Lattice reduction in three dimensions
Author glwhart
2 Stars
Updated Last
11 Months Ago
Started In
March 2021


Stable Runtests codecov

Reduces a basis for a three-dimensional lattice to the basis with the shortest possible basis vectors. Equivalently, the code finds the basis that is as close to orthogonal as possible. This is known as Minkowski reduction. (See Geometrie der Zahlen Hermann Minkowski 1910). In higher dimensions, the famous LLL lattice reduction is commonly used. Lattice reduction is NP-complete in d > 4 dimensions, but a polynomial time algorithm exists for three and four dimensions. (See Phong Nguyen and Damien Stehlé, "Low-Dimensional Lattice Basis Reduction Revisited ") The implementation in this repo for three dimensions (and two dimensions) was inspired by this work.

This code is useful in Density Functional Theory calculations to convert a crystal structure to a compact basis best suited for accurate calculations. It is also useful for reducing a lattice prior to calculating the pointgroup because the symmetry of a reduced lattice can be found in a fixed (relatively small) number of steps. (See pointgroup and spacegroup calculator Spacey.jl)

Example 1: Reduce a slightly skew basis for a simple cubic basis

Consider a simple cubic lattice. The "natural" basis is just the standard basis: (1,0,0), (0,1,0), and (0,0,1) and any equivalent (but skew) basis should reduce to this one. As an example, take this basis: (1,0,0), (0,1,0), and (1,1,1). The first two are orthogonal but the third one is not. Reduce it with the minkReduce function.

julia> a1 = [1.,0.,0.]
julia> a2 = [0.,1.,0.]
julia> a3 = [1.,1.,1.]
julia> minkReduce(a1,a2,a3)
([1.0, 0.0, 0.0], [0.0, 1.0, 0.0], [0.0, 0.0, 1.0])

The output is the standard basis, as expected.

Example 2: Reduce a skew basis for a hexagonal lattice

julia> a1 = [1.,0.,0.]
julia> a2 = [1.5,√3/2,0.]
julia> a3 = [0,0.,√(8/3)]
julia> minkReduce(a1,a2,a3)
([0.5, 0.8660254037844386, 0.0], [0.5, -0.8660254037844386, 0.0], [0.0, 0.0, 1.632993161855452])

Example 3: Reduce a horribly skew basis

julia> bigM = DeviousMat(26) # Matrix whose columns are an extremely skew basis for a simple cubic lattice
3×3 Array{Int64,2}:
  292755045568446  -214311567528244   292755045568445
 -214311567528244   156886956080403  -214311567528244
  292755045568445  -214311567528244   292755045568446
julia> U,V,W = mapslices(x->[x], bigM, dims=2) # Grab columns for input to Minkowski reduction routine
3×1 Array{Array{Int64,1},2}:
 [292755045568446, -214311567528244, 292755045568445]
 [-214311567528244, 156886956080403, -214311567528244]
 [292755045568445, -214311567528244, 292755045568446]
  julia> u,v,w = minkReduce(U,V,W) # Reduce the basis and store in new variables
  ([-1.0, 0.0, 0.0], [0.0, 0.0, -1.0], [0.0, -1.0, 0.0], 15)

The output is the standard basis again (modulo negative signs and ordering). The fourth return value (an integer) is the number of steps required to reduce the basis. In this extreme case, 15 steps are required.

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