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On Tue, Sep 1, 2009 at 10:11 AM, John W Higgins <wishdev / gmail.com> wrote:

> Morning Max,
>
> On Tue, Sep 1, 2009 at 9:41 AM, Max Williams <toastkid.williams / gmail.com
> >wrote:
>
> > Little ruby algorithm puzzle...
> >
> > I have a situation where i have an array of 12 items.  If someone
> > chooses to have n of them (where n can be between 3 and 12) then i want
> > to always include the first and last, and then 'spread' the others out
> > as evenly as possible between the rest.
> >
> > So, lets say for the sake of argument that the array holds the numbers 1
> > to 12.
> >
> > >> arr  1..12).to_a
> > [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
> >
> > I would get results back like this
> >
> > [1,6,12] (or [1,7,12], either is fine)
> >
> > [1, 5, 9, 12]  (or [1,4,8,12] or [1, 5, 8, 12])
> >
> > It feels like there should be a simple solution for this but i can't
> > think of a nice way.  Anyone?
> >
> > thanks
> > max
> > --
> > Posted via http://ww <http://www.ruby-forum.com/>RijndaelManaged<
> http://www.ruby-forum.com/>
> > w.ruby-forum.com/ <http://www.ruby-forum.com/>.
> >
> >
> So here is my silly attempt - simple but probably not the best performing
> answer you will get here.
>

small correction (should use floor instead of ceil)

class Array
ret  rray.new
#We want sections that total 1 less then the number of members we want
back - for example if we
#want 3 members in the spread we want 2 equal groups. We use .to_f
because we care about fractions here.
dist  ount.to_f/(members-1)
#Now we simply run thru the array and take the members at each spread
point.
(members-1).times{ |slot|
ret << at((slot*dist).floor)
}
#Add the last member which is always the last member of the array
ret << at(-1)
end
end

Also, here is a gist which tends to read better.
http://gist.github.com/179241

John

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