[EM] This might be the method we've been looking for:

[fsimmons at pcc.edu]

Chris, 

you're right that it is very close to MinMax(margins). Let's compare and contrast:

In both MinMax versions a matrix M is used to determine the winner in the same way:  if the least
number in row i is greater than the least number in any other row of the matrix M, then candidate i is 
elected.  [By convention each negative number is less than every positive number, and among several 
negative numbers the "most negative" is the least.  So  -6 < -3  <  2 < 5, etc.]

In both methods each entry of the matrix M is the difference betwee two numbers (minuend minus 
subtrahend).

The subrtrahend in this difference is exactly the same in both methods.  The subtrahend in row i column 
j of M is the number of ballots on which candidate j is rated or ranked strictly above candidate i.

It's in the minuend that the two methods part company:

In MinMax(margins) the minuend of the (i, j) entry is the number of ballots on which candidate i is rated 
or ranked strictly above candidate j.

In MinMax(TopTierPairwiseRule) the minuend of the (i, j) entry is the number of ballots on which 
candidate i is ranked (or rated) strictly above candidate j plus the number of ballots on which candidate i 
is rated or ranked equal top with candidate j.

A third method that I call MinMax(EqualRankPairwiseRule) uses the same subtrahend but defines the 
minuend as the number of ballots on which candidate i is ranked both above bottom AND above or equal 
to candidate j.

This last method MinMax(ERPR) also satisfies the FBC, and furthermore it nevers gives incentive for 
insincere order reversal.

Both MinMax(ERPR) and MinMax(TTPR) satisfy the mono-add-equal-top criterion: if additional ballots 
are added with the previous winner ranked top or equal top, the winner is unchanged.

Furthermore, suppose that candidate i is the winner under MinMax(ERPR), and that the least number in 
row i of matrix M is -7.  Suppose that this number -7 appears only in columns  3, 9, and 15 of row i.  If a 
new ballot ranks candidate i above or equal to candidates 3, 9, and 15, then the method will still elect 
candidate i when the new ballot is counted along with the old ones.

Note that in the case of MinMax(ERPR) the diagonal entries (i, i) in the matrix M are the respective 
implicit approvals of the candidates, since ranked candidates are ranked equal to themselves but not 
above themselves.

All three of these MinMax methods are monotone, but fail clone independence in the same sense that 
MinMax(wv) does.  The equal ranking option mitigates this failure.  Perhaps further modifications could 
mitigate it more, if not altogether remove it.  For example, incorporating some version of the Cardinal 
Weighted Pairwise idea might restore clone independence to the same degree enjoyed by Approval and 
other Cardinal Ratings methods.

We can deal with that later.  Meanwhile, with a three slot method, clones tend to get equal ranked a lot, 
so the clone dependence is not much worse than it is in Approval.

We need a popular name that can catch on with the public. Any ideas?

Forest

----- Original Message -----
From: "C.Benham" 
Date: Saturday, December 3, 2011 0:24 am
Subject: This might be the method we've been looking for:
To: em 
Cc: Forest W Simmons 

> Forest,
> 
> I don't understand the algorithm's definition. It seems to be 
> saying 
> that it's MinMax(Margins), only computing X's gross pairwise 
> score 
> against Y by giving X 2 points for every ballot on which X is 
> both 
> top-rated and voted strictly above Y, and otherwise giving X 1 
> point for 
> every ballot on which X is top-rated *or* voted strictly above Y.
> 
> But from trying that on the first example it's obvious that 
> isn't it. 
> Can someone please explain it to me?
> 
> Chris Benham
> 
> 
> Forest Simmons wrote (2 Dec 2011):
> 
> Here’s a method that seems to have the important properties that 
> we have 
> been worrying about lately:
> 
> (1) For each ballot beta, construct two matrices M1 and M2:
> In row X and column Y of matrix M1, enter a one if ballot beta 
> rates X 
> above Y or if beta gives a top
> rating to X. Otherwise enter a zero.
> IN row X and column y of matrix M2, enter a 1 if y is rated 
> strictly 
> above x on beta. Otherwise enter a
> zero.
> 
> (2) Sum the matrices M1 and M2 over all ballots beta.
> 
> (3) Let M be the difference of these respective sums
> .
> (4) Elect the candidate who has the (algebraically) greatest 
> minimum row 
> value in matrix M.
> 
> Consider the scenario
> 49 C
> 27 A>B
> 24 B>A
> Since there are no equal top ratings, the method elects the same 
> candidate A as minmax margins
> would.
> 
> In the case
> 49 C
> 27 A>B
> 24 B
> There are no equal top ratings, so the method gives the same 
> result as 
> minmax margins, namely C wins
> (by the tie breaking rule based on second lowest row value 
> between B and C).
> 
> Now for
> 49 C
> 27 A=B
> 24 B
> In this case B wins, so the A supporters have a way of stopping 
> C from 
> being elected when they know
> that the B voters really are indifferent between A and C.
> 
> The equal top rule for matrix M1 essentially transforms minmax 
> into a 
> method satisfying the FBC.
> 
> Thoughts?
> 
> 
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> 
> 
> 
>