TP
Toby Pereira
Sun, Dec 12, 2021 10:04 AM
If you look at the example here: https://www.rangevoting.org/TobyCondParadox.html having a tie cycle in a Condorcet election can change the result. The ballots are essentially:
1 voter: A>B>C2 voters: B>A>C
Plus
2 voters: A>B>C2 voters: B>C>A2 voters: C>A>B
A is the Condorcet winner in this election, beating both B and C by 5 votes to 4. However, in the top section of ballots, B is the Condorcet winner, and the bottom section is a three-way tie as it is a perfect cycle. Adding in tie cycles to change the result could be seen as a failure of "independence of cycles" (or perhaps there is a better name). This is also a specific case of failure of consistency. Or maybe a weak failure because B doesn't win both the sub-elections. B would have an outright win from the top ballots and tie from the bottom ballots, but loses to A when all are combined.
This example, by the way, is a simplified version of something that Donald Saari used to promote the Borda Count.
The problem is that in an election with more than three candidates, you couldn't simply remove the cycles and calculate the result. Ballots and candidates would potentially be involved in many intertwined cycles, so there would be no straightforward way of doing it.
But what you can do is compare every possible triplet of candidates (like Condorcet methods compare pairs). For each triplet, all tie cycles are removed and you look at the head-to-heads. In the above example it would be:
B beats A by 2 votes to 1B beats C by 2 votes to 0A beats C by 2 votes to 0
If a candidate wins every head-to-head in every triplet, they are elected (the equivalent of a Condorcet winner for this method).
I will presume it's possible for there to be no such winning candidate, and if so, there would need to be some sort of tie-break. But we could simply base it on existing Condorcet methods - e.g. the Ranked Pairs method. Find the single biggest head-to-head win across all the triplets, and lock that in. Then find the next biggest that won't contradict a locked in position and lock that in and so on.
I think the idea of removing tie cycles like this has some merit, although it would be interesting to see what you all think.
Toby
If you look at the example here: https://www.rangevoting.org/TobyCondParadox.html having a tie cycle in a Condorcet election can change the result. The ballots are essentially:
1 voter: A>B>C2 voters: B>A>C
Plus
2 voters: A>B>C2 voters: B>C>A2 voters: C>A>B
A is the Condorcet winner in this election, beating both B and C by 5 votes to 4. However, in the top section of ballots, B is the Condorcet winner, and the bottom section is a three-way tie as it is a perfect cycle. Adding in tie cycles to change the result could be seen as a failure of "independence of cycles" (or perhaps there is a better name). This is also a specific case of failure of consistency. Or maybe a weak failure because B doesn't win both the sub-elections. B would have an outright win from the top ballots and tie from the bottom ballots, but loses to A when all are combined.
This example, by the way, is a simplified version of something that Donald Saari used to promote the Borda Count.
The problem is that in an election with more than three candidates, you couldn't simply remove the cycles and calculate the result. Ballots and candidates would potentially be involved in many intertwined cycles, so there would be no straightforward way of doing it.
But what you can do is compare every possible triplet of candidates (like Condorcet methods compare pairs). For each triplet, all tie cycles are removed and you look at the head-to-heads. In the above example it would be:
B beats A by 2 votes to 1B beats C by 2 votes to 0A beats C by 2 votes to 0
If a candidate wins every head-to-head in every triplet, they are elected (the equivalent of a Condorcet winner for this method).
I will presume it's possible for there to be no such winning candidate, and if so, there would need to be some sort of tie-break. But we could simply base it on existing Condorcet methods - e.g. the Ranked Pairs method. Find the single biggest head-to-head win across all the triplets, and lock that in. Then find the next biggest that won't contradict a locked in position and lock that in and so on.
I think the idea of removing tie cycles like this has some merit, although it would be interesting to see what you all think.
Toby
KM
Kristofer Munsterhjelm
Sun, Dec 12, 2021 11:38 AM
On 12/12/21 11:04 AM, Toby Pereira wrote:
If you look at the example here:
https://www.rangevoting.org/TobyCondParadox.html
https://www.rangevoting.org/TobyCondParadox.html having a tie cycle in
a Condorcet election can change the result. The ballots are essentially:
1 voter: A>B>C
2 voters: B>A>C
Plus
2 voters: A>B>C
2 voters: B>C>A
2 voters: C>A>B
A is the Condorcet winner in this election, beating both B and C by 5
votes to 4. However, in the top section of ballots, B is the Condorcet
winner, and the bottom section is a three-way tie as it is a perfect
cycle. Adding in tie cycles to change the result could be seen as a
failure of "independence of cycles" (or perhaps there is a better name).
This is also a specific case of failure of consistency. Or maybe a weak
failure because B doesn't win both the sub-elections. B would have an
outright win from the top ballots and tie from the bottom ballots, but
loses to A when all are combined.
This example, by the way, is a simplified version of something that
Donald Saari used to promote the Borda Count.
The problem is that in an election with more than three candidates, you
couldn't simply remove the cycles and calculate the result. Ballots and
candidates would potentially be involved in many intertwined cycles, so
there would be no straightforward way of doing it.
But what you can do is compare every possible triplet of candidates
(like Condorcet methods compare pairs). For each triplet, all tie cycles
are removed and you look at the head-to-heads.
You could do this, but as I understand the example, the method would no
longer be a Condorcet method. You could also define the irrelevance of
cycles criterion, perhaps something like:
Removing a constant number of voters who together form an exact tied
Condorcet cycle should not modify the output.
Though I'm not sure what the implications would be - or if it's possible
to pass by any method that fails IIA.
As for using triplets instead of pairs, I think Stensholt suggested that
doing so might be a way to generalize his BPW method, which is only
defined for three candiates. Similarly, it might be a way of
generalizing my fpA-fpC, though I'm again unsure how to do so and
preserve the desired properties of DMTBR and monotonicity.
-km
On 12/12/21 11:04 AM, Toby Pereira wrote:
> If you look at the example here:
> https://www.rangevoting.org/TobyCondParadox.html
> <https://www.rangevoting.org/TobyCondParadox.html> having a tie cycle in
> a Condorcet election can change the result. The ballots are essentially:
>
>
> 1 voter: A>B>C
> 2 voters: B>A>C
>
> Plus
>
> 2 voters: A>B>C
> 2 voters: B>C>A
> 2 voters: C>A>B
>
> A is the Condorcet winner in this election, beating both B and C by 5
> votes to 4. However, in the top section of ballots, B is the Condorcet
> winner, and the bottom section is a three-way tie as it is a perfect
> cycle. Adding in tie cycles to change the result could be seen as a
> failure of "independence of cycles" (or perhaps there is a better name).
> This is also a specific case of failure of consistency. Or maybe a weak
> failure because B doesn't win both the sub-elections. B would have an
> outright win from the top ballots and tie from the bottom ballots, but
> loses to A when all are combined.
>
> This example, by the way, is a simplified version of something that
> Donald Saari used to promote the Borda Count.
>
> The problem is that in an election with more than three candidates, you
> couldn't simply remove the cycles and calculate the result. Ballots and
> candidates would potentially be involved in many intertwined cycles, so
> there would be no straightforward way of doing it.
>
> But what you can do is compare every possible triplet of candidates
> (like Condorcet methods compare pairs). For each triplet, all tie cycles
> are removed and you look at the head-to-heads.
You could do this, but as I understand the example, the method would no
longer be a Condorcet method. You could also define the irrelevance of
cycles criterion, perhaps something like:
Removing a constant number of voters who together form an exact tied
Condorcet cycle should not modify the output.
Though I'm not sure what the implications would be - or if it's possible
to pass by any method that fails IIA.
As for using triplets instead of pairs, I think Stensholt suggested that
doing so might be a way to generalize his BPW method, which is only
defined for three candiates. Similarly, it might be a way of
generalizing my fpA-fpC, though I'm again unsure how to do so and
preserve the desired properties of DMTBR and monotonicity.
-km
CC
Colin Champion
Sun, Dec 12, 2021 4:35 PM
Is "independence of cycles" not the same as Saari's "neutral Condorcet
requirement"? (mentioned in Risse's paper "Why the Count de Borda cannot
beat the Marquis de Condorcet"). The matter is discussed in Pacuit's
Stanford Encyclopedia article. He refers to ballots "cancelling
properly" and mentions a proof that "there is no Condorcet consistent
voting method that cancels properly".
CJC
On 12/12/2021 11:38, Kristofer Munsterhjelm wrote:
On 12/12/21 11:04 AM, Toby Pereira wrote:
If you look at the example here:
https://www.rangevoting.org/TobyCondParadox.html
https://www.rangevoting.org/TobyCondParadox.html having a tie cycle
in a Condorcet election can change the result. The ballots are
essentially:
1 voter: A>B>C
2 voters: B>A>C
Plus
2 voters: A>B>C
2 voters: B>C>A
2 voters: C>A>B
A is the Condorcet winner in this election, beating both B and C by 5
votes to 4. However, in the top section of ballots, B is the
Condorcet winner, and the bottom section is a three-way tie as it is
a perfect cycle. Adding in tie cycles to change the result could be
seen as a failure of "independence of cycles" (or perhaps there is a
better name). This is also a specific case of failure of consistency.
Or maybe a weak failure because B doesn't win both the sub-elections.
B would have an outright win from the top ballots and tie from the
bottom ballots, but loses to A when all are combined.
This example, by the way, is a simplified version of something that
Donald Saari used to promote the Borda Count.
The problem is that in an election with more than three candidates,
you couldn't simply remove the cycles and calculate the result.
Ballots and candidates would potentially be involved in many
intertwined cycles, so there would be no straightforward way of doing
it.
But what you can do is compare every possible triplet of candidates
(like Condorcet methods compare pairs). For each triplet, all tie
cycles are removed and you look at the head-to-heads.
You could do this, but as I understand the example, the method would
no longer be a Condorcet method. You could also define the irrelevance
of cycles criterion, perhaps something like:
Removing a constant number of voters who together form an exact tied
Condorcet cycle should not modify the output.
Though I'm not sure what the implications would be - or if it's
possible to pass by any method that fails IIA.
As for using triplets instead of pairs, I think Stensholt suggested
that doing so might be a way to generalize his BPW method, which is
only defined for three candiates. Similarly, it might be a way of
generalizing my fpA-fpC, though I'm again unsure how to do so and
preserve the desired properties of DMTBR and monotonicity.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
Is "independence of cycles" not the same as Saari's "neutral Condorcet
requirement"? (mentioned in Risse's paper "Why the Count de Borda cannot
beat the Marquis de Condorcet"). The matter is discussed in Pacuit's
Stanford Encyclopedia article. He refers to ballots "cancelling
properly" and mentions a proof that "there is no Condorcet consistent
voting method that cancels properly".
CJC
On 12/12/2021 11:38, Kristofer Munsterhjelm wrote:
> On 12/12/21 11:04 AM, Toby Pereira wrote:
>> If you look at the example here:
>> https://www.rangevoting.org/TobyCondParadox.html
>> <https://www.rangevoting.org/TobyCondParadox.html> having a tie cycle
>> in a Condorcet election can change the result. The ballots are
>> essentially:
>>
>>
>> 1 voter: A>B>C
>> 2 voters: B>A>C
>>
>> Plus
>>
>> 2 voters: A>B>C
>> 2 voters: B>C>A
>> 2 voters: C>A>B
>>
>> A is the Condorcet winner in this election, beating both B and C by 5
>> votes to 4. However, in the top section of ballots, B is the
>> Condorcet winner, and the bottom section is a three-way tie as it is
>> a perfect cycle. Adding in tie cycles to change the result could be
>> seen as a failure of "independence of cycles" (or perhaps there is a
>> better name). This is also a specific case of failure of consistency.
>> Or maybe a weak failure because B doesn't win both the sub-elections.
>> B would have an outright win from the top ballots and tie from the
>> bottom ballots, but loses to A when all are combined.
>>
>> This example, by the way, is a simplified version of something that
>> Donald Saari used to promote the Borda Count.
>>
>> The problem is that in an election with more than three candidates,
>> you couldn't simply remove the cycles and calculate the result.
>> Ballots and candidates would potentially be involved in many
>> intertwined cycles, so there would be no straightforward way of doing
>> it.
>>
>> But what you can do is compare every possible triplet of candidates
>> (like Condorcet methods compare pairs). For each triplet, all tie
>> cycles are removed and you look at the head-to-heads.
>
> You could do this, but as I understand the example, the method would
> no longer be a Condorcet method. You could also define the irrelevance
> of cycles criterion, perhaps something like:
>
> Removing a constant number of voters who together form an exact tied
> Condorcet cycle should not modify the output.
>
> Though I'm not sure what the implications would be - or if it's
> possible to pass by any method that fails IIA.
>
> As for using triplets instead of pairs, I think Stensholt suggested
> that doing so might be a way to generalize his BPW method, which is
> only defined for three candiates. Similarly, it might be a way of
> generalizing my fpA-fpC, though I'm again unsure how to do so and
> preserve the desired properties of DMTBR and monotonicity.
>
> -km
> ----
> Election-Methods mailing list - see https://electorama.com/em for list
> info
KM
Kristofer Munsterhjelm
Sun, Dec 12, 2021 4:55 PM
On 12.12.2021 17:35, Colin Champion wrote:
Is "independence of cycles" not the same as Saari's "neutral Condorcet
requirement"? (mentioned in Risse's paper "Why the Count de Borda cannot
beat the Marquis de Condorcet"). The matter is discussed in Pacuit's
Stanford Encyclopedia article. He refers to ballots "cancelling
properly" and mentions a proof that "there is no Condorcet consistent
voting method that cancels properly".
I wasn't aware of that paper but yes, now that I've looked at the
definition, you're right.
-km
On 12.12.2021 17:35, Colin Champion wrote:
> Is "independence of cycles" not the same as Saari's "neutral Condorcet
> requirement"? (mentioned in Risse's paper "Why the Count de Borda cannot
> beat the Marquis de Condorcet"). The matter is discussed in Pacuit's
> Stanford Encyclopedia article. He refers to ballots "cancelling
> properly" and mentions a proof that "there is no Condorcet consistent
> voting method that cancels properly".
I wasn't aware of that paper but yes, now that I've looked at the
definition, you're right.
-km
TP
Toby Pereira
Sun, Dec 12, 2021 6:14 PM
I suppose I shouldn't be surprised that Saari named the criterion already. As Kristofer pointed out, my proposed method would fail Condorcet, but deliberately. I'm not sure what criteria it might pass or fail generally, but I think it's interesting to look at other methods that might pass Saari's criterion, since the Borda Count is pretty terrible.
Toby
On Sunday, 12 December 2021, 16:56:58 GMT, Kristofer Munsterhjelm km_elmet@t-online.de wrote:
On 12.12.2021 17:35, Colin Champion wrote:
Is "independence of cycles" not the same as Saari's "neutral Condorcet
requirement"? (mentioned in Risse's paper "Why the Count de Borda cannot
beat the Marquis de Condorcet"). The matter is discussed in Pacuit's
Stanford Encyclopedia article. He refers to ballots "cancelling
properly" and mentions a proof that "there is no Condorcet consistent
voting method that cancels properly".
I wasn't aware of that paper but yes, now that I've looked at the
definition, you're right.
-km
Election-Methods mailing list - see https://electorama.com/em for list info
I suppose I shouldn't be surprised that Saari named the criterion already. As Kristofer pointed out, my proposed method would fail Condorcet, but deliberately. I'm not sure what criteria it might pass or fail generally, but I think it's interesting to look at other methods that might pass Saari's criterion, since the Borda Count is pretty terrible.
Toby
On Sunday, 12 December 2021, 16:56:58 GMT, Kristofer Munsterhjelm <km_elmet@t-online.de> wrote:
On 12.12.2021 17:35, Colin Champion wrote:
> Is "independence of cycles" not the same as Saari's "neutral Condorcet
> requirement"? (mentioned in Risse's paper "Why the Count de Borda cannot
> beat the Marquis de Condorcet"). The matter is discussed in Pacuit's
> Stanford Encyclopedia article. He refers to ballots "cancelling
> properly" and mentions a proof that "there is no Condorcet consistent
> voting method that cancels properly".
I wasn't aware of that paper but yes, now that I've looked at the
definition, you're right.
-km
----
Election-Methods mailing list - see https://electorama.com/em for list info
FS
Forest Simmons
Mon, Dec 13, 2021 7:35 AM
For what it's worth, three way tie of the opposite cyclic order would
cancel the above tied cycle and reverse it for x>2.
...
x:B>A>C
x: A>C>B
x: C>B> A
A perfectly balanced cycle is not neutral in its influence ... it exerts
torsion.
The torsion from the original "tied" cycle reinforces the smaller faction
order, but goes against the order of the larger faction ... weakening its
influence.
If we were electing cyclic social orders instead of individual candidates
this would make more sense ... add two of these neutral cycles together ...
if they were absolutely neutral they wouldn't be able to affect each other.
Let's have an election to decide democratically if rock/paper/scissors is
the true cyclic order or its reverse. Some people may believe that rocks
are stronger than paper and that paper can cover scissors and that scissors
can chip away at rocks, for example.
El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 12/12/21 11:04 AM, Toby Pereira wrote:
a Condorcet election can change the result. The ballots are essentially:
1 voter: A>B>C
2 voters: B>A>C
Plus
2 voters: A>B>C
2 voters: B>C>A
2 voters: C>A>B
A is the Condorcet winner in this election, beating both B and C by 5
votes to 4. However, in the top section of ballots, B is the Condorcet
winner, and the bottom section is a three-way tie as it is a perfect
cycle. Adding in tie cycles to change the result could be seen as a
failure of "independence of cycles" (or perhaps there is a better name).
This is also a specific case of failure of consistency. Or maybe a weak
failure because B doesn't win both the sub-elections. B would have an
outright win from the top ballots and tie from the bottom ballots, but
loses to A when all are combined.
This example, by the way, is a simplified version of something that
Donald Saari used to promote the Borda Count.
The problem is that in an election with more than three candidates, you
couldn't simply remove the cycles and calculate the result. Ballots and
candidates would potentially be involved in many intertwined cycles, so
there would be no straightforward way of doing it.
But what you can do is compare every possible triplet of candidates
(like Condorcet methods compare pairs). For each triplet, all tie cycles
are removed and you look at the head-to-heads.
You could do this, but as I understand the example, the method would no
longer be a Condorcet method. You could also define the irrelevance of
cycles criterion, perhaps something like:
Removing a constant number of voters who together form an exact tied
Condorcet cycle should not modify the output.
Though I'm not sure what the implications would be - or if it's possible
to pass by any method that fails IIA.
As for using triplets instead of pairs, I think Stensholt suggested that
doing so might be a way to generalize his BPW method, which is only
defined for three candiates. Similarly, it might be a way of
generalizing my fpA-fpC, though I'm again unsure how to do so and
preserve the desired properties of DMTBR and monotonicity.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
For what it's worth, three way tie of the opposite cyclic order would
cancel the above tied cycle and reverse it for x>2.
...
x:B>A>C
x: A>C>B
x: C>B> A
A perfectly balanced cycle is not neutral in its influence ... it exerts
torsion.
The torsion from the original "tied" cycle reinforces the smaller faction
order, but goes against the order of the larger faction ... weakening its
influence.
If we were electing cyclic social orders instead of individual candidates
this would make more sense ... add two of these neutral cycles together ...
if they were absolutely neutral they wouldn't be able to affect each other.
Let's have an election to decide democratically if rock/paper/scissors is
the true cyclic order or its reverse. Some people may believe that rocks
are stronger than paper and that paper can cover scissors and that scissors
can chip away at rocks, for example.
El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
> On 12/12/21 11:04 AM, Toby Pereira wrote:
> > If you look at the example here:
> > https://www.rangevoting.org/TobyCondParadox.html
> > <https://www.rangevoting.org/TobyCondParadox.html> having a tie cycle
> in
> > a Condorcet election can change the result. The ballots are essentially:
> >
> >
> > 1 voter: A>B>C
> > 2 voters: B>A>C
> >
> > Plus
> >
> > 2 voters: A>B>C
> > 2 voters: B>C>A
> > 2 voters: C>A>B
> >
> > A is the Condorcet winner in this election, beating both B and C by 5
> > votes to 4. However, in the top section of ballots, B is the Condorcet
> > winner, and the bottom section is a three-way tie as it is a perfect
> > cycle. Adding in tie cycles to change the result could be seen as a
> > failure of "independence of cycles" (or perhaps there is a better name).
> > This is also a specific case of failure of consistency. Or maybe a weak
> > failure because B doesn't win both the sub-elections. B would have an
> > outright win from the top ballots and tie from the bottom ballots, but
> > loses to A when all are combined.
> >
> > This example, by the way, is a simplified version of something that
> > Donald Saari used to promote the Borda Count.
> >
> > The problem is that in an election with more than three candidates, you
> > couldn't simply remove the cycles and calculate the result. Ballots and
> > candidates would potentially be involved in many intertwined cycles, so
> > there would be no straightforward way of doing it.
> >
> > But what you can do is compare every possible triplet of candidates
> > (like Condorcet methods compare pairs). For each triplet, all tie cycles
> > are removed and you look at the head-to-heads.
>
> You could do this, but as I understand the example, the method would no
> longer be a Condorcet method. You could also define the irrelevance of
> cycles criterion, perhaps something like:
>
> Removing a constant number of voters who together form an exact tied
> Condorcet cycle should not modify the output.
>
> Though I'm not sure what the implications would be - or if it's possible
> to pass by any method that fails IIA.
>
> As for using triplets instead of pairs, I think Stensholt suggested that
> doing so might be a way to generalize his BPW method, which is only
> defined for three candiates. Similarly, it might be a way of
> generalizing my fpA-fpC, though I'm again unsure how to do so and
> preserve the desired properties of DMTBR and monotonicity.
>
> -km
> ----
> Election-Methods mailing list - see https://electorama.com/em for list
> info
>
TP
Toby Pereira
Tue, Dec 14, 2021 12:55 PM
A perfectly balanced cycle isn't neutral in its influence, but it is neutral in terms of deciding which is the "best" candidate when viewed alone. So arguably it should be neutral in its influence. But then we can add another set of ballots where one candidate would clearly be the winner (A in my example), but the winner of all ballots together would be a different candidate (B in my example).
If I was just looking at the ballots afresh without any method in mind, I might decide that candidate A should be the winner. But Condorcet dictates B should be the winner. I don't subscribe to the view (that others might) that Condorcet methods are the best by default and not to be questioned, although if you devise a method in order to get the result you want in one particular election, it may have glaring faults elsewhere.
The method I devised was in any case arguably over-simplistic as it only deals with three-candiate cycles. I looked up "Neutral Condorcet Criterion" and found it discussed by Mathias Risse - e.g. here http://www.math.buffalo.edu/~dhemmer/UE141F08/Projectpossibilities/RisseBordaCount.pdf
The criterion requires neutrality for larger cycles as well - e.g. with the following four ballots:
A>B>C>DB>C>D>AC>D>A>BD>A>B>C
It might be that in order to pass exactly it requires the Borda Count, but another method that might go a decent way to passing it would be to eliminate any cycles involving all remaining candidates, then eliminate the "worst" candidate (e.g. ranked pairs loser) and continue until you have a winner.
So if there are four candidates, take out any four-way cycles as above (and any going the other way) and then eliminate the last placed candidate. Then with the three remaining candidates, eliminate all three way cycles, and eliminate the losing candidate. Then pick the winner.
Toby
On Monday, 13 December 2021, 07:35:19 GMT, Forest Simmons <forest.simmons21@gmail.com> wrote:
For what it's worth, three way tie of the opposite cyclic order would cancel the above tied cycle and reverse it for x>2....x:B>A>Cx: A>C>Bx: C>B> A
A perfectly balanced cycle is not neutral in its influence ... it exerts torsion.
The torsion from the original "tied" cycle reinforces the smaller faction order, but goes against the order of the larger faction ... weakening its influence.
If we were electing cyclic social orders instead of individual candidates this would make more sense ... add two of these neutral cycles together ... if they were absolutely neutral they wouldn't be able to affect each other.
Let's have an election to decide democratically if rock/paper/scissors is the true cyclic order or its reverse. Some people may believe that rocks are stronger than paper and that paper can cover scissors and that scissors can chip away at rocks, for example.
El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm km_elmet@t-online.de escribió:
On 12/12/21 11:04 AM, Toby Pereira wrote:
If you look at the example here:
https://www.rangevoting.org/TobyCondParadox.html
https://www.rangevoting.org/TobyCondParadox.html having a tie cycle in
a Condorcet election can change the result. The ballots are essentially:
1 voter: A>B>C
2 voters: B>A>C
Plus
2 voters: A>B>C
2 voters: B>C>A
2 voters: C>A>B
A is the Condorcet winner in this election, beating both B and C by 5
votes to 4. However, in the top section of ballots, B is the Condorcet
winner, and the bottom section is a three-way tie as it is a perfect
cycle. Adding in tie cycles to change the result could be seen as a
failure of "independence of cycles" (or perhaps there is a better name).
This is also a specific case of failure of consistency. Or maybe a weak
failure because B doesn't win both the sub-elections. B would have an
outright win from the top ballots and tie from the bottom ballots, but
loses to A when all are combined.
This example, by the way, is a simplified version of something that
Donald Saari used to promote the Borda Count.
The problem is that in an election with more than three candidates, you
couldn't simply remove the cycles and calculate the result. Ballots and
candidates would potentially be involved in many intertwined cycles, so
there would be no straightforward way of doing it.
But what you can do is compare every possible triplet of candidates
(like Condorcet methods compare pairs). For each triplet, all tie cycles
are removed and you look at the head-to-heads.
You could do this, but as I understand the example, the method would no
longer be a Condorcet method. You could also define the irrelevance of
cycles criterion, perhaps something like:
Removing a constant number of voters who together form an exact tied
Condorcet cycle should not modify the output.
Though I'm not sure what the implications would be - or if it's possible
to pass by any method that fails IIA.
As for using triplets instead of pairs, I think Stensholt suggested that
doing so might be a way to generalize his BPW method, which is only
defined for three candiates. Similarly, it might be a way of
generalizing my fpA-fpC, though I'm again unsure how to do so and
preserve the desired properties of DMTBR and monotonicity.
-km
Election-Methods mailing list - see https://electorama.com/em for list info
A perfectly balanced cycle isn't neutral in its influence, but it is neutral in terms of deciding which is the "best" candidate when viewed alone. So arguably it should be neutral in its influence. But then we can add another set of ballots where one candidate would clearly be the winner (A in my example), but the winner of all ballots together would be a different candidate (B in my example).
If I was just looking at the ballots afresh without any method in mind, I might decide that candidate A should be the winner. But Condorcet dictates B should be the winner. I don't subscribe to the view (that others might) that Condorcet methods are the best by default and not to be questioned, although if you devise a method in order to get the result you want in one particular election, it may have glaring faults elsewhere.
The method I devised was in any case arguably over-simplistic as it only deals with three-candiate cycles. I looked up "Neutral Condorcet Criterion" and found it discussed by Mathias Risse - e.g. here http://www.math.buffalo.edu/~dhemmer/UE141F08/Projectpossibilities/RisseBordaCount.pdf
The criterion requires neutrality for larger cycles as well - e.g. with the following four ballots:
A>B>C>DB>C>D>AC>D>A>BD>A>B>C
It might be that in order to pass exactly it requires the Borda Count, but another method that might go a decent way to passing it would be to eliminate any cycles involving all remaining candidates, then eliminate the "worst" candidate (e.g. ranked pairs loser) and continue until you have a winner.
So if there are four candidates, take out any four-way cycles as above (and any going the other way) and then eliminate the last placed candidate. Then with the three remaining candidates, eliminate all three way cycles, and eliminate the losing candidate. Then pick the winner.
Toby
On Monday, 13 December 2021, 07:35:19 GMT, Forest Simmons <forest.simmons21@gmail.com> wrote:
For what it's worth, three way tie of the opposite cyclic order would cancel the above tied cycle and reverse it for x>2....x:B>A>Cx: A>C>Bx: C>B> A
A perfectly balanced cycle is not neutral in its influence ... it exerts torsion.
The torsion from the original "tied" cycle reinforces the smaller faction order, but goes against the order of the larger faction ... weakening its influence.
If we were electing cyclic social orders instead of individual candidates this would make more sense ... add two of these neutral cycles together ... if they were absolutely neutral they wouldn't be able to affect each other.
Let's have an election to decide democratically if rock/paper/scissors is the true cyclic order or its reverse. Some people may believe that rocks are stronger than paper and that paper can cover scissors and that scissors can chip away at rocks, for example.
El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm <km_elmet@t-online.de> escribió:
On 12/12/21 11:04 AM, Toby Pereira wrote:
> If you look at the example here:
> https://www.rangevoting.org/TobyCondParadox.html
> <https://www.rangevoting.org/TobyCondParadox.html> having a tie cycle in
> a Condorcet election can change the result. The ballots are essentially:
>
>
> 1 voter: A>B>C
> 2 voters: B>A>C
>
> Plus
>
> 2 voters: A>B>C
> 2 voters: B>C>A
> 2 voters: C>A>B
>
> A is the Condorcet winner in this election, beating both B and C by 5
> votes to 4. However, in the top section of ballots, B is the Condorcet
> winner, and the bottom section is a three-way tie as it is a perfect
> cycle. Adding in tie cycles to change the result could be seen as a
> failure of "independence of cycles" (or perhaps there is a better name).
> This is also a specific case of failure of consistency. Or maybe a weak
> failure because B doesn't win both the sub-elections. B would have an
> outright win from the top ballots and tie from the bottom ballots, but
> loses to A when all are combined.
>
> This example, by the way, is a simplified version of something that
> Donald Saari used to promote the Borda Count.
>
> The problem is that in an election with more than three candidates, you
> couldn't simply remove the cycles and calculate the result. Ballots and
> candidates would potentially be involved in many intertwined cycles, so
> there would be no straightforward way of doing it.
>
> But what you can do is compare every possible triplet of candidates
> (like Condorcet methods compare pairs). For each triplet, all tie cycles
> are removed and you look at the head-to-heads.
You could do this, but as I understand the example, the method would no
longer be a Condorcet method. You could also define the irrelevance of
cycles criterion, perhaps something like:
Removing a constant number of voters who together form an exact tied
Condorcet cycle should not modify the output.
Though I'm not sure what the implications would be - or if it's possible
to pass by any method that fails IIA.
As for using triplets instead of pairs, I think Stensholt suggested that
doing so might be a way to generalize his BPW method, which is only
defined for three candiates. Similarly, it might be a way of
generalizing my fpA-fpC, though I'm again unsure how to do so and
preserve the desired properties of DMTBR and monotonicity.
-km
----
Election-Methods mailing list - see https://electorama.com/em for list info
FS
Forest Simmons
Tue, Dec 14, 2021 9:11 PM
How does this idea relate to "Break Cycle"?
El mar., 14 de dic. de 2021 4:55 a. m., Toby Pereira tdp201b@yahoo.co.uk
escribió:
A perfectly balanced cycle isn't neutral in its influence, but it is
neutral in terms of deciding which is the "best" candidate when viewed
alone. So arguably it should be neutral in its influence. But then we can
add another set of ballots where one candidate would clearly be the winner
(A in my example), but the winner of all ballots together would be a
different candidate (B in my example).
If I was just looking at the ballots afresh without any method in mind, I
might decide that candidate A should be the winner. But Condorcet dictates
B should be the winner. I don't subscribe to the view (that others might)
that Condorcet methods are the best by default and not to be questioned,
although if you devise a method in order to get the result you want in one
particular election, it may have glaring faults elsewhere.
The method I devised was in any case arguably over-simplistic as it only
deals with three-candiate cycles. I looked up "Neutral Condorcet Criterion"
and found it discussed by Mathias Risse - e.g. here
http://www.math.buffalo.edu/~dhemmer/UE141F08/Projectpossibilities/RisseBordaCount.pdf
The criterion requires neutrality for larger cycles as well - e.g. with
the following four ballots:
A>B>C>D
B>C>D>A
C>D>A>B
D>A>B>C
It might be that in order to pass exactly it requires the Borda Count, but
another method that might go a decent way to passing it would be to
eliminate any cycles involving all remaining candidates, then eliminate the
"worst" candidate (e.g. ranked pairs loser) and continue until you have a
winner.
So if there are four candidates, take out any four-way cycles as above
(and any going the other way) and then eliminate the last placed candidate.
Then with the three remaining candidates, eliminate all three way cycles,
and eliminate the losing candidate. Then pick the winner.
Toby
On Monday, 13 December 2021, 07:35:19 GMT, Forest Simmons <
forest.simmons21@gmail.com> wrote:
For what it's worth, three way tie of the opposite cyclic order would
cancel the above tied cycle and reverse it for x>2.
...
x:B>A>C
x: A>C>B
x: C>B> A
A perfectly balanced cycle is not neutral in its influence ... it exerts
torsion.
The torsion from the original "tied" cycle reinforces the smaller faction
order, but goes against the order of the larger faction ... weakening its
influence.
If we were electing cyclic social orders instead of individual candidates
this would make more sense ... add two of these neutral cycles together ...
if they were absolutely neutral they wouldn't be able to affect each other.
Let's have an election to decide democratically if rock/paper/scissors is
the true cyclic order or its reverse. Some people may believe that rocks
are stronger than paper and that paper can cover scissors and that scissors
can chip away at rocks, for example.
El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 12/12/21 11:04 AM, Toby Pereira wrote:
a Condorcet election can change the result. The ballots are essentially:
1 voter: A>B>C
2 voters: B>A>C
Plus
2 voters: A>B>C
2 voters: B>C>A
2 voters: C>A>B
A is the Condorcet winner in this election, beating both B and C by 5
votes to 4. However, in the top section of ballots, B is the Condorcet
winner, and the bottom section is a three-way tie as it is a perfect
cycle. Adding in tie cycles to change the result could be seen as a
failure of "independence of cycles" (or perhaps there is a better name).
This is also a specific case of failure of consistency. Or maybe a weak
failure because B doesn't win both the sub-elections. B would have an
outright win from the top ballots and tie from the bottom ballots, but
loses to A when all are combined.
This example, by the way, is a simplified version of something that
Donald Saari used to promote the Borda Count.
The problem is that in an election with more than three candidates, you
couldn't simply remove the cycles and calculate the result. Ballots and
candidates would potentially be involved in many intertwined cycles, so
there would be no straightforward way of doing it.
But what you can do is compare every possible triplet of candidates
(like Condorcet methods compare pairs). For each triplet, all tie cycles
are removed and you look at the head-to-heads.
You could do this, but as I understand the example, the method would no
longer be a Condorcet method. You could also define the irrelevance of
cycles criterion, perhaps something like:
Removing a constant number of voters who together form an exact tied
Condorcet cycle should not modify the output.
Though I'm not sure what the implications would be - or if it's possible
to pass by any method that fails IIA.
As for using triplets instead of pairs, I think Stensholt suggested that
doing so might be a way to generalize his BPW method, which is only
defined for three candiates. Similarly, it might be a way of
generalizing my fpA-fpC, though I'm again unsure how to do so and
preserve the desired properties of DMTBR and monotonicity.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
How does this idea relate to "Break Cycle"?
El mar., 14 de dic. de 2021 4:55 a. m., Toby Pereira <tdp201b@yahoo.co.uk>
escribió:
> A perfectly balanced cycle isn't neutral in its influence, but it is
> neutral in terms of deciding which is the "best" candidate when viewed
> alone. So arguably it should be neutral in its influence. But then we can
> add another set of ballots where one candidate would clearly be the winner
> (A in my example), but the winner of all ballots together would be a
> different candidate (B in my example).
>
> If I was just looking at the ballots afresh without any method in mind, I
> might decide that candidate A should be the winner. But Condorcet dictates
> B should be the winner. I don't subscribe to the view (that others might)
> that Condorcet methods are the best by default and not to be questioned,
> although if you devise a method in order to get the result you want in one
> particular election, it may have glaring faults elsewhere.
>
> The method I devised was in any case arguably over-simplistic as it only
> deals with three-candiate cycles. I looked up "Neutral Condorcet Criterion"
> and found it discussed by Mathias Risse - e.g. here
> http://www.math.buffalo.edu/~dhemmer/UE141F08/Projectpossibilities/RisseBordaCount.pdf
>
> The criterion requires neutrality for larger cycles as well - e.g. with
> the following four ballots:
>
> A>B>C>D
> B>C>D>A
> C>D>A>B
> D>A>B>C
>
> It might be that in order to pass exactly it requires the Borda Count, but
> another method that might go a decent way to passing it would be to
> eliminate any cycles involving all remaining candidates, then eliminate the
> "worst" candidate (e.g. ranked pairs loser) and continue until you have a
> winner.
>
> So if there are four candidates, take out any four-way cycles as above
> (and any going the other way) and then eliminate the last placed candidate.
> Then with the three remaining candidates, eliminate all three way cycles,
> and eliminate the losing candidate. Then pick the winner.
>
> Toby
>
>
> On Monday, 13 December 2021, 07:35:19 GMT, Forest Simmons <
> forest.simmons21@gmail.com> wrote:
>
>
> For what it's worth, three way tie of the opposite cyclic order would
> cancel the above tied cycle and reverse it for x>2.
> ...
> x:B>A>C
> x: A>C>B
> x: C>B> A
>
> A perfectly balanced cycle is not neutral in its influence ... it exerts
> torsion.
>
> The torsion from the original "tied" cycle reinforces the smaller faction
> order, but goes against the order of the larger faction ... weakening its
> influence.
>
> If we were electing cyclic social orders instead of individual candidates
> this would make more sense ... add two of these neutral cycles together ...
> if they were absolutely neutral they wouldn't be able to affect each other.
>
> Let's have an election to decide democratically if rock/paper/scissors is
> the true cyclic order or its reverse. Some people may believe that rocks
> are stronger than paper and that paper can cover scissors and that scissors
> can chip away at rocks, for example.
>
> El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm <
> km_elmet@t-online.de> escribió:
>
> On 12/12/21 11:04 AM, Toby Pereira wrote:
> > If you look at the example here:
> > https://www.rangevoting.org/TobyCondParadox.html
> > <https://www.rangevoting.org/TobyCondParadox.html> having a tie cycle
> in
> > a Condorcet election can change the result. The ballots are essentially:
> >
> >
> > 1 voter: A>B>C
> > 2 voters: B>A>C
> >
> > Plus
> >
> > 2 voters: A>B>C
> > 2 voters: B>C>A
> > 2 voters: C>A>B
> >
> > A is the Condorcet winner in this election, beating both B and C by 5
> > votes to 4. However, in the top section of ballots, B is the Condorcet
> > winner, and the bottom section is a three-way tie as it is a perfect
> > cycle. Adding in tie cycles to change the result could be seen as a
> > failure of "independence of cycles" (or perhaps there is a better name).
> > This is also a specific case of failure of consistency. Or maybe a weak
> > failure because B doesn't win both the sub-elections. B would have an
> > outright win from the top ballots and tie from the bottom ballots, but
> > loses to A when all are combined.
> >
> > This example, by the way, is a simplified version of something that
> > Donald Saari used to promote the Borda Count.
> >
> > The problem is that in an election with more than three candidates, you
> > couldn't simply remove the cycles and calculate the result. Ballots and
> > candidates would potentially be involved in many intertwined cycles, so
> > there would be no straightforward way of doing it.
> >
> > But what you can do is compare every possible triplet of candidates
> > (like Condorcet methods compare pairs). For each triplet, all tie cycles
> > are removed and you look at the head-to-heads.
>
> You could do this, but as I understand the example, the method would no
> longer be a Condorcet method. You could also define the irrelevance of
> cycles criterion, perhaps something like:
>
> Removing a constant number of voters who together form an exact tied
> Condorcet cycle should not modify the output.
>
> Though I'm not sure what the implications would be - or if it's possible
> to pass by any method that fails IIA.
>
> As for using triplets instead of pairs, I think Stensholt suggested that
> doing so might be a way to generalize his BPW method, which is only
> defined for three candiates. Similarly, it might be a way of
> generalizing my fpA-fpC, though I'm again unsure how to do so and
> preserve the desired properties of DMTBR and monotonicity.
>
> -km
>
> ----
> Election-Methods mailing list - see https://electorama.com/em for list
> info
>
>
TP
Toby Pereira
Tue, Dec 14, 2021 11:10 PM
I'm not sure I understand the question.
On Tuesday, 14 December 2021, 21:11:39 GMT, Forest Simmons forest.simmons21@gmail.com wrote:
How does this idea relate to "Break Cycle"?
El mar., 14 de dic. de 2021 4:55 a. m., Toby Pereira tdp201b@yahoo.co.uk escribió:
A perfectly balanced cycle isn't neutral in its influence, but it is neutral in terms of deciding which is the "best" candidate when viewed alone. So arguably it should be neutral in its influence. But then we can add another set of ballots where one candidate would clearly be the winner (A in my example), but the winner of all ballots together would be a different candidate (B in my example).
If I was just looking at the ballots afresh without any method in mind, I might decide that candidate A should be the winner. But Condorcet dictates B should be the winner. I don't subscribe to the view (that others might) that Condorcet methods are the best by default and not to be questioned, although if you devise a method in order to get the result you want in one particular election, it may have glaring faults elsewhere.
The method I devised was in any case arguably over-simplistic as it only deals with three-candiate cycles. I looked up "Neutral Condorcet Criterion" and found it discussed by Mathias Risse - e.g. here http://www.math.buffalo.edu/~dhemmer/UE141F08/Projectpossibilities/RisseBordaCount.pdf
The criterion requires neutrality for larger cycles as well - e.g. with the following four ballots:
A>B>C>DB>C>D>AC>D>A>BD>A>B>C
It might be that in order to pass exactly it requires the Borda Count, but another method that might go a decent way to passing it would be to eliminate any cycles involving all remaining candidates, then eliminate the "worst" candidate (e.g. ranked pairs loser) and continue until you have a winner.
So if there are four candidates, take out any four-way cycles as above (and any going the other way) and then eliminate the last placed candidate. Then with the three remaining candidates, eliminate all three way cycles, and eliminate the losing candidate. Then pick the winner.
Toby
On Monday, 13 December 2021, 07:35:19 GMT, Forest Simmons <forest.simmons21@gmail.com> wrote:
For what it's worth, three way tie of the opposite cyclic order would cancel the above tied cycle and reverse it for x>2....x:B>A>Cx: A>C>Bx: C>B> A
A perfectly balanced cycle is not neutral in its influence ... it exerts torsion.
The torsion from the original "tied" cycle reinforces the smaller faction order, but goes against the order of the larger faction ... weakening its influence.
If we were electing cyclic social orders instead of individual candidates this would make more sense ... add two of these neutral cycles together ... if they were absolutely neutral they wouldn't be able to affect each other.
Let's have an election to decide democratically if rock/paper/scissors is the true cyclic order or its reverse. Some people may believe that rocks are stronger than paper and that paper can cover scissors and that scissors can chip away at rocks, for example.
El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm km_elmet@t-online.de escribió:
On 12/12/21 11:04 AM, Toby Pereira wrote:
If you look at the example here:
https://www.rangevoting.org/TobyCondParadox.html
https://www.rangevoting.org/TobyCondParadox.html having a tie cycle in
a Condorcet election can change the result. The ballots are essentially:
1 voter: A>B>C
2 voters: B>A>C
Plus
2 voters: A>B>C
2 voters: B>C>A
2 voters: C>A>B
A is the Condorcet winner in this election, beating both B and C by 5
votes to 4. However, in the top section of ballots, B is the Condorcet
winner, and the bottom section is a three-way tie as it is a perfect
cycle. Adding in tie cycles to change the result could be seen as a
failure of "independence of cycles" (or perhaps there is a better name).
This is also a specific case of failure of consistency. Or maybe a weak
failure because B doesn't win both the sub-elections. B would have an
outright win from the top ballots and tie from the bottom ballots, but
loses to A when all are combined.
This example, by the way, is a simplified version of something that
Donald Saari used to promote the Borda Count.
The problem is that in an election with more than three candidates, you
couldn't simply remove the cycles and calculate the result. Ballots and
candidates would potentially be involved in many intertwined cycles, so
there would be no straightforward way of doing it.
But what you can do is compare every possible triplet of candidates
(like Condorcet methods compare pairs). For each triplet, all tie cycles
are removed and you look at the head-to-heads.
You could do this, but as I understand the example, the method would no
longer be a Condorcet method. You could also define the irrelevance of
cycles criterion, perhaps something like:
Removing a constant number of voters who together form an exact tied
Condorcet cycle should not modify the output.
Though I'm not sure what the implications would be - or if it's possible
to pass by any method that fails IIA.
As for using triplets instead of pairs, I think Stensholt suggested that
doing so might be a way to generalize his BPW method, which is only
defined for three candiates. Similarly, it might be a way of
generalizing my fpA-fpC, though I'm again unsure how to do so and
preserve the desired properties of DMTBR and monotonicity.
-km
Election-Methods mailing list - see https://electorama.com/em for list info
I'm not sure I understand the question.
On Tuesday, 14 December 2021, 21:11:39 GMT, Forest Simmons <forest.simmons21@gmail.com> wrote:
How does this idea relate to "Break Cycle"?
El mar., 14 de dic. de 2021 4:55 a. m., Toby Pereira <tdp201b@yahoo.co.uk> escribió:
A perfectly balanced cycle isn't neutral in its influence, but it is neutral in terms of deciding which is the "best" candidate when viewed alone. So arguably it should be neutral in its influence. But then we can add another set of ballots where one candidate would clearly be the winner (A in my example), but the winner of all ballots together would be a different candidate (B in my example).
If I was just looking at the ballots afresh without any method in mind, I might decide that candidate A should be the winner. But Condorcet dictates B should be the winner. I don't subscribe to the view (that others might) that Condorcet methods are the best by default and not to be questioned, although if you devise a method in order to get the result you want in one particular election, it may have glaring faults elsewhere.
The method I devised was in any case arguably over-simplistic as it only deals with three-candiate cycles. I looked up "Neutral Condorcet Criterion" and found it discussed by Mathias Risse - e.g. here http://www.math.buffalo.edu/~dhemmer/UE141F08/Projectpossibilities/RisseBordaCount.pdf
The criterion requires neutrality for larger cycles as well - e.g. with the following four ballots:
A>B>C>DB>C>D>AC>D>A>BD>A>B>C
It might be that in order to pass exactly it requires the Borda Count, but another method that might go a decent way to passing it would be to eliminate any cycles involving all remaining candidates, then eliminate the "worst" candidate (e.g. ranked pairs loser) and continue until you have a winner.
So if there are four candidates, take out any four-way cycles as above (and any going the other way) and then eliminate the last placed candidate. Then with the three remaining candidates, eliminate all three way cycles, and eliminate the losing candidate. Then pick the winner.
Toby
On Monday, 13 December 2021, 07:35:19 GMT, Forest Simmons <forest.simmons21@gmail.com> wrote:
For what it's worth, three way tie of the opposite cyclic order would cancel the above tied cycle and reverse it for x>2....x:B>A>Cx: A>C>Bx: C>B> A
A perfectly balanced cycle is not neutral in its influence ... it exerts torsion.
The torsion from the original "tied" cycle reinforces the smaller faction order, but goes against the order of the larger faction ... weakening its influence.
If we were electing cyclic social orders instead of individual candidates this would make more sense ... add two of these neutral cycles together ... if they were absolutely neutral they wouldn't be able to affect each other.
Let's have an election to decide democratically if rock/paper/scissors is the true cyclic order or its reverse. Some people may believe that rocks are stronger than paper and that paper can cover scissors and that scissors can chip away at rocks, for example.
El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm <km_elmet@t-online.de> escribió:
On 12/12/21 11:04 AM, Toby Pereira wrote:
> If you look at the example here:
> https://www.rangevoting.org/TobyCondParadox.html
> <https://www.rangevoting.org/TobyCondParadox.html> having a tie cycle in
> a Condorcet election can change the result. The ballots are essentially:
>
>
> 1 voter: A>B>C
> 2 voters: B>A>C
>
> Plus
>
> 2 voters: A>B>C
> 2 voters: B>C>A
> 2 voters: C>A>B
>
> A is the Condorcet winner in this election, beating both B and C by 5
> votes to 4. However, in the top section of ballots, B is the Condorcet
> winner, and the bottom section is a three-way tie as it is a perfect
> cycle. Adding in tie cycles to change the result could be seen as a
> failure of "independence of cycles" (or perhaps there is a better name).
> This is also a specific case of failure of consistency. Or maybe a weak
> failure because B doesn't win both the sub-elections. B would have an
> outright win from the top ballots and tie from the bottom ballots, but
> loses to A when all are combined.
>
> This example, by the way, is a simplified version of something that
> Donald Saari used to promote the Borda Count.
>
> The problem is that in an election with more than three candidates, you
> couldn't simply remove the cycles and calculate the result. Ballots and
> candidates would potentially be involved in many intertwined cycles, so
> there would be no straightforward way of doing it.
>
> But what you can do is compare every possible triplet of candidates
> (like Condorcet methods compare pairs). For each triplet, all tie cycles
> are removed and you look at the head-to-heads.
You could do this, but as I understand the example, the method would no
longer be a Condorcet method. You could also define the irrelevance of
cycles criterion, perhaps something like:
Removing a constant number of voters who together form an exact tied
Condorcet cycle should not modify the output.
Though I'm not sure what the implications would be - or if it's possible
to pass by any method that fails IIA.
As for using triplets instead of pairs, I think Stensholt suggested that
doing so might be a way to generalize his BPW method, which is only
defined for three candiates. Similarly, it might be a way of
generalizing my fpA-fpC, though I'm again unsure how to do so and
preserve the desired properties of DMTBR and monotonicity.
-km
----
Election-Methods mailing list - see https://electorama.com/em for list info
