[EM] Approval-Sorted Margins(Ranking) Elimination
Chris Benham
chrisjbenham at optusnet.com.au
Tue Apr 17 08:41:21 PDT 2007
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
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