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).
Is traveling salesman NP-hard?
In Chapter 15 we introduced the Traveling Salesman Problem (TSP) and showed that it is NP-hard (Theorem 15.42). The TSP is perhaps the best-studied NP-hard combinatorial optimization problem, and there are many techniques which have been applied.
What is the Travelling sales problem is it NP-complete prove?
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.
Why is travel salesman NP-complete?
Why TSP Is Not NP-complete
The simple answer is that it’s NP-hard, but
it’s not in NP
. Since it’s not in NP, it can’t be NP-complete. … Since it takes exponential time to solve NP, the solution cannot be checked in polynomial time. Thus this problem is NP-hard, but not in NP.
Has traveling salesman problem been solved?
Scientists in Japan
have solved a more complex traveling salesman problem than ever before. The previous standard for instant solving was 16 “cities,” and these scientists have used a new kind of processor to solve 22 cities. They say it would have taken a traditional von Neumann CPU 1,200 years to do the same task.
Is Travelling salesman problem minimum spanning tree?
The Minimum Spanning Tree problem asks you to build a tree that connects all cities and
has minimum total weight
, while the Travelling Salesman Problem asks you to find a trip that visits all cities with minimum total weight (and possibly coming back to your starting point).
Is Travelling salesman backtracking?
Travelling Salesman Problem (TSP): Given a set of cities and distance between every pair of cities, the problem is to find the shortest possible route that visits every city exactly once and
returns back
to the starting point.
Why is NP-hard not in NP?
An NP-hard problem can be
beyond NP
. The polynomial-time reduction from your X to any problem in NP does not necessarily have a polynomial-time inverse. If the inverse is harder, then the verification is harder.
Is longest common subsequence NP?
The general longest common subsequence problem (LCS) over a binary alphabet is
NP-complete
.
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.
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.
Are NP problems solvable?
The short answer is that
if a problem is in NP, it is indeed solvable
.
Is Floyd warshall NP-hard?
It is not NP-complete
, because it is not a decision problem. In weighted complete graphs with non-negative edge weights, the weighted longest path problem is the same as the Travelling salesman path problem, because the longest path always includes all vertices.
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 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).
NP is set of problems that
can be solved
by a Non-deterministic Turing Machine in Polynomial time. P is subset of NP (any problem that can be solved by deterministic machine in polynomial time can also be solved by non-deterministic machine in polynomial time) but P≠NP.
