overlapping sets

[Gerard Flanagan]

Gerard Flanagan wrote:

> kpp9c wrote:
>
> > I have a question... and ... whew ... i am gonna be honest, i haven't
> > the slightest clue how to even start ... i am not sure if i used up all
> > my good will here or can take a mulligan.. i love to try to at least
> > post some lame broken code of my own at first... but like i said, not
> > being a math person i am not even sure how to start or if it is even
> > possible.
> >
> > here's the deal... i have a dictionary that defines some collections..
> > like so:
> >
> > sets = { ('one') : [0, 4, 7, 9 ],
> > ('two') : [0, 3, 7, 9 ],
> > ('three') : [0, 4, 7, 11],
> > ('four') : [0, 3, 7, 10 ],
> > ('five') : [0, 4, 7, 10 ],
> > ('six') : [0, 4, 8, 10 ],
> > ('seven') : [0, 3, 6, 10],
> > ('eight') : [0, 3, 6, 9 ],
> > ('nine') : [0, 3, 7, 11 ],
> > ('ten') : [0, 5, 7, 10 ] }
> >
> > I every time i call this function i would like like it to return a
> > collection at random, any collection, so long as it has all but one
> > element that is the same. So if i grab [0, 4, 7, 9 ] as my first set

>
> # k is the length of the required output lists, 0<k<n
> # n is the number of lists to output
>
> import random
>
> alphabet = [ 0, 3, 4, 5, 6, 7, 8, 9, 10, 11 ]
>
> def randomiser( a_list, n, k ):
>     d = len( a_list )
>     choice = range(d)
>     vector = []
>     for i in range(k):
>         vector.append( choice.pop(random.randint(0,d-i-1)) )
>     yield [ a_list[s] for s in vector ]
>     #yield sorted( a_list[s] for s in vector )
>     for _ in range(n):
>         rand_vector_idx = random.randint(0,k-1)
>         rand_choice_idx = random.randint(0,d-k-1)
>         rand_vector_val = vector[rand_vector_idx]  #remember old value
>         vector[rand_vector_idx] = choice.pop( rand_choice_idx )
>         choice.append( rand_vector_val ) #add old value back to choice
>         yield [ a_list[t] for t in vector ]
>         #yield sorted( a_list[t] for t in vector )
>


Sorry, I realise this doesn't answer your question - it is the
collections themselves which are your 'alphabet', not the set of their
elements.  Looks like it's Graph Theory for you!

Apologies.

Gerard