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Computing Standard-Deviation-to-Mean and Variance-to-Mean Ratios under Interval Uncertainty Is NP-Hard

机译:在区间不确定性下计算标准差 - 均值和均值 - 均值比是Np难

摘要

Once we have a collection of values x1, ..., ,xn corresponding a class of objects, a usual way to decide whether a new object with the value x of the corresponding property belongs to this class is to check whether the value x belongs to interval [E - k0 * s, E + k0 * s], where E = (1/n) * (x1 + ... + xn) is the sample mean, V = s^2 = (1/n) * ((x1 - E)^2 + ... + (xn - E)^2) is the sample variance, and the parameter k0 is determined by the degree of confidence with which we want to make the decision. For each value x, the degree of confidence that x belongs to the class depends on the smallest value k for which x belongs to the interval [E - k * s, E + k * s], i.e., on the ratio r = 1/k = s/(E - x). In practice, we often only know the intervals [xi] that contain the actual values xi. Different values xi from these intervals lead, in general, to different values of r, so it is desirable to compute the range [r] of corresponding values of r. Polynomial-time algorithms are known for computing [r] under certain conditions; whether it is possible that [r] can be computed in polynomial time was unknown. In this paper, we prove that the problem of computing [r] is NP-hard. A similar NP-hardness result is proven for a similar ratio V/E that is used in clustering.
机译:一旦有了对应于一类对象的值x1,...,xn的集合,确定具有对应属性的值x的新对象是否属于此类的通常方法是检查值x是否属于到区间[E-k0 * s,E + k0 * s],其中E =(1 / n)*(x1 + ... + xn)是样本均值,V = s ^ 2 =(1 / n) *((x1-E)^ 2 + ... +(xn-E)^ 2)是样本方差,参数k0由我们要进行决策的置信度确定。对于每个值x,x属于类别的置信度取决于x属于区间[E-k * s,E + k * s]的最小值k,即比率r = 1 / k = s /(E-x)。在实践中,我们通常仅知道包含实际值xi的间隔xi。通常,来自这些间隔的不同值xi导致r的值不同,因此希望计算r的相应值的范围[r]。已知多项式时间算法可以在某些条件下进行计算[r]。未知是否可以在多项式时间内计算[r]。在本文中,我们证明了[r]的计算问题是NP难的。对于在聚类中使用的比率V / E,证明了类似的NP硬度结果。

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    Lo Sio-Long;

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