[EM] Improved Copeland (was "A New Spinoff of Our Recent Discussions")
Forest Simmons
fsimmons at pcc.edu
Thu Jun 13 19:50:50 PDT 2019
Sorry to take so long to reply.
First I want to give another formulation of Imprroved copeland:
It takes two passes (barring some miracle approach). The first pass is to
construct the pairwise win matrix in which the element in row i and column
j is a one if i beats j pairwise, negative one if j beats i, and zero
otherwise.
Based on this matrix construct a directed graph ffor each ballot, with an
arrow from each candidate to every candidate he pairwise beats. If any
arrow points from a lower ranked candidate to an higher ranked one, then
delete it along with all of the other arrows pointing to the higher
candidate and all of the arrows emanating from the lower one. The score of
each candidate is the total (over all ballots) of the indegrees minus the
outdegrees of the remaining arrows impinging on that candidate.
In matrix terms, for each ballot the matrix is constructed (with reference
to the pairwise win matrix) as follows. Initially If candidate i is ranked
ahead of j on
the ballot then put a one in the (i, j) position in the matrix, zero
otherwise. Subtract the transpose of this matrix from itself to get a new
matrix. Then mark all of the elements that are not identical in the two
matrices, and then zero out the rows and columns of the marked elements.
The resultingg matrix is the contribution of the ballot to the total
matrix.
The row with the largest sum (or the column with the smallest sum)
represents the winner.
Now if we just zeroed outt the offending elements without zeroing out their
entire rows and columns the sum would be something we could do sneakily:
Average the Margins matrix with n copies of the win/loss matrix where n is
the total number of ballots. That is the matrix you would get iff you only
removed the arrows pointing in the wrong direction without removing the
other associated witth the affected indegrees and outdegrees.
o that related, but inferior< method is precinct summable.
Inferior because it definitely still has all of the clone problems off
Borda and copeland, while ffailing Landau. The zeroing out off the rrows
and columns is crucial to preserving Landau, and (I hope it tkes care of
the clone dependence, too, since it makes the remaining vertices more like
the first place vertices that we were trying to limit our count to in ourr
first attempt).
Thinking in terms of the (modified) directed graphs correspondinig to the
matrices with the rows and columns zeroed out, it is easy to see how
monotonicitty is preserrved, and almost as easy to see how Landau is
preserved.
That's all I have time for now.
Best,
Forest
On Thu, Jun 6, 2019 at 2:11 PM Forest Simmons <fsimmons at pcc.edu> wrote:
> Here's another slightly simpler approach aimed at the lay voter:
>
> Tell the audience that the Condorcet ideal is a candidate that is not
> pairwise beaten by any other candidate.
>
> When that is not possible, it is natural to consider a candidate that is
> beaten pairwise by the fewest other candidates. This idea is the basis of
> the Copeland Method.
>
> There are two problems with the Copeland method: (1) It has a strong
> tendency to produce ties, and (2) More subtle problems created by cloning
> certain candidates to increase the number of defeats suffered by certain
> other candidates without increasing the number of defeats of the cloned
> candidates.
>
> Because of these two problems, Copeland is not considered a serious
> contender for use in public elections.
>
> But what if there were a simple modification of Copeland that would
> totally resolve these two problems in one fell swoop?
>
> There is; instead of counting the number of candidates that defeat
> candidate X, (and electing the candidate with the smallest count), we add
> up all of the first place votes of all of the candidates that defeat X, and
> elect the candidate with the smallest sum.
>
> This solves the first problem because in any moderate to large sized
> election, it would be rare for two candidates to have the same minimum sum.
>
> It also solves the second problem because if a candidate is cloned, the
> first place votes of the cloned candidate are divided up among the clones.
>
> [End of Introduction to Improved Copeland for the lay voter.]
>
> Now, as mentioned in my last post, it is more general to replace the
> phrase "first place votes" with "random candidate probabilities," i.e.
> benchmark lottery probabilities. Even other suitable lottery
> probabilities could be used.
>
> This method (at least under the top rank counts or benchmark lottery)
> always elects from Landau, since if X covers Y, then only a subset of the
> candidates that beat Y will beat X, yielding X a smaller probability sum
> than Y.
>
> Also since if candidate X is raised on a ballot it can only decrease the
> benchmark probability of any other candidate, and the set of candidates
> that now beat X will be a (possibly proper) subset of the candidates that
> did before raising X on the ballot; i.e. this method is monotonous (if not
> monogamous).
>
> And it seems tp satisfy the Chicken Defense criterion:
>
> 49 C
> 26 A>B
> 25 B (sincere is B>A)
>
> C>A is the only pairwise defeat of A, so the A sum is 49.
> A>B is the only pairwise defeat of B, so the B sum is 26.
> B>C is the only pairwise defeat of C, so the C sum is 25.
>
> Candidate C (with the smallest sum) is elected, thus thwarting the
> threatened chicken attack
>
> What's not to like?
>
> Now think in terms of "Yee BoLson Diagrams":
>
> A candidate's score is the sum of the Dirichlet Cell probabilities (i.e.
> Voronoi Polygon probabilities). These are the Dirichlet/Voronoicells of
> the candidates that are closer to the center of the distribution than the
> given candidate. [the respective cells represent the voters that top rank
> the respective candidates.]
>
> So the winning candidate is the candidate for which the mass of cells
> closer to the center than the candidate has the smallest total probability.
>
> In the case of the standard centrally symmetric distribution used in Yee
> Bolson diagrams, the candidate closest to the center will be the winner
> with no defeats, so the "mass of defeating cells" will be empty.
>
> Not bad!
>
> Is it good enough and simple enough to propose?
> .
> Forest
>
>
>
>
>> ----------------------------------------------------------------------
>>
>> Message: 1
>> Date: Wed, 5 Jun 2019 19:28:48 -0700
>> From: Forest Simmons <fsimmons at pcc.edu>
>> To: Kristofer Munsterhjelm <km_elmet at t-online.de>, EM
>> <election-methods at lists.electorama.com>, "C.Benham"
>> <cbenham at adam.com.au>, Kevin Venzke <stepjak at yahoo.fr>
>> Subject: [EM] A New Spinoff of Our Recent Discussions
>> Message-ID:
>> <
>> CAP29onc__B876PaCWDT-d_T7BwQi5eR3yDR6dV8PDGOYOto4ew at mail.gmail.com>
>> Content-Type: text/plain; charset="utf-8"
>>
>> I was mulling over Kristofer's ideas abbout using first place counts in
>> new
>> ways.
>>
>> It reminded me about what we called "Borda Done Right" where we decloned
>> Borda by use of the first place counts.
>>
>> Then a light bulb turned on: Why not de-clone Copeland in rthe same way?
>> After all, Copeland always chooses from the Landau set, the set of
>> uncovered candidates. [I first realized this a few years ago when Marcus
>> expressed doubt that there was a monotonic method that would always choose
>> from Landau. After a little thought Copeland was the most obvious example
>> to clear up the question.]
>>
>> So here is Improved Copeland:
>>
>> For each candidate X, let S(X) be the sum of first place votes of the
>> candidates that do not pairwise beat X.
>>
>> [Note that this sum includes the number of first place votes received by
>> X,
>> since X does not pairwise beat X.]
>>
>> Elect from argmax(S(X)).
>>
>> Note that in large public elections argmax(S(X)) will almost surely
>> consist
>> of only one candidate.
>>
>> That version works best when the ballots have have easily discerned
>> favorites.
>>
>> Here's a version that works better for a greater variety of ballots,
>> especially where equal top votes are allowed:
>>
>> For each candidate Y let P(Y) be the probability that Y would be chosen by
>> a random ballot lottery. [Actually, any other decent lottery would work
>> just as well.]
>>
>> For each candidate X, let S(X) be the sum (over all candidates Y that do
>> not pairwise beat X) of P(Y).
>>
>> Elect from argmax(S(X)).
>>
>> Note that if Z covers X, then S(Z) is greater than or equal to S(X),
>> because every P(Y) in the sum for S(X) will also be a term in the sum
>> defining S(Z).
>>
>> Therefore the max(S(X)) candidate is uncovered.
>>
>> Like Copeland the method is also monotone, and unlike Copeland the method
>> is clone proof.
>>
>> Since Copeland is one of the most familiar Condorcet methods, and has an
>> obvious appeal until the clone dependence probblem is pointed out, ithis
>> new method can be presented as a simple , easily understandable solution
>> to
>> that problem.
>>
>> How does it hold up on our favorite examples?
>>
>> Try
>>
>> 49 C
>> 26 A>B
>> 25 B
>>
>> S(C)=49+26=75
>> S(A)=26+25=51
>> S(B)=25+49=74
>>
>> Arrgmax(S(X))={C}
>>
>> The method passes the CD criterion.
>>
>> Is this too good to be true?
>> ***************************
>>
>
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