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<div class="moz-cite-prefix">Luděk,<br>
<br>
One of the methods I like for public elections is a Condorcet-IRV
hybrid (the simplest of several) that has been dubbed "Benham".<br>
<br>
Voters strictly rank from the top however many candidates they
wish. If a Condorcet winner exists, he/she wins. Otherwise
continue<br>
with IRV, checking before each elimination for a CW among
remaining candidates and electing the first one to appear.<br>
<br>
A few others I like use "approval" information, interpreting
above-bottom ranking as approval (unless the ballot rules
allow/invite<br>
voters to explicitly specify an approval threshold in their
rankings). <br>
<br>
One of these methods that works here is Approval Margins Sort:
Line up the candidates in order of approval from highest to lowest<br>
as the first tentative ordering of the candidates (with the
Approval winner highest in the order). Look for adjacent pairs of
candidates<br>
where the candidate lower in the order pairwise beats the one
higher in the order. If there are none then the order is confirmed
and<br>
candidate highest in the order wins.<br>
<br>
If there are adjacent pairs of candidate who are "out of pairwise
order", flip the order of the one (of those pairs) with the
smallest approval-score gap.<br>
Repeat until each candidate not bottom in the order pairwise beats
the candidate just below it. Then elect the highest-ordered
candidate.<br>
<br>
<a class="moz-txt-link-freetext" href="http://wiki.electorama.com/wiki/Approval_Sorted_Margins">http://wiki.electorama.com/wiki/Approval_Sorted_Margins</a><br>
<br>
<br>
1: A>B<br>
2: B>C<br>
1: C>A<br>
1: C>A<br>
<br>
Approval Scores: C4 > A3 = B3<br>
<br>
Here B and A have the same approval scores but are out of order
pairwise, so the next tentative order is C > A > B. Now no
adjacent<br>
pair of candidates (C>A or A>B) is out of pairwise order
so that order is final and C wins.<br>
<br>
Another one of the methods I like for public office is
Smith//Approval: Elect the most approved member of the Smith set.
The "Smith<br>
set" is the smallest set of candidates that pairwise-beat all (if
any) of the candidates outside the set.<br>
<br>
Here all the candidates are in the Smith set, and the most
approved of them is C, so again C wins.<br>
<br>
Something a bit more elegant that might give a different result in
a complicated example with more than 3 candidates in the Smith set<br>
is a suggestion of Forrest Simmons: Construct a "chain" of
candidates thus: begin with the most approved candidate. If no
candidate<br>
"covers" this candidate A (i.e. pairwise beats A and also pairwise
beats every candidate that A pairwise beats and doesn't pairwise
lose<br>
to any candidate A pairwise ties with) then A wins. Otherwise add
to the chain the most approved candidate that covers A. For as
long<br>
as it is possible, keep adding to the chain the most approved
candidate that pairwise beats the last candidate added. When no
new candidate<br>
can be added, elect the last added candidate.<br>
<br>
In this example all the voters have submitted a full ranking, so
maybe they (and/or you) think that interpreting above-bottom
ranking as<br>
approval is arbitrary and too much focused on the bottoms of the
ballots. <br>
<br>
In that case a possible alternative is Borda Margins-Sort
Elimination: Using Borda scores (on each ballot each candidate
gets a point-score<br>
equal to the number of candidates ranked below it. Candidates
ranked the same should perhaps be handled in a away that meets
"Symmetric<br>
Completion", so that say a "A=B > C" ballot should give 1.5
points each to A and B and zero points to C.) instead of Approval
scores, order<br>
the candidates as in Approval-Margins Sort. <br>
<br>
Then eliminate the lowest ordered candidate, recalculate new Borda
scores without the eliminated candidate. Repeat until the last
remaining 3 candidates<br>
are finally ordered and then elect the one highest in that order.<br>
<br>
In your example the candidates' Borda scores are the same (5), so
here this method is also a tie.<br>
<br>
Chris Benham<br>
<br>
<br>
<br>
On 9/14/2016 5:27 PM, Luděk Belán wrote:<br>
</div>
<blockquote cite="mid:C7N.Bgss.4azFvSyh4rf.1NsGBE@seznam.cz"
type="cite">Chris,
<div>thank you for confirmation. Do you know any variant of
Schulze method (or of other Condorcet method) to determine the
winner in this example?</div>
<div>Thank you.</div>
<div><br>
</div>
<div>Luděk Belán</div>
<div><br>
</div>
<div><br>
<p>---------- Původní zpráva ----------<br>
Od: C.Benham <a class="moz-txt-link-rfc2396E" href="mailto:cbenham@adam.com.au"><cbenham@adam.com.au></a><br>
Komu: <a class="moz-txt-link-abbreviated" href="mailto:election-methods@lists.electorama.com">election-methods@lists.electorama.com</a><br>
Datum: 12. 9. 2016 18:36:36<br>
Předmět: Re: [EM] Example with tie in Schulze method but not
in IRV</p>
<br>
<blockquote>Luděk,<br>
<br>
Yes, you are right.<br>
<br>
<br>
Chris Benham<br>
<br>
<br>
<br>
On 9/12/2016 7:47 PM, Luděk Belán wrote:<br>
> Dear all,<br>
><br>
> excuse my bad English.<br>
><br>
> Example for discusion:<br>
> Candidates: A, B, C<br>
> Ballots (count: order):<br>
> 1: A>B>C<br>
> 2: B>C>A<br>
> 1: C>A>B<br>
> 1: A>C>B<br>
><br>
> In Instant Runoff Voting wins candidate A, but in Schulze
method in my opinion is result tie.<br>
> It's true, please?<br>
><br>
> Best regards<br>
><br>
> Luděk Belán<br>
> ----<br>
> Election-Methods mailing list - see
<a class="moz-txt-link-freetext" href="http://electorama.com/em">http://electorama.com/em</a> for list info<br>
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