# Np Complete Vs Np Hardclevelandmultifiles

- Np Complete Vs Np Hardclevelandmultifiles Exam
- Np Complete Vs Np Hardclevelandmultifiles Nurse Practitioner
- Np Complete Vs Np Hardclevelandmultifiles Nursing

Nurse practitioner programs are often part of a master's degree program and prepare graduates to perform advanced levels of medical care. Those who earn a Master of Science in Nursing may focus on. NP deals with the gap between computers being able to quickly solve problems vs. Just being able to test proposed solutions for correctness. As such, the P vs. NP problem is the search for a way to solve problems that require the trying of millions, billions, or trillions of combinations without actually having to try each one. NP-Complete- The group of problems which are both in NP and NP-hard are known as NP-Complete problem. Now suppose we have a NP-Complete problem R and it is reducible to Q then Q is at least as hard as R and since R is an NP-hard problem. Therefore Q will also be at least NP-hard, it may be NP-complete also.

*Science and technology journalists pride themselves on the ability to explain complicated ideas in accessible ways, but there are some technical principles that we encounter so often in our reporting that paraphrasing them or writing around them begins to feel like missing a big part of the story. So in a new series of articles called 'Explained,' MIT News Office staff will explain some of the core ideas in the areas they cover, as reference points for future reporting on MIT research.*

In the 1995 Halloween episode of *The Simpsons*, Homer Simpson finds a portal to the mysterious Third Dimension behind a bookcase, and desperate to escape his in-laws, he plunges through. He finds himself wandering across a dark surface etched with green gridlines and strewn with geometric shapes, above which hover strange equations. One of these is the deceptively simple assertion that P = NP.

In fact, in a 2002 poll, 61 mathematicians and computer scientists said that they thought P probably didn’t equal NP, to only nine who thought it did — and of those nine, several told the pollster that they took the position just to be contrary. But so far, no one’s been able to decisively answer the question one way or the other. Frequently called the most important outstanding question in theoretical computer science, the equivalency of P and NP is one of the seven problems that the Clay Mathematics Institute will give you a million dollars for proving — or disproving. Roughly speaking, P is a set of relatively easy problems, and NP is a set that includes what seem to be very, very hard problems, so P = NP would imply that the apparently hard problems actually have relatively easy solutions. But the details are more complicated.

Computer science is largely concerned with a single question: How long does it take to execute a given algorithm? But computer scientists don’t give the answer in minutes or milliseconds; they give it relative to the number of elements the algorithm has to manipulate.

Imagine, for instance, that you have an unsorted list of numbers, and you want to write an algorithm to find the largest one. The algorithm has to look at all the numbers in the list: there’s no way around that. But if it simply keeps a record of the largest number it’s seen so far, it has to look at each entry only once. The algorithm’s execution time is thus directly proportional to the number of elements it’s handling — which computer scientists designate N. Of course, most algorithms are more complicated, and thus less efficient, than the one for finding the largest number in a list; but many common algorithms have execution times proportional to N^{2}, or N times the logarithm of N, or the like.

A mathematical expression that involves N’s and N^{2}s and N’s raised to other powers is called a polynomial, and that’s what the “P” in “P = NP” stands for. P is the set of problems whose solution times are proportional to polynomials involving N's.

Obviously, an algorithm whose execution time is proportional to N^{3} is slower than one whose execution time is proportional to N. But such differences dwindle to insignificance compared to another distinction, between polynomial expressions — where N is the number being raised to a power — and expressions where a number is raised to the Nth power, like, say, 2^{N}.

If an algorithm whose execution time is proportional to N takes a second to perform a computation involving 100 elements, an algorithm whose execution time is proportional to N^{3} takes almost three hours. But an algorithm whose execution time is proportional to 2^{N} takes 300 quintillion years. And that discrepancy gets much, much worse the larger N grows.

NP (which stands for nondeterministic polynomial time) is the set of problems whose solutions can be verified in polynomial time. But as far as anyone can tell, many of those problems take exponential time to solve. Perhaps the most famous exponential-time problem in NP, for example, is finding prime factors of a large number. Verifying a solution just requires multiplication, but solving the problem seems to require systematically trying out lots of candidates.

So the question “Does P equal NP?” means “If the solution to a problem can be verified in polynomial time, can it be found in polynomial time?” Part of the question’s allure is that the vast majority of NP problems whose solutions seem to require exponential time are what’s called NP-complete, meaning that a polynomial-time solution to one can be adapted to solve all the others. And in real life, NP-complete problems are fairly common, especially in large scheduling tasks. The most famous NP-complete problem, for instance, is the so-called traveling-salesman problem: given N cities and the distances between them, can you find a route that hits all of them but is shorter than … whatever limit you choose to set?

Given that P probably doesn’t equal NP, however — that efficient solutions to NP problems will probably never be found — what’s all the fuss about? Michael Sipser, the head of the MIT Department of Mathematics and a member of the Computer Science and Artificial Intelligence Lab’s Theory of Computation Group (TOC), says that the P-versus-NP problem is important for deepening our understanding of computational complexity.

“A major application is in the cryptography area,” Sipser says, where the security of cryptographic codes is often ensured by the complexity of a computational task. The RSA cryptographic scheme, which is commonly used for secure Internet transactions — and was invented at MIT — “is really an outgrowth of the study of the complexity of doing certain number-theoretic computations,” Sipser says.

Similarly, Sipser says, “the excitement around quantum computation really boiled over when Peter Shor” — another TOC member — “discovered a method for factoring numbers on a quantum computer. Peter's breakthrough inspired an enormous amount of research both in the computer science community and in the physics community.” Indeed, for a while, Shor’s discovery sparked the hope that quantum computers, which exploit the counterintuitive properties of extremely small particles of matter, could solve NP-complete problems in polynomial time. But that now seems unlikely: the factoring problem is actually one of the few hard NP problems that is not known to be NP-complete.

Sipser also says that “the P-versus-NP problem has become broadly recognized in the mathematical community as a mathematical question that is fundamental and important and beautiful. I think it has helped bridge the mathematics and computer science communities.”

But if, as Sipser says, “complexity adds a new wrinkle on old problems” in mathematics, it’s changed the questions that computer science asks. “When you’re faced with a new computational problem,” Sipser says, “what the theory of NP-completeness offers you is, instead of spending all of your time looking for a fast algorithm, you can spend half your time looking for a fast algorithm and the other half of your time looking for a proof of NP-completeness.”

Sipser points out that some algorithms for NP-complete problems exhibit exponential complexity only in the worst-case scenario and that, in the average case, they can be more efficient than polynomial-time algorithms. But even there, NP-completeness “tells you something very specific,” Sipser says. “It tells you that if you’re going to look for an algorithm that’s going to work in every case and give you the best solution, you’re doomed: don’t even try. That’s useful information.”

- Design and Analysis of Algorithms

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A problem is in the class NPC if it is in NP and is as **hard** as any problem in NP. 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**. The phenomenon of NP-completeness is important for both theoretical and practical reasons.

## Definition of NP-Completeness

A language **B** is ** NP-complete** if it satisfies two conditions

**B**is in NPEvery

**A**in NP is polynomial time reducible to**B**.

If a language satisfies the second property, but not necessarily the first one, the language **B** is known as **NP-Hard**. Informally, a search problem **B** is **NP-Hard** if there exists some **NP-Complete** problem **A** that Turing reduces to **B**.

The problem in NP-Hard cannot be solved in polynomial time, until **P = NP**. If a problem is proved to be NPC, there is no need to waste time on trying to find an efficient algorithm for it. Instead, we can focus on design approximation algorithm.

## NP-Complete Problems

Following are some NP-Complete problems, for which no polynomial time algorithm is known.

- Determining whether a graph has a Hamiltonian cycle
- Determining whether a Boolean formula is satisfiable, etc.

## NP-Hard Problems

The following problems are NP-Hard

- The circuit-satisfiability problem
- Set Cover
- Vertex Cover
- Travelling Salesman Problem

### Np Complete Vs Np Hardclevelandmultifiles Exam

In this context, now we will discuss TSP is NP-Complete

## TSP is NP-Complete

The traveling salesman problem consists of a salesman and a set of cities. The salesman has to visit each one of the cities starting from a certain one and returning to the same city. The challenge of the problem is that the traveling salesman wants to minimize the total length of the trip

## Proof

To prove ** TSP is NP-Complete**, first we have to prove that

**. In TSP, we find a tour and check that the tour contains each vertex once. Then the total cost of the edges of the tour is calculated. Finally, we check if the cost is minimum. This can be completed in polynomial time. Thus**

*TSP belongs to NP***.**

*TSP belongs to NP*Secondly, we have to prove that ** TSP is NP-hard**. To prove this, one way is to show that

**(as we know that the Hamiltonian cycle problem is NPcomplete).**

*Hamiltonian cycle ≤*_{p}TSPAssume ** G = (V, E)** to be an instance of Hamiltonian cycle.

Hence, an instance of TSP is constructed. We create the complete graph ** G^{'} = (V, E^{'})**, where

$$E^{'}=lbrace(i, j)colon i, j in V ::and:ineq j$$

Thus, the cost function is defined as follows −

$$t(i,j)=begin{cases}0 & if: (i, j): in E1 & otherwiseend{cases}$$

Now, suppose that a Hamiltonian cycle ** h** exists in

**. It is clear that the cost of each edge in**

*G***is**

*h***0**in

**as each edge belongs to**

*G*^{'}**. Therefore,**

*E***has a cost of**

*h***0**in

**. Thus, if graph**

*G*^{'}**has a Hamiltonian cycle, then graph**

*G***has a tour of**

*G*^{'}**0**cost.

### Np Complete Vs Np Hardclevelandmultifiles Nurse Practitioner

Conversely, we assume that ** G^{'}** has a tour

**of cost at most**

*h*^{'}**0**. The cost of edges in

**are**

*E*^{'}**0**and

**1**by definition. Hence, each edge must have a cost of

**0**as the cost of

**is**

*h*^{'}**0**. We therefore conclude that

**contains only edges in**

*h*^{'}**.**

*E*### Np Complete Vs Np Hardclevelandmultifiles Nursing

We have thus proven that ** G** has a Hamiltonian cycle, if and only if

**has a tour of cost at most**

*G*^{'}**0**. TSP is NP-complete.