from __future__ import print_function, division

import random
from collections import defaultdict

from sympy.core import Basic
from sympy.core.compatibility import is_sequence, reduce, range, as_int
from sympy.utilities.iterables import (flatten, has_variety, minlex,
    has_dups, runs)
from sympy.polys.polytools import lcm
from sympy.matrices import zeros
from mpmath.libmp.libintmath import ifac


def _af_rmul(a, b):
    """
    Return the product b*a; input and output are array forms. The ith value
    is a[b[i]].

    Examples
    ========

    >>> from sympy.combinatorics.permutations import _af_rmul, Permutation
    >>> Permutation.print_cyclic = False

    >>> a, b = [1, 0, 2], [0, 2, 1]
    >>> _af_rmul(a, b)
    [1, 2, 0]
    >>> [a[b[i]] for i in range(3)]
    [1, 2, 0]

    This handles the operands in reverse order compared to the ``*`` operator:

    >>> a = Permutation(a)
    >>> b = Permutation(b)
    >>> list(a*b)
    [2, 0, 1]
    >>> [b(a(i)) for i in range(3)]
    [2, 0, 1]

    See Also
    ========
    rmul, _af_rmuln
    """
    return [a[i] for i in b]


def _af_rmuln(*abc):
    """
    Given [a, b, c, ...] return the product of ...*c*b*a using array forms.
    The ith value is a[b[c[i]]].

    Examples
    ========

    >>> from sympy.combinatorics.permutations import _af_rmul, Permutation
    >>> Permutation.print_cyclic = False

    >>> a, b = [1, 0, 2], [0, 2, 1]
    >>> _af_rmul(a, b)
    [1, 2, 0]
    >>> [a[b[i]] for i in range(3)]
    [1, 2, 0]

    This handles the operands in reverse order compared to the ``*`` operator:

    >>> a = Permutation(a); b = Permutation(b)
    >>> list(a*b)
    [2, 0, 1]
    >>> [b(a(i)) for i in range(3)]
    [2, 0, 1]

    See Also
    ========
    rmul, _af_rmul
    """
    a = abc
    m = len(a)
    if m == 3:
        p0, p1, p2 = a
        return [p0[p1[i]] for i in p2]
    if m == 4:
        p0, p1, p2, p3 = a
        return [p0[p1[p2[i]]] for i in p3]
    if m == 5:
        p0, p1, p2, p3, p4 = a
        return [p0[p1[p2[p3[i]]]] for i in p4]
    if m == 6:
        p0, p1, p2, p3, p4, p5 = a
        return [p0[p1[p2[p3[p4[i]]]]] for i in p5]
    if m == 7:
        p0, p1, p2, p3, p4, p5, p6 = a
        return [p0[p1[p2[p3[p4[p5[i]]]]]] for i in p6]
    if m == 8:
        p0, p1, p2, p3, p4, p5, p6, p7 = a
        return [p0[p1[p2[p3[p4[p5[p6[i]]]]]]] for i in p7]
    if m == 1:
        return a[0][:]
    if m == 2:
        a, b = a
        return [a[i] for i in b]
    if m == 0:
        raise ValueError("String must not be empty")
    p0 = _af_rmuln(*a[:m//2])
    p1 = _af_rmuln(*a[m//2:])
    return [p0[i] for i in p1]


def _af_parity(pi):
    """
    Computes the parity of a permutation in array form.

    The parity of a permutation reflects the parity of the
    number of inversions in the permutation, i.e., the
    number of pairs of x and y such that x > y but p[x] < p[y].

    Examples
    ========

    >>> from sympy.combinatorics.permutations import _af_parity
    >>> _af_parity([0, 1, 2, 3])
    0
    >>> _af_parity([3, 2, 0, 1])
    1

    See Also
    ========

    Permutation
    """
    n = len(pi)
    a = [0] * n
    c = 0
    for j in range(n):
        if a[j] == 0:
            c += 1
            a[j] = 1
            i = j
            while pi[i] != j:
                i = pi[i]
                a[i] = 1
    return (n - c) % 2


def _af_invert(a):
    """
    Finds the inverse, ~A, of a permutation, A, given in array form.

    Examples
    ========

    >>> from sympy.combinatorics.permutations import _af_invert, _af_rmul
    >>> A = [1, 2, 0, 3]
    >>> _af_invert(A)
    [2, 0, 1, 3]
    >>> _af_rmul(_, A)
    [0, 1, 2, 3]

    See Also
    ========

    Permutation, __invert__
    """
    inv_form = [0] * len(a)
    for i, ai in enumerate(a):
        inv_form[ai] = i
    return inv_form


def _af_pow(a, n):
    """
    Routine for finding powers of a permutation.

    Examples
    ========

    >>> from sympy.combinatorics.permutations import Permutation, _af_pow
    >>> Permutation.print_cyclic = False
    >>> p = Permutation([2, 0, 3, 1])
    >>> p.order()
    4
    >>> _af_pow(p._array_form, 4)
    [0, 1, 2, 3]
    """
    if n == 0:
        return list(range(len(a)))
    if n < 0:
        return _af_pow(_af_invert(a), -n)
    if n == 1:
        return a[:]
    elif n == 2:
        b = [a[i] for i in a]
    elif n == 3:
        b = [a[a[i]] for i in a]
    elif n == 4:
        b = [a[a[a[i]]] for i in a]
    else:
        # use binary multiplication
        b = list(range(len(a)))
        while 1:
            if n & 1:
                b = [b[i] for i in a]
                n -= 1
                if not n:
                    break
            if n % 4 == 0:
                a = [a[a[a[i]]] for i in a]
                n = n // 4
            elif n % 2 == 0:
                a = [a[i] for i in a]
                n = n // 2
    return b


def _af_commutes_with(a, b):
    """
    Checks if the two permutations with array forms
    given by ``a`` and ``b`` commute.

    Examples
    ========

    >>> from sympy.combinatorics.permutations import _af_commutes_with
    >>> _af_commutes_with([1, 2, 0], [0, 2, 1])
    False

    See Also
    ========

    Permutation, commutes_with
    """
    return not any(a[b[i]] != b[a[i]] for i in range(len(a) - 1))


class Cycle(dict):
    """
    Wrapper around dict which provides the functionality of a disjoint cycle.

    A cycle shows the rule to use to move subsets of elements to obtain
    a permutation. The Cycle class is more flexible than Permutation in
    that 1) all elements need not be present in order to investigate how
    multiple cycles act in sequence and 2) it can contain singletons:

    >>> from sympy.combinatorics.permutations import Perm, Cycle

    A Cycle will automatically parse a cycle given as a tuple on the rhs:

    >>> Cycle(1, 2)(2, 3)
    (1 3 2)

    The identity cycle, Cycle(), can be used to start a product:

    >>> Cycle()(1, 2)(2, 3)
    (1 3 2)

    The array form of a Cycle can be obtained by calling the list
    method (or passing it to the list function) and all elements from
    0 will be shown:

    >>> a = Cycle(1, 2)
    >>> a.list()
    [0, 2, 1]
    >>> list(a)
    [0, 2, 1]

    If a larger (or smaller) range is desired use the list method and
    provide the desired size -- but the Cycle cannot be truncated to
    a size smaller than the largest element that is out of place:

    >>> b = Cycle(2, 4)(1, 2)(3, 1, 4)(1, 3)
    >>> b.list()
    [0, 2, 1, 3, 4]
    >>> b.list(b.size + 1)
    [0, 2, 1, 3, 4, 5]
    >>> b.list(-1)
    [0, 2, 1]

    Singletons are not shown when printing with one exception: the largest
    element is always shown -- as a singleton if necessary:

    >>> Cycle(1, 4, 10)(4, 5)
    (1 5 4 10)
    >>> Cycle(1, 2)(4)(5)(10)
    (1 2)(10)

    The array form can be used to instantiate a Permutation so other
    properties of the permutation can be investigated:

    >>> Perm(Cycle(1, 2)(3, 4).list()).transpositions()
    [(1, 2), (3, 4)]

    Notes
    =====

    The underlying structure of the Cycle is a dictionary and although
    the __iter__ method has been redefined to give the array form of the
    cycle, the underlying dictionary items are still available with the
    such methods as items():

    >>> list(Cycle(1, 2).items())
    [(1, 2), (2, 1)]

    See Also
    ========

    Permutation
    """
    def __missing__(self, arg):
        """Enter arg into dictionary and return arg."""
        arg = as_int(arg)
        self[arg] = arg
        return arg

    def __iter__(self):
        for i in self.list():
            yield i

    def __call__(self, *other):
        """Return product of cycles processed from R to L.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Cycle as C
        >>> from sympy.combinatorics.permutations import Permutation as Perm
        >>> C(1, 2)(2, 3)
        (1 3 2)

        An instance of a Cycle will automatically parse list-like
        objects and Permutations that are on the right. It is more
        flexible than the Permutation in that all elements need not
        be present:

        >>> a = C(1, 2)
        >>> a(2, 3)
        (1 3 2)
        >>> a(2, 3)(4, 5)
        (1 3 2)(4 5)

        """
        rv = Cycle(*other)
        for k, v in zip(list(self.keys()), [rv[self[k]] for k in self.keys()]):
            rv[k] = v
        return rv

    def list(self, size=None):
        """Return the cycles as an explicit list starting from 0 up
        to the greater of the largest value in the cycles and size.

        Truncation of trailing unmoved items will occur when size
        is less than the maximum element in the cycle; if this is
        desired, setting ``size=-1`` will guarantee such trimming.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Cycle
        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> p = Cycle(2, 3)(4, 5)
        >>> p.list()
        [0, 1, 3, 2, 5, 4]
        >>> p.list(10)
        [0, 1, 3, 2, 5, 4, 6, 7, 8, 9]

        Passing a length too small will trim trailing, unchanged elements
        in the permutation:

        >>> Cycle(2, 4)(1, 2, 4).list(-1)
        [0, 2, 1]
        """
        if not self and size is None:
            raise ValueError('must give size for empty Cycle')
        if size is not None:
            big = max([i for i in self.keys() if self[i] != i] + [0])
            size = max(size, big + 1)
        else:
            size = self.size
        return [self[i] for i in range(size)]

    def __repr__(self):
        """We want it to print as a Cycle, not as a dict.

        Examples
        ========

        >>> from sympy.combinatorics import Cycle
        >>> Cycle(1, 2)
        (1 2)
        >>> print(_)
        (1 2)
        >>> list(Cycle(1, 2).items())
        [(1, 2), (2, 1)]
        """
        if not self:
            return 'Cycle()'
        cycles = Permutation(self).cyclic_form
        s = ''.join(str(tuple(c)) for c in cycles)
        big = self.size - 1
        if not any(i == big for c in cycles for i in c):
            s += '(%s)' % big
        return 'Cycle%s' % s

    def __str__(self):
        """We want it to be printed in a Cycle notation with no
        comma in-between.

        Examples
        ========

        >>> from sympy.combinatorics import Cycle
        >>> Cycle(1, 2)
        (1 2)
        >>> Cycle(1, 2, 4)(5, 6)
        (1 2 4)(5 6)
        """
        if not self:
            return '()'
        cycles = Permutation(self).cyclic_form
        s = ''.join(str(tuple(c)) for c in cycles)
        big = self.size - 1
        if not any(i == big for c in cycles for i in c):
            s += '(%s)' % big
        s = s.replace(',', '')
        return s

    def __init__(self, *args):
        """Load up a Cycle instance with the values for the cycle.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Cycle
        >>> Cycle(1, 2, 6)
        (1 2 6)
        """

        if not args:
            return
        if len(args) == 1:
            if isinstance(args[0], Permutation):
                for c in args[0].cyclic_form:
                    self.update(self(*c))
                return
            elif isinstance(args[0], Cycle):
                for k, v in args[0].items():
                    self[k] = v
                return
        args = [as_int(a) for a in args]
        if any(i < 0 for i in args):
            raise ValueError('negative integers are not allowed in a cycle.')
        if has_dups(args):
            raise ValueError('All elements must be unique in a cycle.')
        for i in range(-len(args), 0):
            self[args[i]] = args[i + 1]

    @property
    def size(self):
        if not self:
            return 0
        return max(self.keys()) + 1

    def copy(self):
        return Cycle(self)


class Permutation(Basic):
    """
    A permutation, alternatively known as an 'arrangement number' or 'ordering'
    is an arrangement of the elements of an ordered list into a one-to-one
    mapping with itself. The permutation of a given arrangement is given by
    indicating the positions of the elements after re-arrangement [2]_. For
    example, if one started with elements [x, y, a, b] (in that order) and
    they were reordered as [x, y, b, a] then the permutation would be
    [0, 1, 3, 2]. Notice that (in SymPy) the first element is always referred
    to as 0 and the permutation uses the indices of the elements in the
    original ordering, not the elements (a, b, etc...) themselves.

    >>> from sympy.combinatorics import Permutation
    >>> Permutation.print_cyclic = False

    Permutations Notation
    =====================

    Permutations are commonly represented in disjoint cycle or array forms.

    Array Notation and 2-line Form
    ------------------------------------

    In the 2-line form, the elements and their final positions are shown
    as a matrix with 2 rows:

    [0    1    2     ... n-1]
    [p(0) p(1) p(2)  ... p(n-1)]

    Since the first line is always range(n), where n is the size of p,
    it is sufficient to represent the permutation by the second line,
    referred to as the "array form" of the permutation. This is entered
    in brackets as the argument to the Permutation class:

    >>> p = Permutation([0, 2, 1]); p
    Permutation([0, 2, 1])

    Given i in range(p.size), the permutation maps i to i^p

    >>> [i^p for i in range(p.size)]
    [0, 2, 1]

    The composite of two permutations p*q means first apply p, then q, so
    i^(p*q) = (i^p)^q which is i^p^q according to Python precedence rules:

    >>> q = Permutation([2, 1, 0])
    >>> [i^p^q for i in range(3)]
    [2, 0, 1]
    >>> [i^(p*q) for i in range(3)]
    [2, 0, 1]

    One can use also the notation p(i) = i^p, but then the composition
    rule is (p*q)(i) = q(p(i)), not p(q(i)):

    >>> [(p*q)(i) for i in range(p.size)]
    [2, 0, 1]
    >>> [q(p(i)) for i in range(p.size)]
    [2, 0, 1]
    >>> [p(q(i)) for i in range(p.size)]
    [1, 2, 0]

    Disjoint Cycle Notation
    -----------------------

    In disjoint cycle notation, only the elements that have shifted are
    indicated. In the above case, the 2 and 1 switched places. This can
    be entered in two ways:

    >>> Permutation(1, 2) == Permutation([[1, 2]]) == p
    True

    Only the relative ordering of elements in a cycle matter:

    >>> Permutation(1,2,3) == Permutation(2,3,1) == Permutation(3,1,2)
    True

    The disjoint cycle notation is convenient when representing
    permutations that have several cycles in them:

    >>> Permutation(1, 2)(3, 5) == Permutation([[1, 2], [3, 5]])
    True

    It also provides some economy in entry when computing products of
    permutations that are written in disjoint cycle notation:

    >>> Permutation(1, 2)(1, 3)(2, 3)
    Permutation([0, 3, 2, 1])
    >>> _ == Permutation([[1, 2]])*Permutation([[1, 3]])*Permutation([[2, 3]])
    True

        Caution: when the cycles have common elements
        between them then the order in which the
        permutations are applied matters. The
        convention is that the permutations are
        applied from *right to left*. In the following, the
        transposition of elements 2 and 3 is followed
        by the transposition of elements 1 and 2:

        >>> Permutation(1, 2)(2, 3) == Permutation([(1, 2), (2, 3)])
        True
        >>> Permutation(1, 2)(2, 3).list()
        [0, 3, 1, 2]

        If the first and second elements had been
        swapped first, followed by the swapping of the second
        and third, the result would have been [0, 2, 3, 1].
        If, for some reason, you want to apply the cycles
        in the order they are entered, you can simply reverse
        the order of cycles:

        >>> Permutation([(1, 2), (2, 3)][::-1]).list()
        [0, 2, 3, 1]

    Entering a singleton in a permutation is a way to indicate the size of the
    permutation. The ``size`` keyword can also be used.

    Array-form entry:

    >>> Permutation([[1, 2], [9]])
    Permutation([0, 2, 1], size=10)
    >>> Permutation([[1, 2]], size=10)
    Permutation([0, 2, 1], size=10)

    Cyclic-form entry:

    >>> Permutation(1, 2, size=10)
    Permutation([0, 2, 1], size=10)
    >>> Permutation(9)(1, 2)
    Permutation([0, 2, 1], size=10)

    Caution: no singleton containing an element larger than the largest
    in any previous cycle can be entered. This is an important difference
    in how Permutation and Cycle handle the __call__ syntax. A singleton
    argument at the start of a Permutation performs instantiation of the
    Permutation and is permitted:

    >>> Permutation(5)
    Permutation([], size=6)

    A singleton entered after instantiation is a call to the permutation
    -- a function call -- and if the argument is out of range it will
    trigger an error. For this reason, it is better to start the cycle
    with the singleton:

    The following fails because there is is no element 3:

    >>> Permutation(1, 2)(3)
    Traceback (most recent call last):
    ...
    IndexError: list index out of range

    This is ok: only the call to an out of range singleton is prohibited;
    otherwise the permutation autosizes:

    >>> Permutation(3)(1, 2)
    Permutation([0, 2, 1, 3])
    >>> Permutation(1, 2)(3, 4) == Permutation(3, 4)(1, 2)
    True


    Equality testing
    ----------------

    The array forms must be the same in order for permutations to be equal:

    >>> Permutation([1, 0, 2, 3]) == Permutation([1, 0])
    False


    Identity Permutation
    --------------------

    The identity permutation is a permutation in which no element is out of
    place. It can be entered in a variety of ways. All the following create
    an identity permutation of size 4:

    >>> I = Permutation([0, 1, 2, 3])
    >>> all(p == I for p in [
    ... Permutation(3),
    ... Permutation(range(4)),
    ... Permutation([], size=4),
    ... Permutation(size=4)])
    True

    Watch out for entering the range *inside* a set of brackets (which is
    cycle notation):

    >>> I == Permutation([range(4)])
    False


    Permutation Printing
    ====================

    There are a few things to note about how Permutations are printed.

    1) If you prefer one form (array or cycle) over another, you can set that
    with the print_cyclic flag.

    >>> Permutation(1, 2)(4, 5)(3, 4)
    Permutation([0, 2, 1, 4, 5, 3])
    >>> p = _

    >>> Permutation.print_cyclic = True
    >>> p
    (1 2)(3 4 5)
    >>> Permutation.print_cyclic = False

    2) Regardless of the setting, a list of elements in the array for cyclic
    form can be obtained and either of those can be copied and supplied as
    the argument to Permutation:

    >>> p.array_form
    [0, 2, 1, 4, 5, 3]
    >>> p.cyclic_form
    [[1, 2], [3, 4, 5]]
    >>> Permutation(_) == p
    True

    3) Printing is economical in that as little as possible is printed while
    retaining all information about the size of the permutation:

    >>> Permutation([1, 0, 2, 3])
    Permutation([1, 0, 2, 3])
    >>> Permutation([1, 0, 2, 3], size=20)
    Permutation([1, 0], size=20)
    >>> Permutation([1, 0, 2, 4, 3, 5, 6], size=20)
    Permutation([1, 0, 2, 4, 3], size=20)

    >>> p = Permutation([1, 0, 2, 3])
    >>> Permutation.print_cyclic = True
    >>> p
    (3)(0 1)
    >>> Permutation.print_cyclic = False

    The 2 was not printed but it is still there as can be seen with the
    array_form and size methods:

    >>> p.array_form
    [1, 0, 2, 3]
    >>> p.size
    4

    Short introduction to other methods
    ===================================

    The permutation can act as a bijective function, telling what element is
    located at a given position

    >>> q = Permutation([5, 2, 3, 4, 1, 0])
    >>> q.array_form[1] # the hard way
    2
    >>> q(1) # the easy way
    2
    >>> {i: q(i) for i in range(q.size)} # showing the bijection
    {0: 5, 1: 2, 2: 3, 3: 4, 4: 1, 5: 0}

    The full cyclic form (including singletons) can be obtained:

    >>> p.full_cyclic_form
    [[0, 1], [2], [3]]

    Any permutation can be factored into transpositions of pairs of elements:

    >>> Permutation([[1, 2], [3, 4, 5]]).transpositions()
    [(1, 2), (3, 5), (3, 4)]
    >>> Permutation.rmul(*[Permutation([ti], size=6) for ti in _]).cyclic_form
    [[1, 2], [3, 4, 5]]

    The number of permutations on a set of n elements is given by n! and is
    called the cardinality.

    >>> p.size
    4
    >>> p.cardinality
    24

    A given permutation has a rank among all the possible permutations of the
    same elements, but what that rank is depends on how the permutations are
    enumerated. (There are a number of different methods of doing so.) The
    lexicographic rank is given by the rank method and this rank is used to
    increment a permutation with addition/subtraction:

    >>> p.rank()
    6
    >>> p + 1
    Permutation([1, 0, 3, 2])
    >>> p.next_lex()
    Permutation([1, 0, 3, 2])
    >>> _.rank()
    7
    >>> p.unrank_lex(p.size, rank=7)
    Permutation([1, 0, 3, 2])

    The product of two permutations p and q is defined as their composition as
    functions, (p*q)(i) = q(p(i)) [6]_.

    >>> p = Permutation([1, 0, 2, 3])
    >>> q = Permutation([2, 3, 1, 0])
    >>> list(q*p)
    [2, 3, 0, 1]
    >>> list(p*q)
    [3, 2, 1, 0]
    >>> [q(p(i)) for i in range(p.size)]
    [3, 2, 1, 0]

    The permutation can be 'applied' to any list-like object, not only
    Permutations:

    >>> p(['zero', 'one', 'four', 'two'])
     ['one', 'zero', 'four', 'two']
    >>> p('zo42')
    ['o', 'z', '4', '2']

    If you have a list of arbitrary elements, the corresponding permutation
    can be found with the from_sequence method:

    >>> Permutation.from_sequence('SymPy')
    Permutation([1, 3, 2, 0, 4])

    See Also
    ========

    Cycle

    References
    ==========

    .. [1] Skiena, S. 'Permutations.' 1.1 in Implementing Discrete Mathematics
           Combinatorics and Graph Theory with Mathematica.  Reading, MA:
           Addison-Wesley, pp. 3-16, 1990.

    .. [2] Knuth, D. E. The Art of Computer Programming, Vol. 4: Combinatorial
           Algorithms, 1st ed. Reading, MA: Addison-Wesley, 2011.

    .. [3] Wendy Myrvold and Frank Ruskey. 2001. Ranking and unranking
           permutations in linear time. Inf. Process. Lett. 79, 6 (September 2001),
           281-284. DOI=10.1016/S0020-0190(01)00141-7

    .. [4] D. L. Kreher, D. R. Stinson 'Combinatorial Algorithms'
           CRC Press, 1999

    .. [5] Graham, R. L.; Knuth, D. E.; and Patashnik, O.
           Concrete Mathematics: A Foundation for Computer Science, 2nd ed.
           Reading, MA: Addison-Wesley, 1994.

    .. [6] http://en.wikipedia.org/wiki/Permutation#Product_and_inverse

    .. [7] http://en.wikipedia.org/wiki/Lehmer_code

    """

    is_Permutation = True

    _array_form = None
    _cyclic_form = None
    _cycle_structure = None
    _size = None
    _rank = None

    def __new__(cls, *args, **kwargs):
        """
        Constructor for the Permutation object from a list or a
        list of lists in which all elements of the permutation may
        appear only once.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False

        Permutations entered in array-form are left unaltered:

        >>> Permutation([0, 2, 1])
        Permutation([0, 2, 1])

        Permutations entered in cyclic form are converted to array form;
        singletons need not be entered, but can be entered to indicate the
        largest element:

        >>> Permutation([[4, 5, 6], [0, 1]])
        Permutation([1, 0, 2, 3, 5, 6, 4])
        >>> Permutation([[4, 5, 6], [0, 1], [19]])
        Permutation([1, 0, 2, 3, 5, 6, 4], size=20)

        All manipulation of permutations assumes that the smallest element
        is 0 (in keeping with 0-based indexing in Python) so if the 0 is
        missing when entering a permutation in array form, an error will be
        raised:

        >>> Permutation([2, 1])
        Traceback (most recent call last):
        ...
        ValueError: Integers 0 through 2 must be present.

        If a permutation is entered in cyclic form, it can be entered without
        singletons and the ``size`` specified so those values can be filled
        in, otherwise the array form will only extend to the maximum value
        in the cycles:

        >>> Permutation([[1, 4], [3, 5, 2]], size=10)
        Permutation([0, 4, 3, 5, 1, 2], size=10)
        >>> _.array_form
        [0, 4, 3, 5, 1, 2, 6, 7, 8, 9]
        """
        size = kwargs.pop('size', None)
        if size is not None:
            size = int(size)

        #a) ()
        #b) (1) = identity
        #c) (1, 2) = cycle
        #d) ([1, 2, 3]) = array form
        #e) ([[1, 2]]) = cyclic form
        #f) (Cycle) = conversion to permutation
        #g) (Permutation) = adjust size or return copy
        ok = True
        if not args:  # a
            return cls._af_new(list(range(size or 0)))
        elif len(args) > 1:  # c
            return cls._af_new(Cycle(*args).list(size))
        if len(args) == 1:
            a = args[0]
            if isinstance(a, cls):  # g
                if size is None or size == a.size:
                    return a
                return cls(a.array_form, size=size)
            if isinstance(a, Cycle):  # f
                return cls._af_new(a.list(size))
            if not is_sequence(a):  # b
                return cls._af_new(list(range(a + 1)))
            if has_variety(is_sequence(ai) for ai in a):
                ok = False
        else:
            ok = False
        if not ok:
            raise ValueError("Permutation argument must be a list of ints, "
                             "a list of lists, Permutation or Cycle.")

        # safe to assume args are valid; this also makes a copy
        # of the args
        args = list(args[0])

        is_cycle = args and is_sequence(args[0])
        if is_cycle:  # e
            args = [[int(i) for i in c] for c in args]
        else:  # d
            args = [int(i) for i in args]

        # if there are n elements present, 0, 1, ..., n-1 should be present
        # unless a cycle notation has been provided. A 0 will be added
        # for convenience in case one wants to enter permutations where
        # counting starts from 1.

        temp = flatten(args)
        if has_dups(temp) and not is_cycle:
            raise ValueError('there were repeated elements.')
        temp = set(temp)

        if not is_cycle and \
                any(i not in temp for i in range(len(temp))):
            raise ValueError("Integers 0 through %s must be present." %
                             max(temp))

        if is_cycle:
            # it's not necessarily canonical so we won't store
            # it -- use the array form instead
            c = Cycle()
            for ci in args:
                c = c(*ci)
            aform = c.list()
        else:
            aform = list(args)
        if size and size > len(aform):
            # don't allow for truncation of permutation which
            # might split a cycle and lead to an invalid aform
            # but do allow the permutation size to be increased
            aform.extend(list(range(len(aform), size)))

        return cls._af_new(aform)

    @classmethod
    def _af_new(cls, perm):
        """A method to produce a Permutation object from a list;
        the list is bound to the _array_form attribute, so it must
        not be modified; this method is meant for internal use only;
        the list ``a`` is supposed to be generated as a temporary value
        in a method, so p = Perm._af_new(a) is the only object
        to hold a reference to ``a``::

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Perm
        >>> Perm.print_cyclic = False
        >>> a = [2,1,3,0]
        >>> p = Perm._af_new(a)
        >>> p
        Permutation([2, 1, 3, 0])

        """
        p = Basic.__new__(cls, perm)
        p._array_form = perm
        p._size = len(perm)
        return p

    def _hashable_content(self):
        # the array_form (a list) is the Permutation arg, so we need to
        # return a tuple, instead
        return tuple(self.array_form)

    @property
    def array_form(self):
        """
        Return a copy of the attribute _array_form
        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> p = Permutation([[2, 0], [3, 1]])
        >>> p.array_form
        [2, 3, 0, 1]
        >>> Permutation([[2, 0, 3, 1]]).array_form
        [3, 2, 0, 1]
        >>> Permutation([2, 0, 3, 1]).array_form
        [2, 0, 3, 1]
        >>> Permutation([[1, 2], [4, 5]]).array_form
        [0, 2, 1, 3, 5, 4]
        """
        return self._array_form[:]

    def __repr__(self):
        from sympy.combinatorics.permutations import Permutation, Cycle
        if Permutation.print_cyclic:
            if not self.size:
                return 'Permutation()'
            # before taking Cycle notation, see if the last element is
            # a singleton and move it to the head of the string
            s = Cycle(self)(self.size - 1).__repr__()[len('Cycle'):]
            last = s.rfind('(')
            if not last == 0 and ',' not in s[last:]:
                s = s[last:] + s[:last]
            return 'Permutation%s' %s
        else:
            s = self.support()
            if not s:
                if self.size < 5:
                    return 'Permutation(%s)' % str(self.array_form)
                return 'Permutation([], size=%s)' % self.size
            trim = str(self.array_form[:s[-1] + 1]) + ', size=%s' % self.size
            use = full = str(self.array_form)
            if len(trim) < len(full):
                use = trim
            return 'Permutation(%s)' % use

    def list(self, size=None):
        """Return the permutation as an explicit list, possibly
        trimming unmoved elements if size is less than the maximum
        element in the permutation; if this is desired, setting
        ``size=-1`` will guarantee such trimming.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> p = Permutation(2, 3)(4, 5)
        >>> p.list()
        [0, 1, 3, 2, 5, 4]
        >>> p.list(10)
        [0, 1, 3, 2, 5, 4, 6, 7, 8, 9]

        Passing a length too small will trim trailing, unchanged elements
        in the permutation:

        >>> Permutation(2, 4)(1, 2, 4).list(-1)
        [0, 2, 1]
        >>> Permutation(3).list(-1)
        []
        """
        if not self and size is None:
            raise ValueError('must give size for empty Cycle')
        rv = self.array_form
        if size is not None:
            if size > self.size:
                rv.extend(list(range(self.size, size)))
            else:
                # find first value from rhs where rv[i] != i
                i = self.size - 1
                while rv:
                    if rv[-1] != i:
                        break
                    rv.pop()
                    i -= 1
        return rv

    @property
    def cyclic_form(self):
        """
        This is used to convert to the cyclic notation
        from the canonical notation. Singletons are omitted.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> p = Permutation([0, 3, 1, 2])
        >>> p.cyclic_form
        [[1, 3, 2]]
        >>> Permutation([1, 0, 2, 4, 3, 5]).cyclic_form
        [[0, 1], [3, 4]]

        See Also
        ========

        array_form, full_cyclic_form
        """
        if self._cyclic_form is not None:
            return list(self._cyclic_form)
        array_form = self.array_form
        unchecked = [True] * len(array_form)
        cyclic_form = []
        for i in range(len(array_form)):
            if unchecked[i]:
                cycle = []
                cycle.append(i)
                unchecked[i] = False
                j = i
                while unchecked[array_form[j]]:
                    j = array_form[j]
                    cycle.append(j)
                    unchecked[j] = False
                if len(cycle) > 1:
                    cyclic_form.append(cycle)
                    assert cycle == list(minlex(cycle, is_set=True))
        cyclic_form.sort()
        self._cyclic_form = cyclic_form[:]
        return cyclic_form

    @property
    def full_cyclic_form(self):
        """Return permutation in cyclic form including singletons.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation([0, 2, 1]).full_cyclic_form
        [[0], [1, 2]]
        """
        need = set(range(self.size)) - set(flatten(self.cyclic_form))
        rv = self.cyclic_form
        rv.extend([[i] for i in need])
        rv.sort()
        return rv

    @property
    def size(self):
        """
        Returns the number of elements in the permutation.

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> Permutation([[3, 2], [0, 1]]).size
        4

        See Also
        ========

        cardinality, length, order, rank
        """
        return self._size

    def support(self):
        """Return the elements in permutation, P, for which P[i] != i.

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> p = Permutation([[3, 2], [0, 1], [4]])
        >>> p.array_form
        [1, 0, 3, 2, 4]
        >>> p.support()
        [0, 1, 2, 3]
        """
        a = self.array_form
        return [i for i, e in enumerate(a) if a[i] != i]

    def __add__(self, other):
        """Return permutation that is other higher in rank than self.

        The rank is the lexicographical rank, with the identity permutation
        having rank of 0.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> I = Permutation([0, 1, 2, 3])
        >>> a = Permutation([2, 1, 3, 0])
        >>> I + a.rank() == a
        True

        See Also
        ========

        __sub__, inversion_vector

        """
        rank = (self.rank() + other) % self.cardinality
        rv = self.unrank_lex(self.size, rank)
        rv._rank = rank
        return rv

    def __sub__(self, other):
        """Return the permutation that is other lower in rank than self.

        See Also
        ========

        __add__
        """
        return self.__add__(-other)

    @staticmethod
    def rmul(*args):
        """
        Return product of Permutations [a, b, c, ...] as the Permutation whose
        ith value is a(b(c(i))).

        a, b, c, ... can be Permutation objects or tuples.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import _af_rmul, Permutation
        >>> Permutation.print_cyclic = False

        >>> a, b = [1, 0, 2], [0, 2, 1]
        >>> a = Permutation(a); b = Permutation(b)
        >>> list(Permutation.rmul(a, b))
        [1, 2, 0]
        >>> [a(b(i)) for i in range(3)]
        [1, 2, 0]

        This handles the operands in reverse order compared to the ``*`` operator:

        >>> a = Permutation(a); b = Permutation(b)
        >>> list(a*b)
        [2, 0, 1]
        >>> [b(a(i)) for i in range(3)]
        [2, 0, 1]

        Notes
        =====

        All items in the sequence will be parsed by Permutation as
        necessary as long as the first item is a Permutation:

        >>> Permutation.rmul(a, [0, 2, 1]) == Permutation.rmul(a, b)
        True

        The reverse order of arguments will raise a TypeError.

        """
        rv = args[0]
        for i in range(1, len(args)):
            rv = args[i]*rv
        return rv

    @classmethod
    def rmul_with_af(cls, *args):
        """
        same as rmul, but the elements of args are Permutation objects
        which have _array_form
        """
        a = [x._array_form for x in args]
        rv = cls._af_new(_af_rmuln(*a))
        return rv

    def mul_inv(self, other):
        """
        other*~self, self and other have _array_form
        """
        a = _af_invert(self._array_form)
        b = other._array_form
        return self._af_new(_af_rmul(a, b))

    def __rmul__(self, other):
        """This is needed to coerce other to Permutation in rmul."""
        cls = type(self)
        return cls(other)*self

    def __mul__(self, other):
        """
        Return the product a*b as a Permutation; the ith value is b(a(i)).

        Examples
        ========

        >>> from sympy.combinatorics.permutations import _af_rmul, Permutation
        >>> Permutation.print_cyclic = False

        >>> a, b = [1, 0, 2], [0, 2, 1]
        >>> a = Permutation(a); b = Permutation(b)
        >>> list(a*b)
        [2, 0, 1]
        >>> [b(a(i)) for i in range(3)]
        [2, 0, 1]

        This handles operands in reverse order compared to _af_rmul and rmul:

        >>> al = list(a); bl = list(b)
        >>> _af_rmul(al, bl)
        [1, 2, 0]
        >>> [al[bl[i]] for i in range(3)]
        [1, 2, 0]

        It is acceptable for the arrays to have different lengths; the shorter
        one will be padded to match the longer one:

        >>> b*Permutation([1, 0])
        Permutation([1, 2, 0])
        >>> Permutation([1, 0])*b
        Permutation([2, 0, 1])

        It is also acceptable to allow coercion to handle conversion of a
        single list to the left of a Permutation:

        >>> [0, 1]*a # no change: 2-element identity
        Permutation([1, 0, 2])
        >>> [[0, 1]]*a # exchange first two elements
        Permutation([0, 1, 2])

        You cannot use more than 1 cycle notation in a product of cycles
        since coercion can only handle one argument to the left. To handle
        multiple cycles it is convenient to use Cycle instead of Permutation:

        >>> [[1, 2]]*[[2, 3]]*Permutation([]) # doctest: +SKIP
        >>> from sympy.combinatorics.permutations import Cycle
        >>> Cycle(1, 2)(2, 3)
        (1 3 2)

        """
        a = self.array_form
        # __rmul__ makes sure the other is a Permutation
        b = other.array_form
        if not b:
            perm = a
        else:
            b.extend(list(range(len(b), len(a))))
            perm = [b[i] for i in a] + b[len(a):]
        return self._af_new(perm)

    def commutes_with(self, other):
        """
        Checks if the elements are commuting.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> a = Permutation([1, 4, 3, 0, 2, 5])
        >>> b = Permutation([0, 1, 2, 3, 4, 5])
        >>> a.commutes_with(b)
        True
        >>> b = Permutation([2, 3, 5, 4, 1, 0])
        >>> a.commutes_with(b)
        False
        """
        a = self.array_form
        b = other.array_form
        return _af_commutes_with(a, b)

    def __pow__(self, n):
        """
        Routine for finding powers of a permutation.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> p = Permutation([2,0,3,1])
        >>> p.order()
        4
        >>> p**4
        Permutation([0, 1, 2, 3])
        """
        if isinstance(n, Permutation):
            raise NotImplementedError(
                'p**p is not defined; do you mean p^p (conjugate)?')
        n = int(n)
        return self._af_new(_af_pow(self.array_form, n))

    def __rxor__(self, i):
        """Return self(i) when ``i`` is an int.

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> p = Permutation(1, 2, 9)
        >>> 2^p == p(2) == 9
        True
        """
        if int(i) == i:
            return self(i)
        else:
            raise NotImplementedError(
                "i^p = p(i) when i is an integer, not %s." % i)

    def __xor__(self, h):
        """Return the conjugate permutation ``~h*self*h` `.

        If ``a`` and ``b`` are conjugates, ``a = h*b*~h`` and
        ``b = ~h*a*h`` and both have the same cycle structure.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = True
        >>> p = Permutation(1, 2, 9)
        >>> q = Permutation(6, 9, 8)
        >>> p*q != q*p
        True

        Calculate and check properties of the conjugate:

        >>> c = p^q
        >>> c == ~q*p*q and p == q*c*~q
        True

        The expression q^p^r is equivalent to q^(p*r):

        >>> r = Permutation(9)(4, 6, 8)
        >>> q^p^r == q^(p*r)
        True

        If the term to the left of the conjugate operator, i, is an integer
        then this is interpreted as selecting the ith element from the
        permutation to the right:

        >>> all(i^p == p(i) for i in range(p.size))
        True

        Note that the * operator as higher precedence than the ^ operator:

        >>> q^r*p^r == q^(r*p)^r == Permutation(9)(1, 6, 4)
        True

        Notes
        =====

        In Python the precedence rule is p^q^r = (p^q)^r which differs
        in general from p^(q^r)

        >>> q^p^r
        (9)(1 4 8)
        >>> q^(p^r)
        (9)(1 8 6)

        For a given r and p, both of the following are conjugates of p:
        ~r*p*r and r*p*~r. But these are not necessarily the same:

        >>> ~r*p*r == r*p*~r
        True

        >>> p = Permutation(1, 2, 9)(5, 6)
        >>> ~r*p*r == r*p*~r
        False

        The conjugate ~r*p*r was chosen so that ``p^q^r`` would be equivalent
        to ``p^(q*r)`` rather than ``p^(r*q)``. To obtain r*p*~r, pass ~r to
        this method:

        >>> p^~r == r*p*~r
        True
        """

        if self.size != h.size:
            raise ValueError("The permutations must be of equal size.")
        a = [None]*self.size
        h = h._array_form
        p = self._array_form
        for i in range(self.size):
            a[h[i]] = h[p[i]]
        return self._af_new(a)

    def transpositions(self):
        """
        Return the permutation decomposed into a list of transpositions.

        It is always possible to express a permutation as the product of
        transpositions, see [1]

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([[1, 2, 3], [0, 4, 5, 6, 7]])
        >>> t = p.transpositions()
        >>> t
        [(0, 7), (0, 6), (0, 5), (0, 4), (1, 3), (1, 2)]
        >>> print(''.join(str(c) for c in t))
        (0, 7)(0, 6)(0, 5)(0, 4)(1, 3)(1, 2)
        >>> Permutation.rmul(*[Permutation([ti], size=p.size) for ti in t]) == p
        True

        References
        ==========

        1. http://en.wikipedia.org/wiki/Transposition_%28mathematics%29#Properties

        """
        a = self.cyclic_form
        res = []
        for x in a:
            nx = len(x)
            if nx == 2:
                res.append(tuple(x))
            elif nx > 2:
                first = x[0]
                for y in x[nx - 1:0:-1]:
                    res.append((first, y))
        return res

    @classmethod
    def from_sequence(self, i, key=None):
        """Return the permutation needed to obtain ``i`` from the sorted
        elements of ``i``. If custom sorting is desired, a key can be given.

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> Permutation.print_cyclic = True

        >>> Permutation.from_sequence('SymPy')
        (4)(0 1 3)
        >>> _(sorted("SymPy"))
        ['S', 'y', 'm', 'P', 'y']
        >>> Permutation.from_sequence('SymPy', key=lambda x: x.lower())
        (4)(0 2)(1 3)
        """
        ic = list(zip(i, list(range(len(i)))))
        if key:
            ic.sort(key=lambda x: key(x[0]))
        else:
            ic.sort()
        return ~Permutation([i[1] for i in ic])

    def __invert__(self):
        """
        Return the inverse of the permutation.

        A permutation multiplied by its inverse is the identity permutation.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([[2,0], [3,1]])
        >>> ~p
        Permutation([2, 3, 0, 1])
        >>> _ == p**-1
        True
        >>> p*~p == ~p*p == Permutation([0, 1, 2, 3])
        True
        """
        return self._af_new(_af_invert(self._array_form))

    def __iter__(self):
        """Yield elements from array form.

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> list(Permutation(range(3)))
        [0, 1, 2]
        """
        for i in self.array_form:
            yield i

    def __call__(self, *i):
        """
        Allows applying a permutation instance as a bijective function.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([[2, 0], [3, 1]])
        >>> p.array_form
        [2, 3, 0, 1]
        >>> [p(i) for i in range(4)]
        [2, 3, 0, 1]

        If an array is given then the permutation selects the items
        from the array (i.e. the permutation is applied to the array):

        >>> from sympy.abc import x
        >>> p([x, 1, 0, x**2])
        [0, x**2, x, 1]
        """
        # list indices can be Integer or int; leave this
        # as it is (don't test or convert it) because this
        # gets called a lot and should be fast
        if len(i) == 1:
            i = i[0]
            try:
                # P(1)
                return self._array_form[i]
            except TypeError:
                try:
                    # P([a, b, c])
                    return [i[j] for j in self._array_form]
                except Exception:
                    raise TypeError('unrecognized argument')
        else:
            # P(1, 2, 3)
            return self*Permutation(Cycle(*i), size=self.size)

    def atoms(self):
        """
        Returns all the elements of a permutation

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> Permutation([0, 1, 2, 3, 4, 5]).atoms()
        {0, 1, 2, 3, 4, 5}
        >>> Permutation([[0, 1], [2, 3], [4, 5]]).atoms()
        {0, 1, 2, 3, 4, 5}
        """
        return set(self.array_form)

    def next_lex(self):
        """
        Returns the next permutation in lexicographical order.
        If self is the last permutation in lexicographical order
        it returns None.
        See [4] section 2.4.


        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([2, 3, 1, 0])
        >>> p = Permutation([2, 3, 1, 0]); p.rank()
        17
        >>> p = p.next_lex(); p.rank()
        18

        See Also
        ========

        rank, unrank_lex
        """
        perm = self.array_form[:]
        n = len(perm)
        i = n - 2
        while perm[i + 1] < perm[i]:
            i -= 1
        if i == -1:
            return None
        else:
            j = n - 1
            while perm[j] < perm[i]:
                j -= 1
            perm[j], perm[i] = perm[i], perm[j]
            i += 1
            j = n - 1
            while i < j:
                perm[j], perm[i] = perm[i], perm[j]
                i += 1
                j -= 1
        return self._af_new(perm)

    @classmethod
    def unrank_nonlex(self, n, r):
        """
        This is a linear time unranking algorithm that does not
        respect lexicographic order [3].

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> Permutation.unrank_nonlex(4, 5)
        Permutation([2, 0, 3, 1])
        >>> Permutation.unrank_nonlex(4, -1)
        Permutation([0, 1, 2, 3])

        See Also
        ========

        next_nonlex, rank_nonlex
        """
        def _unrank1(n, r, a):
            if n > 0:
                a[n - 1], a[r % n] = a[r % n], a[n - 1]
                _unrank1(n - 1, r//n, a)

        id_perm = list(range(n))
        n = int(n)
        r = r % ifac(n)
        _unrank1(n, r, id_perm)
        return self._af_new(id_perm)

    def rank_nonlex(self, inv_perm=None):
        """
        This is a linear time ranking algorithm that does not
        enforce lexicographic order [3].


        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 2, 3])
        >>> p.rank_nonlex()
        23

        See Also
        ========

        next_nonlex, unrank_nonlex
        """
        def _rank1(n, perm, inv_perm):
            if n == 1:
                return 0
            s = perm[n - 1]
            t = inv_perm[n - 1]
            perm[n - 1], perm[t] = perm[t], s
            inv_perm[n - 1], inv_perm[s] = inv_perm[s], t
            return s + n*_rank1(n - 1, perm, inv_perm)

        if inv_perm is None:
            inv_perm = (~self).array_form
        if not inv_perm:
            return 0
        perm = self.array_form[:]
        r = _rank1(len(perm), perm, inv_perm)
        return r

    def next_nonlex(self):
        """
        Returns the next permutation in nonlex order [3].
        If self is the last permutation in this order it returns None.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> p = Permutation([2, 0, 3, 1]); p.rank_nonlex()
        5
        >>> p = p.next_nonlex(); p
        Permutation([3, 0, 1, 2])
        >>> p.rank_nonlex()
        6

        See Also
        ========

        rank_nonlex, unrank_nonlex
        """
        r = self.rank_nonlex()
        if r == ifac(self.size) - 1:
            return None
        return self.unrank_nonlex(self.size, r + 1)

    def rank(self):
        """
        Returns the lexicographic rank of the permutation.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 2, 3])
        >>> p.rank()
        0
        >>> p = Permutation([3, 2, 1, 0])
        >>> p.rank()
        23

        See Also
        ========

        next_lex, unrank_lex, cardinality, length, order, size
        """
        if not self._rank is None:
            return self._rank
        rank = 0
        rho = self.array_form[:]
        n = self.size - 1
        size = n + 1
        psize = int(ifac(n))
        for j in range(size - 1):
            rank += rho[j]*psize
            for i in range(j + 1, size):
                if rho[i] > rho[j]:
                    rho[i] -= 1
            psize //= n
            n -= 1
        self._rank = rank
        return rank

    @property
    def cardinality(self):
        """
        Returns the number of all possible permutations.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 2, 3])
        >>> p.cardinality
        24

        See Also
        ========

        length, order, rank, size
        """
        return int(ifac(self.size))

    def parity(self):
        """
        Computes the parity of a permutation.

        The parity of a permutation reflects the parity of the
        number of inversions in the permutation, i.e., the
        number of pairs of x and y such that ``x > y`` but ``p[x] < p[y]``.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 2, 3])
        >>> p.parity()
        0
        >>> p = Permutation([3, 2, 0, 1])
        >>> p.parity()
        1

        See Also
        ========

        _af_parity
        """
        if self._cyclic_form is not None:
            return (self.size - self.cycles) % 2

        return _af_parity(self.array_form)

    @property
    def is_even(self):
        """
        Checks if a permutation is even.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 2, 3])
        >>> p.is_even
        True
        >>> p = Permutation([3, 2, 1, 0])
        >>> p.is_even
        True

        See Also
        ========

        is_odd
        """
        return not self.is_odd

    @property
    def is_odd(self):
        """
        Checks if a permutation is odd.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 2, 3])
        >>> p.is_odd
        False
        >>> p = Permutation([3, 2, 0, 1])
        >>> p.is_odd
        True

        See Also
        ========

        is_even
        """
        return bool(self.parity() % 2)

    @property
    def is_Singleton(self):
        """
        Checks to see if the permutation contains only one number and is
        thus the only possible permutation of this set of numbers

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> Permutation([0]).is_Singleton
        True
        >>> Permutation([0, 1]).is_Singleton
        False

        See Also
        ========

        is_Empty
        """
        return self.size == 1

    @property
    def is_Empty(self):
        """
        Checks to see if the permutation is a set with zero elements

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> Permutation([]).is_Empty
        True
        >>> Permutation([0]).is_Empty
        False

        See Also
        ========

        is_Singleton
        """
        return self.size == 0

    @property
    def is_Identity(self):
        """
        Returns True if the Permutation is an identity permutation.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([])
        >>> p.is_Identity
        True
        >>> p = Permutation([[0], [1], [2]])
        >>> p.is_Identity
        True
        >>> p = Permutation([0, 1, 2])
        >>> p.is_Identity
        True
        >>> p = Permutation([0, 2, 1])
        >>> p.is_Identity
        False

        See Also
        ========

        order
        """
        af = self.array_form
        return not af or all(i == af[i] for i in range(self.size))

    def ascents(self):
        """
        Returns the positions of ascents in a permutation, ie, the location
        where p[i] < p[i+1]

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([4, 0, 1, 3, 2])
        >>> p.ascents()
        [1, 2]

        See Also
        ========

        descents, inversions, min, max
        """
        a = self.array_form
        pos = [i for i in range(len(a) - 1) if a[i] < a[i + 1]]
        return pos

    def descents(self):
        """
        Returns the positions of descents in a permutation, ie, the location
        where p[i] > p[i+1]

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([4, 0, 1, 3, 2])
        >>> p.descents()
        [0, 3]

        See Also
        ========

        ascents, inversions, min, max
        """
        a = self.array_form
        pos = [i for i in range(len(a) - 1) if a[i] > a[i + 1]]
        return pos

    def max(self):
        """
        The maximum element moved by the permutation.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([1, 0, 2, 3, 4])
        >>> p.max()
        1

        See Also
        ========

        min, descents, ascents, inversions
        """
        max = 0
        a = self.array_form
        for i in range(len(a)):
            if a[i] != i and a[i] > max:
                max = a[i]
        return max

    def min(self):
        """
        The minimum element moved by the permutation.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 4, 3, 2])
        >>> p.min()
        2

        See Also
        ========

        max, descents, ascents, inversions
        """
        a = self.array_form
        min = len(a)
        for i in range(len(a)):
            if a[i] != i and a[i] < min:
                min = a[i]
        return min

    def inversions(self):
        """
        Computes the number of inversions of a permutation.

        An inversion is where i > j but p[i] < p[j].

        For small length of p, it iterates over all i and j
        values and calculates the number of inversions.
        For large length of p, it uses a variation of merge
        sort to calculate the number of inversions.

        References
        ==========

        [1] http://www.cp.eng.chula.ac.th/~piak/teaching/algo/algo2008/count-inv.htm

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 2, 3, 4, 5])
        >>> p.inversions()
        0
        >>> Permutation([3, 2, 1, 0]).inversions()
        6

        See Also
        ========

        descents, ascents, min, max
        """
        inversions = 0
        a = self.array_form
        n = len(a)
        if n < 130:
            for i in range(n - 1):
                b = a[i]
                for c in a[i + 1:]:
                    if b > c:
                        inversions += 1
        else:
            k = 1
            right = 0
            arr = a[:]
            temp = a[:]
            while k < n:
                i = 0
                while i + k < n:
                    right = i + k * 2 - 1
                    if right >= n:
                        right = n - 1
                    inversions += _merge(arr, temp, i, i + k, right)
                    i = i + k * 2
                k = k * 2
        return inversions

    def commutator(self, x):
        """Return the commutator of self and x: ``~x*~self*x*self``

        If f and g are part of a group, G, then the commutator of f and g
        is the group identity iff f and g commute, i.e. fg == gf.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> p = Permutation([0, 2, 3, 1])
        >>> x = Permutation([2, 0, 3, 1])
        >>> c = p.commutator(x); c
        Permutation([2, 1, 3, 0])
        >>> c == ~x*~p*x*p
        True

        >>> I = Permutation(3)
        >>> p = [I + i for i in range(6)]
        >>> for i in range(len(p)):
        ...     for j in range(len(p)):
        ...         c = p[i].commutator(p[j])
        ...         if p[i]*p[j] == p[j]*p[i]:
        ...             assert c == I
        ...         else:
        ...             assert c != I
        ...

        References
        ==========

        http://en.wikipedia.org/wiki/Commutator
        """

        a = self.array_form
        b = x.array_form
        n = len(a)
        if len(b) != n:
            raise ValueError("The permutations must be of equal size.")
        inva = [None]*n
        for i in range(n):
            inva[a[i]] = i
        invb = [None]*n
        for i in range(n):
            invb[b[i]] = i
        return self._af_new([a[b[inva[i]]] for i in invb])

    def signature(self):
        """
        Gives the signature of the permutation needed to place the
        elements of the permutation in canonical order.

        The signature is calculated as (-1)^<number of inversions>

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 2])
        >>> p.inversions()
        0
        >>> p.signature()
        1
        >>> q = Permutation([0,2,1])
        >>> q.inversions()
        1
        >>> q.signature()
        -1

        See Also
        ========

        inversions
        """
        if self.is_even:
            return 1
        return -1

    def order(self):
        """
        Computes the order of a permutation.

        When the permutation is raised to the power of its
        order it equals the identity permutation.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> p = Permutation([3, 1, 5, 2, 4, 0])
        >>> p.order()
        4
        >>> (p**(p.order()))
        Permutation([], size=6)

        See Also
        ========

        identity, cardinality, length, rank, size
        """

        return reduce(lcm, [len(cycle) for cycle in self.cyclic_form], 1)

    def length(self):
        """
        Returns the number of integers moved by a permutation.

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> Permutation([0, 3, 2, 1]).length()
        2
        >>> Permutation([[0, 1], [2, 3]]).length()
        4

        See Also
        ========

        min, max, support, cardinality, order, rank, size
        """

        return len(self.support())

    @property
    def cycle_structure(self):
        """Return the cycle structure of the permutation as a dictionary
        indicating the multiplicity of each cycle length.

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> Permutation.print_cyclic = True
        >>> Permutation(3).cycle_structure
        {1: 4}
        >>> Permutation(0, 4, 3)(1, 2)(5, 6).cycle_structure
        {2: 2, 3: 1}
        """
        if self._cycle_structure:
            rv = self._cycle_structure
        else:
            rv = defaultdict(int)
            singletons = self.size
            for c in self.cyclic_form:
                rv[len(c)] += 1
                singletons -= len(c)
            if singletons:
                rv[1] = singletons
            self._cycle_structure = rv
        return dict(rv)  # make a copy

    @property
    def cycles(self):
        """
        Returns the number of cycles contained in the permutation
        (including singletons).

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> Permutation([0, 1, 2]).cycles
        3
        >>> Permutation([0, 1, 2]).full_cyclic_form
        [[0], [1], [2]]
        >>> Permutation(0, 1)(2, 3).cycles
        2

        See Also
        ========
        sympy.functions.combinatorial.numbers.stirling
        """
        return len(self.full_cyclic_form)

    def index(self):
        """
        Returns the index of a permutation.

        The index of a permutation is the sum of all subscripts j such
        that p[j] is greater than p[j+1].

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([3, 0, 2, 1, 4])
        >>> p.index()
        2
        """
        a = self.array_form

        return sum([j for j in range(len(a) - 1) if a[j] > a[j + 1]])

    def runs(self):
        """
        Returns the runs of a permutation.

        An ascending sequence in a permutation is called a run [5].


        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([2, 5, 7, 3, 6, 0, 1, 4, 8])
        >>> p.runs()
        [[2, 5, 7], [3, 6], [0, 1, 4, 8]]
        >>> q = Permutation([1,3,2,0])
        >>> q.runs()
        [[1, 3], [2], [0]]
        """
        return runs(self.array_form)

    def inversion_vector(self):
        """Return the inversion vector of the permutation.

        The inversion vector consists of elements whose value
        indicates the number of elements in the permutation
        that are lesser than it and lie on its right hand side.

        The inversion vector is the same as the Lehmer encoding of a
        permutation.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([4, 8, 0, 7, 1, 5, 3, 6, 2])
        >>> p.inversion_vector()
        [4, 7, 0, 5, 0, 2, 1, 1]
        >>> p = Permutation([3, 2, 1, 0])
        >>> p.inversion_vector()
        [3, 2, 1]

        The inversion vector increases lexicographically with the rank
        of the permutation, the -ith element cycling through 0..i.

        >>> p = Permutation(2)
        >>> while p:
        ...     print('%s %s %s' % (p, p.inversion_vector(), p.rank()))
        ...     p = p.next_lex()
        ...
        Permutation([0, 1, 2]) [0, 0] 0
        Permutation([0, 2, 1]) [0, 1] 1
        Permutation([1, 0, 2]) [1, 0] 2
        Permutation([1, 2, 0]) [1, 1] 3
        Permutation([2, 0, 1]) [2, 0] 4
        Permutation([2, 1, 0]) [2, 1] 5

        See Also
        ========
        from_inversion_vector
        """
        self_array_form = self.array_form
        n = len(self_array_form)
        inversion_vector = [0] * (n - 1)

        for i in range(n - 1):
            val = 0
            for j in range(i + 1, n):
                if self_array_form[j] < self_array_form[i]:
                    val += 1
            inversion_vector[i] = val
        return inversion_vector

    def rank_trotterjohnson(self):
        """
        Returns the Trotter Johnson rank, which we get from the minimal
        change algorithm. See [4] section 2.4.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 1, 2, 3])
        >>> p.rank_trotterjohnson()
        0
        >>> p = Permutation([0, 2, 1, 3])
        >>> p.rank_trotterjohnson()
        7

        See Also
        ========

        unrank_trotterjohnson, next_trotterjohnson
        """
        if self.array_form == [] or self.is_Identity:
            return 0
        if self.array_form == [1, 0]:
            return 1
        perm = self.array_form
        n = self.size
        rank = 0
        for j in range(1, n):
            k = 1
            i = 0
            while perm[i] != j:
                if perm[i] < j:
                    k += 1
                i += 1
            j1 = j + 1
            if rank % 2 == 0:
                rank = j1*rank + j1 - k
            else:
                rank = j1*rank + k - 1
        return rank

    @classmethod
    def unrank_trotterjohnson(cls, size, rank):
        """
        Trotter Johnson permutation unranking. See [4] section 2.4.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.unrank_trotterjohnson(5, 10)
        Permutation([0, 3, 1, 2, 4])

        See Also
        ========

        rank_trotterjohnson, next_trotterjohnson
        """
        perm = [0]*size
        r2 = 0
        n = ifac(size)
        pj = 1
        for j in range(2, size + 1):
            pj *= j
            r1 = (rank * pj) // n
            k = r1 - j*r2
            if r2 % 2 == 0:
                for i in range(j - 1, j - k - 1, -1):
                    perm[i] = perm[i - 1]
                perm[j - k - 1] = j - 1
            else:
                for i in range(j - 1, k, -1):
                    perm[i] = perm[i - 1]
                perm[k] = j - 1
            r2 = r1
        return cls._af_new(perm)

    def next_trotterjohnson(self):
        """
        Returns the next permutation in Trotter-Johnson order.
        If self is the last permutation it returns None.
        See [4] section 2.4. If it is desired to generate all such
        permutations, they can be generated in order more quickly
        with the ``generate_bell`` function.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> p = Permutation([3, 0, 2, 1])
        >>> p.rank_trotterjohnson()
        4
        >>> p = p.next_trotterjohnson(); p
        Permutation([0, 3, 2, 1])
        >>> p.rank_trotterjohnson()
        5

        See Also
        ========

        rank_trotterjohnson, unrank_trotterjohnson, sympy.utilities.iterables.generate_bell
        """
        pi = self.array_form[:]
        n = len(pi)
        st = 0
        rho = pi[:]
        done = False
        m = n-1
        while m > 0 and not done:
            d = rho.index(m)
            for i in range(d, m):
                rho[i] = rho[i + 1]
            par = _af_parity(rho[:m])
            if par == 1:
                if d == m:
                    m -= 1
                else:
                    pi[st + d], pi[st + d + 1] = pi[st + d + 1], pi[st + d]
                    done = True
            else:
                if d == 0:
                    m -= 1
                    st += 1
                else:
                    pi[st + d], pi[st + d - 1] = pi[st + d - 1], pi[st + d]
                    done = True
        if m == 0:
            return None
        return self._af_new(pi)

    def get_precedence_matrix(self):
        """
        Gets the precedence matrix. This is used for computing the
        distance between two permutations.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation.josephus(3, 6, 1)
        >>> p
        Permutation([2, 5, 3, 1, 4, 0])
        >>> p.get_precedence_matrix()
        Matrix([
        [0, 0, 0, 0, 0, 0],
        [1, 0, 0, 0, 1, 0],
        [1, 1, 0, 1, 1, 1],
        [1, 1, 0, 0, 1, 0],
        [1, 0, 0, 0, 0, 0],
        [1, 1, 0, 1, 1, 0]])

        See Also
        ========

        get_precedence_distance, get_adjacency_matrix, get_adjacency_distance
        """
        m = zeros(self.size)
        perm = self.array_form
        for i in range(m.rows):
            for j in range(i + 1, m.cols):
                m[perm[i], perm[j]] = 1
        return m

    def get_precedence_distance(self, other):
        """
        Computes the precedence distance between two permutations.

        Suppose p and p' represent n jobs. The precedence metric
        counts the number of times a job j is preceded by job i
        in both p and p'. This metric is commutative.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([2, 0, 4, 3, 1])
        >>> q = Permutation([3, 1, 2, 4, 0])
        >>> p.get_precedence_distance(q)
        7
        >>> q.get_precedence_distance(p)
        7

        See Also
        ========

        get_precedence_matrix, get_adjacency_matrix, get_adjacency_distance
        """
        if self.size != other.size:
            raise ValueError("The permutations must be of equal size.")
        self_prec_mat = self.get_precedence_matrix()
        other_prec_mat = other.get_precedence_matrix()
        n_prec = 0
        for i in range(self.size):
            for j in range(self.size):
                if i == j:
                    continue
                if self_prec_mat[i, j] * other_prec_mat[i, j] == 1:
                    n_prec += 1
        d = self.size * (self.size - 1)//2 - n_prec
        return d

    def get_adjacency_matrix(self):
        """
        Computes the adjacency matrix of a permutation.

        If job i is adjacent to job j in a permutation p
        then we set m[i, j] = 1 where m is the adjacency
        matrix of p.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation.josephus(3, 6, 1)
        >>> p.get_adjacency_matrix()
        Matrix([
        [0, 0, 0, 0, 0, 0],
        [0, 0, 0, 0, 1, 0],
        [0, 0, 0, 0, 0, 1],
        [0, 1, 0, 0, 0, 0],
        [1, 0, 0, 0, 0, 0],
        [0, 0, 0, 1, 0, 0]])
        >>> q = Permutation([0, 1, 2, 3])
        >>> q.get_adjacency_matrix()
        Matrix([
        [0, 1, 0, 0],
        [0, 0, 1, 0],
        [0, 0, 0, 1],
        [0, 0, 0, 0]])

        See Also
        ========

        get_precedence_matrix, get_precedence_distance, get_adjacency_distance
        """
        m = zeros(self.size)
        perm = self.array_form
        for i in range(self.size - 1):
            m[perm[i], perm[i + 1]] = 1
        return m

    def get_adjacency_distance(self, other):
        """
        Computes the adjacency distance between two permutations.

        This metric counts the number of times a pair i,j of jobs is
        adjacent in both p and p'. If n_adj is this quantity then
        the adjacency distance is n - n_adj - 1 [1]

        [1] Reeves, Colin R. Landscapes, Operators and Heuristic search, Annals
        of Operational Research, 86, pp 473-490. (1999)


        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 3, 1, 2, 4])
        >>> q = Permutation.josephus(4, 5, 2)
        >>> p.get_adjacency_distance(q)
        3
        >>> r = Permutation([0, 2, 1, 4, 3])
        >>> p.get_adjacency_distance(r)
        4

        See Also
        ========

        get_precedence_matrix, get_precedence_distance, get_adjacency_matrix
        """
        if self.size != other.size:
            raise ValueError("The permutations must be of the same size.")
        self_adj_mat = self.get_adjacency_matrix()
        other_adj_mat = other.get_adjacency_matrix()
        n_adj = 0
        for i in range(self.size):
            for j in range(self.size):
                if i == j:
                    continue
                if self_adj_mat[i, j] * other_adj_mat[i, j] == 1:
                    n_adj += 1
        d = self.size - n_adj - 1
        return d

    def get_positional_distance(self, other):
        """
        Computes the positional distance between two permutations.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> p = Permutation([0, 3, 1, 2, 4])
        >>> q = Permutation.josephus(4, 5, 2)
        >>> r = Permutation([3, 1, 4, 0, 2])
        >>> p.get_positional_distance(q)
        12
        >>> p.get_positional_distance(r)
        12

        See Also
        ========

        get_precedence_distance, get_adjacency_distance
        """
        a = self.array_form
        b = other.array_form
        if len(a) != len(b):
            raise ValueError("The permutations must be of the same size.")
        return sum([abs(a[i] - b[i]) for i in range(len(a))])

    @classmethod
    def josephus(cls, m, n, s=1):
        """Return as a permutation the shuffling of range(n) using the Josephus
        scheme in which every m-th item is selected until all have been chosen.
        The returned permutation has elements listed by the order in which they
        were selected.

        The parameter ``s`` stops the selection process when there are ``s``
        items remaining and these are selected by continuing the selection,
        counting by 1 rather than by ``m``.

        Consider selecting every 3rd item from 6 until only 2 remain::

            choices    chosen
            ========   ======
              012345
              01 345   2
              01 34    25
              01  4    253
              0   4    2531
              0        25314
                       253140

        Examples
        ========

        >>> from sympy.combinatorics import Permutation
        >>> Permutation.josephus(3, 6, 2).array_form
        [2, 5, 3, 1, 4, 0]

        References
        ==========

        1. http://en.wikipedia.org/wiki/Flavius_Josephus
        2. http://en.wikipedia.org/wiki/Josephus_problem
        3. http://www.wou.edu/~burtonl/josephus.html

        """
        from collections import deque
        m -= 1
        Q = deque(list(range(n)))
        perm = []
        while len(Q) > max(s, 1):
            for dp in range(m):
                Q.append(Q.popleft())
            perm.append(Q.popleft())
        perm.extend(list(Q))
        return cls(perm)

    @classmethod
    def from_inversion_vector(cls, inversion):
        """
        Calculates the permutation from the inversion vector.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> Permutation.from_inversion_vector([3, 2, 1, 0, 0])
        Permutation([3, 2, 1, 0, 4, 5])

        """
        size = len(inversion)
        N = list(range(size + 1))
        perm = []
        try:
            for k in range(size):
                val = N[inversion[k]]
                perm.append(val)
                N.remove(val)
        except IndexError:
            raise ValueError("The inversion vector is not valid.")
        perm.extend(N)
        return cls._af_new(perm)

    @classmethod
    def random(cls, n):
        """
        Generates a random permutation of length ``n``.

        Uses the underlying Python pseudo-random number generator.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.random(2) in (Permutation([1, 0]), Permutation([0, 1]))
        True

        """
        perm_array = list(range(n))
        random.shuffle(perm_array)
        return cls._af_new(perm_array)

    @classmethod
    def unrank_lex(cls, size, rank):
        """
        Lexicographic permutation unranking.

        Examples
        ========

        >>> from sympy.combinatorics.permutations import Permutation
        >>> Permutation.print_cyclic = False
        >>> a = Permutation.unrank_lex(5, 10)
        >>> a.rank()
        10
        >>> a
        Permutation([0, 2, 4, 1, 3])

        See Also
        ========

        rank, next_lex
        """
        perm_array = [0] * size
        psize = 1
        for i in range(size):
            new_psize = psize*(i + 1)
            d = (rank % new_psize) // psize
            rank -= d*psize
            perm_array[size - i - 1] = d
            for j in range(size - i, size):
                if perm_array[j] > d - 1:
                    perm_array[j] += 1
            psize = new_psize
        return cls._af_new(perm_array)

    # global flag to control how permutations are printed
    # when True, Permutation([0, 2, 1, 3]) -> Cycle(1, 2)
    # when False, Permutation([0, 2, 1, 3]) -> Permutation([0, 2, 1])
    print_cyclic = True


def _merge(arr, temp, left, mid, right):
    """
    Merges two sorted arrays and calculates the inversion count.

    Helper function for calculating inversions. This method is
    for internal use only.
    """
    i = k = left
    j = mid
    inv_count = 0
    while i < mid and j <= right:
        if arr[i] < arr[j]:
            temp[k] = arr[i]
            k += 1
            i += 1
        else:
            temp[k] = arr[j]
            k += 1
            j += 1
            inv_count += (mid -i)
    while i < mid:
        temp[k] = arr[i]
        k += 1
        i += 1
    if j <= right:
        k += right - j + 1
        j += right - j + 1
        arr[left:k + 1] = temp[left:k + 1]
    else:
        arr[left:right + 1] = temp[left:right + 1]
    return inv_count

Perm = Permutation
_af_new = Perm._af_new
