def exercise_is_redundant():
edge_list = double_banana_edge_list
for n in [2,3]:
perm = [0,1,2]
while True:
el = [edge_list[i] for i in perm]
rrfrm = row_reduced_float_rigidity_matrix(
n_dim=3, n_vertices=3, edge_list=el[:n])
if (n == 2):
assert rrfrm.dof() == 7
else:
assert rrfrm.dof() == 6
for edge in el[:n]:
assert rrfrm.is_redundant(edge=edge)
if (n == 2):
assert not rrfrm.is_redundant(el[2])
update_repeat_log(nr=rrfrm.repeats)
if (not next_permutation(perm)):
break
#
rrfrm = row_reduced_float_rigidity_matrix(
n_dim=3, n_vertices=5, edge_list=edge_list[:9])
assert rrfrm.dof() == 6
for edge in edge_list[:9]:
assert rrfrm.is_redundant(edge=edge)
update_repeat_log(nr=rrfrm.repeats)
#
rrfrm = row_reduced_float_rigidity_matrix(
n_dim=3, n_vertices=8, edge_list=edge_list)
assert rrfrm.dof() == 7
for edge in edge_list:
assert rrfrm.is_redundant(edge=edge)
assert rrfrm.is_redundant(edge=(3,4))
for i in [0,1,2]:
for j in [5,6,7]:
assert not rrfrm.is_redundant(edge=(i,j))
update_repeat_log(nr=rrfrm.repeats)
#
edge_list = [ # octahedron
(0,1),(0,2),(0,3),(0,4),
(1,2),(2,3),(3,4),(4,1),
(5,1),(5,2),(5,3),(5,4)]
rrfrm = row_reduced_float_rigidity_matrix(
n_dim=3, n_vertices=6, edge_list=edge_list)
assert rrfrm.dof() == 6
assert rrfrm.is_redundant(edge=(0,5))
assert rrfrm.is_redundant(edge=(1,3))
assert rrfrm.is_redundant(edge=(1,3))
update_repeat_log(nr=rrfrm.repeats)
#
del edge_list[3]
rrfrm = row_reduced_float_rigidity_matrix(
n_dim=3, n_vertices=6, edge_list=edge_list)
assert rrfrm.dof() == 7
assert not rrfrm.is_redundant(edge=(0,5))
assert not rrfrm.is_redundant(edge=(1,3))
assert not rrfrm.is_redundant(edge=(1,3))
update_repeat_log(nr=rrfrm.repeats)
def exercise_next_permutation():
from libtbx.math_utils import next_permutation
seq = []
assert next_permutation(seq) is False
seq = [0]
assert next_permutation(seq) is False
seq = [0, 1]
assert next_permutation(seq)
assert seq == [1, 0]
assert not next_permutation(seq)
assert seq == [0, 1]
seq = [0, 1, 2]
result = []
while True:
result.append(tuple(seq))
if (not next_permutation(seq)):
break
assert result == [(0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1),
(2, 1, 0)]
assert seq == [0, 1, 2]
expected_n = 1
for m in xrange(1, 7):
expected_n *= m
seq = range(m)
n = 0
while True:
n += 1
if (not next_permutation(seq)):
break
assert seq == range(m)
assert n == expected_n
def exercise_next_permutation():
from libtbx.math_utils import next_permutation
seq = []
assert next_permutation(seq) is False
seq = [0]
assert next_permutation(seq) is False
seq = [0,1]
assert next_permutation(seq)
assert seq == [1, 0]
assert not next_permutation(seq)
assert seq == [0, 1]
seq = [0,1,2]
result = []
while True:
result.append(tuple(seq))
if (not next_permutation(seq)):
break
assert result == [
(0, 1, 2),
(0, 2, 1),
(1, 0, 2),
(1, 2, 0),
(2, 0, 1),
(2, 1, 0)]
assert seq == [0,1,2]
expected_n = 1
for m in xrange(1,7):
expected_n *= m
seq = range(m)
n = 0
while True:
n += 1
if (not next_permutation(seq)):
break
assert seq == range(m)
assert n == expected_n
