[Edu-sig] Group Theory for Girl Scouts (and Boys too)

[Kirby Urner]

> There's some playfulness here, with duck typing, in that the parameter
> name suggests a set, yet we pass a string, then work with listifications.
> This allows more types of input, yet the output should always be:  a dict,
> a mapping, a permutation.  Running it now, I get:
>
> P1:  {'1': '6', '0': '1', '3': '8', '2': '5', '5': '3', '4': '0', '7':
> '2', '6': '4', '9': '9', '8': '7'}
> P2:  {'1': '0', '0': '5', '3': '6', '2': '4', '5': '8', '4': '3', '7':
> '2', '6': '7', '9': '9', '8': '1'}
>
> So what's cyclic notation again?  I hadn't gotten that far, as we needed
> an easy way to get permutations.  Thanks Python.
>
> Lets work on P1.  You start a tuple going (1, 6, 4, 0).  See what I did?
> I started with 1, arbitrarily and found it maps to 6.  Then I looked up 6
> and found it mapped to 4.  I keep following that trail until I either get
> back to where I started, meaning the last element is deemed to connect to
> the first.  It's a circle, a cycle.  But maybe we're not done.  2 is not
> mentioned so lets start there.  2 maps to 5 maps to 3 maps to 8 maps to 7
> maps to 2.  Another circle:  (2, 5, 3, 8, 7).  Anyone missing?  9, where's
> 9.  9 maps to itself, so (9,) -- using Python notation.  And we're done:
> P1 may be expressed as ((0, 1, 6, 4), (2, 5, 3, 8, 7), (9,)).  I can start
> with the lowest number in each cycle and sort the cycles by lowest
> leftmost, if I want a canonical order.  The above is canonical and unique
> for P1.
>
>
P1 =  {'1': '6', '0': '1', '3': '8', '2': '5', '5': '3', '4': '0', '7':
'2', '6': '4', '9': '9', '8': '7'}
P2 =  {'1': '0', '0': '5', '3': '6', '2': '4', '5': '8', '4': '3', '7':
'2', '6': '7', '9': '9', '8': '1'}


def get_cyclic(codedict):
    """
    Return Permutation dict as a tuple of cyclic tuples, e.g.
    (('A','R','C'),('D','Q')...)
    """
    dictkeys = list(codedict.keys())
    result = []
    for i in list(dictkeys):  # using copy of dictkeys
        newtuple = ()
        if i in dictkeys:
            initval,nextval = i,i
            while True:
               newtuple += (nextval,)
               dictkeys.remove(nextval)
               nextval = codedict[nextval]
               if nextval==initval:  # cycle complete
                   break
            result.append(newtuple)
    return tuple(result)

print(get_cyclic(P1))
print(get_cyclic(P2))

Running....

(('1', '6', '4', '0'), ('3', '8', '7', '2', '5'), ('9',))
(('1', '0', '5', '8'), ('3', '6', '7', '2', '4'), ('9',))

Process finished with exit code 0

Kirby
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.python.org/pipermail/edu-sig/attachments/20130719/ebc860c4/attachment.html>