<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<meta content="text/html;charset=ISO-8859-1" http-equiv="Content-Type">
<title></title>
</head>
<body bgcolor="#ffffff" text="#000000">
<br>
<br>
Brian Olson wrote:<br>
<blockquote type="cite">
<pre wrap="">I'm trying to understand the details of this procedure.
On Apr 16, 2007, at 12:03 PM, Chris Benham wrote:
</pre>
<blockquote type="cite">
<pre wrap="">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).
</pre>
</blockquote>
<pre wrap=""><!---->
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?</pre>
</blockquote>
<br>
If there are no other candidates then for the initial ordering (or
"seeding" as the electowiki<br>
ASM entry puts it) yes.<br>
<blockquote type="cite">
<pre wrap="">And I'd think A>B>C=D would mean I approve of A and B,..</pre>
</blockquote>
Same answer. If there are no other candidates then C and D are ranked
"equal-bottom".<br>
The "above" in step 2 applies to both "bottom" and "equal-bottom".<br>
<br>
<blockquote type="cite">
<pre wrap="">...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</pre>
</blockquote>
<br>
Brian, is this exactly what you meant to write?<br>
<br>
<br>
<blockquote type="cite">
<blockquote type="cite">
<pre wrap="">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.
</pre>
</blockquote>
<pre wrap=""><!---->
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?)</pre>
</blockquote>
Yes, ASM is a Condorcet method. And so of course is ASM(R)E. As I put
it in my Apr.16 post:<br>
<br>
<blockquote type="cite">At some point in the process all except the
candidates in
the top-cycle will be eliminated,..</blockquote>
The "top-cycle" is the Smith set.<br>
<br>
<blockquote type="cite">
<pre wrap="">4. Eliminate and drop from the ballots the (now) lowest ordered
candidate.
5. Repeat steps 2-4 until one candidate (the winner) remains.</pre>
</blockquote>
<br>
Thanks for taking an interest,<br>
<br>
Chris Benham<br>
<br>
<br>
</body>
</html>