Table of Contents
Is P NP a solvable problem?
P is the set of all decision problems that are efficiently solvable. P is a subset of NP. P is the set of all decision problems that are efficiently solvable and is a subset of NP. Basic Arithmetic is solvable in Polynomial-time, thus belongs to P.
What is the fundamental problem of computer science?
Put another way, the fundamental question of computer science is simply What can be computed? process of examining algorithms and problems mathematically. Computer scientists have shown that some seemingly simple problems are not solvable by any algorithm. Other problems are intractable.
Are NP complete problems reducible?
Every problem in NP is reducible to. in polynomial time.
What is P vs NP problem in Computer Science?
The P versus NP problem is a major unsolved problem in computer science. It asks whether every problem whose solution can be quickly verified can also be solved quickly. Every problem as defined in theoretical computer science has sizes and the notion of quickness is defined relative to the size of the problem.
What is the difference between NP-hard problems and NP-complete problems?
That is, any NP problem can be transformed into any of the NP-complete problems. Informally, an NP-complete problem is an NP problem that is at least as “tough” as any other problem in NP. NP-hard problems are those at least as hard as NP problems, i.e., all NP problems can be reduced (in polynomial time) to them.
What is the proof that P = NP?
If P = NP, then all of these problems will be proven to have an efficient (polynomial time) solution. Most scientists believe that P != NP. However, no proof has yet been established for either P = NP or P != NP. If anyone provides a proof for either conjecture, they will win US $1 million. To give the simplest answer I can think of:
What is a yes-or-no problem in NP?
A yes-or-no problem is in P ( P olynomial time) if the answer can be computed in polynomial time. A yes-or-no problem is in NP ( N on-deterministic P olynomial time) if a yes answer can be verified in polynomial time. Intuitively, we can see that if a problem is in P, then it is in NP.