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Permutation Generator

P: n/a
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
Aug 12 '05 #1
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10 Replies


P: n/a
In article <ma***************************************@python. org>,
Talin <ta***@acm.org> 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
Aug 12 '05 #2

P: n/a
Talin <ta***@acm.org> 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.
Aug 12 '05 #3

P: n/a
"Talin" <ta***@acm.org> wrote in message
news:ma***************************************@pyt hon.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
Aug 13 '05 #4

P: n/a
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,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]

An interesting thing about the factoradic method is its reversibility.

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
def factoradic(anInt,order=0):
"""calculate the factoradic on anInt
factoradic(859) [1, 1, 0, 3, 0, 1, 0]
factoradic(11233111122213455539988899978655326328) [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.append(int(anInt % z))
anInt /= z
factoradic.reverse()
if order:
while len(factoradic) < order:
factoradic.insert(0,0)

return factoradic

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
def unfactoradic(aList):
"""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]
"""
f = factoradic(anInt,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(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)
factoradic = []
while len(data) > 0:
r = data.pop(0)
factoradic.append(r-1)
for idx,val in enumerate(data):
if val >= r:
data[idx] = val -1
return unfactoradic(factoradic)

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))):
b = factoradic(k,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)
Aug 13 '05 #5

P: n/a
On Fri, Aug 12, 2005 at 03:48:38PM -0400, Michael J. Fromberger wrote:
In article <ma***************************************@python. org>,
Talin <ta***@acm.org> 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
Aug 14 '05 #6

P: n/a
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
Aug 14 '05 #7

P: n/a

"Casey Hawthorne" <ca***************@istar.ca> wrote in message
news:s1********************************@4ax.com...
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.
Aug 14 '05 #8

P: n/a
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
<Mi******************@Clothing.Dartmouth.EDU> wrote:
In article <ma***************************************@python. org>,
Talin <ta***@acm.org> 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.
Aug 15 '05 #9

P: n/a
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(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.
Aug 15 '05 #10

P: n/a
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


As it happens, I've just started learning Python and my 'Hello World'
was a class which generates all permutations of the set {1,2,...n}.
It's a translation of some Pascal code from the book "Programming for
Mathematicians" by Raymond Seroul (2000 Springer-Verlag), and produces
a (non-lexicographic) total ordering of the set using Johnson's
Algorithm.

To explain the algorithm briefly. Each element is 'adorned' with a flag
( either 1 or -1 ) which determines the direction that an element is
'looking' - so, given the sequence

[ [-1,3], [1,2], [1,1], [-1,4] ]

1 sees 4
2 sees 1
3 sees nothing
4 sees 1

(with the convention that -1 means 'looking left')

An element is said to be 'mobile' if it is looking at a smaller number.
eg. in the sequence above, both 2 and 4 are mobile.

Then the algorithm is:

- find the highest mobile
- swap this mobile with the element it sees, but don't swap their
direction flags
- reverse the direction flags of any element larger than the mobile

In coding the algorithm, sentinels with value 1 bigger than n are added
at either end of the sequence, this helps when checking which elements
are mobile. Also, 1 and -1 aren't arbitrary flags, these values are
used because you can find which element another element 'sees' by using
'index + flag'.

Here's the code:
def factorial( n ):
if n <= 1:
return 1
else:
return n * factorial( n-1 )

class Permutation:

def __init__( self, items ):
seq = list( items[:] )
n = len( seq )
self.sequence = seq
self.length = n
self.count = factorial( n )

def __getitem__( self, key ):
result = []
sequence = self.sequence[:]
N = self.count
index = key
for i in range( self.length, 0, -1):
N = N / i
choice, index = index // N, index % N
result += [ sequence.pop(choice) ]
return result

class NPermutation( Permutation ):

def __init__( self, n ):
Permutation.__init__( self, range( 1, n+1 ) )
self.reset()

def reset( self ):
list = [ [-1,i] for i in range(self.length+2) ]
list[0][1] = self.length+1
#eg. n=3 -> list = [[-1,4], [-1,1], [-1,2], [-1,3], [-1,4]]
self.__current = list
self.__index = 0

def __iter__( self ):
return self

def next( self ):
if self.__index == self.count:
self.reset()
raise StopIteration
elif self.__index > 0:
j = self.__get_index_of_highest_mobile()
#remember the mobile itself before you move it
mobile = self.__current[j][1]
#swap the mobile with the element it 'sees'
self.__move_mobile_at_index( j )
#switch the direction of elements greater than mobile
if mobile < self.length:
self.__reorient_list_elements( mobile )
self.__index += 1
return [ a[1] for a in self.__current[ 1:self.length+1 ] ]

def __get_index_of_highest_mobile( self ):
high_value = 0
high_index = 0
for i in range( 1, self.length+1 ):
direction = self.__current[i][0]
value = self.__current[i][1]
if value > high_value and value >
self.__current[i+direction][1]:
high_value = value
high_index = i
return high_index

def __move_mobile_at_index( self, index ):
direction = self.__current[index][0]
value = self.__current[index][1]
self.__current[index] = self.__current[index+direction]
self.__current[index+direction] = [direction,value]

def __reorient_list_elements( self, mobile ):
for i in range( 1, self.length+1 ):
if self.__current[i][1] > mobile:
self.__current[i][0] = -self.__current[i][0]


x = NPermutation( 6 )

print 'loop with __getitem__'
print '---------------'
for i in range( x.count ):
print x[i]

print 'loop with __iter__'
print '---------------'
for perm in x:
print perm
Gerard Flanagan

15/8/05

Aug 15 '05 #11

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