# Source code for sympy.combinatorics.permutations

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
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]

========
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]

========
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

========

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]

========

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

========

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

[docs]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.""" 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
[docs] 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]) 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 = [int(a) for a in args] 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): return max(self.keys()) + 1 def copy(self): return Cycle(self)
[docs]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 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 >>> dict([(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 _af_new(list(range(size or 0))) elif len(args) > 1: # c return _af_new(Cycle(*args).list(size)) if len(args) == 1: a = args[0] if isinstance(a, Perm): # g if size is None or size == a.size: return a return Perm(a.array_form, size=size) if isinstance(a, Cycle): # f return _af_new(a.list(size)) if not is_sequence(a): # b return _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): if is_cycle: raise ValueError('there were repeated elements; to resolve ' 'cycles use Cycle%s.' % ''.join([str(tuple(c)) for c in args])) else: 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))) size = len(aform) obj = Basic.__new__(cls, aform) obj._array_form = aform obj._size = size return obj @staticmethod def _af_new(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__(Perm, 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
[docs] 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
[docs] 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
[docs] 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
[docs] 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
[docs] 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
[docs] 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 = Perm.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
[docs] 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
@staticmethod
[docs] def rmul_with_af(*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 = _af_new(_af_rmuln(*a)) return rv
[docs] def mul_inv(self, other): """ other*~self, self and other have _array_form """ a = _af_invert(self._array_form) b = other._array_form return _af_new(_af_rmul(a, b))
def __rmul__(self, other): """This is needed to coerse other to Permutation in rmul.""" return Perm(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 _af_new(perm)
[docs] 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 type(n) == Perm: raise NotImplementedError( 'p**p is not defined; do you mean p^p (conjugate)?') n = int(n) return _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 _af_new(a)
[docs] 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
[docs] 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 _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)
[docs] def atoms(self): """ Returns all the elements of a permutation Examples ======== >>> from sympy.combinatorics import Permutation >>> Permutation([0, 1, 2, 3, 4, 5]).atoms() set([0, 1, 2, 3, 4, 5]) >>> Permutation([[0, 1], [2, 3], [4, 5]]).atoms() set([0, 1, 2, 3, 4, 5]) """ return set(self.array_form)
[docs] 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 _af_new(perm)
@classmethod
[docs] 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 _af_new(id_perm)
[docs] 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
[docs] 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 Perm.unrank_nonlex(self.size, r + 1)
[docs] 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
[docs] 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))
[docs] 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
[docs] 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
[docs] 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
[docs] 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
[docs] 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
[docs] 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))
[docs] 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
[docs] 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
[docs] 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
[docs] 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
[docs] 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
[docs] 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 _af_new([a[b[inva[i]]] for i in invb])
[docs] 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
[docs] 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)
[docs] 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
[docs] 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
[docs] 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)
[docs] 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]])
[docs] 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)
[docs] 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
[docs] 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
[docs] def unrank_trotterjohnson(self, 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 _af_new(perm)
[docs] 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 _af_new(pi)
[docs] 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
[docs] 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
[docs] 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
[docs] 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
[docs] 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
[docs] def josephus(self, 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 Perm(perm)
@classmethod
[docs] def from_inversion_vector(self, 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 _af_new(perm)
@classmethod
[docs] def random(self, 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 _af_new(perm_array)
@classmethod
[docs] def unrank_lex(self, 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 _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