[EM] Kemeny challenge

[Peter Zbornik]

Dear all,

Some weeks ago, I showed how to complete incomplete ballots by adding
the null candidate (i.e. the option "none of the above") with lower
rank than the ranked candidates and equally ranking the unranked
candidates below the null candidate.  This ballot completion method
gives the same result for winning votes and margins for Condorcet
elections,

I guess the comment the Kemeny method not being able to handle
incomplete ballots, does not apply if the ballot completion method
above is used.

Best regards
Peter Zbornik

On 9/14/11, Richard Fobes <ElectionMethods at votefair.org> wrote:
> The Condorcet-Kemeny method does allow candidates to be ranked at the
> same preference level, and no special calculations are needed to handle
> these ballots.  Such "ties" can occur at any combination of preference
> levels.  The interactive ballots at VoteFair.org allow such "ties" and,
> more broadly, allow any one oval to be marked for each choice.  (On a
> paper-based version, if a voter marks more than one oval, only the
> left-most marked oval is used.)
>
> I've addressed the "clone dependence" issue previously, yet I'll repeat
> the important points:  Exact clones (which is what clone dependence
> assumes) are very rare in real elections, and circular ambiguity (that
> includes the winner) is not common (because Condorcet winners are more
> common), so the combination of these two events -- which is what must
> occur in order to fail the clone independence criteria -- is extremely rare.
>
> When I get time to reply to Warren's other message I'll address the
> "computational intractability" misconception.
>
> Richard Fobes
>
> On 9/13/2011 2:39 PM, fsimmons at pcc.edu wrote:
>> The problems with Kemeny are the same as the problems with Dodgson:
>> (1) computational intractability
>> (2) clone dependence
>> (3) they require completely ordered ballots (no truncations or equal
>> ranking), so they do not readily adapt to Approval ballots, for example.
>> In my posting several weeks ago under the title "Dodgson done right" I
>> showed how to overcome these three problems. (The same modifications do
>> the trick for both methods.) However, much of the simplicity of the
>> statements of these two methods (Dodgson and Kemeny) gets lost in the
>> translation.
>
>
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