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All NP complete problems are NP hard problems when solved in a different way.
But, all NP hard problems are not NP complete.
Ex:
1. Traveling salesman problem. It is both NP hard and NP complete. We can find that whether the solution is correct or not in the given period of time. In this way, it is NP complete.
But, to find the shortest path, i.e. optimization of Traveling Salesman problem is NP hard. If there will be changing costs, then every time when the salesperson returns to the source node, then he will be having different shortest path. In this way, it is hard to solve. It cannot be solved in the polynomial time. In this way, it is NP hard problem.
2. Halting problem.
3. Sum of subset problem.
What else can I help you with?
Is -225.4 a rational number?
Yes
237 93 np 4 9 he plus 233 pa?
91
If the original price of a computer was reduced by 250 dollars and the new discounted price is 1499 dollars what was the original price of the computer?
$1749 RP-DA = NP where RP is regular price minus discount amount (DA) is equal to New price (NP) Change the formula to solve for RP you get NP + DA = RP 1,499 + 250 = RP 1,749 = RP The Regular price was one thousand seven hundred and forty nine
What is the mean variance and standard deviation for the number of heads when 20 coins are tossed?
mean = np = 20*0.5 = 10 Variance = npq = 20*0.5*0.5 = 5 Std dev = sqrt(npq) = sqrt(20*0.5*0.5) = 2.23
How can you approximate a binomial distribution to a poison distribution when the number of binomial trials became large enough?
The Poisson distribution with parameter np will be a good approximation for the binomial distribution with parameters n and p when n is large and p is small. For more details See related link below
Difference between np and np complete?
A problem is 'in NP' if there exists a polynomial time complexity algorithm which runs on a Non-Deterministic Turing Machine that solves it. A problem is 'NP Hard' if all problems in NP can be reduced to it in polynomial time, or equivalently if there is a polynomial-time reduction of any other NP Hard problem to it. A problem is NP Complete if it is both in NP and NP hard.
Can you prove that the problem is NP-complete?
Proving that a problem is NP-complete involves demonstrating that it is both in the NP complexity class and that it is at least as hard as any other problem in NP. This typically involves reducing a known NP-complete problem to the problem in question, showing that a solution to the problem in question can be used to solve the known NP-complete problem efficiently.
How can one demonstrate that a problem is NP-complete?
One can demonstrate that a problem is NP-complete by showing that it belongs to the NP complexity class and that it is at least as hard as any other problem in NP. This can be done by reducing a known NP-complete problem to the problem in question through a polynomial-time reduction.
What is the difference between NP and NP Complete problems?
- a problem in NP means that it can be solved in polynomial time with a non-deterministic turing machine - a problem that is NP-hard means that all problems in NP are "easier" than this problem - a problem that is NP-complete means that it is in NP and it is NP-hard example - Hamiltonian path in a graph: The problem is: given a graph as input, an algorithm must say whether there is a hamiltonian path in it or not. in NP: here is an algorithm that works in polynomial time on a non-deterministic turing machine: guess a path in the graph. Check that it is really a hamiltonian path. NP-hard: we use reduction from a problem that is NP-comlete (SAT for example). Given an input for the other problem we construct a graph for the hamiltonian-path problem. The graph should have a path iff the original problem should return "true". Therefore, if there is an algorithm that executes in polynomial time, we solve all the problems in NP in polynomial time.j
Is there a way to demonstrate that the problem of double-sat is NP-complete?
Yes, the problem of double-sat can be demonstrated to be NP-complete by reducing it to a known NP-complete problem, such as 3-SAT. This reduction shows that solving double-sat is at least as hard as solving 3-SAT, which is a known NP-complete problem.
What is the proof that the Clique Problem is NP-complete?
The proof that the Clique Problem is NP-complete involves showing that it is both in the NP complexity class and that it is as hard as any problem in NP. This is typically done by reducing a known NP-complete problem, such as the SAT problem, to the Clique Problem in polynomial time. This reduction demonstrates that if a polynomial-time algorithm exists for the Clique Problem, then one also exists for the known NP-complete problem, which implies that the Clique Problem is NP-complete.
Is there a way to demonstrate that the problem of determining whether a given path exists in a graph is NP-complete?
Yes, the problem of determining whether a given path exists in a graph can be demonstrated as NP-complete by reducing it to a known NP-complete problem, such as the Hamiltonian path problem. This reduction shows that the path existence problem is at least as hard as the known NP-complete problem, making it NP-complete as well.
Is prime factorization an NP-complete problem?
Yes, prime factorization is not an NP-complete problem. It is in fact in the complexity class NP, but it is not known to be NP-complete.
How can NP completeness reductions be used to demonstrate the complexity of a computational problem?
NP completeness reductions are used to show that a computational problem is at least as hard as the hardest problems in the NP complexity class. By reducing a known NP-complete problem to a new problem, it demonstrates that the new problem is also NP-complete. This helps in understanding the complexity of the new problem by showing that it is as difficult to solve as the known NP-complete problem.
Is the Knapsack Problem NP-complete?
The Knapsack Problem is NP-complete. This means that it is a problem in computational complexity theory that belongs to the NP complexity class and is at least as hard as the hardest problems in NP. It is a classic optimization problem where the goal is to maximize the total value of items placed into a knapsack without exceeding the knapsack's capacity. The NP-completeness of the Knapsack Problem has been proven through reductions from other NP-complete problems such as the Boolean Satisfiability Problem.
Is the clique problem NP-complete?
Yes, the clique problem is NP-complete.
Is the partition problem NP-complete?
Yes, the partition problem is NP-complete.
