Is set cover NP-complete?

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.

What does NP stand for in business?

No protest (NP) is a term used when a bank receives instructions from another bank not to protest items in the event that a negotiable instrument is not paid or accepted.

Which problems are NP-hard?

In computational complexity theory, NP-hardness (non-deterministic polynomial-time hardness) is the defining property of a class of problems that are informally “at least as hard as the hardest problems in NP”. A simple example of an NP-hard problem is the subset sum problem.

What is meant by NP-hard problem?

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.

What does NP stand for in school?

NP in Education

2 NP National Percentile School, Educational, Score
1 NP No Pass Student, Course, Grading
1 NP Notary Public Genealogy, Jurisprudence, Slang
1 NP Nuclear Physics Politics, Physics, University
1 NP Nurse Practioner Practitioner, Medical, Health

Is chess an NP problem?

Generalized chess may be NP-hard. Chess has an 8×8 board, generalized chess has an nxn board with many pieces. There may be a “yes” answer and the certificate for NP might be a list of perfect moves for both players, but it’s intractable to check if those moves by black are actually perfect.

Is vertex cover NP-complete?

Its decision version, the vertex cover problem, was one of Karp’s 21 NP-complete problems and is therefore a classical NP-complete problem in computational complexity theory.

Is subset sum NP-hard?

Subset sum can also be regarded as an optimization problem: find a subset whose sum is as close as possible to T. It is NP-hard, but there are several algorithms that can solve it reasonably quickly in practice.

How do you prove a problem is NP?

The easiest way to prove some problem is in NP is using the certificate definiiton of NP mentioned in other answers. The nondeterministic definition of NP is usually not very useful for showing a problem belongs to NP.

Does every set have an open cover?

The answer to your question is yes. In a metric space X, X is open. Since (very reduntantly) every subset of X is a subset of X, then X functions as an open cover for each of its subsets. A set K⊂X of a metric space X is compact iff every open cover of K has a finite subcover.

What is difference between NP-hard and NP-complete?

A problem X is NP-Complete if there is an NP problem Y, such that Y is reducible to X in polynomial time….Difference between NP-Hard and NP-Complete:

NP-hard NP-Complete
To solve this problem, do not have to be in NP . To solve this problem, it must be both NP and NP-hard problems.
Do not have to be a Decision problem. It is exclusively a Decision problem.

What set cover?

Buildings insurance – Replacement of sets cover is only available with Home Insurance Ultimate. Where there is a valid claim, this covers the changing or replacement of any associated undamaged items that form part of a matching set or suite that is lost or damaged where a reasonable match cannot be obtained.

How many NP-complete problems are there?

This list is in no way comprehensive (there are more than 3000 known NP-complete problems). Most of the problems in this list are taken from Garey and Johnson’s seminal book Computers and Intractability: A Guide to the Theory of NP-Completeness, and are here presented in the same order and organization.

Can P be reduced to NP-complete?

Quick reply: No, it does not. Recall the definition of NP-hard problems. A problem X is NP-Hard if every problem in NP can be polynomially reduced to X. Q: If a Problem X lying in P or NP is such that an NP-Complete problem can be reduced to it, is that problem X automatically an NP-Hard problem?

Are NP-hard problems solvable?

A problem is NP-hard if all problems in NP are polynomial time reducible to it, even though it may not be in NP itself. If a polynomial time algorithm exists for any of these problems, all problems in NP would be polynomial time solvable. These problems are called NP-complete.

Is traveling salesman NP-hard?

The travelling salesman problem (also called the traveling salesperson problem or TSP) asks the following question: “Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city exactly once and returns to the origin city?” It is an NP-hard problem in …

What does P vs NP stand for?

nondeterministic polynomial time

What does the acronym NP stand for?

Acronym Definition
NP Noun Phrase
NP New Patient
NP No Problem
NP Not Possible

Is clique problem NP-complete?

In computer science, the clique problem is the computational problem of finding cliques (subsets of vertices, all adjacent to each other, also called complete subgraphs) in a graph. Most versions of the clique problem are hard. The clique decision problem is NP-complete (one of Karp’s 21 NP-complete problems).

Is set cover NP-hard?

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.

How do you prove that vertex cover is NP-complete?

Thus, we can say that there is a clique of size k in graph G if and only if there is a vertex cover of size |V| – k in G’, and hence, any instance of the clique problem can be reduced to an instance of the vertex cover problem. Thus, vertex cover is NP Hard.

Is P equal to NP?

The statement P=NP means that if a problem takes polynomial time on a non-deterministic TM, then one can build a deterministic TM which would solve the same problem also in polynomial time.

Why do we need to prove NP completeness?

Proving a problem NP-Complete is a research success because it frees you from having to search for an efficient and exact solution for the general problem you are studying.

What does NP stand for CS?

In computational complexity theory, NP (nondeterministic polynomial time) is a complexity class used to classify decision problems. NP is the set of decision problems for which the problem instances, where the answer is “yes”, have proofs verifiable in polynomial time by a deterministic Turing machine.

What is NP price?

Net price is the value at which a product or service is sold after all taxes and other costs are added and all discounts subtracted. Amounts subtracted from a list price to arrive at a net price include sales and other discount amounts, rebates and any amounts negotiated between the customer and the seller.

How do you solve NP-hard problems?

The first and most obvious approach to getting “pretty good” solutions to NP-Hard problems is to devise greedy algorithms. There are usually many possible greedy choices, some of which, like Dijkstra’s algorithm, can actually be quite sophisticated.

What is NP-complete with example?

NP-complete problem, any of a class of computational problems for which no efficient solution algorithm has been found. Many significant computer-science problems belong to this class—e.g., the traveling salesman problem, satisfiability problems, and graph-covering problems.

What is a Subcover?

Noun. subcover (plural subcovers) (topology) A cover which is a subset of another cover. The open intervals cover the real numbers; the open intervals of the form (x, x+1) are a subcover.

What is Open cover in topology?

A collection of open sets of a topological space whose union contains a given subset. For example, an open cover of the real line, with respect to the Euclidean topology, is the set of all open intervals , where . The set of all intervals , where , is an open cover of the open interval .