Can NP-hard problems be solved in polynomial time?

If P ≠ NP, then NP-hard problems could not be solved in polynomial time. Some NP-hard optimization problems can be polynomial-time approximated up to some constant approximation ratio (in particular, those in APX) or even up to any approximation ratio (those in PTAS or FPTAS).

Is it true that if an NP-complete problem can be solved in polynomial time then P NP justify your answer?

The answer is B (no NP-Complete problem can be solved in polynomial time). Because, if one NP-Complete problem can be solved in polynomial time, then all NP problems can solved in polynomial time. If that is the case, then NP and P set become same which contradicts the given condition.

Is set of problems that can be solved in polynomial time?

Problems that can be solved in polynomial time are called Tractable problems.

Is there a problem in NP that has a polynomial time algorithm?

Strictly speaking, as the other answers explain, no. A polynomial-time algorithm for an NP-hard problem is not known nor expected to exist.

Can an NP problem be solved?

When an NP-complete problem must be solved, one approach is to use a polynomial algorithm to approximate the solution; the answer thus obtained will not necessarily be optimal but will be reasonably close. The Editors of Encyclopaedia Britannica This article was most recently revised and updated by Erik Gregersen.

Is TSP 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).

Can P vs NP be solved?

Informally, they are the “hardest” of the NP problems. Thus if any one NP-Complete problem can be solved in polynomial time, then every NP-Complete problem can be solved in polynomial time, and every problem in NP can be solved in polynomial time (i.e. P=NP).

Are NP hard problems NP-complete?

A problem X is NP-Complete if there is an NP problem Y, such that Y is reducible to X in polynomial time. NP-Complete problems are as hard as NP problems.
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Difference between NP-Hard and NP-Complete:
NP-hard NP-Complete
To solve this problem, it do not have to be in NP . To solve this problem, it must be both NP and NP-hard problems.
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Oct 4, 2021

Which is the class of decision problems that can be solved by non deterministic polynomial algorithms?

Explanation: NP problems are called as non-deterministic polynomial problems. They are a class of decision problems that can be solved using NP algorithms.

Is chess a NP?

For two-player games, one encounters a similar phenomenon at a higher level of complexity. … For this reason games like chess cannot themselves be NP-complete, as they only have a finite (albeit unthinkably large) number of possible positions.

Is NP harder than P?

Roughly speaking, P is a set of relatively easy problems, and NP is a set that includes what seem to be very, very hard problems, so P = NP would imply that the apparently hard problems actually have relatively easy solutions.

Who Solved P versus NP problem?

In 2000 American mathematician Stephen Smale devised an influential list of 18 important mathematical problems for solving in the 21st century. The third problem on his list was the P versus NP problem.

Is GO NP-hard?

It’s a common misconception that chess, GO is NP-hard. … Chess and Go are both EXPTIME complete. IIRC, Go has more possible moves, so I think it a higher multiple of that complexity class than chess.

Is Sokoban NP-complete?

SOKOBAN(1,0) is NP-hard. As mentioned, SOKOBAN is similar to a motion planning problem with ‘movable obstacles’ studied by Wilfong [13].

Is Rush Hour NP-complete?

Each of the 24 puzzles below have all been shown to be NP-Complete (with the exception of Rush Hour, which is PSPACE-Complete). However, we should emphasize that any problem that has a finite problem space cannot be NP-Complete as we could solve the problem in constant time.

Is 8×8 chess hard?

It’s a common misconception that chess is NP-hard. Generalized chess may be NP-hard. Chess has an 8×8 board, generalized chess has an nxn board with many pieces. … I reason that it’s not NP-complete; not because it’s easier than NP-complete problems but because it’s harder.

Why is chess not in Pspace?

Some other generalized games, such as chess, checkers (draughts), and Go are EXPTIME-complete because a game between two perfect players can be very long, so they are unlikely to be in PSPACE. … It is also possible for puzzles played by a single player to be PSPACE-complete.

Is chess a polynomial time?

8×8 chess is in P of course because we have only a polynomial number of possible positions, albeit enormously large but still polynomial. We have to notice that chess can’t be in NP because there is no polynomial time verifier for the decision problem “is e4 the best initial move for white?”.

Is chess an Exptime?

For many generalized games which may last for a number of moves exponential in the size of the board, the problem of determining if there is a win for the first player in a given position is EXPTIME-complete. Generalized chess, go (with Japanese ko rules), Quixo, and checkers are EXPTIME-complete.

Is NP a PSPACE?

Since, PSPACE is closed under reductions and NP is contained in PSPACE, then we have that NP = PSPACE. The P versus NP problem is a major unsolved problem in computer science. This problem was introduced in 1971 by Stephen Cook [1]. It is considered by many to be the most important open problem in the field [2].

Is chess PSPACE-complete?

On the complexity of chess shows that deciding which player wins is PSPACE-complete, if the rules only permit a polynomial number of moves (such as the 50-move drawing rule.) “Real” chess is in P because it’s of finite size so all positions can be (in a theoretical, computational-complexity sense) looked up in a table.

What is NP problem?

That is, co-NP is the set of decision problems where there exists a polynomial p(n) and a polynomial-time bounded Turing machine M such that for every instance x, x is a no-instance if and only if: for some possible certificate c of length bounded by p(n), the Turing machine M accepts the pair (x, c). …