Mono-switch-Plump criterion compliance claims corrected
The Mono-switch-plump criterion is much stronger than I previously
thought, and is probably simply incompatible with the
Condorcet criterion.
I used to think that its met by two of my favourite Condorcet methods,
Margins-Sorted Losing Votes (erw) Elimination (equivalent in the 3
candidate case
to the "MMLV(erw)M" I discuss in the May 2014 post) and Approval Sorted
Margins. Consider this election under MSLVerwE :
40: A
29: C>A
03: B
28: B>C
A>B 69-31, B>C 31-29, C>A 57-40. LV(erw) scores: A40 > B31 >
C29. No adjacent pair is out-of-order pairwise, so MSLV(erw)E elects A.
But if we switch the 3 B plumping ballots to A then C becomes the
Condorcet winner (C>B 29-28, C>A 57-43).
43: A
29: C>A
28: B>C
And now this election under Approval Sorted Margins:
30: C
04: C>A
33: A>B
32: B
A>B 37-32, B>C 64-34, C>A 34-33. (Implicit) Approval scores: B64
A37 > C34. The adjacent pair with the smallest (absolute margin)
difference
in their scores (A > C) is pairwise out of order so we flip that to give
B > C > A. Now neither adjacent pair is pairwise out-of-order, so the
order is
final and so Margins Sorted Approval elects B.
But if we switch two of the 32 B plumping ballots to A then A becomes
the Condorcet winner (A>B 39-34, A>C 35-34).
30: C
04: C>A
33: A>B
02: A
30: B
I doubt that IBIFA meets the criterion.
But I remain sure that it's met by Bucklin (and similar methods like MTA
and MCA and QLTD).
Chris Benham
On 11 May 2014 Chris Benham posted to EM:
Mono-switch-plump:
The probability of candidate X winning must not be reduced if one or
more ballots that
plump for any not-X are replaced by an equal number of ballots that
plump for X.
Previously I showed that this is failed by the following methods:
Schulze (aka Beatpath), Ranked Pairs, River, MinMax (all equivalent
with 3 candidates) if they use Winning Votes to weigh pairwise defeats.
IRV and the Condorcet methods based on IRV (such as Benham and Woodall)
Total Approval Chain Climbing.
I claim that it is met by Margins, any positional method, IBIFA,
Bucklin and Bucklin-like methods like Median Ratings and MCA and MTA.
And also it is met by MMLV(erw)M. To support that claim I'll just
talk about the Margins Sort version with 3 candidates.
Plumping ballots for any X always contribute to X's score and
switching plumping ballots to X might get rid of one of X's pairwise
defeats.
If X has no pairwise defeats then that will always be still the case
after switching some plumping ballots to X and so X will still win. X
can't
be a winner with all pairwise defeats so we are only concerned about
the case when X has just one (and so will the other 2 candidates).
Say we designate the candidate with the highest score 1, the
second-highest 2 and and the lowest 3. The algorithm in this
3-candidate cycle
situation elects 1 unless 2 both pairwise beats 1 and has a score
that is closer to 1's than to 3's.
If winning candidate X is in position 2 then the effect of plumping
ballots being switched from 1 to 2 will be to just make 2 still
closer to 1,
and the effect of plumping ballots being switched from 3 to 2 will
have the same effect (and make 3 further away).
If winning candidate X is 1 and pairwise beats 2 and loses to 3,
then the only hope of making 1 lose is to switch some plumping ballots
from
2 to 1 sufficient for 2 and 3 to change places but that won't work
because then 2 and 3 will be adjacent candidates that are out of pairwise
order and will be much closer together score-wise than the other such
pair and they'll be switched back to give the final order 1>2>3.
And if X is 1 and losing to 2 then it means that 1's distance
(scorewise) from 2 is such that 2 and 3 are switched in the order, and
switching
any plumping ballots to 1 will only increase that distance.
I hope that (almost confused) waffle is not too confusing or opaque.
Chris Benham
Mono-switch-plump:
The probability of candidate X winning must not be reduced if one or
more ballots that
plump for any not-X are replaced by an equal number of ballots that
plump for X.
Mono-raise is the traditional monotonicity criterion, but I don't see
why anyone would
see failure of Mono-switch-plump as less embarrassing than failing
Mono-raise.
25 A>B
26 B>C
23 C>A
22 C
04 A
B>C 51-45 C>A 71-29 A>B 52-26
Top Preferences: C45 > A29 > B26
When there are three candidates the MinMax , Beatpath (aka Schulze),
Ranked Pairs and River algorithms
are all equivalent. When they use Winning Votes as the measure of
defeat strength they all elect C.
IRV (aka the Alternative Vote) and Benham (and Woodall) also elect
C. But if we replace the 4A ballots
with 4C ballots the winner with all these methods changes from C to B.
25 A>B
26 B>C
23 C>A
26 C
B>C 51-49 C>A 71-29 A>B 48-26
Top Preferences: C45 > B26 > A25
Total Approval Chain Climbing also fails.
25 A>B
06 A>C
32 B>C
27 C>A
08 C
02 B
C>A>B>C, Approvals C73 > B59 > A58
TACC elects C, but if the 2B ballots are changed to 2C, then the
winner changes to A.
25 A>B
06 A>C
32 B>C
27 C>A
10 C
C>A>B>C, Approvals C75 > A58 > B57
Forest has pointed out that my supposed example of Approval Sorted
Margins failing mono-switch-plump is nonsense.
So (at least for the time being) I am not able to show that Approval
Sorted Margins fails the mono-switch-plump criterion.
(BTW I am still very happy with the Margins-Sorted Losing Votes (erw)
Elimination method.)
Chris Benham
On 10/24/2016 10:28 PM, C.Benham wrote:
The Mono-switch-plump criterion is much stronger than I previously
thought, and is probably simply incompatible with the
Condorcet criterion.
I used to think that its met by two of my favourite Condorcet
methods, Margins-Sorted Losing Votes (erw) Elimination (equivalent in
the 3 candidate case
to the "MMLV(erw)M" I discuss in the May 2014 post) and Approval
Sorted Margins. Consider this election under MSLVerwE :
40: A
29: C>A
03: B
28: B>C
A>B 69-31, B>C 31-29, C>A 57-40. LV(erw) scores: A40 > B31 >
C29. No adjacent pair is out-of-order pairwise, so MSLV(erw)E elects A.
But if we switch the 3 B plumping ballots to A then C becomes the
Condorcet winner (C>B 29-28, C>A 57-43).
43: A
29: C>A
28: B>C
And now this election under Approval Sorted Margins:
30: C
04: C>A
33: A>B
32: B
A>B 37-32, B>C 64-34, C>A 34-33. (Implicit) Approval scores:
B64 > A37 > C34. The adjacent pair with the smallest (absolute
margin) difference
in their scores (A > C) is pairwise out of order so we flip that to
give B > C > A. Now neither adjacent pair is pairwise out-of-order,
so the order is
final and so Margins Sorted Approval elects B.
But if we switch two of the 32 B plumping ballots to A then A becomes
the Condorcet winner (A>B 39-34, A>C 35-34).
30: C
04: C>A
33: A>B
02: A
30: B
I doubt that IBIFA meets the criterion.
But I remain sure that it's met by Bucklin (and similar methods like
MTA and MCA and QLTD).
Chris Benham
On 11 May 2014 Chris Benham posted to EM:
Mono-switch-plump:
The probability of candidate X winning must not be reduced if one
or more ballots that
plump for any not-X are replaced by an equal number of ballots that
plump for X.
Previously I showed that this is failed by the following methods:
Schulze (aka Beatpath), Ranked Pairs, River, MinMax (all equivalent
with 3 candidates) if they use Winning Votes to weigh pairwise defeats.
IRV and the Condorcet methods based on IRV (such as Benham and Woodall)
Total Approval Chain Climbing.
I claim that it is met by Margins, any positional method, IBIFA,
Bucklin and Bucklin-like methods like Median Ratings and MCA and MTA.
And also it is met by MMLV(erw)M. To support that claim I'll just
talk about the Margins Sort version with 3 candidates.
Plumping ballots for any X always contribute to X's score and
switching plumping ballots to X might get rid of one of X's pairwise
defeats.
If X has no pairwise defeats then that will always be still the case
after switching some plumping ballots to X and so X will still win. X
can't
be a winner with all pairwise defeats so we are only concerned about
the case when X has just one (and so will the other 2 candidates).
Say we designate the candidate with the highest score 1, the
second-highest 2 and and the lowest 3. The algorithm in this
3-candidate cycle
situation elects 1 unless 2 both pairwise beats 1 and has a score
that is closer to 1's than to 3's.
If winning candidate X is in position 2 then the effect of plumping
ballots being switched from 1 to 2 will be to just make 2 still
closer to 1,
and the effect of plumping ballots being switched from 3 to 2 will
have the same effect (and make 3 further away).
If winning candidate X is 1 and pairwise beats 2 and loses to 3,
then the only hope of making 1 lose is to switch some plumping
ballots from
2 to 1 sufficient for 2 and 3 to change places but that won't work
because then 2 and 3 will be adjacent candidates that are out of
pairwise
order and will be much closer together score-wise than the other such
pair and they'll be switched back to give the final order 1>2>3.
And if X is 1 and losing to 2 then it means that 1's distance
(scorewise) from 2 is such that 2 and 3 are switched in the order,
and switching
any plumping ballots to 1 will only increase that distance.
I hope that (almost confused) waffle is not too confusing or opaque.
Chris Benham
Mono-switch-plump:
The probability of candidate X winning must not be reduced if one or
more ballots that
plump for any not-X are replaced by an equal number of ballots that
plump for X.
Mono-raise is the traditional monotonicity criterion, but I don't see
why anyone would
see failure of Mono-switch-plump as less embarrassing than failing
Mono-raise.
25 A>B
26 B>C
23 C>A
22 C
04 A
B>C 51-45 C>A 71-29 A>B 52-26
Top Preferences: C45 > A29 > B26
When there are three candidates the MinMax , Beatpath (aka Schulze),
Ranked Pairs and River algorithms
are all equivalent. When they use Winning Votes as the measure of
defeat strength they all elect C.
IRV (aka the Alternative Vote) and Benham (and Woodall) also elect
C. But if we replace the 4A ballots
with 4C ballots the winner with all these methods changes from C to B.
25 A>B
26 B>C
23 C>A
26 C
B>C 51-49 C>A 71-29 A>B 48-26
Top Preferences: C45 > B26 > A25
Total Approval Chain Climbing also fails.
25 A>B
06 A>C
32 B>C
27 C>A
08 C
02 B
C>A>B>C, Approvals C73 > B59 > A58
TACC elects C, but if the 2B ballots are changed to 2C, then the
winner changes to A.
25 A>B
06 A>C
32 B>C
27 C>A
10 C
C>A>B>C, Approvals C75 > A58 > B57
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Can you please post an explanation of "Margins-Sorted Losing Votes (erw)
Elimination" on electowiki?
2016-10-26 0:08 GMT-04:00 C.Benham cbenham@adam.com.au:
Forest has pointed out that my supposed example of Approval Sorted
Margins failing mono-switch-plump is nonsense.
So (at least for the time being) I am not able to show that Approval
Sorted Margins fails the mono-switch-plump criterion.
(BTW I am still very happy with the Margins-Sorted Losing Votes (erw)
Elimination method.)
Chris Benham
On 10/24/2016 10:28 PM, C.Benham wrote:
The Mono-switch-plump criterion is much stronger than I previously
thought, and is probably simply incompatible with the
Condorcet criterion.
I used to think that its met by two of my favourite Condorcet methods,
Margins-Sorted Losing Votes (erw) Elimination (equivalent in the 3
candidate case
to the "MMLV(erw)M" I discuss in the May 2014 post) and Approval Sorted
Margins. Consider this election under MSLVerwE :
40: A
29: C>A
03: B
28: B>C
A>B 69-31, B>C 31-29, C>A 57-40. LV(erw) scores: A40 > B31 >
C29. No adjacent pair is out-of-order pairwise, so MSLV(erw)E elects A.
But if we switch the 3 B plumping ballots to A then C becomes the
Condorcet winner (C>B 29-28, C>A 57-43).
43: A
29: C>A
28: B>C
And now this election under Approval Sorted Margins:
30: C
04: C>A
33: A>B
32: B
A>B 37-32, B>C 64-34, C>A 34-33. (Implicit) Approval scores: B64 >
A37 > C34. The adjacent pair with the smallest (absolute margin) difference
in their scores (A > C) is pairwise out of order so we flip that to give B
C > A. Now neither adjacent pair is pairwise out-of-order, so the order
is
final and so Margins Sorted Approval elects B.
But if we switch two of the 32 B plumping ballots to A then A becomes the
Condorcet winner (A>B 39-34, A>C 35-34).
30: C
04: C>A
33: A>B
02: A
30: B
I doubt that IBIFA meets the criterion.
But I remain sure that it's met by Bucklin (and similar methods like MTA
and MCA and QLTD).
Chris Benham
On 11 May 2014 Chris Benham posted to EM:
Mono-switch-plump:
The probability of candidate X winning must not be reduced if one or more
ballots that
plump for any not-X are replaced by an equal number of ballots that plump
for X.
Previously I showed that this is failed by the following methods:
Schulze (aka Beatpath), Ranked Pairs, River, MinMax (all equivalent with 3
candidates) if they use Winning Votes to weigh pairwise defeats.
IRV and the Condorcet methods based on IRV (such as Benham and Woodall)
Total Approval Chain Climbing.
I claim that it is met by Margins, any positional method, IBIFA, Bucklin
and Bucklin-like methods like Median Ratings and MCA and MTA.
And also it is met by MMLV(erw)M. To support that claim I'll just talk
about the Margins Sort version with 3 candidates.
Plumping ballots for any X always contribute to X's score and switching
plumping ballots to X might get rid of one of X's pairwise defeats.
If X has no pairwise defeats then that will always be still the case after
switching some plumping ballots to X and so X will still win. X can't
be a winner with all pairwise defeats so we are only concerned about the
case when X has just one (and so will the other 2 candidates).
Say we designate the candidate with the highest score 1, the
second-highest 2 and and the lowest 3. The algorithm in this 3-candidate
cycle
situation elects 1 unless 2 both pairwise beats 1 and has a score that is
closer to 1's than to 3's.
If winning candidate X is in position 2 then the effect of plumping
ballots being switched from 1 to 2 will be to just make 2 still closer to
1,
and the effect of plumping ballots being switched from 3 to 2 will have
the same effect (and make 3 further away).
If winning candidate X is 1 and pairwise beats 2 and loses to 3, then
the only hope of making 1 lose is to switch some plumping ballots from
2 to 1 sufficient for 2 and 3 to change places but that won't work because
then 2 and 3 will be adjacent candidates that are out of pairwise
order and will be much closer together score-wise than the other such pair
and they'll be switched back to give the final order 1>2>3.
And if X is 1 and losing to 2 then it means that 1's distance (scorewise)
from 2 is such that 2 and 3 are switched in the order, and switching
any plumping ballots to 1 will only increase that distance.
I hope that (almost confused) waffle is not too confusing or opaque.
Chris Benham
Mono-switch-plump:
The probability of candidate X winning must not be reduced if one or more
ballots that
plump for any not-X are replaced by an equal number of ballots that plump
for X.
Mono-raise is the traditional monotonicity criterion, but I don't see why
anyone would
see failure of Mono-switch-plump as less embarrassing than failing
Mono-raise.
25 A>B
26 B>C
23 C>A
22 C
04 A
B>C 51-45 C>A 71-29 A>B 52-26
Top Preferences: C45 > A29 > B26
When there are three candidates the MinMax , Beatpath (aka Schulze),
Ranked Pairs and River algorithms
are all equivalent. When they use Winning Votes as the measure of defeat
strength they all elect C.
IRV (aka the Alternative Vote) and Benham (and Woodall) also elect C.
But if we replace the 4A ballots
with 4C ballots the winner with all these methods changes from C to B.
25 A>B
26 B>C
23 C>A
26 C
B>C 51-49 C>A 71-29 A>B 48-26
Top Preferences: C45 > B26 > A25
Total Approval Chain Climbing also fails.
25 A>B
06 A>C
32 B>C
27 C>A
08 C
02 B
C>A>B>C, Approvals C73 > B59 > A58
TACC elects C, but if the 2B ballots are changed to 2C, then the winner
changes to A.
25 A>B
06 A>C
32 B>C
27 C>A
10 C
C>A>B>C, Approvals C75 > A58 > B57
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The Approval Sorted Margins example I gave earlier didn't work, so below
I've substituted one that does.
Chris Benham
On 10/24/2016 10:28 PM, C.Benham wrote:
The Mono-switch-plump criterion is much stronger than I previously
thought, and is probably simply incompatible with the
Condorcet criterion.
I used to think that its met by two of my favourite Condorcet
methods, Margins-Sorted Losing Votes (erw) Elimination (equivalent
in the 3 candidate case
to the "MMLV(erw)M" I discuss in the May 2014 post) and Approval
Sorted Margins. Consider this election under MSLVerwE :
40: A
29: C>A
03: B
28: B>C
A>B 69-31, B>C 31-29, C>A 57-40. LV(erw) scores: A40 > B31 >
C29. No adjacent pair is out-of-order pairwise, so MSLV(erw)E elects A.
But if we switch the 3 B plumping ballots to A then C becomes the
Condorcet winner (C>B 29-28, C>A 57-43).
43: A
29: C>A
28: B>C
And now this election under Approval Sorted Margins:
43: A
04: A>C
19: B>C
07: B
27: C>B
B>A 53-47, A>C 47-46, C>B 31-26. (Implicit) Approval scores:
B53 > C50 > A47.
Both adjacent pairs are out-of-order pairwise and the approval score
differences are the same (3) in both
cases so we flip the order of the lower-ordered pair to give B>A>C.
Now no adjacent pair is pairwise out-of-order
so that order is final and B wins.
Now say we change two of the A-plumping ballots into B-plumping
ballots. Then C will be the Condorcet winner.
41: A
04: A>C
19: B>C
09: B
27: C>B
C>A 46-45, C>B 31-28, B>A 55-45
I doubt that IBIFA meets the criterion.
But I remain sure that it's met by Bucklin (and similar methods like
MTA and MCA and QLTD).
Chris Benham
On 11 May 2014 Chris Benham posted to EM:
Mono-switch-plump:
The probability of candidate X winning must not be reduced if one
or more ballots that
plump for any not-X are replaced by an equal number of ballots
that plump for X.
Previously I showed that this is failed by the following methods:
Schulze (aka Beatpath), Ranked Pairs, River, MinMax (all equivalent
with 3 candidates) if they use Winning Votes to weigh pairwise defeats.
IRV and the Condorcet methods based on IRV (such as Benham and
Woodall)
Total Approval Chain Climbing.
I claim that it is met by Margins, any positional method, IBIFA,
Bucklin and Bucklin-like methods like Median Ratings and MCA and MTA.
And also it is met by MMLV(erw)M. To support that claim I'll
just talk about the Margins Sort version with 3 candidates.
Plumping ballots for any X always contribute to X's score and
switching plumping ballots to X might get rid of one of X's pairwise
defeats.
If X has no pairwise defeats then that will always be still the case
after switching some plumping ballots to X and so X will still win.
X can't
be a winner with all pairwise defeats so we are only concerned about
the case when X has just one (and so will the other 2 candidates).
Say we designate the candidate with the highest score 1, the
second-highest 2 and and the lowest 3. The algorithm in this
3-candidate cycle
situation elects 1 unless 2 both pairwise beats 1 and has a score
that is closer to 1's than to 3's.
If winning candidate X is in position 2 then the effect of plumping
ballots being switched from 1 to 2 will be to just make 2 still
closer to 1,
and the effect of plumping ballots being switched from 3 to 2 will
have the same effect (and make 3 further away).
If winning candidate X is 1 and pairwise beats 2 and loses to 3,
then the only hope of making 1 lose is to switch some plumping
ballots from
2 to 1 sufficient for 2 and 3 to change places but that won't work
because then 2 and 3 will be adjacent candidates that are out of
pairwise
order and will be much closer together score-wise than the other
such pair and they'll be switched back to give the final order 1>2>3.
And if X is 1 and losing to 2 then it means that 1's distance
(scorewise) from 2 is such that 2 and 3 are switched in the order,
and switching
any plumping ballots to 1 will only increase that distance.
I hope that (almost confused) waffle is not too confusing or opaque.
Chris Benham
Mono-switch-plump:
The probability of candidate X winning must not be reduced if one
or more ballots that
plump for any not-X are replaced by an equal number of ballots that
plump for X.
Mono-raise is the traditional monotonicity criterion, but I don't
see why anyone would
see failure of Mono-switch-plump as less embarrassing than failing
Mono-raise.
25 A>B
26 B>C
23 C>A
22 C
04 A
B>C 51-45 C>A 71-29 A>B 52-26
Top Preferences: C45 > A29 > B26
When there are three candidates the MinMax , Beatpath (aka Schulze),
Ranked Pairs and River algorithms
are all equivalent. When they use Winning Votes as the measure of
defeat strength they all elect C.
IRV (aka the Alternative Vote) and Benham (and Woodall) also elect
C. But if we replace the 4A ballots
with 4C ballots the winner with all these methods changes from C to B.
25 A>B
26 B>C
23 C>A
26 C
B>C 51-49 C>A 71-29 A>B 48-26
Top Preferences: C45 > B26 > A25
Total Approval Chain Climbing also fails.
25 A>B
06 A>C
32 B>C
27 C>A
08 C
02 B
C>A>B>C, Approvals C73 > B59 > A58
TACC elects C, but if the 2B ballots are changed to 2C, then the
winner changes to A.
25 A>B
06 A>C
32 B>C
27 C>A
10 C
C>A>B>C, Approvals C75 > A58 > B57