[EM] My Matrix for Kemeny's Rule, n=3
barnes99 at vaxa.cis.uwosh.edu
Thu Jan 2 14:04:22 PST 2003
Thanks for confirming that my matrix for determining the Kemeny outcome for 3
candidates is correct. Now, is Kemeny's Rule, as defined by my matrix for 3
candidate tallies, the same as the Condorcet method which is described on the
EM website <http://electionmethods.org/>?
--- In election-methods-list at yahoogroups.com, Forest Simmons <fsimmons at p...>
> Your example is correctly done.
> Despite the intractability of the method for large numbers of candidates,
> it seems like an ideal method for some situations.
> One application could be in choosing between several orders that have been
> found by other means.
> [The main computational difficulty of finding the Kemeny order is not in
> calculating the mean Kemeny distance to any particular order, but in the
> sheer magnitude of the number of possible orders.]
> Here's a homely example:
> Start with the candidates ranked by Approval, Borda, Ranked Pairs, etc.
> Do both a sink sort and a bubble sort to each of these preliminary orders.
> [These sortings are ways of achieving "local Keminization." Ranked Pairs
> is already locally Kemeny optimal.]
> See which of the resulting locally optimum orders is best (among those
> considered) globally by calculating the mean distance of each to the
> preference orders on the ballots.
> On Fri, 20 Dec 2002, barnes99 wrote:
> > Here is my matrix for Kemeny's Rule with 3 candidates. Please let me know
> > got it right or wrong.
> > Here is an example of a Kemeny Rule tally for profile p, p=(1,1,0,0,0,0),
> > where the first thru sixth columns represent, respectively, the number of
> > ACB, CAB, CBA, BCA, and BAC voters.
> > Voting Vector:
> > p=[ 1 1 0 0 0 0 ]
> > Matrix (M) for Kemeny's Rule:
> > [[ 0 1 2 3 2 1 ]
> > [ 1 0 1 2 3 2 ]
> > [ 2 1 0 1 2 3 ]
> > [ 3 2 1 0 1 2 ]
> > [ 2 3 2 1 0 1 ]
> > [ 1 2 3 2 1 0 ]]
> > The KR tally is:
> > p(M)=[ 1 1 3 5 5 3 ], where
> > ABC=[0+1+0+0+0+0]=1
> > ACB=[1+0+0+0+0+0]=1
> > CAB=[2+1+0+0+0+0]=3
> > CBA=[3+2+0+0+0+0]=5
> > BCA=[2+3+0+0+0+0]=5
> > BAC=[1+2+0+0+0+0]=3
> > This is the measure of the "distance" from unanimity, so the lower the
> > the better. In this example, we have a tie between the ABC and ACB
> > in which case I guess the final outcome must be A>B~C. This may not be a
> > interesting example, but the point is that I believe this is how to do a
> > tally with 3 candidates. Please correct me, if I'm wrong.
> > Thank you,
> > SB
Richard M. Hare, 1919 - 2002, In Memoriam: <http://www.petersingerlinks.com/hare.htm>.
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