What are Hitting sets?
A hitting set for a collection of sets is a set that has a non- empty intersection with each set in the collection; the hitting set problem is to find a hitting set of minimum cardinality.
Is hitting set NP hard?
Hitting Set is NP-Hard: In order to prove Hitting Set is NP-Hard, we will perform a reduction from which vertex cover problem can be reduced to the Hitting Set problem.
Is vertex cover a NP?
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. It is often used in computational complexity theory as a starting point for NP-hardness proofs.
Is set cover in P?
Since our contruction takes polynomial time, and we have shown that Set Cover is in NP, we can conclude that Set Cover is NP-Complete.
Is independent set NP?
The independent set decision problem is NP-complete, and hence it is not believed that there is an efficient algorithm for solving it. The maximum independent set problem is NP-hard and it is also hard to approximate.
What is the ρ approximation?
ρ-approximation algorithm (algorithmic technique) Definition: An approximation algorithm guaranteed to find a solution at most (or at least, as appropriate) ρ times the optimum. The ratio ρ is the performance ratio or relative performance guarantee of the algorithm.
Is knapsack NP-complete?
Theorem 1 Knapsack is NP-complete. Proof: First of all, Knapsack is NP. The proof is the set S of items that are chosen and the verification process is to compute ∑i∈S si and ∑i∈S vi, which takes polynomial time in the size of input.
Is Edge cover NP-complete?
Computing total edge covers. We now consider the t-Total Edge Cover problem. For this problem becomes the Edge Cover problem, which has long been known to be solvable in polynomial time [8]. The problem is NP-complete for each fixed 2 ≤ t ≤ k [9, Theorem 3].
Why Is Set cover 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.
Is K independent set NP-complete?
Independent Set is NP-complete. k-Independent Set is in P for every k∈ℕ.
What is C approximation algorithm?
We say that A is a c-approximation for P if for every input x of P: OPT(x) ≤ A(x) ≤ c · OPT(x) 2. If P is a maximization problem. We say that A is a c-approximation for P if for every input x of P: OPT(x) c ≤ A(x) ≤ OPT(x) In both cases, we call c the approximation ratio of A.
What is P N approximation algorithm?
If an algorithm reaches an approximation ratio of P(n), then we call it a P(n)-approximation algorithm. For a maximization problem, 0< C < C*, and the ratio of C*/C gives the factor by which the cost of an optimal solution is larger than the cost of the approximate algorithm.
What is k-approximation of k- hitting set?
In computer science, k-approximation of k-hitting set is an approximation algorithm for weighted hitting set. The input is a collection S of subsets of some universe T and a mapping W from T to non-negative numbers called the weights of the elements of T.
What is k-hitting set?
The input is a collection S of subsets of some universe T and a mapping W from T to non-negative numbers called the weights of the elements of T. In k-hitting set the size of the sets in S cannot be larger than k.
Why does it return a hitting set when it exits?
It returns a hitting set, because when the loop exits, all sets in S contain a tight element from T, and the set of these tight elements are returned. . This follows from the invariant property.