How do I prove NP hard reduction?
To prove that problem A is NP-hard, reduce a known NP-hard problem to A. In other words, to prove that your problem is hard, you need to describe an ecient algorithm to solve a dierent problem, which you already know is hard, using an hypothetical ecient algorithm for your problem as a black-box subroutine.
Is min vertex cover NP-complete?
Computational problem. The minimum vertex cover problem is the optimization problem of finding a smallest vertex cover in a given graph. The vertex cover problem is an NP-complete problem: it was one of Karp’s 21 NP-complete problems.
How do you prove set cover is NP-complete?
To prove a problem X is NP-complete, you need to show that it is both in NP and that it is at least as hard any other problem in NP. This last step is typically done by showing that Y ≤P X for some problem Y that you already know to be NP-Complete.
How do you find the optimal vertex cover?
Now, we want to solve the optimal version of the vertex cover problem, i.e., we want to find a minimum size vertex cover of a given graph….Vertex Cover
- Approx-Vertex-Cover (G = (V, E))
- {
- C = empty-set;
- E’= E;
- While E’ is not empty do.
- {
- Let (u, v) be any edge in E’: (*)
- Add u and v to C;
How do you know if you have a NP-hard problem?
NP-Hard problems(say X) can be solved if and only if there is a NP-Complete problem(say Y) that can be reducible into X in polynomial time. NP-Complete problems can be solved by a non-deterministic Algorithm/Turing Machine in polynomial time. To solve this problem, it do not have to be in NP .
Is vertex cover NP hard or NP complete?
Thus, vertex cover is NP Hard. Since vertex cover is in both NP and NP Hard classes, it is NP Complete.
Is set cover NP-hard?
The decision version of set covering is NP-complete, and the optimization/search version of set cover is NP-hard. If each set is assigned a cost, it becomes a weighted set cover problem.
Is NP-hard the same as NP-complete?
A problem is said to be NP-hard if everything in NP can be transformed in polynomial time into it even though it may not be in NP. Conversely, a problem is NP-complete if it is both in NP and NP-hard. The NP-complete problems represent the hardest problems in NP.
What is vertex cover problem prove that vertex cover problem is NP complete?
Proof that vertex cover is in NP – If any problem is in NP, then, given a ‘certificate’ (a solution) to the problem and an instance of the problem (a graph G and a positive integer k, in this case), we will be able to verify (check whether the solution given is correct or not) the certificate in polynomial time.
Can P reduce to NP?
(For example, solving Sudoku is NP, because you can show me the solution, and I can verify it is correct in polynomial time). All P problems are NP. So, “polynomial time reduction of a P problem to an NP problem” is pretty trivial, and doesn’t prove anything very interesting. We already know that P is a subset of NP.
Is vertex cover an NP-hard problem?
Vertex Cover problem is defined as follows: Given a Graph G and integer k. Does G have a subset S of vertices, such that: |S| = k, and every edge in G has at least one of the end points in S. Prove that Vertex Cover is an NP-hard problem by showing a reduction from one of the known NP-complete problems.
What is the certificate for the vertex cover problem?
The certificate for the vertex cover problem is a subset V’ of V, which contains the vertices in the vertex cover. We can check whether the set V’ is a vertex cover of size k using the following strategy (for a graph G (V, E)):
Is the vertex cover problem in polynomial time?
We can check whether the set V’ is a vertex cover of size k using the following strategy (for a graph G (V, E)): It is plain to see that this can be done in polynomial time. Thus the vertex cover problem is in the class NP.
Is the set cover problem NP-complete?
Thus, it can be verified that the set cover problem is NP-Complete using the following propositions: