本文整理汇总了Python中sympy.combinatorics.permutations.Permutation.josephus方法的典型用法代码示例。如果您正苦于以下问题：Python Permutation.josephus方法的具体用法？Python Permutation.josephus怎么用？Python Permutation.josephus使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类sympy.combinatorics.permutations.Permutation的用法示例。
在下文中一共展示了Permutation.josephus方法的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的Python代码示例。
示例1: test_Permutation
# 需要导入模块: from sympy.combinatorics.permutations import Permutation [as 别名]
# 或者: from sympy.combinatorics.permutations.Permutation import josephus [as 别名]
#.........这里部分代码省略.........
assert Permutation.from_inversion_vector(q.inversion_vector()).array_form\
== q.array_form
raises(ValueError, lambda: Permutation.from_inversion_vector([0, 2]))
assert Permutation([i for i in range(500, -1, -1)]).inversions() == 125250
s = Permutation([0, 4, 1, 3, 2])
assert s.parity() == 0
_ = s.cyclic_form # needed to create a value for _cyclic_form
assert len(s._cyclic_form) != s.size and s.parity() == 0
assert not s.is_odd
assert s.is_even
assert Permutation([0, 1, 4, 3, 2]).parity() == 1
assert _af_parity([0, 4, 1, 3, 2]) == 0
assert _af_parity([0, 1, 4, 3, 2]) == 1
s = Permutation([0])
assert s.is_Singleton
assert Permutation([]).is_Empty
r = Permutation([3, 2, 1, 0])
assert (r**2).is_Identity
assert rmul(~p, p).is_Identity
assert (~p)**13 == Permutation([5, 2, 0, 4, 6, 1, 3])
assert ~(r**2).is_Identity
assert p.max() == 6
assert p.min() == 0
q = Permutation([[6], [5], [0, 1, 2, 3, 4]])
assert q.max() == 4
assert q.min() == 0
p = Permutation([1, 5, 2, 0, 3, 6, 4])
q = Permutation([[1, 2, 3, 5, 6], [0, 4]])
assert p.ascents() == [0, 3, 4]
assert q.ascents() == [1, 2, 4]
assert r.ascents() == []
assert p.descents() == [1, 2, 5]
assert q.descents() == [0, 3, 5]
assert Permutation(r.descents()).is_Identity
assert p.inversions() == 7
# test the merge-sort with a longer permutation
big = list(p) + list(range(p.max() + 1, p.max() + 130))
assert Permutation(big).inversions() == 7
assert p.signature() == -1
assert q.inversions() == 11
assert q.signature() == -1
assert rmul(p, ~p).inversions() == 0
assert rmul(p, ~p).signature() == 1
assert p.order() == 6
assert q.order() == 10
assert (p**(p.order())).is_Identity
assert p.length() == 6
assert q.length() == 7
assert r.length() == 4
assert p.runs() == [[1, 5], [2], [0, 3, 6], [4]]
assert q.runs() == [[4], [2, 3, 5], [0, 6], [1]]
assert r.runs() == [[3], [2], [1], [0]]
assert p.index() == 8
assert q.index() == 8
assert r.index() == 3
assert p.get_precedence_distance(q) == q.get_precedence_distance(p)
assert p.get_adjacency_distance(q) == p.get_adjacency_distance(q)
assert p.get_positional_distance(q) == p.get_positional_distance(q)
p = Permutation([0, 1, 2, 3])
q = Permutation([3, 2, 1, 0])
assert p.get_precedence_distance(q) == 6
assert p.get_adjacency_distance(q) == 3
assert p.get_positional_distance(q) == 8
p = Permutation([0, 3, 1, 2, 4])
q = Permutation.josephus(4, 5, 2)
assert p.get_adjacency_distance(q) == 3
raises(ValueError, lambda: p.get_adjacency_distance(Permutation([])))
raises(ValueError, lambda: p.get_positional_distance(Permutation([])))
raises(ValueError, lambda: p.get_precedence_distance(Permutation([])))
a = [Permutation.unrank_nonlex(4, i) for i in range(5)]
iden = Permutation([0, 1, 2, 3])
for i in range(5):
for j in range(i + 1, 5):
assert a[i].commutes_with(a[j]) == \
(rmul(a[i], a[j]) == rmul(a[j], a[i]))
if a[i].commutes_with(a[j]):
assert a[i].commutator(a[j]) == iden
assert a[j].commutator(a[i]) == iden
a = Permutation(3)
b = Permutation(0, 6, 3)(1, 2)
assert a.cycle_structure == {1: 4}
assert b.cycle_structure == {2: 1, 3: 1, 1: 2}示例2: test_josephus
# 需要导入模块: from sympy.combinatorics.permutations import Permutation [as 别名]
# 或者: from sympy.combinatorics.permutations.Permutation import josephus [as 别名]
def test_josephus():
assert Permutation.josephus(4, 6, 1) == Permutation([3, 1, 0, 2, 5, 4])
assert Permutation.josephus(1, 5, 1).is_Identity