Proof that the halting problem is NP-hard? - Stack Overflow In particular, I cannot seem to find a polynomial-time many-to-one mapping from instances of every problem in NP (or at least, any known NP-complete problem) onto the halting problem Is there a straightforward proof that the halting problem is NP-hard?
Example of a problem that is NP-Hard but not NP-Complete An example of a decidable, NP-Hard but not NP-Complete problem is the Towers of Hanoi problem Intuitively, the T of H problem is not in NP because the exploration of all possible configurations cannot be done in polynomial time, on a non-deterministic machine
NP-hardness - Wikipedia In computational complexity theory, a computational problem H is called NP-hard if, for every problem L which can be solved in non-deterministic polynomial-time, there is a polynomial-time reduction from L to H That is, assuming a solution for H takes 1 unit time, H ' s solution can be used to solve L in polynomial time
Does the halting problem belong to NP class of problems? If it doesn't halt, then the answer to the NP problem is "No" So deciding that the program halts allows us to find the solution to the NP problem Any problem in NP can be reduced to the question whether my "hint-guesser" halts or not
NP-Hard Class - GeeksforGeeks NP-Hard is a computational complexity theory that acts as a defining property for the class of problems that are "at least as hard as the hardest problems in NP" P's solution can be used to solve 'Q' in polynomial time For example, the Subset Sum Problem, and Turing Halting problem is an NP-Hard problem All NP-hard problems are not in NP
The Halting Problem P vs. NP - Kalamazoo College It turns out there is (at least) one problem that all problems in NP reduce to: SAT – SAT is the problem of checking to see if a boolean expression is satisfiable Therefore an efficient solution to SAT would yield an efficient solution to every problem in NP – SAT is NPComplete
Why the Halting Problem is a Poor Example of NP-Harder Problems The Halting Problem is often cited as an example of an NP-Harder problem, but this classification is debated for several reasons: Verification of "Yes" Instances: In NP problems, a "yes" answer can be verified in polynomial time given a witness
Why are NP problems called that way (and NP-hard and NP-complete)? Saying that a problem Q is "NP-hard" means that any problem in NP can be reduced to problem Q (in polynomial time) Since the relation "can be reduced to" between problems is an order relation, you can think of "NP-hard" as meaning "at least as hard as all NP problems"
Difference between NP hard and NP complete problem NP Hard: A problem is NP-Hard if it obeys Property 2 (all NP problems can be reduced to it) of NP Complete and need not obey Property 1 (It is in NP) Therefore, a problem is NP-complete if it is both NP and NP-hard