Velocity Reviews > Permutation Generator

# Permutation Generator

Talin
Guest
Posts: n/a

 08-12-2005
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

Michael J. Fromberger
Guest
Posts: n/a

 08-12-2005
In article <(E-Mail Removed)>,
Talin <(E-Mail Removed)> 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

You're right that you're not the first person to do this: Many others
have also posted incorrect permutation generators.

Have you tried your code on some simple test cases?

list(permute([1, 2, 3]))
==> [[1, 2, 3], [2, 3, 1], [1, 3, 2], [3, 2, 1]]

Notably absent from this list are [2, 1, 3] and [2, 3, 1]. The problem
gets worse with longer lists. The basic problem is that x needs to be
able to occur in ALL positions, not just the beginning and the end.

Cheers,
-M

--
Michael J. Fromberger | Lecturer, Dept. of Computer Science
http://www.dartmouth.edu/~sting/ | Dartmouth College, Hanover, NH, USA

Paul Rubin
Guest
Posts: n/a

 08-12-2005
Talin <(E-Mail Removed)> writes:
> 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

Hmm:

>>> for p in permute([1,2,3]):

print p

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

Oops.

David Isaac
Guest
Posts: n/a

 08-13-2005
"Talin" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> I wanted to share
> this: a generator which returns all permutations of a list:

def permuteg(lst): return ([lst[i]]+x
for i in range(len(lst))
for x in permute(lst[:i]+lst[i+1:])) \
or [[]]

Alan Isaac

Jim Washington
Guest
Posts: n/a

 08-13-2005
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/de...rmutations.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 = [1]
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)

Jack Diederich
Guest
Posts: n/a

 08-14-2005
On Fri, Aug 12, 2005 at 03:48:38PM -0400, Michael J. Fromberger wrote:
> In article <(E-Mail Removed)>,
> Talin <(E-Mail Removed)> 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:

>
> You're right that you're not the first person to do this: Many others
> have also posted incorrect permutation generators.
>

Amen, combinatorics are so popular they should be in the FAQ.
groups.google.com can show you many pure python recipies and benchmarks,
but I'll give my ususal response:
http://probstat.sourceforge.net/

I'm not just the author, I'm a client-ly,
-jackdied

Casey Hawthorne
Guest
Posts: n/a

 08-14-2005
It's hard to make "complete" permutation generators, Knuth has a whole
fascicle on it - "The Art of Computer Programming - Volume 4 Fascicle
2 - Generating All Tuples and Permutations" - 2005
--
Regards,
Casey

David Isaac
Guest
Posts: n/a

 08-14-2005

"Casey Hawthorne" <(E-Mail Removed)> wrote in message
news:(E-Mail Removed)...
> It's hard to make "complete" permutation generators, Knuth has a whole
> fascicle on it - "The Art of Computer Programming - Volume 4 Fascicle
> 2 - Generating All Tuples and Permutations" - 2005

Can you elaborate a bit on what you mean?
Given a list of unique elements, it is easy enough to produce a
complete permutation generator in Python,
in the sense that it yields every possible permuation.
(See my previous post.) So you must mean
something else?

Cheers,
Alan Isaac

PS If the elements are not unique, that is easy enough to
deal with too, as long as you say what you want the
outcome to be.

Matt Hammond
Guest
Posts: n/a

 08-15-2005
Just satisfied my curiosity wrt this problem, so I might as well share

>>> def permute(list):

.... if len(list) <= 1:
.... yield list
.... else:
.... for i in xrange(0,len(list)):
.... for tail in permute( list[:i] + list[i+1:] ):
.... yield [ list[i] ] + tail
....
>>> for o in permute([1,2,3]):

.... print o
....
[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 1, 2]
[3, 2, 1]

regards

Matt

On Fri, 12 Aug 2005 20:48:38 +0100, Michael J. Fromberger
<(E-Mail Removed)> wrote:

> In article <(E-Mail Removed)>,
> Talin <(E-Mail Removed)> 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

>
> You're right that you're not the first person to do this: Many others
> have also posted incorrect permutation generators.
>
> Have you tried your code on some simple test cases?
>
> list(permute([1, 2, 3]))
> ==> [[1, 2, 3], [2, 3, 1], [1, 3, 2], [3, 2, 1]]
>
> Notably absent from this list are [2, 1, 3] and [2, 3, 1]. The problem
> gets worse with longer lists. The basic problem is that x needs to be
> able to occur in ALL positions, not just the beginning and the end.
>
> Cheers,
> -M
>

--

| Matt Hammond
| R&D Engineer, BBC Research and Development, Tadworth, Surrey, UK.

Tom Anderson
Guest
Posts: n/a

 08-15-2005
On Mon, 15 Aug 2005, Matt Hammond wrote:

> Just satisfied my curiosity wrt this problem, so I might as well share
>
>>>> def permute(list):

def permutation(l, i):
"Makes the ith permutation of the sequence l."
# leave out the reverses if you don't care about the order of the permutations
l_ = []; l_[:] = l; l_.reverse()
m = []
for j in xrange(len(l_)):
m.append(l_.pop((i % len(l_))))
i = i / (len(l_) + 1)
m.reverse()
return m

def factorial(n):
if (n == 1): return 1
else: return n * factorial((n - 1))

def permute(l):
for i in xrange(factorial(len(l))):
yield permutation(l, i)

>>> for o in permute([1,2,3]): print o

....
[1, 2, 3]
[1, 3, 2]
[2, 3, 1]
[2, 1, 3]
[3, 1, 2]
[3, 2, 1]

The thing i like about doing it this way is that you can use
permutation(l, i) to make arbitrary permutations on their own, should you
ever need to do that.

Also, it gives you an easy way to make only the even permutations of a
list - just feed even numbers into permutation(l, i) (i think!). This
could be useful if you wanted to build an alternating group for some
obscure discrete mathematics purpose.

tom

--
The future is still out there, somewhere.