[EM] Condorcet Matrix Decomposition
Dan Bishop
daniel-j-bishop at neo.tamu.edu
Fri Nov 25 14:54:58 PST 2005
Rob Brown wrote:
> Why do you need to break ties? Wouldn't it make more sense to consider ties to
> be ranked equally?
Perhaps you're trying to approximate an election method that requires
fully-ranked ballots, in which case it makes sense to avoid ties
whenever possible.
More importantly, always ranking ties equally causes the method to fail.
Take a look at my example. Once the matrix [[0, 4, 4], [4, 0, 4], [4,
4, 0]] is reached, you can't just create A=B=C ballots from then on,
because that would create an infinite loop.
> The problem I see is that using borda scores is just rather arbitrary,
True, but it has the advantage up being computationally cheap.
> but worse, randomly breaking ties is non-deterministic.
I agree that this is a problem, but it's one that could easily be fixed.
Recomputing my example by removing Borda scores and using your
suggestion of breaking ties according to the previous ballot gives the CMD:
Ballot #1: A>B>C
Ballot #2: A>B>C
Ballot #3: A>B>C (A=B>C with A=B broken by previous A>B)
Ballot #4: B>A>C (B>A=C with A=C broken my previous A>C)
Ballot #5: B>A>C (A=B=C with ties broken by previous ballot)
Ballot #6: C>B>A (C>A=B with A=B broken by previous B>A)
Ballot #7: C>A>B (A=C>B with A=C broken by previous C>A)
Ballot #8: A>C>B (A>B=C with B=C broken by previous C>B)
Ballot #9: A>B>C (A=B>C with A=B broken by previous A>B)
Ballot #10: B>A>C (B>A=C with A=C broken by previous A>C)
Ballot #11: B>C>A (B=C>A with B=C broken by previous B>C)
Unfortunately, your "previous ballot" rule (which would give C>B>A)
fails for the final ballot, because the B>A preferences have been
depleted. Fortunately, the remaining preferences do form a ballot,
which is C>A>B.
In summary, the CMD under these revised rules is:
4: A>B>C
1: A>C>B
3: B>A>C
1: B>C>A
2: C>A>B
1: C>B>A
which happens to be identical to the one in my previous e-mail.
> In any case this is a fascinating exercise and I could see it being useful. I
> would be very interested to see it applied to a real IRV election....see how
> different the results are if you first "compress" it into a pairwise matrix then
> reexpand (deconstruct) it into ballots and tabulate it again.
How about that San Francisco election? Are the ballots still available
online somewhere?
Also, like Smith, I'm curious as to whether there's a counterexample for
my method, i.e., a Condorcet matrix for which a CMD exists, but my
method doesn't find one.
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