[EM] Approval-Sorted Margins(Ranking) Elimination

[Chris Benham]

Brian Olson wrote:

>I'm trying to understand the details of this procedure.
>
>On Apr 16, 2007, at 12:03 PM, Chris Benham wrote:
>
>  
>
>>My current favourite plain ranked-ballot method is  "Approval- 
>>Sorted Margins(Ranking) Elimination":
>>
>>1. Voters rank candidates, truncation and equal-ranking allowed.
>>
>>2. Interpreting ranking above bottom or equal-bottom as 'approval',  
>>initially order the candidates
>>according to their approval scores from the most approved (highest  
>>ordered) to the least approved
>>(lowest ordered).
>>    
>>
>
>I'm a little fuzzy on this step, it sounds like that reverse-IRV  
>method of disqualifying the most-last-placed choice.
>Does A>B>C>D mean I approve of all but D?
>

If there are no other candidates then for the initial ordering  (or 
"seeding" as the electowiki
ASM entry puts it)  yes.

>And I'd think A>B>C=D would mean I approve of A and B,..
>
Same answer. If there are no other candidates then C and D are ranked 
"equal-bottom".
The "above" in step 2 applies to both "bottom" and "equal-bottom".

>...but this  
>statement seems to imply approval for all of A-D, unless perhaps  
>there's E and F left unranked then it would approve A-D and not E,F
>

Brian, is this exactly what you meant to write?


>>3. If any candidate Y pairwise beats the candidate next highest in  
>>the order (X) , then modify the order
>>by switching  the order of the X>Y  pair  (to Y>X) that are closest  
>>in approval score.
>>Repeat until all the candidates not ordered top are pairwise beaten  
>>by the next highest-ordered candidate.
>>    
>>
>
>So, said another way, if the intermediate total order is
>A>B>C>D>E>F as ordered by approval counts, but more ballots rank C>B than B>C,  
>and more ballots rank E>D than D>E, then if the approval count [difference] of B-C  
>is less than D-E, then flop B and C in the intermediate order.
>Repeat fixing up the intermediate order, always with the closest  
>approval count difference, until no neighbors in the intermediate  
>order violate pairwise ranking winner. (This seems to be very much  
>like a condorcet process, actually, is it ever different unless  
>there's a tie?)
>
Yes, ASM is a Condorcet method.  And so of course is ASM(R)E. As I put 
it in my Apr.16 post:

> At some point in the process all except the candidates in the 
> top-cycle will be eliminated,..

The "top-cycle" is the Smith set.

>4. Eliminate and drop from the ballots the (now) lowest ordered  
>candidate.
>
>5. Repeat steps 2-4 until one candidate (the winner) remains.
>

Thanks for taking an interest,

Chris Benham


-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20070418/74421449/attachment-0003.htm>