recursive function

[cesco]

Hendrik van Rooyen wrote:
> "cesco" <fd.calabrese at gmail.com> wrote:
>
> >
> > Neil Cerutti wrote:
> > > On 2007-01-08, cesco <fd.calabrese at gmail.com> wrote:
> > > > Hi,
> > > >
> > > > I have a dictionary of lists of tuples like in the following example:
> > > > dict = {1: [(3, 4), (5, 8)],
> > > >         2: [(5, 4), (21, 3), (19, 2)],
> > > >         3: [(16, 1), (0, 2), (1, 2), (3, 4)]]
> > > >
> > > > In this case I have three lists inside the dict but this number
> > > > is known only at runtime. I have to write a function that
> > > > considers all the possible combinations of tuples belonging to
> > > > the different lists and return a list of tuples of tuples for
> > > > which the sum of the first element of the most inner tuple is
> > > > equal to N.
> > > >
> > > > For example, assuming N = 24, in this case it should return:
> > > > [((3, 4), (5, 4), (16, 1)), ((3, 4), (21, 3), (0, 2)), ((5, 8), (19,
> > > > 2), (0, 2))]
> > >
> > > What do you mean by "most inner tuple"?
> > >
> > > > A simple list comprehension would be enough if only I knew the
> > > > number of keys/lists beforehand
> > >
> > > len(dict.keys()).
> >
> > What I mean is that the number of keys/lists is not known until runtime
> > and it changes randomly from time to time which means I would have to
> > write every time a different list comprehension (or a different number
> > of nested loops) to accomplish the same thing.
> > In the example the result of the search should be a list containing
> > three tuples each of which contains again three tuple (the most inner
> > tuples). If you consider the first element of each tuple the sum is 24
> > (like in 3+5+16, or 3+21+0 or 5+19+0).
> >
> > Any other help will be appreciated
>
>
> Is there any reliable structure in the data?
> - for instance in your example, the first list
> has two tuples, the second one three, and the
> third one four - Is this a pattern?

Hi Hendrik,

there is no such a pattern (I just happened to insert that ascending
number of lists). Anyway the answer from bearophileH... at lycos.com was
quite satisfactory:-)

regards
Cesco