[EM] Has this idea been considered?

[Russ Paielli]

On Tue, Jul 5, 2011 at 2:14 AM, Juho Laatu <juho4880 at yahoo.co.uk> wrote:

> On 5.7.2011, at 11.19, Russ Paielli wrote:
>
> If one wants to simplify the inheritance rules even more then we might end
>> up using a tree method (I seem to mention it in every mail I send:). In that
>> approach there is no risk of having loops in the candidate transfer order.
>> Votes would be counted right away for each branch, and the candidate of the
>> largest brach of the largest branch of the ... would win.
>>
>
> That sounds interesting, but I'm not sure I understand what you mean. Can
> you give an example?
>
>
> Here's one example.
>
> Tree of candidates + number of personal votes + sum of votes of candidates
> of each branch:
>
> Branch1 (13)
>     Branch1.1 (7)
>         A (4)
>         B (3)
>     Branch1.2 (6)
>         C (6)
> Branch2 (18)
>     Branch2.1 (12)
>         D (5)
>         E (7)
>     Branch2.2 (5)
>         F (3)
>         G (2)
>     Branch2.3 (1)
>         H (1)
>
> - Branch2 has more votes than Branch1 => Branch2 wins
> - Branch2.1 has more votes than Branch2.2 and Branch2.3 => Branch2.1 wins
> - candidate E has more votes than candidate D => candidate E wins
>
> The tree approach thus forces the order of transfer to be non-cyclic. The
> transfer order of candidate E is E > D > {F, G, H}.
>
> The tree format can be printed on paper and it is easy to grasp. The ballot
> sheet may also follow the same tree format. Branches may have names (e.g.
> party names) or be unnamed. Left wing parties could join forces under one
> branch. Candidates of one party could be divided in smaller groups. Or maybe
> the branches have no party names and party affiliations, maybe just
> descriptive names, maybe no branch names at all.
>
>
Thanks for the example, but I don't understand. Who decides what the
branches are, and based on what? Why is E transferring votes if E has the
most votes? And what are the counts after each transfer? Sorry if those are
dumb questions.

--Russ P.

-- 
http://RussP.us
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