[EM] Example with tie in Schulze method but not in IRV (small correction)
cbenham at adam.com.au
Thu Sep 15 00:52:30 PDT 2016
In defining Forest's algorithm I wrote:
> "For as long as it is possible, keep adding to the chain the most
> approved candidate that pairwise beats the last candidate added."
The "pairwise beats" is wrong. It should be ' *covers*'.
Also I omitted the detail in the Margins-Sort algorithm of what to do
when score gaps between pairwise out-of-order candidates are the same.
These mistakes are fixed in text below.
On 9/15/2016 3:25 AM, C.Benham wrote:
> One of the methods I like for public elections is a Condorcet-IRV
> hybrid (the simplest of several) that has been dubbed "Benham".
> Voters strictly rank from the top however many candidates they wish.
> If a Condorcet winner exists, he/she wins. Otherwise continue
> with IRV, checking before each elimination for a CW among remaining
> candidates and electing the first one to appear.
> A few others I like use "approval" information, interpreting
> above-bottom ranking as approval (unless the ballot rules allow/invite
> voters to explicitly specify an approval threshold in their rankings).
> One of these methods that works here is Approval Margins Sort: Line up
> the candidates in order of approval from highest to lowest
> as the first tentative ordering of the candidates (with the Approval
> winner highest in the order). Look for adjacent pairs of candidates
> 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
> candidate highest in the order wins.
> 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.
> If among the out-of-pairwise-order adjacent pairs there is gap-size
> tie, flip the order of lowest ordered pair.
> Repeat until each candidate not bottom in the order pairwise beats the
> candidate just below it. Then elect the highest-ordered candidate.
> 1: A>B
> 2: B>C
> 1: C>A
> 1: C>A
> Approval Scores: C4 > A3 = B3
> 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
> pair of candidates (C>A or A>B) is out of pairwise order so that
> order is final and C wins.
> Another one of the methods I like for public office is
> Smith//Approval: Elect the most approved member of the Smith set. The
> set" is the smallest set of candidates that pairwise-beat all (if any)
> of the candidates outside the set.
> Here all the candidates are in the Smith set, and the most approved of
> them is C, so again C wins.
> Something a bit more elegant that might give a different result in a
> complicated example with more than 3 candidates in the Smith set
> is a suggestion of Forrest Simmons: Construct a "chain" of candidates
> thus: begin with the most approved candidate. If no candidate
> "covers" this candidate A (i.e. pairwise beats A and also pairwise
> beats every candidate that A pairwise beats and doesn't pairwise lose
> 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
> as it is possible, keep adding to the chain the most approved
> candidate that covers the last candidate added. When no new candidate
> can be added, elect the last added candidate.
> In this example all the voters have submitted a full ranking, so maybe
> they (and/or you) think that interpreting above-bottom ranking as
> approval is arbitrary and too much focused on the bottoms of the ballots.
> In that case a possible alternative is Borda Margins-Sort
> Elimination: Using Borda scores (on each ballot each candidate gets a
> equal to the number of candidates ranked below it. Candidates ranked
> the same should perhaps be handled in a away that meets "Symmetric
> 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
> the candidates as in Approval-Margins Sort.
> Then eliminate the lowest ordered candidate, recalculate new Borda
> scores without the eliminated candidate. Repeat until the last
> remaining 3 candidates
> are finally ordered and then elect the one highest in that order.
> In your example the candidates' Borda scores are the same (5), so
> here this method is also a tie.
> Chris Benham
> On 9/14/2016 5:27 PM, Luděk Belán wrote:
>> thank you for confirmation. Do you know any variant of Schulze method
>> (or of other Condorcet method) to determine the winner in this example?
>> Thank you.
>> Luděk Belán
>> ---------- Původní zpráva ----------
>> Od: C.Benham <cbenham at adam.com.au>
>> Komu: election-methods at lists.electorama.com
>> Datum: 12. 9. 2016 18:36:36
>> Předmět: Re: [EM] Example with tie in Schulze method but not in IRV
>> Yes, you are right.
>> Chris Benham
>> On 9/12/2016 7:47 PM, Luděk Belán wrote:
>> > Dear all,
>> > excuse my bad English.
>> > Example for discusion:
>> > Candidates: A, B, C
>> > Ballots (count: order):
>> > 1: A>B>C
>> > 2: B>C>A
>> > 1: C>A>B
>> > 1: A>C>B
>> > In Instant Runoff Voting wins candidate A, but in Schulze
>> method in my opinion is result tie.
>> > It's true, please?
>> > Best regards
>> > Luděk Belán
>> > ----
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