[EM] RE: Improved Condorcet Approval (ICA)
simmonfo at up.edu
Fri Sep 16 14:33:32 PDT 2005
you are right in all of your comments. The mix up with PM and CM and rows and columns was due to me being in an extreme hurry.
Here's another possibility:
A candidate's score is the ratio of the minimum entry in his row of PM and the maximum entry in his row of CM.
The candidate with the highest score wins.
In other words, the candidate with the highest value of L/G where L is his least score against another candidate, and G is the greatest score some other candidate scores against him.
[But it has to be understood that any first place vote is counted as a score for you against every other candidate in the computation of L, and that every truncated vote is counted as a score against you by every other candidate in the computation of G.]
I think that this satisfies the FBC, but optimal strategy might be to rank equal all approved candidates, and truncate all others. I'm not sure.
Kevin replied to the following:
--- "Simmons, Forest " <simmonfo at up.edu> a écrit :
> Kevin, your ICA method interests me. In particular, your creative use of "equal ranked top"
> might be called "power top" analogous to what Mike Ossipoff recently called "power truncation"
> for equal (non)ranking at the bottom.
Maybe. The "tied at the top" rule can be used with just one pairwise matrix
if you only consider "votes for" and don't find the winners of pairwise contests.
For ICA you need a second (symmetrical) matrix to keep track of how many voters
vote each pair of candidates tied at the top.
You can ditch ICA's second matrix if you only respect majority-strength wins
(i.e., the MDDA method). But I feel like this method is much less sensitive
to the ranking data.
> I suggest that we consider methods that sum two modified pairwise matrices in addition to the
> basic pairwise matrix:
> (This description is at the ballot level)
> In the ordinary pairwise matrix M, the (i,j) entry is a one or a zero depending on whether or
> not candidate i is ranked ahead of candidate j on the ballot.
> In the "Pro modification" PM, if candidate i is ranked equal first, then row i is filled in
> with ones.
This sounds right. (ICA isn't done in this way, though.)
> In the "Con modification" CM, if candidate k is truncated, then column k is filled in with
> ones. This is Ossipoff's "power truncation" matrix.
> We pit the candidate with the strongest offense against the candidate with the strongest
> The offensive winner is the candidate for whom the minimal row element of the PM matrix is
> maximal, i.e. the MMPO winner with power truncation.
"Row element" and "offensive" mean "votes for," don't they? That doesn't sound
like MMPO, but rather Woodall's MinGS, under which if there is some candidate
against whom you have no pairwise votes, then your score is 0.
> The defensive winner is the candidate for whom the maximal element of the CM matrix is minimal.
This sounds like MMPO to me.
> If these two winners are different, then the ordinary pairwise matrix M decides between them.
> It seems like this method might satisfy the FBC: the process of picking the offensive winner
> must satisfy the FBC for the same reason that MMPO does. And it seems to me that the process of
> picking the defensive winner satisfies the FBC for the same reason that ICA does.
> What do you think?
I think the pairwise comparison on the M matrix forfeits FBC. I don't know a
way of doing pairwise comparisons and preserving FBC without either 1) having
a "tied at the top" matrix in addition to the pairwise matrix, or 2) only
respecting wins with more than half of the voters on the winning side.
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