[EM] De-Cloned Kendall-tau Example

[Forest Simmons]

Once you have a good metric,  like de-cloned Kendall-tau, on the ballots,
you can cancel diameter endpoints until all of the remaining ballots are
identical.

If the ballot distribution is essentially one dimensional, it is easy to
see that this cancellation process finds the median ballot, and
consequently elects the Condorcet candidate.

So this cancellation method is a natural choice for Condorcet completion.
It seems to me that it would have robust resistance against manipulation,
because the geometry has more substance than its mere face value.

A natural two candidate runoff would be between the two winners of method X
... one based on the explicit "face value" ballot preferences, and the
other based on the preferences deduced from geometric proximity based
affinities...no runoff necessary if the two method X counts yield the same
winner.

That kind of runoff would be highly immune to manipulation.

Another application of the cancellation idea (having nothing else in common
with our Kendall-tau cancellation method) is an alternate way of completing
single winner asset voting:

After the asset transfer stage stalls, while there remains more than one
candidate with any assets left, the candidate with the fewest remaining
assets uses all of them to cancel that many assets from one of the other
candidates.

-Forest



El lun., 7 de mar. de 2022 9:22 p. m., Forest Simmons <
forest.simmons21 at gmail.com> escribió:

> I appreciate very much Kevin Venzke's recent tour de force towards
> classification of methods by their criteria compliances ... finding seven
> 3-candidate, 4-faction profiles that distinguish all of the well known
> methods by their basic compliances.
>
> I see this as an opportunity to put the de-cloned Kendall-tau metric
> through its paces.
>
> First a simple example with factions suitable for hand computation:
>
> 4ABC3+3BCA+2CAB
>
> We need favorite and anti-favorite counts for the de-cloning:
>
> The respective favorite counts f, are 4, 3, and 2. The respective a
> anti-favorite counts f', are 3, 2, and 4.
>
> The respective costs of the basic decloned order swaps are ...
> AB to BA 4×2 from f(A)×f'(B)
> BA to AB 3×3 from f(B)×f'(A)
> BC to CB 3×4 from f(B)×f'(C)
> CB to BC 2×2 from f(C)×f'(B)
> CA to AC 2×3 from f(C)× f'(A)
> AC to CA 4×4 from f(A)×f'(C)
>
> The respective costs from faction ...
> ABC to BCA 8+16
> BCA to ABC 6+9
> [Round trip 39]
> BCA to CAB 12+9
> CAB to BCA 8+4
> [Round trip 33]
> CAB to ABC 6+4
> ABC to CAB 12+16
> [Round trip 38]
>
> Our method is to remove pairs of ballots that are as far apart as possible
> until only one faction remains:
>
> ABC and BCA are the furthest apart ballots with a round trip distance of
> 39.
>
> Removing 3  ballots from each of these factions leaves ...
>
> ABC+2CAB
>
> After removing one ballot from each of these two remaining factions, we
> see that the winning faction is CAB.
>
> So this is one way to use the de-cloned Kendall-tau distance.
>
> To do Kevin's seven examples I will need some triple A batteries for my
> calculator.
>
> The most interesting part for me is that the geometrically derived
> preferences would be ...
>
> The 4 members of the ABC faction should geometrically prefer the CAB
> faction over the BCA faction because they are closer to it ... distance 38
> verses 39.
>
> The 3 members of the BCA faction should prefer the CAB faction over the
> ABC faction ... distance 33 versus 39.
>
> The two member CAB faction should prefer BCA to ABC .... distance 33 vs 39.
>
> So geometric preferences are
>
> 2 CAB>BCA>ABC
> 3 BCA>CAB>ABC
> 4 ABC>CAB>BCA
>
> The CAB faction is the Condorcet faction.
>
> In other words sincere preferences seem to be
>
> 4 A>C>B
> 3 B>C>A
> 2 C>B>A
>
> It looks like insincere or mistaken burials of C by A and A by C.
>
> Any other thoughts?
>
>
>
>
>
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