Why do you need to break ties? Wouldn't it make more sense to consider ties to
be ranked equally?
If you are trying to do it in as few ballots as possible, you might also
consider leaving tied candidates in their previous relative sort position (from
the last time through the loop).
The problem
RLSuter at aol.com writes:
It's a nice article and nicely formatted.
Thanks. And yes, I cleaned up the title. :)
What do you plan to
link to the underlined word in the bottom box for people wanting
to learn more about tabulating ranked ballots?
Hopefully an article I write myself,
Hi Rob.
Your little essay about how political parties form aka movie night
started out nice but got lame at the end.
(You also exhibit some high class knowledge of how to create web pages...
my web pages use old technology and I think simply cannot do the stuff you
did...)
Anyhow. To answer
I have examined this issue before in an unpublished paper whch I can
tell you about in separate email.
Anyhow, the thing is that some, but not other, Condorcet matrices are
actually achieveable as arising from actual sets of ballots.
Which ones are achievable? Well, you can tell by solving an
Warren Smith wds at math.temple.edu writes:
Your little essay about how political parties form aka movie night
started out nice but got lame at the end.
You mean the happily ever after part? (I'll agree that was lame and was kind
of a joke. I would like to find a better ending)
Or is there
Rob Brown wrote:
Why do you need to break ties? Wouldn't it make more sense to consider ties
to
be ranked equally?
Perhaps you're trying to approximate an election method that requires
fully-ranked ballots, in which case it makes sense to avoid ties
whenever possible.
More importantly,
Suppose that after the ballots come in and after the pairwise matrix has been
published, any and all are allowed to submit single elimination tournament
schedules in the form of binary trees of minimum possible depth (approximately
log base two of the number of candidates).
All of these
