1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
|
import numpy as np
from ase.cell import Cell
def test_bravais_eps():
# This tests a BCT cell which would be mischaracterized as MCLC
# depending on comparson's precision (fix: c432fd52ecfdca).
# The cell should actually be MCLC for small tolerances,
# and BCT with larger ones. But it would always come out MCLC.
#
# The solution is that the Niggli reduction must run with a more
# coarse precision than the lattice recognition algorithm.
#
# Danger: Since the two mechanisms (Niggli, lattice recognition)
# define their precisions differently, it is not certain whether this
# problem is entirely gone.
cellpar = np.array([3.42864, 3.42864, 3.42864, 125.788, 125.788, 80.236])
cell = Cell.new(cellpar)
mclc = cell.get_bravais_lattice(eps=1e-4)
bct = cell.get_bravais_lattice(eps=1e-3)
print(mclc)
print(bct)
assert mclc.name == 'MCLC'
assert bct.name == 'BCT'
# Original cell is not perfect (rounding).
perfect_bct_cell = bct.tocell()
# perfect_bct_cellpar = bct.cellpar()
assert perfect_bct_cell.get_bravais_lattice().name == 'BCT'
def test_mclc_eps():
a = 6.41
c = 5.87
alpha = 76.7
beta = 103.3
gamma = 152.2
# This lattice maps to something within tolerance of an MCLC lattice,
# but the candidate was discarded due to being unconventional (b > c).
# This test verifies that the problem was fixed.
#
# There could be similar problems for other kinds of lattice. It
# could perhaps cause the algorithm to find MCL/MCLC/TRI lattices
# with higher-than-necessary orthogonality defect if some
# candidates are discarded for this reason.
cell = Cell.new([a, a, c, alpha, beta, gamma])
lat = cell.get_bravais_lattice(eps=1e-2)
print(lat)
assert lat.name == 'MCLC'
|
