Python Permutation.rmul方法代码示例

# 需要导入模块: from sympy.combinatorics.permutations import Permutation [as 别名]
# 或者: from sympy.combinatorics.permutations.Permutation import rmul [as 别名]
def test_coset_factor():
    a = Permutation([0, 2, 1])
    G = PermutationGroup([a])
    c = Permutation([2, 1, 0])
    assert not G.coset_factor(c)
    assert G.coset_rank(c) is None

    a = Permutation([2, 0, 1, 3, 4, 5])
    b = Permutation([2, 1, 3, 4, 5, 0])
    g = PermutationGroup([a, b])
    assert g.order() == 360
    d = Permutation([1, 0, 2, 3, 4, 5])
    assert not g.coset_factor(d.array_form)
    assert not g.contains(d)
    assert Permutation(2) in G
    c = Permutation([1, 0, 2, 3, 5, 4])
    v = g.coset_factor(c, True)
    tr = g.basic_transversals
    p = Permutation.rmul(*[tr[i][v[i]] for i in range(len(g.base))])
    assert p == c
    v = g.coset_factor(c)
    p = Permutation.rmul(*v)
    assert p == c
    assert g.contains(c)
    G = PermutationGroup([Permutation([2, 1, 0])])
    p = Permutation([1, 0, 2])
    assert G.coset_factor(p) == []

# 需要导入模块: from sympy.combinatorics.permutations import Permutation [as 别名]
# 或者: from sympy.combinatorics.permutations.Permutation import rmul [as 别名]
def test_mul():
    a, b = [0, 2, 1, 3], [0, 1, 3, 2]
    assert _af_rmul(a, b) == [0, 2, 3, 1]
    assert _af_rmuln(a, b, range(4)) == [0, 2, 3, 1]
    assert rmul(Permutation(a), Permutation(b)).array_form == [0, 2, 3, 1]

    a = Permutation([0, 2, 1, 3])
    b = (0, 1, 3, 2)
    c = (3, 1, 2, 0)
    assert Permutation.rmul(a, b, c) == Permutation([1, 2, 3, 0])
    assert Permutation.rmul(a, c) == Permutation([3, 2, 1, 0])
    raises(TypeError, lambda: Permutation.rmul(b, c))

    n = 6
    m = 8
    a = [Permutation.unrank_nonlex(n, i).array_form for i in range(m)]
    h = range(n)
    for i in range(m):
        h = _af_rmul(h, a[i])
        h2 = _af_rmuln(*a[:i + 1])
        assert h == h2

# 需要导入模块: from sympy.combinatorics.permutations import Permutation [as 别名]
# 或者: from sympy.combinatorics.permutations.Permutation import rmul [as 别名]
def test_Permutation():
    # don't auto fill 0
    raises(ValueError, lambda: Permutation([1]))
    p = Permutation([0, 1, 2, 3])
    # call as bijective
    assert [p(i) for i in range(p.size)] == list(p)
    # call as operator
    assert p(range(p.size)) == list(p)
    # call as function
    assert list(p(1, 2)) == [0, 2, 1, 3]
    # conversion to list
    assert list(p) == range(4)
    # cycle form with size
    assert Permutation([[1, 2]], size=4) == Permutation([[1, 2], [0], [3]])
    # random generation
    assert Permutation.random(2) in (Permutation([1, 0]), Permutation([0, 1]))

    p = Permutation([2, 5, 1, 6, 3, 0, 4])
    q = Permutation([[1], [0, 3, 5, 6, 2, 4]])
    assert len(set([p, p])) == 1
    r = Permutation([1, 3, 2, 0, 4, 6, 5])
    ans = Permutation(_af_rmuln(*[w.array_form for w in (p, q, r)])).array_form
    assert rmul(p, q, r).array_form == ans
    # make sure no other permutation of p, q, r could have given
    # that answer
    for a, b, c in permutations((p, q, r)):
        if (a, b, c) == (p, q, r):
            continue
        assert rmul(a, b, c).array_form != ans

    assert p.support() == range(7)
    assert q.support() == [0, 2, 3, 4, 5, 6]
    assert Permutation(p.cyclic_form).array_form == p.array_form
    assert p.cardinality == 5040
    assert q.cardinality == 5040
    assert q.cycles == 2
    assert rmul(q, p) == Permutation([4, 6, 1, 2, 5, 3, 0])
    assert rmul(p, q) == Permutation([6, 5, 3, 0, 2, 4, 1])
    assert _af_rmul(p.array_form, q.array_form) == \
        [6, 5, 3, 0, 2, 4, 1]

    assert rmul(Permutation([[1, 2, 3], [0, 4]]),
                Permutation([[1, 2, 4], [0], [3]])).cyclic_form == \
        [[0, 4, 2], [1, 3]]
    assert q.array_form == [3, 1, 4, 5, 0, 6, 2]
    assert q.cyclic_form == [[0, 3, 5, 6, 2, 4]]
    assert q.full_cyclic_form == [[0, 3, 5, 6, 2, 4], [1]]
    assert p.cyclic_form == [[0, 2, 1, 5], [3, 6, 4]]
    t = p.transpositions()
    assert t == [(0, 5), (0, 1), (0, 2), (3, 4), (3, 6)]
    assert Permutation.rmul(*[Permutation(Cycle(*ti)) for ti in (t)])
    assert Permutation([1, 0]).transpositions() == [(0, 1)]

    assert p**13 == p
    assert q**0 == Permutation(range(q.size))
    assert q**-2 == ~q**2
    assert q**2 == Permutation([5, 1, 0, 6, 3, 2, 4])
    assert q**3 == q**2*q
    assert q**4 == q**2*q**2

    a = Permutation(1, 3)
    b = Permutation(2, 0, 3)
    I = Permutation(3)
    assert ~a == a**-1
    assert a*~a == I
    assert a*b**-1 == a*~b

    ans = Permutation(0, 5, 3, 1, 6)(2, 4)
    assert (p + q.rank()).rank() == ans.rank()
    assert (p + q.rank())._rank == ans.rank()
    assert (q + p.rank()).rank() == ans.rank()
    raises(TypeError, lambda: p + Permutation(range(10)))

    assert (p - q.rank()).rank() == Permutation(0, 6, 3, 1, 2, 5, 4).rank()
    assert p.rank() - q.rank() < 0  # for coverage: make sure mod is used
    assert (q - p.rank()).rank() == Permutation(1, 4, 6, 2)(3, 5).rank()

    assert p*q == Permutation(_af_rmuln(*[list(w) for w in (q, p)]))
    assert p*Permutation([]) == p
    assert Permutation([])*p == p
    assert p*Permutation([[0, 1]]) == Permutation([2, 5, 0, 6, 3, 1, 4])
    assert Permutation([[0, 1]])*p == Permutation([5, 2, 1, 6, 3, 0, 4])

    pq = p^q
    assert pq == Permutation([5, 6, 0, 4, 1, 2, 3])
    assert pq == rmul(q, p, ~q)
    qp = q^p
    assert qp == Permutation([4, 3, 6, 2, 1, 5, 0])
    assert qp == rmul(p, q, ~p)
    raises(ValueError, lambda: p^Permutation([]))

    assert p.commutator(q) == Permutation(0, 1, 3, 4, 6, 5, 2)
    assert q.commutator(p) == Permutation(0, 2, 5, 6, 4, 3, 1)
    assert p.commutator(q) == ~q.commutator(p)
    raises(ValueError, lambda: p.commutator(Permutation([])))

    assert len(p.atoms()) == 7
    assert q.atoms() == set([0, 1, 2, 3, 4, 5, 6])

    assert p.inversion_vector() == [2, 4, 1, 3, 1, 0]
#.........这里部分代码省略.........