<div dir="ltr">Can you please post an explanation of "<span style="font-size:12.8px">Margins-Sorted Losing Votes (erw) Elimination" on electowiki?</span></div><div class="gmail_extra"><br><div class="gmail_quote">2016-10-26 0:08 GMT-04:00 C.Benham <span dir="ltr"><<a href="mailto:cbenham@adam.com.au" target="_blank">cbenham@adam.com.au</a>></span>:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
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<div class="m_-6885124151727990076moz-cite-prefix">Forest has pointed out that my supposed
example of Approval Sorted Margins failing mono-switch-plump is
nonsense.<br>
<br>
So (at least for the time being) I am not able to show that
Approval Sorted Margins fails the mono-switch-plump criterion.<br>
<br>
(BTW I am still very happy with the Margins-Sorted Losing Votes
(erw) Elimination method.)<br>
<br>
Chris Benham<div><div class="h5"><br>
<br>
<br>
On 10/24/2016 10:28 PM, C.Benham wrote:<br>
</div></div></div>
<blockquote type="cite"><div><div class="h5">
<p>The Mono-switch-plump criterion is much stronger than I
previously thought, and is probably simply incompatible with the
<br>
Condorcet criterion.<br>
</p>
<p>I used to think that its met by two of my favourite Condorcet
methods, Margins-Sorted Losing Votes (erw) Elimination
(equivalent in the 3 candidate case<br>
to the "MMLV(erw)M" I discuss in the May 2014 post) and Approval
Sorted Margins. Consider this election under MSLVerwE :<br>
</p>
<p>40: A<br>
29: C>A<br>
03: B<br>
28: B>C<br>
</p>
<p>A>B 69-31, B>C 31-29, C>A 57-40. LV(erw)
scores: A40 > B31 > C29. No adjacent pair is
out-of-order pairwise, so MSLV(erw)E elects A.<br>
<br>
But if we switch the 3 B plumping ballots to A then C becomes
the Condorcet winner (C>B 29-28, C>A 57-43).<br>
<br>
43: A<br>
29: C>A<br>
28: B>C<br>
</p>
<p>And now this election under Approval Sorted Margins:<br>
<br>
30: C<br>
04: C>A<br>
33: A>B<br>
32: B<br>
<br>
A>B 37-32, B>C 64-34, C>A 34-33. (Implicit)
Approval scores: B64 > A37 > C34. The adjacent pair with
the smallest (absolute margin) difference<br>
in their scores (A > C) is pairwise out of order so we flip
that to give B > C > A. Now neither adjacent pair is
pairwise out-of-order, so the order is<br>
final and so Margins Sorted Approval elects B.<br>
</p>
<p>But if we switch two of the 32 B plumping ballots to A then A
becomes the Condorcet winner (A>B 39-34, A>C 35-34).<br>
</p>
<p>30: C<br>
04: C>A<br>
33: A>B<br>
02: A<br>
30: B<br>
<br>
I doubt that IBIFA meets the criterion. <br>
<br>
But I remain sure that it's met by Bucklin (and similar methods
like MTA and MCA and QLTD).<br>
<br>
Chris Benham<br>
<br>
</p>
<p>On 11 May 2014 Chris Benham posted to EM:<br>
</p>
<blockquote type="cite">
<div class="m_-6885124151727990076moz-text-flowed" style="font-family:-moz-fixed;font-size:14px" lang="x-western"> <br>
<blockquote type="cite" style="color:#000000"> Mono-switch-plump:
<br>
<br>
*The probability of candidate X winning must not be reduced
if one or more ballots that <br>
plump for any not-X are replaced by an equal number of
ballots that plump for X.* <br>
</blockquote>
<br>
Previously I showed that this is failed by the following
methods: <br>
<br>
Schulze (aka Beatpath), Ranked Pairs, River, MinMax (all
equivalent with 3 candidates) if they use Winning Votes to
weigh pairwise defeats. <br>
<br>
IRV and the Condorcet methods based on IRV (such as Benham
and Woodall) <br>
<br>
Total Approval Chain Climbing. <br>
<br>
I claim that it is met by Margins, any positional method,
IBIFA, Bucklin and Bucklin-like methods like Median Ratings
and MCA and MTA. <br>
<br>
And also it is met by MMLV(erw)M. To support that claim
I'll just talk about the Margins Sort version with 3
candidates. <br>
<br>
Plumping ballots for any X always contribute to X's score
and switching plumping ballots to X might get rid of one of
X's pairwise defeats. <br>
<br>
If X has no pairwise defeats then that will always be still
the case after switching some plumping ballots to X and so X
will still win. X can't <br>
be a winner with all pairwise defeats so we are only concerned
about the case when X has just one (and so will the other 2
candidates). <br>
<br>
Say we designate the candidate with the highest score 1, the
second-highest 2 and and the lowest 3. The algorithm in this
3-candidate cycle <br>
situation elects 1 unless 2 both pairwise beats 1 and has a
score that is closer to 1's than to 3's. <br>
<br>
If winning candidate X is in position 2 then the effect of
plumping ballots being switched from 1 to 2 will be to just
make 2 still closer to 1, <br>
and the effect of plumping ballots being switched from 3 to 2
will have the same effect (and make 3 further away). <br>
<br>
If winning candidate X is 1 and pairwise beats 2 and loses
to 3, then the only hope of making 1 lose is to switch some
plumping ballots from <br>
2 to 1 sufficient for 2 and 3 to change places but that won't
work because then 2 and 3 will be adjacent candidates that are
out of pairwise <br>
order and will be much closer together score-wise than the
other such pair and they'll be switched back to give the final
order 1>2>3. <br>
<br>
And if X is 1 and losing to 2 then it means that 1's distance
(scorewise) from 2 is such that 2 and 3 are switched in the
order, and switching <br>
any plumping ballots to 1 will only increase that distance. <br>
<br>
I hope that (almost confused) waffle is not too confusing or
opaque. <br>
<br>
Chris Benham <br>
<br>
<br>
<br>
<br>
<br>
Mono-switch-plump: <br>
<br>
*The probability of candidate X winning must not be reduced if
one or more ballots that <br>
plump for any not-X are replaced by an equal number of
ballots that plump for X.* <br>
<br>
Mono-raise is the traditional monotonicity criterion, but I
don't see why anyone would <br>
see failure of Mono-switch-plump as less embarrassing than
failing Mono-raise. <br>
<br>
<br>
25 A>B <br>
26 B>C <br>
23 C>A <br>
22 C <br>
04 A <br>
<br>
B>C 51-45 C>A 71-29 A>B 52-26 <br>
<br>
Top Preferences: C45 > A29 > B26 <br>
<br>
When there are three candidates the MinMax , Beatpath (aka
Schulze), Ranked Pairs and River algorithms <br>
are all equivalent. When they use Winning Votes as the measure
of defeat strength they all elect C. <br>
<br>
IRV (aka the Alternative Vote) and Benham (and Woodall) also
elect C. But if we replace the 4A ballots <br>
with 4C ballots the winner with all these methods changes from
C to B. <br>
<br>
25 A>B <br>
26 B>C <br>
23 C>A <br>
26 C <br>
<br>
B>C 51-49 C>A 71-29 A>B 48-26 <br>
<br>
Top Preferences: C45 > B26 > A25 <br>
<br>
Total Approval Chain Climbing also fails. <br>
<br>
25 A>B <br>
06 A>C <br>
32 B>C <br>
27 C>A <br>
08 C <br>
02 B <br>
<br>
C>A>B>C, Approvals C73 > B59 > A58 <br>
<br>
TACC elects C, but if the 2B ballots are changed to 2C, then
the winner changes to A. <br>
<br>
25 A>B <br>
06 A>C <br>
32 B>C <br>
27 C>A <br>
10 C <br>
<br>
C>A>B>C, Approvals C75 > A58 > B57 <br>
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</blockquote>
<br>
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