De-Cloned Kendall-tau Example
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?
El lun., 7 de mar. de 2022 9:22 p. m., Forest Simmons <
forest.simmons21@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]
At this point we can insert the respective decloned Kemeny-Young costs:
Converting all ballots to ABC:
3×15+2×10=65
Converting all ballots to BCA:
4×24+2×12=120
Converting all ballots to CAB:
4×28+3×21=175
So the de-cloned K-Y finish order is ABC, the same as clone dependent K-Y
... not particularly interesting.
This method ignores the preference information implicit in the geometrc
proximities ... blissfully taking the ballot preferences at face value
despite the geometric evidence to the contrary.
So the distant pair removal method is a more interesting application of
de-cloned Kendall-tau.
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?
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@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?