Is set cover problem NP?
The set cover problem is a classical question in combinatorics, computer science, operations research, and complexity theory. It is one of Karp’s 21 NP-complete problems shown to be NP-complete in 1972. The decision version of set covering is NP-complete, and the optimization/search version of set cover is NP-hard.
Why set cover problem is NP-complete?
Theorem: Set Cover is NP-Complete. Proof: First, we argue that Set Cover is in NP, since given a collection of sets C, a certifier can efficiently check that C indeed contains at most k elements, and that the union of all sets listed in C does include all elements from the ground set U.
What are set covering constraints?
Variants are discussed in Section 4.2. Constraints (2) ensure trip coverage by at least one of the duties of the final schedule. The covering constraints “ ≥ ” allow over-covering of trips, i.e. a crew works on the train while others are allowed to travel on the same trip without fulfilling working tasks.
Is vertex cover a NP?
The vertex cover problem is an NP-complete problem: it was one of Karp’s 21 NP-complete problems.
What is meant by NP-hard?
A problem is NP-hard if an algorithm for solving it can be translated into one for solving any NP- problem (nondeterministic polynomial time) problem. NP-hard therefore means “at least as hard as any NP-problem,” although it might, in fact, be harder.
How do you prove a problem is NP?
A problem is “in NP” if, given a potential solution, you can verify that it is correct or incorrect in polynomial time. For instance if the problem is sorting lists, if you can verify that one list is the sorted version of another list in polynomial time, then sorting is in NP.
Can we reduce the problem of vertex cover to the problem of clique?
Since VERTEX-COVER can be reduced to CLIQUE in polynomial time, CLIQUE ∈ NP and VERTEX-COVER is NP-Complete, CLIQUE is also NP-Complete.
Can set cover be reduced to vertex cover?
i.e. The Vertex Cover problem was reduced to the Set Cover problem.) Question: Does this reduced instance of the Set Cover problem have a set cover of size at most k? of the graph. Hence, the answer to the Vertex Cover instance is also “yes”.
What is a set covering model?
The set covering problem is a specific type of a discrete location model. In this model, a facility can serve all demand nodes that are within a given coverage distance Dc from the facility. The problem is the place the minimum number of facilities so as to ensure that all demand nodes can be served.
What is greedy set cover algorithm?
When the greedy algorithm chooses a set s, imagine that it charges the price per element for that iteration to each element newly covered by s. Then the total weight of the sets chosen by the algorithm equals the total amount charged, and each element is charged once.
Is the set cover problem NP-complete?
Thus, it can be verified that the set cover problem is NP-Complete using the following propositions:
How to prove set cover is NP-hard?
Set Cover is NP-Hard: In order to prove set cover is NP Hard, we will perform a reduction from a known NP-Hard problem, that is, vertex cover to set cover problem.For the vertex cover problem, we have as input a graph G = (V, E) and an integer k. Now, let the ground set Subset S u for every vertex u in V, contains the edges incident to u.
What is set cover problem in Computer Science?
The set cover problem is a classical question in combinatorics, computer science and complexity theory. It is one of Karp’s 21 NP-complete problems shown to be NP-complete in 1972. It is a problem “whose study has led to the development of fundamental techniques for the entire field” of approximation algorithms.
How do you prove a problem is NP-complete?
An instance of the Set Cover problem is a Ground set X, an integer k and a collection of subsets Si formed out of X. Since an NP-complete problem, by definition, is a problem which is both NP and NP-Hard, the proof or statement that a problem is NP-Complete consists of two parts: Attention reader!