E) is a set of edges such that no distinct pair of edges {a, b} and {c, d} have endpoints that are equal: {a, b} ∩ {c, d} = ∅. Say that a strict matching is matching with the property that no pair of distinct edges have endpoints that are connected by an edge: {a, c} ̸∈ E, {a, d} ̸∈ E, {b, c} ̸∈ E, and {b, d} ̸∈ E. (Since a strict matching is also a matching, we also require {a, b} ∩ {c, d} = ∅.) The problem Strict-Matching is then given a graph G and an integer k, does G contain a strict matching with at least k edges. Prove that Strict-Matching is NP-complete.
Consider the following generalization of the maximum matching problem, which we call
Strict-Matching. Recall that a matching in an undirected graph G = (V, E) is a set
of edges such that no distinct pair of edges {a, b} and {c, d} have endpoints that are
equal: {a, b} ∩ {c, d} = ∅. Say that a strict matching is matching with the property
that no pair of distinct edges have endpoints that are connected by an edge: {a, c} ̸∈ E,
{a, d} ̸∈ E, {b, c} ̸∈ E, and {b, d} ̸∈ E. (Since a strict matching is also a matching, we
also require {a, b} ∩ {c, d} = ∅.) The problem Strict-Matching is then given a graph
G and an integer k, does G contain a strict matching with at least k edges.
Prove that Strict-Matching is NP-complete.
Strict-Matching is NP-complete because it is at least as hard as the maximum matching problem, which is known to be NP-complete, and it can be reduced to the maximum matching problem in polynomial time.
To show that Strict-Matching is NP-complete, we need to prove two things:
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Strict-Matching is in NP: This means that given a proposed solution, we can check its validity in polynomial time. In this case, a proposed solution would be a set of edges that form a strict matching. To verify that a solution is valid, we can simply check that every vertex has degree 1, that there are no self-loops, and that no two edges share endpoints.
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Strict-Matching is NP-hard: This means that we can reduce an NP-hard problem to Strict-Matching in polynomial time. In this case, we will reduce the maximum matching problem to Strict-Matching.
Reducing Maximum Matching to Strict-Matching:
Given an instance of the maximum matching problem, represented by an undirected graph G = (V, E), we can construct a new graph G' = (V', E') in polynomial time as follows:
- Create a new vertex for each vertex in G and call this set of vertices V'.
- For each vertex v in V, connect the two vertices in V' that correspond to v with a new edge (v1, v2).
- Add all the edges in E to E'.
Now, the maximum matching in G' will be a strict matching in G if and only if the number of edges in the maximum matching is equal to the number of vertices in G.
Since the maximum matching problem is NP-complete, it follows that Strict-Matching is also NP-complete.
In conclusion, Strict-Matching is NP-complete because it can be reduced to the maximum matching problem, which is known to be NP-complete, and because it is in NP. This means that Strict-Matching is as hard as the maximum matching problem and that finding an exact solution for Strict-Matching is at least as hard as finding an exact solution for the maximum matching problem.
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