Is timetabling NP complete?

tl;dr Automatic scheduling(timetabling) comes under a special set of mathematical problems named as NP Complete which makes it impossible to generate and validate solutions ‘quickly’. Imagine a magical way of solving a Sudoku puzzle without having to check whether it is correct.

Why a problem is NP-hard?

A problem is NP-hard if an algorithm for solving it can be translated into one for solving any NP-problem (nondeterministic polynomial time) problem. NP-hard therefore means “at least as hard as any NP-problem,” although it might, in fact, be harder.

Are scheduling problems NP-hard?

Each task takes a certain amount of time and gives a certain amount of profit (gain). Moreover, to get the benefit of the task it has to finish by its deadline. When stated in this way, the Scheduling problem is NP-hard, and we do not know of any efficient, that is, polynomial-time algorithm for it.

What is NP scheduling problem?

All jobs require one time unit. All jobs require one or two time units, and there are only two processor resolving (in the negative a conjecture of R. L. Graham, Proc. As a consequence, the general preemptive scheduling problem is also NP-complete. …

What is NP and NP-hard problem?

A problem is NP-hard if an algorithm for solving it can be translated into one for solving any NP- problem (nondeterministic polynomial time) problem. NP-hard therefore means “at least as hard as any NP-problem,” although it might, in fact, be harder.

What is NP-hard problem in DAA?

A problem is NP-hard if all problems in NP are polynomial time reducible to it, even though it may not be in NP itself. If a polynomial time algorithm exists for any of these problems, all problems in NP would be polynomial time solvable. These problems are called NP-complete.

Is NP-complete problem NP-hard?

A problem is said to be NP-hard if everything in NP can be transformed in polynomial time into it even though it may not be in NP. Conversely, a problem is NP-complete if it is both in NP and NP-hard. The NP-complete problems represent the hardest problems in NP.