[EM] Schulze ("Approval-Domination prioritised Margins")

C.Benham cbenhamau at yahoo.com.au
Tue Sep 21 18:10:04 PDT 2010

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"

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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