Permutation Generator

[Talin]

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:
head = lst[:1]
for x in permute( lst[1:] ):
yield head + x
yield x + head
return

-- Talin

[Michael J. Fromberger]

In article <ma************ *************** ************@py thon.org>,
Talin <ta***@acm.or g> 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:
head = lst[:1]
for x in permute( lst[1:] ):
yield head + x
yield x + head
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]

Talin <ta***@acm.or g> 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:
head = lst[:1]
for x in permute( lst[1:] ):
yield head + x
yield x + head
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]

"Talin" <ta***@acm.or g> wrote in message
news:ma******** *************** *************** *@python.org...

I wanted to share
this: a generator which returns all permutations of a list:

Try this instead:
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]

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:
head = lst[:1]
for x in permute( lst[1:] ):
yield head + x
yield x + head
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

""" ***Reversible** * Permutations using factoradics.

factoradic concept at:

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,N one])

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]

An interesting thing about the factoradic method is its reversibility.

If you have a list: ['a','Fred',23,N one]

and you are presented with an ordering: [23,'a',None,'Fr ed']
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,N one])
ix = foo.getPermutat ionIndex([23,'a',None,'Fr ed'])

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
def factoradic(anIn t,order=0):
"""calculat e the factoradic on anInt

factoradic(859) [1, 1, 0, 3, 0, 1, 0]
factoradic(1123 311112221345553 998889997865532 6328) [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]
factoradic(0,4) [0, 0, 0, 0]
factoradic(1) [1, 0]
factoradic(1047 ) [1, 2, 3, 2, 1, 1, 0]
factoradic(5,4) [0, 2, 1, 0]
"""

factoradic = []

z = 0
while anInt > 0:
z += 1
factoradic.appe nd(int(anInt % z))
anInt /= z
factoradic.reve rse()
if order:
while len(factoradic) < order:
factoradic.inse rt(0,0)

return factoradic

def factorial(anInt ):
"""factoria l
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
def unfactoradic(aL ist):
"""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(obj ect):
"""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,d ata):
self.data = data

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

def permutationIndi ces(self,anInt) :
"""calculat e the permutation indices of self from anInt
z = Permutation([1,2,3,4,5,6,7])
z.permutationIn dices(1047) [1, 3, 5, 4, 2, 6, 0] z = Permutation([0,1,2,3])
z.permutationIn dices(5) [0, 3, 2, 1]
"""
f = factoradic(anIn t,self.order)
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(sel f,anInt):
"""return a list of permutated items
z = Permutation([1,2,3,4,5,6,7])
z.permutation(1 047) [2, 4, 6, 5, 3, 7, 1]

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

def randomPermutati on(self):
"""just get one of them, randomly"""
r = random.randint( 0,factorial(sel f.order))
return self.permutatio n(r)

def getPermutationI ndex(self,aPerm utation):
"""presumin g a unique list, get the permutation index of the given
permutation list.
d = [1,2,3,4,5,6,7]
z = Permutation(d)
z.getPermutatio nIndex([2, 4, 6, 5, 3, 7, 1])

1047
"""
indexkey = []
for k in aPermutation:
indexkey.append (self.data.inde x(k))
data = []
for k in indexkey:
data.append(k+1 )
factoradic = []
while len(data) > 0:
r = data.pop(0)
factoradic.appe nd(r-1)
for idx,val in enumerate(data) :
if val >= r:
data[idx] = val -1
return unfactoradic(fa ctoradic)

order = property(getOrd er)

def listAll(anInt):
theList = []
for k in range(anInt):
theList.append( k)
z = Permutation(the List)
for k in range(factorial (len(z.data))):
b = factoradic(k,le n(z.data))
c = z.permutation(k )
d = z.getPermutatio nIndex(c)
print "%s\t%s\t%s\t%s " % (k,b,c,d)
def _test():
import doctest,permuta tion
return doctest.testmod (permutation)
if __name__ == '__main__':
_test()
listAll(4)

[Jack Diederich]

On Fri, Aug 12, 2005 at 03:48:38PM -0400, Michael J. Fromberger wrote:

In article <ma************ *************** ************@py thon.org>,
Talin <ta***@acm.or g> 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.c om 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]

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]

"Casey Hawthorne" <ca************ ***@istar.ca> wrote in message
news:s1******** *************** *********@4ax.c om...

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]

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(li st)):
.... 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
<Mi************ ******@Clothing .Dartmouth.EDU> wrote:
In article <ma************ *************** ************@py thon.org>,
Talin <ta***@acm.or g> 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:
head = lst[:1]
for x in permute( lst[1:] ):
yield head + x
yield x + head
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]

On Mon, 15 Aug 2005, Matt Hammond wrote:

Just satisfied my curiosity wrt this problem, so I might as well share :-)

def permute(list):
How about:

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(factoria l(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.