# Permutation Generator

Jim Washington jwashin at vt.edu
Sun Aug 14 01:21:30 CEST 2005

```On Fri, 12 Aug 2005 12:39:08 -0700, Talin wrote:

> I'm sure I am not the first person to do this, but I wanted to share
> this: a generator which returns all permutations of a list:
>
> def permute( lst ):
>     if len( lst ) == 1:
>         yield lst
>     else:
>         for x in permute( lst[1:] ):
>     return
>
> -- Talin

If we are sharing permutation algorithms today, here's one.

The following likes to be in a file called "permutation.py" for __main__
to work. A couple of lines went over 80 characters, so you might have to
put those back together.

-Jim Washington

http://msdn.microsoft.com/library/default.asp?url=/library/en-us/dnnetsec/html/permutations.asp

Why permutations?  Sometimes, you need to list your objects in a different order.
Maybe, when you are dealing with something persistent like Zope, you wish
your users to access things in a different order than other users.  Think
quizzes or photo galleries.

You think you want randomness, but what you really want is that different users
get different orderings of things, so that the first item is likely different
for each individual.  But you do not really want randomness; you want a
particular user always to get the same ordering.

One way would be to store for each individual the complete list in order,
This is another way that allows you to just store an index that refers to a
particular ordering.

For a list of n items, there are n factorial (n!) possible permutations.  So,
any number from 0 to n!-1  is a valid index to a unique ordering.

If you have

foo = Permutation(['a','Fred',23,None])

the possible indices are numbered 0 to 23 (0 to 4!-1)

sam = foo.permutation(10)
mary = foo.permutation(4)

sam is ['Fred', None, 'a', 23]
mary is ['a', None,'Fred', 23]

If you have a list: ['a','Fred',23,None]

and you are presented with an ordering: [23,'a',None,'Fred']
the factoradic method can algorithmically determine that this ordering is
index 13 of 24 of the possible permutations, without going forward through
your generating algorithm to get there.

foo = Permutation(['a','Fred',23,None])
ix = foo.getPermutationIndex([23,'a',None,'Fred'])

ix is 13.

For the above example, I used a list of mixed items; you probably will not.
Reversibility does not work if items are repeated, since it cannot know the
original positions of repeated items.  If you have duplicated items, use their
list index instead of the items themselves.

"""
try:
import psyco
psyco.full()
except:
pass

import random

[1, 1, 0, 3, 0, 1, 0]

[1, 9, 22, 2, 20, 20, 7, 14, 0, 19, 2, 13, 2, 5, 14, 18, 2, 0, 10, 1, 9, 3, 11, 9, 9, 4, 1, 4, 0, 0, 1, 1, 0, 0]

[0, 0, 0, 0]

[1, 0]

[1, 2, 3, 2, 1, 1, 0]

[0, 2, 1, 0]

"""

z = 0
while anInt > 0:
z += 1
anInt /= z

if order:

def factorial(anInt):
"""factorial

>>> factorial(3)
6
>>> factorial(0)
1
>>> factorial(1)
1
"""
if anInt == 0:
return 1
if anInt < 0:
raise ValueError, "Cannot factorialize negative numbers"
result = 1

while anInt > 1:
result = result * anInt
anInt -= 1
return result

"""from a factoradic list, calculate the integer

>>> unfactoradic([1, 1, 0, 3, 0, 1, 0])
859

"""
aList.reverse()
result = 0
for idx,val in enumerate(aList):
result += factorial(idx) * val
return result

class Permutation(object):
"""Base object for doing permutations.  Generally initialized with a list
of the items to do permutations on.  Works by the factoradic method,
which provides reversibility."""

_order = None

def __init__(self,data):
self.data = data

def getOrder(self):
if not self._order:
self._order = len(self.data)
return self._order

def permutationIndices(self,anInt):
"""calculate the permutation indices of self from anInt

>>> z = Permutation([1,2,3,4,5,6,7])
>>> z.permutationIndices(1047)
[1, 3, 5, 4, 2, 6, 0]
>>> z = Permutation([0,1,2,3])
>>> z.permutationIndices(5)
[0, 3, 2, 1]

"""
temp = []
for k in f:
temp.append(k + 1)

data = 
temp.reverse()
for k in temp[1:]:
data.insert(0,k)
for idx,val in enumerate(data[1:]):
if val >= k:
data[idx+1] = val + 1
for idx,val in enumerate(data):
data[idx] = val-1
return data

def permutation(self,anInt):
"""return a list of permutated items

>>> z = Permutation([1,2,3,4,5,6,7])
>>> z.permutation(1047)
[2, 4, 6, 5, 3, 7, 1]

"""
indices = self.permutationIndices(anInt)
newlist = []
for k in indices:
newlist.append(self.data[k])
return newlist

def randomPermutation(self):
"""just get one of them, randomly"""
r = random.randint(0,factorial(self.order))
return self.permutation(r)

def getPermutationIndex(self,aPermutation):
"""presuming a unique list, get the permutation index of the given
permutation list.

>>> d = [1,2,3,4,5,6,7]
>>> z = Permutation(d)
>>> z.getPermutationIndex([2, 4, 6, 5, 3, 7, 1])
1047
"""
indexkey = []
for k in aPermutation:
indexkey.append(self.data.index(k))
data = []
for k in indexkey:
data.append(k+1)
while len(data) > 0:
r = data.pop(0)
for idx,val in enumerate(data):
if val >= r:
data[idx] = val -1

order = property(getOrder)

def listAll(anInt):
theList = []
for k in range(anInt):
theList.append(k)
z = Permutation(theList)
for k in range(factorial(len(z.data))):
c = z.permutation(k)
d = z.getPermutationIndex(c)
print "%s\t%s\t%s\t%s" % (k,b,c,d)

def _test():
import doctest,permutation
return doctest.testmod(permutation)

if __name__ == '__main__':
_test()
listAll(4)

```