Has anyone solved NP or P?
Now, a German man named Norbert Blum has claimed to have solved the above riddle, which is properly known as the P vs NP problem. Unfortunately, his purported solution doesn’t bear good news. Blum, who is from the University of Bonn, claims in his recently published 38-page paper that P does not equal NP.
What happens if P vs NP is solved?
If P=NP, then all of the NP problems can be solved deterministically in Polynomial time. If you could solve clique with a polynomial time algorithm, this would prove that P=NP, and then you could also use your method for solving clique to solve all of the other problems on that wiki-list, as an implication.
What is NP-hard problem with example?
Examples. An example of an NP-hard problem is the decision subset sum problem: given a set of integers, does any non-empty subset of them add up to zero? That is a decision problem and happens to be 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 …
Can all NP problems be reduced to NP-complete?
Yes. By definition any NP problem can be reduced to an NP-complete problem in polynomial time. Since NP-complete problems are themselves NP problems, all NP-complete problems can be reduced to each other in polynomial time.
Why do researchers believe P is not equal to NP?
According to polls, most computer scientists believe that P ≠ NP. A key reason for this belief is that after decades of studying these problems no one has been able to find a polynomial-time algorithm for any of more than 3000 important known NP-complete problems (see List of NP-complete problems).
What is N and P NP-complete problems?
What are NP, P, NP-complete and NP-Hard problems? P is set of problems that can be solved by a deterministic Turing machine in Polynomial time. NP is set of decision problems that can be solved by a Non-deterministic Turing Machine in Polynomial time. NP-completeness applies to the realm of decision problems.
What does it mean if Q is NP-hard?
3)A problem is called NP-hard if any other problem in NP can be reduced to it in polynomial time. 4)It follows that a problem P is NP-hard if there exists a problem Q known to be NP-complete, such that Q can be reduced to P in polynomial time.
What are hard problems?
A problem is “hard” if it requires (or we think it requires) “large” computational resources to solve, and “easy” if it doesn’t. “Large” depends on context but, in most contexts, a problem that can be solved in polynomial time is considered “easy”.
What is the difference between P and NP problems?
P = the set of problems that are solvable in polynomial time by a Deterministic Turing Machine. NP = the set of decision problems (answer is either yes or no) that are solvable in nondeterministic polynomial time i.e can be solved in polynomial time by a Nondeterministic Turing Machine[4].
Are all NP problems Decidable?
Note that every NP problem is decidable. This is a key concept. Remember P problems also fit the definition of NP, so…. There are certain NP-Hard problems that also exist in NP.
Are all decision problems in NP?
Decision Problems, Optimization Problems P and NP only contains decision problems. But many decision problems can be transformed in polynomial time to and from a corresponding optimization problem.
Are all NP languages Decidable?
All languages in P and NP are decidable. Any undecidable language, such as the halting problem, cannot be in NP. The nondeterministic time hierarchy theorem can be used to show that there are languages in NEXP that are not in NP.
Can NP-hard problems be verified in polynomial time?
An NP-Hard problem is one that is not solvable in polynomial time but can be verified in polynomial time.
Is Hamiltonian cycle NP-complete?
A Hamiltonian path is a simple open path that contains each vertex in a graph exactly once. The Hamiltonian Path problem is the problem to determine whether a given graph contains a Hamiltonian path. Hamiltonian Cycle is NP-complete, so we may try to reduce this problem to Hamiltonian Path.
What makes a problem 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.
How do you deal with NP completeness?
One relaxation of these problems is to find an approximate solution with desired approximation guarantee. Theoreticians proved that efficient algorithms exist for certain NP-hard problems, and simple greedy strategy usually provides good approximation.
Is shortest path NP-complete?
The Shortest Path Problem with Edge Information Reuse is 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.
Why is longest path NP hard?
-G has a negative cycle, in which case the shortest path in -G cannot be found. -G does not have a negative cycle, in which case Floy-Warshall or Bellman-Ford algorithm can find the shortest path in -G in polynomial time.