Np Complete Vs Np Hardclevelandmultifiles

  1. Np Complete Vs Np Hardclevelandmultifiles Practice
  2. Np Complete Vs Np Hardclevelandmultifiles Exam

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Dreds more problems NP-complete (see 17). An e cient solution to any NP-complete problem would imply P = NP and an e cient solution to every NP-complete problem. Most computer scientists quickly came to believe P 6= NP and trying to prove it quickly became the single most impor-tant question in all of theoretical computer science and one. This discussion is deliberately fuzzy, since it is supposed to be addressed to a child as per the question. They have years to learn the right details. To keep things simple, let's just talk about problems with "yes/no" answers. Oct 29, 2009 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.

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Np Complete Vs Np Hardclevelandmultifiles Practice

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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 NP

  • Every 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 Vs Np Hardclevelandmultifiles Exam

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

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

TSP is NP-Complete

Np complete vs np hardclevelandmultifiles nursing

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


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

Np Complete Vs Np Hardclevelandmultifiles

Secondly, we have to prove that TSP is NP-hard. To prove this, one way is to show that Hamiltonian cycle ≤p TSP (as we know that the Hamiltonian cycle problem is NPcomplete).

Assume 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 G. It is clear that the cost of each edge in h is 0 in G' as each edge belongs to E. Therefore, h has a cost of 0 in G'. Thus, if graph G has a Hamiltonian cycle, then graph G' has a tour of 0 cost.

Conversely, we assume that G' has a tour h' of cost at most 0. The cost of edges in E' are 0 and 1 by definition. Hence, each edge must have a cost of 0 as the cost of h' is 0. We therefore conclude that h' contains only edges in E.

We have thus proven that G has a Hamiltonian cycle, if and only if G' has a tour of cost at most 0. TSP is NP-complete.