[EM] Condorcet Matrix Decomposition
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
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:
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
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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