maximin.EquidistantLHD

maximin.EquidistantLHD(N, method=1)

Generate an Equidistant Latin Hypercube

Parameters

Name	Type	Description	Default
N	int	An odd integer	required
method	int	Specify construction method, can either be 1 or 2. Defaults to 1.	1

Returns

Type	Description
numpy.numpy.ndarray	If method=1, given an odd integer \(N=(2m+1)\), return an \((m \times m)\) equidistant LHD. This design, is a cyclic Latin square, with each level occuring once in each row and once in each column. It is also a maximin distance LHD in terms of \(L_1\)-distance

Notes

If method=1, construction method is based on “OPTIMAL MAXIMIN L1-DISTANCE LATIN HYPERCUBE DESIGNS BASED ON GOOD LATTICE POINT DESIGNS” by LIN WANG QIAN XIAO AND HONGQUAN XU

If method=2, constuction method is based on “A CONSTRUCTION METHOD FOR MAXIMIN L1-DISTANCE LATIN HYPERCUBE DESIGNS” by Ru Yuan, Yuhao Yin, Hongquan Xu, Min-Qian Liu

Example:

import pyLHD
N = 11
sample = pyLHD.EquidistantLHD(N = N)
sample

array([[1, 2, 3, 4, 5],
       [2, 4, 5, 3, 1],
       [3, 5, 2, 1, 4],
       [4, 3, 1, 5, 2],
       [5, 1, 4, 2, 3]])

l1 = pyLHD.LqDistance(sample,q=1)
l1.pairwise()

array([10., 10., 10., 10., 10., 10., 10., 10., 10., 10.])

l1.design()

10.0