Contents
Why k-means NP-hard?
Abstract We show that k-means clustering is an NP-hard optimization problem, even for instances in the plane. Specifically, the hardness holds for k = Θ(nϵ), for any ϵ > 0, where n is the number of points in the instance, and k is the number of clusters.
What does it mean if problem B is NP-hard?
Definition of NP-Completeness If a language satisfies the second property, but not necessarily the first one, the language B is known as NP-Hard. Informally, a search problem B is NP-Hard if there exists some NP-Complete problem A that Turing reduces to B.
Are NP-Hard problems solvable?
This is known as Cook’s theorem. What makes NP-complete problems important is that if a deterministic polynomial time algorithm can be found to solve one of them, every NP problem is solvable in polynomial time (one problem to rule them all).
Is the planar k-means problem NP hard?
While clustering problems generally tend to be NP-hard even in the plane, there are surprising exceptions – the problem of covering a set of points by k balls so as to minimize the sum of the radii of the balls admits a polynomial time algorithm if we use L
What’s the difference between hard NP and hard NP?
NP-Hard: These are at least as hard as the hardest problems in NP. If we can solve these problems in polynomial time, we can solve any NP problem that can possibly exist. Note that these problems are not necessarily NP problems.
What is the goal of the k mean problem?
First, the goal of K-mean is to produce an optimal (in the sense of Euclidean distance) set of set-vector pairs { ( S, μ) }, where set represents the membership of the data and the vector represents the centroid of the data. This is clearly not a decision problem.
Are there any decision problems that are NP-hard?
There are decision problems that are NP-hard but not NP-complete such as the halting problem. That is the problem which asks “given a program and its input, will it run forever?” That is a yes / no question and so is a decision problem. It is easy to prove that the halting problem is NP-hard but not NP-complete.