On 18 Jan 2009 I proposed a Condorcet method, "Approval-Domination
Prioritised Margins":
I have an idea for a new defeat-strength measure for the Schulze algorithm
(and similar such as Ranked Pairs and River), which I'll call:
"Approval-Domination prioritised Margins":
*Voters rank from the top however many candidates they wish.
Interpreting ranking (in any position, or alternatively above at least
one other
candidate) as approval, candidate A is considered as "approval dominating"
candidate B if A's approval-opposition to B (i.e. A's approval score
on ballots
that don't approve B) is greater than B's total approval score.
All pairwise defeats/victories where the victor "approval dominates"
the loser
are considered as stronger than all the others.
With that sole modification, we use Margins as the measure of defeat
strength.*
This aims to meet SMD (and so Plurality and Minimal Defense,
criteria failed
by regular Margins) and my recently suggested "Smith- Comprehensive 3-slot
Ratings Winner" criterion (failed by Winning Votes).
http://lists.electorama.com/pipermail/election-methods-electorama.com/2008-December/023595.html
Here is an example where the result differs from regular Margins,
Winning Votes
and Schwartz//Approval.
44: A
46: B>C
07: C>A
03: C
A>B 51-46 = 5 *
B>C 46-10 = 36
C>A 56-44 = 12
Plain Margins would consider B's defeat to be the weakest and elect B,
but that is the only
one of the three pairwise results where the victor
"approval-dominates" the loser. A's approval
opposition to B is 51, higher than B's total approval score of 46.
So instead my suggested alternative considers A's defeat (with the
next smallest margin) to be
the weakest and elects A. Looking at it from the point of view of
the Ranked Pairs algorithm
(MinMax, Schulze, Ranked Pairs, River are all equivalent with three
candidates), the A>B result
is considered strongest and so "locked", followed by the B>C result
(with the greatest margin)
to give the final order A>B>C.
Winning Votes considers C's defeat to be weakest and so elects C.
Schwartz//Approval also
elects C.
Margins election of B is a failure of Minimal Defense. Maybe the B
supporters are Burying
against A and A is the sincere Condorcet winner.
I've discovered that this actually fails my suggested "Smith-
Comprehensive 3-slot Ratings Winner"
criterion.
20: A>B
20: A=B
15: B>C
45: C
C>A 60-40 = 20 *
A>B 20-15 = 5
B>C 55-45 = 10
In this example borrowed from Kevin Venzke, C is in the Smith set, has
the highest Top-Ratings score,
the highest Approval score and the lowest Maximum Approval Opposition
score and yet B wins.
So I withdraw my endorsement of this method. I no longer see any real
justification for preferring it to
the much simpler Smith//Approval, which I continue to endorse.
Chris Benham
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