Um, wouldn't that require 10! people = 3,628,800 for ten candidates?

-- Ernie P.

On Nov 17, 2003, at 5:39 PM, Gervase Lam wrote:

Date: Mon, 17 Nov 2003 14:11:01 -0500
From: Dave Ketchum <[EMAIL PROTECTED]>
Subject: Re: [EM] Electronic Voting Bill of Rights?

CONDORCET:  Choices here also:
      Count as in Plurality?  Tempting, since a majority winner would
end the task.

That's it! You've reminded me of a way to do a Condorcet hand count.


An explanation follows, the logistics of which could be improved.  For
brevity, I have not taken into account tied votes nor do I know how to
atm.  I want to get to bed as soon as possible.

Suppose there are 10 candidates.  Create 10 large teams.  Each team
represents a candidate who was given the top rank.  Each of the teams
obtain the relevant ballots.  From this, you are now able to fill in a
large proportion of the pairwise matrix as you know that a top ranked
candidate is a beats all candidate.

Each large team has 9 sub-teams, the 9 being for each 2nd ranked
candidate. Each sub-team receives the ballots from its parent large team.
Again, from this, you are now able to fill in a large proportion of the
pairwise matrix, a part of which has already been filled in by the parent.


The 9 sub-teams each have 8 sub-teams etc, etc...

Therefore, you are able to complete a Condorcet hand count in N - 1
iterations, where N is the number of candidates.

Thanks,
Gervase.
----
Election-methods mailing list - see http://electorama.com/em for list info

---- Election-methods mailing list - see http://electorama.com/em for list info

Reply via email to