[EM] Re: Approval/Condorcet Hybrids

[Jobst Heitzig]

Dear Ted!

You wrote:
> One thing I find is that in elections with lots
> of candidates, any strong method (Ranked Pairs, Schulze, River or
> Short Ranked Pairs) quickly becomes mind-numbingly complicated for
> the average voter, who would lose confidence in the outcome,
> regardless of any Immunity from Majority Complaint (or similar)
> property.

I don't think so, or do you find the River solution to the Baseball
example
(http://lists.electorama.com/htdig.cgi/election-methods-electorama.com/2004-October/014102.html)
too complicated to follow for a voter who is interested enough in how
the process worked? Those who only ask for a justification of the
outcome will not actually have to follow the process but will only have
to be shown that to the resulting tree diagram no defeat can be added
without removing a stronger one. But you are of course right in pointing
out that the immune methods are usually more complicated to apply than
approval or the like.

By the way, I was too quick to announce that Short Ranked Pairs was
cloneproof and monotonic -- unfortunately both proof ideas didn't work
right and both properties are actually not fulfilled. I tried to rescue
the idea by varying the definition but did not succeed and have not much
 hope that the lexicographic approach of RP, River, and Beatpath can be
adjusted to get uncovered candidates only. That was the reason I
returned to Lasliers book to get new inspiration and found the section
on the bipartisan set and the tournament game which I reported on lately.

> I like the first step of your Sprucing Up process, eliminating
> covered candidates.  It quickly de-noises the candidate pool, and
> uncovered candidates have short beatpaths to counter complaints.  But
> it isn't hard to find examples where the set of uncovered candidates
> is clone-free and large enough (5 or more) to still be too large for
> the average voter to follow the remaining steps.

You are quite pessimistic as to what an average voter will accept I
think. However, Forest has made the plausible that in public elections
it will be paramountly probable that there is either a CW or a
three-element covering set, that is, a cycle of three candidates of
which each other beats at most one. Hence I think that he is right in
suggesting that we should not only reduce to the uncovered candidates
but even to the members of the minimal covering set (=Dutta set). I
conjecture that this set can be found in O(n^3) time. Anyway, once
found, it can easily be shown to be the minimal covering set. Once we
have reduced everything to at most three candidates, we can proceed as
we like by either dropping the weakest defeat in the cycle or by drawing
 random ballots or by maximizing approval etc.


> In your Approval/Condorcet hybrid example above, I like choosing the
> winner from U(A).  The winner will almost certainly be in even
> smaller set of candidates, with high voter preference, and with all
> the strong properties of being in the uncovered set.  I can't imagine
> any non-pathological real election example with more than 3 or 4
> candidates in U(A).  And even 3 would be an extreme case.
> 
> But ... your argument that, if W differs from A, this implies "that W
> beat every candidate that A beats head to head" does not follow.  It
> only implies that W has highest approval in U(A).

No, Forest is right, he defined:
>> Let U(A) be the set of uncovered candidates that cover the approval
>> winner A.

Hence every member of U(A) not only defeats A but covers A (or is equal
to A), so it defeats all candidates A defeats, by definition!

> I think a better winner could be found by determining a strong
> clone-proof method winner from U(A) [with 3 candidates it won't
> really make any difference which method, will it?  But I prefer
> Ranked Pairs].  And if approval weighting (a la James
> Green-Armytage's approval-weighted pairwise) is used to rank defeats,
> approval is still part of the process.  Call this winner V.

I tend to agree with you here. But it would be even better (and
simpler!) to just use Random Ballot (as Forest lately suggested) or
"Random Ballot Approval Cutoff" (=draw a random ballot, keep only
candidates approved on that ballot (if any at all), repeat until only
one remains).

Yours, Jobst