[EM] Proposal: Majority Enhanced Approval (MEA)

[fsimmons at pcc.edu]

----- Original Message -----
From: Chris Benham 

> Forest,
> 
> This MEA method you have suggested would nearly always give the 
> same winner as 
> Smith//Approval (ranking), one of the methods I endorse.
> 
> Where they do give different winners, would it be the case that 
> the Smith//Approval winner
> is the more approved?  But outside the Uncovered set?

Yes.  So when the Smith consists entirely of uncovered candidates as in the case of a cycle of three, or 
a cycle of cycles of three, etc. they are the same.

The main difference is that MEA does the job "seamlessly," as Rob LeGrand wouold say, i.e. the 
method doesn't have to say do such and such if there is no Condorcet Winner, and detecting the Smith 
set is not some separate step in the procedure.

> 
> What is the most "realistic" example you can give of the two 
> methods giving different winners?

I haven't thought about this yet because mostly I think of MEA as a simpler way of implementing 
Smith//Approval with a few extra nice properties that are the frosting on the cake.

> 
> A feature of both methods that I am going off is that voters can 
> be punished for failing to truncate
> their least preferred of the viable candidates.
> 
> 49: A1>A2
> 24: B
> 27: C>B>A1
> 
> A1 is uncovered and most approved, so both methods elect A1. But 
> the presence of the weakly
> pareto-dominated clone A2 on the ballot caused the C supporters 
> to not truncate A1. If they had
> done so then B would have won.

It might be better to have an optional "approval cutoff" that voters can rank if they don't want to truncate 
all of the candidates they disapprove of.  I know this takes us ouoside of the standard pairwise matrix, 
but as i mentioned before, it seems like a waste to not use the diagonal entries for something, and why 
not approval?

Here's another proposal.  Let M be the matrix whose (i,j) element is the number of ballots on which 
candidate i is ranked ahead of candidate j.  I think that this is what you mean by the "normal gross 
pairwise matrix" that you mention below.

For each candidate i, let d(i) be the difference of the maximum number in column i and the minimum 
number in row i.   In other words d(i) is the difference is the maximum number of points scored against 
candidate i in a pairwise contest and the minimum number of points that candidate i scored in a pairwise 
contest.

Generally speaking, the smaller d(i), the stronger candidate i.

So list the candidates in increasing order of d(i) instead of the order of decreasing approval, and apply 
the enhancement as before:

Let D1 be the candidate i with the smallest difference d(i).  Elect D1 if uncovered, else let D2 be the 
smallest d(i) candidate among those that cover D1, etc.

This method wastes the diagonal slots of matrix M just like all of the other standard Condorcet 
methods.  But I would be interested if you would run it by your standard test cases.

> 
> This type of example is what motivated me to recently propose 
> that a candidate's 'biggest gross
> pairwise score in a pairwise victory over an uncovered 
> candidate' be used as a quasi-approval score.
> B then wins whether the C voters truncate A1 or not.
> 
> (The other motivation was that I was looking for something that 
> used nothing but the normal gross
> pairwise matrix.)
> 
> I recognize that this can cause failure of mono-raise, but 
> probably only in a complicated not very
> likely example.
> 
> Chris Benham
> 
> 
> 
> Forest Simmons wrote (8 May 2010):
> 
> I have a proposal that uses the same pairwise win/loss/tie 
> information that Copeland is based on, along with 
> the complementary information that Approval is based on.  It’s a 
> simple and powerful Condorcet/Approval 
> hybrid which, like Copeland, always elects an uncovered 
> candidate, but without the indecisiveness or clone 
> dependence of Copeland.
> 
> I used to call it UncAAO, but for better name recognition, I’m 
> changing the name to Majority Enhanced 
> Approval (MEA).
> 
> The method is extremely easy to understand once you get the 
> simple concept of covering.  Candidate X 
> covers candidate Y if candidate X pairwise beats both Y and 
> every candidate that Y beats pairwise.
> 
> MEA elects the candidate A1 that is approved on the greatest 
> number of ballots if A1 is uncovered. 
> Otherwise it elects the highest approval candidate A2 that 
> covers A1 if A2 is uncovered.  Otherwise it elects 
> the highest approval candidate A3 that covers A2 if A3 is 
> uncovered.  Otherwise, etc. until we arrive at an 
> uncovered candidate An, which is elected.
> 
> MEA satisfies Monotonicity, Clone Independence, Independence 
> from Pareto Dominated Alternatives, and 
> Independence from Non-Smith Alternatives, as well as all of the 
> following:
> 1.  It elects the same member of a clone set as the method would 
> when restricted to the clone set.
> 
> 2.  If a candidate that beats the winner is removed, the winner 
> is unchanged.
> 
> 3.  If an added candidate covers the winner, the new candidate 
> becomes the new winner.
> 
> 4.  If the old winner covers an added candidate, the old winner 
> still wins.
> 
> 5.  It always chooses from the uncovered set.
> 
> 6.  It is easy to describe:  Initialize L to be an empty list.  
> While there exists some alternative that covers 
> every member of L,  add to L the one (from among those) with the 
> greatest approval.  Elect the last 
> candidate added to L. 
> 
> What other deterministic method (based on ranked ballots with 
> truncations allowed) satisfies all of these 
> criteria?
> 
> Forest
> 
> 
> 
> 
> 
>