On Tue, 19 Jun 2001, LAYTON Craig wrote in part:
> Forest,
>
> you wrote:
>
> >This fact suggests also using the d'Hondt count within each list to decide
> >which of the list's candidates get to fill the seats won by the list
> >(party).
> >
> >Then if there is only one list, the election will reduce to sequential
> >PAV, maintaining the PR status of the election.
>
> d'Hondt isn't possible in the conventional sense,
True
> because each candidate
> can't win more than one seat,
But the voters who approved that one candidate form a set whose influence
on subsequent choices can be reduced by the d'Hondt rule, as in sequential
PAV.
> although PAV would be a good method to decide
> between the candidates on each list. I'm not sure why you switched to
> sequential PAV. It doesn't appear to me to be an improvement, but you are
> much better qualified than me on this subject, as it's your method and all.
>
It's not an improvement unless a pecking order is desirable. However in
the case of many winners it is more computationally feasible, since
non-sequential PAV requires rating all of the subsets of size N if there
are to be N winners.
> >What about allowing parties to submit multiple lists with the party name
> >on them without actually splitting the party?
>
> I would limit parties to just one list. Keeping the number of lists down
> would be difficult enough. In the last New South Wales state election (my
> state) the upper house (elected by list STV) had over 30 or 40 party lists -
> and that was with each party being allowed only one list. Voters often
> complain about the size of ballot papers (which can be up to a few feet
> across).
I certainly wouldn't want to multiply lists in a situation like that!
I think the simplicity of your original suggestion out weighs any possible
benefits from the tweaks that I suggested, especially in the case of
ordinary public elections.
One question: What if the d'Hondt quota for a party's list exceeds the
number of candidates on the list? Are all of the "extra" votes wasted? Or
are they redistributed proportionally when the list is exhausted?
Forest
