# [EM] Example with tie in Schulze method but not in IRV (small correction)

C.Benham 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.

Chris Benham

On 9/15/2016 3:25 AM, C.Benham wrote:
> Luděk,
>
> 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.
>
> http://wiki.electorama.com/wiki/Approval_Sorted_Margins
>
>
> 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
> "Smith
> 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
> point-score
> 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:
>> Chris,
>> 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
>>
>>
>>     Luděk,
>>
>>     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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