Is The Knapsack Problem NP-complete?

The decision problem form of the knapsack problem (Can a value of at least V be achieved without exceeding the weight W?) is

NP-complete

, thus there is no known algorithm both correct and fast (polynomial-time) in all cases. … There is a pseudo-polynomial time algorithm using dynamic programming.

How do you know if a problem is NP-complete?

A problem is called NP (nondeterministic polynomial) if its solution can be guessed and verified in polynomial time; nondeterministic means that no particular rule is followed to make the guess. If a problem is NP and all other NP problems are

polynomial-time reducible to it

, the problem is NP-complete.

Are NP-hard problems NP-complete?

The complexity class of problems of this form is called NP, an abbreviation for “nondeterministic polynomial time”. 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.

What problems are NP-complete?

Other well-known NP-complete problems are satisfiability (SAT),

traveling salesman, the bin packing problem

, and the knapsack problem. (Strictly the related decision problems are NP-complete.) “NP” comes from the class that a Nondeterministic Turing machine accepts in Polynomial time.

Is subset sum and knapsack problem NP-complete?

For this reason, it is also called the Subset Sum Problem. Clearly, the Knapsack (Subset Sum) Problem re- duces to the 0 -1 Knapsack Problem, and thus the 0 -1 Knapsack Problem is also

NP-complete

.

Is clique and set cover problem NP-complete?

Since VERTEX-COVER can be reduced to CLIQUE in polynomial time, CLIQUE ∈ NP and VERTEX

-COVER is NP-Complete

, CLIQUE is also NP-Complete.

Is it possible for a problem to be in both P and NP?

Is it possible for a problem to be in both P and NP?

Yes

. Since P is a subset of NP, every problem in P is in both P and NP.

Can NP-hard be reduced to NP-complete?

(If P and NP are the same class, then NP-intermediate problems do not exist because in this case every NP-complete problem would fall in P, and by definition,

every problem in NP can be reduced to

an NP-complete problem.)

Are NP problems solvable?

The short answer is that

if a problem is in NP, it is indeed solvable

.

Is traveling salesman NP-complete?

Traveling Salesman Optimization(TSP-OPT) is a NP-hard problem and

Traveling Salesman Search(TSP) is NP-complete

. However, TSP-OPT can be reduced to TSP since if TSP can be solved in polynomial time, then so can TSP-OPT(1).

What does it mean if Q is NP-hard?

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.

Which type of problem may be NP-hard?

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.

Is clique a problem with NP?

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).

What’s the difference between NP hard and NP-complete?

Is shortest path problem NP-complete?

We show that the following variation of the single-source shortest path problem is NP-complete. Let a weighted, directed, acyclic graph G=(V,E,w) with source and sink vertices s and t be given. It is NP-complete by reduction from 3SAT. …

Is NP equal to NP-complete?

What is the point of classifying the two if they are the same? In other words, if we have an NP problem then through (2) this problem can transform into an NP-complete problem. Therefore, the NP problem is now NP-complete, and

NP = NP-complete

. Both classes are equivalent.