MinLV(erw) Sorted Margins Elimination
I made a strange mistake in the working out of my first example (but not
the result). Below it is corrected.
(I actually first made the mistake in 2016 and I soon noticed it then
and posted a corrected version, but what I posted last month
was mostly copied from the initial uncorrected post. )
Chris Benham
My favourite method that meets both Condorcet and Chicken Dilemma is
'Min Losing-Votes (equal-ranking whole) Sorted Margins Elimination':
*Voters rank from the top whatever number of candidates they like.
Equal-ranking and truncation are allowed.
For the purpose of determining candidates' pairwise scores:
a ballot that votes both X and Y above no other (remaining) candidates
contributes nothing to X's pairwise score versus Y and vice versa,
a ballot that ranks X and Y equal and above at least one (remaining)
candidate contributes a whole vote to X's pairwise score versus Y and
vice versa,
a ballot that ranks X above Y contributes a whole vote to X's pairwise
score versus Y and nothing to Y's pairwise score
versus X.
Give each candidate X a score equal to X's smallest losing pairwise score.
Initially order the candidates from highest-scored to lowest scored. If
any adjacent pair is out-of-order pairwise, then swap
the out-of-order pair with the smallest score-difference. If there is a
tie for that then swap the tied pair that is lowest in
the order. Repeat until no adjacent pair is pairwise out-of-order, and
then eliminate the lowest-ordered candidate.
Repeat (disregarding any pairwise scores with eliminated candidates)
until one candidate remains. *
Some examples:
46 A>B
44 B>C (sincere is B or B>A)
05 C>A
05 C>B
A>B 51-49, B>C 90-10, C>A 54-46.
MinLV(erw) scores: B49 > A46 > C10.
Both adjacent pairs (B>A and A>C) are pairwise out of order. The B>A
score-difference is the smallest of the two
(3 versus 36) so we first swap that order to give
A49 > B51 > C10
Now neither pair of adjacent candidates is pairwise out of order so C is
eliminated and A wins.
Winning Votes, Margins, MMPO elect the Burier's candidate.
25 A>B
26 B>C
23 C>A
26 C
C>A 75-25, A>B 48-26, B>C 51-49.
MinLV(erw) scores: C49 > B26 > A25.
Both adjacent pairs (C>B and B>A) are pairwise out-of-order. The B-A
score difference is by
far the smallest, so we swap the B>A order to give
C > A > B. That order is final and C wins. C is the most top ranked
and the most above-bottom ranked
candidate. WV, MMPO, IRV, Benham elect B.
35 A
10 A=B
30 B>C
25 C
C>A 55-45, A>B 45-40 (note 10A=B effect), B>C 40-25.
MinLV(erw) scores: A45 > B40 > C25. Neither adjacent pair is pairwise
out-of-order so the order is final
and A wins.
A both pairwise-beats and positionally dominates B, but WV, Margins,
MMPO all elect B.
Chris Benham
On Thu, May 4, 2023, 1:25 AM C.Benham cbenham@adam.com.au wrote:
I made a strange mistake in the working out of my first example (but not
the result). Below it is corrected.
(I actually first made the mistake in 2016 and I soon noticed it then and
posted a corrected version, but what I posted last month
was mostly copied from the initial uncorrected post. )
Chris Benham
My favourite method that meets both Condorcet and Chicken Dilemma is 'Min
Losing-Votes (equal-ranking whole) Sorted Margins Elimination':
*Voters rank from the top whatever number of candidates they like.
Equal-ranking and truncation are allowed.
For the purpose of determining candidates' pairwise scores:
a ballot that votes both X and Y above no other (remaining) candidates
contributes nothing to X's pairwise score versus Y and vice versa,
a ballot that ranks X and Y equal and above at least one (remaining)
candidate contributes a whole vote to X's pairwise score versus Y and vice
versa,
a ballot that ranks X above Y contributes a whole vote to X's pairwise
score versus Y and nothing to Y's pairwise score
versus X.
Give each candidate X a score equal to X's smallest losing pairwise score.
Initially order the candidates from highest-scored to lowest scored. If
any adjacent pair is out-of-order pairwise, then swap
the out-of-order pair with the smallest score-difference. If there is a
tie for that then swap the tied pair that is lowest in
the order. Repeat until no adjacent pair is pairwise out-of-order, and
then eliminate the lowest-ordered candidate.
Repeat (disregarding any pairwise scores with eliminated candidates) until
one candidate remains. *
Some examples:
46 A>B
44 B>C (sincere is B or B>A)
05 C>A
05 C>B
A>B 51-49, B>C 90-10, C>A 54-46.
MinLV(erw) scores: B49 > A46 > C10.
Both adjacent pairs (B>A and A>C) are pairwise out of order. The B>A
score-difference is the smallest of the two
(3 versus 36) so we first swap that order to give
A49 > B51 > C10
Now neither pair of adjacent candidates is pairwise out of order so C is
eliminated and A wins.
Winning Votes, Margins, MMPO elect the Burier's candidate.
25 A>B
26 B>C
23 C>A
26 C
C>A 75-25, A>B 48-26, B>C 51-49.
MinLV(erw) scores: C49 > B26 > A25.
Both adjacent pairs (C>B and B>A) are pairwise out-of-order. The B-A score
difference is by
far the smallest, so we swap the B>A order to give
C > A > B. That order is final and C wins. C is the most top ranked and
the most above-bottom ranked
candidate. WV, MMPO, IRV, Benham elect B.
35 A
10 A=B
30 B>C
25 C
C>A 55-45, A>B 45-40 (note 10A=B effect), B>C 40-25.
I noticed that A has more losing votes (45) than B has wining votes (40).
It seems to me that this fact (by itself) should disqualify B.
Also B is Ranked on feet ballots (40) than A is ranked Top (45) so
Plurality requires B to los as well.
So how about this as a tournament versión of Plurality:
If B's maxPairwiseSuppoft is less than A's minPS, then B should not win.
Anybody ever proposed this Criterion before?
MinLV(erw) scores: A45 > B40 > C25. Neither adjacent pair is pairwise
out-of-order so the order is final
and A wins.
A both pairwise-beats and positionally dominates B, but WV, Margins, MMPO
all elect B.
Chris Benham
On 5/06/2023 9:16 am, Forest Simmons wrote:
35 A
10 A=B
30 B>C
25 C
C>A 55-45, A>B 45-40 (note 10A=B effect), B>C 40-25.
I noticed that A has more losing votes (45) than B has wining votes (40).
It seems to me that this fact (by itself) should disqualify B.
So how about this as a tournament versión of Plurality:
If B's maxPairwiseSuppoft is less than A's minPS, then B should not win.
Anybody ever proposed this Criterion before?
I and/or Kevin Venzke may have mentioned it in passing. The example is
originally from Kevin.
I agree that is a good criterion and should be a strong standard. I
quite a while ago rejected the idea
that the best Condorcet methods were those that focused purely on
"defeat strengths" with a view to
simply "break the cycle at its weakest link".
Also B is Ranked on fewer ballots (40) than A is ranked Top (45) so
Plurality requires B to lose as well.
The Plurality criterion was coined by Douglas Woodall, who only
discussed ballots with strict ranking from
the top with truncation allowed. So it says that that B isn't allowed
to win if B is voted above bottom on fewer
ballots than A is voted alone above all others.
So it generally accepted that Winning Votes meets the ("normal",
original) Plurality criterion.
Chris Benham
On Thu, May 4, 2023, 1:25 AM C.Benham cbenham@adam.com.au wrote:
My favourite method that meets both Condorcet and Chicken Dilemma
is 'Min Losing-Votes (equal-ranking whole) Sorted Margins
Elimination':
*Voters rank from the top whatever number of candidates they like.
Equal-ranking and truncation are allowed.
For the purpose of determining candidates' pairwise scores:
a ballot that votes both X and Y above no other (remaining)
candidates contributes nothing to X's pairwise score versus Y and
vice versa,
a ballot that ranks X and Y equal and above at least one
(remaining) candidate contributes a whole vote to X's pairwise
score versus Y and vice versa,
a ballot that ranks X above Y contributes a whole vote to X's
pairwise score versus Y and nothing to Y's pairwise score
versus X.
Give each candidate X a score equal to X's smallest losing
pairwise score.
Initially order the candidates from highest-scored to lowest
scored. If any adjacent pair is out-of-order pairwise, then swap
the out-of-order pair with the smallest score-difference. If there
is a tie for that then swap the tied pair that is lowest in
the order. Repeat until no adjacent pair is pairwise out-of-order,
and then eliminate the lowest-ordered candidate.
Repeat (disregarding any pairwise scores with eliminated
candidates) until one candidate remains. *
Some examples:
46 A>B
44 B>C (sincere is B or B>A)
05 C>A
05 C>B
A>B 51-49, B>C 90-10, C>A 54-46.
MinLV(erw) scores: B49 > A46 > C10.
Both adjacent pairs (B>A and A>C) are pairwise out of order. The
B>A score-difference is the smallest of the two
(3 versus 36) so we first swap that order to give
A49 > B51 > C10
Now neither pair of adjacent candidates is pairwise out of order
so C is eliminated and A wins.
Winning Votes, Margins, MMPO elect the Burier's candidate.
25 A>B
26 B>C
23 C>A
26 C
C>A 75-25, A>B 48-26, B>C 51-49.
MinLV(erw) scores: C49 > B26 > A25.
Both adjacent pairs (C>B and B>A) are pairwise out-of-order. The
B-A score difference is by
far the smallest, so we swap the B>A order to give
C > A > B. That order is final and C wins. C is the most top
ranked and the most above-bottom ranked
candidate. WV, MMPO, IRV, Benham elect B.
35 A
10 A=B
30 B>C
25 C
C>A 55-45, A>B 45-40 (note 10A=B effect), B>C 40-25.
I noticed that A has more losing votes (45) than B has wining votes (40).
It seems to me that this fact (by itself) should disqualify B.
Also B is Ranked on feet ballots (40) than A is ranked Top (45) so
Plurality requires B to los as well.
So how about this as a tournament versión of Plurality:
If B's maxPairwiseSuppoft is less than A's minPS, then B should not win.
Anybody ever proposed this Criterion before?
MinLV(erw) scores: A45 > B40 > C25. Neither adjacent pair is
pairwise out-of-order so the order is final
and A wins.
A both pairwise-beats and positionally dominates B, but WV,
Margins, MMPO all elect B.
Chris Benham
On Sun, Jun 4, 2023, 11:03 PM C.Benham cbenham@adam.com.au wrote:
On 5/06/2023 9:16 am, Forest Simmons wrote:
35 A
10 A=B
30 B>C
25 C
C>A 55-45, A>B 45-40 (note 10A=B effect), B>C 40-25.
I noticed that A has more losing votes (45) than B has wining votes (40).
It seems to me that this fact (by itself) should disqualify B.
So how about this as a tournament versión of Plurality:
If B's maxPairwiseSuppoft is less than A's minPS, then B should not win.
Anybody ever proposed this Criterion before?
I and/or Kevin Venzke may have mentioned it in passing. The example is
originally from Kevin.
I agree that is a good criterion and should be a strong standard. I quite
a while ago rejected the idea
that the best Condorcet methods were those that focused purely on "defeat
strengths" with a view to
simply "break the cycle at its weakest link".
Me too. I cannot make that kind of method (weakest link cycle breaking)
sufficiently burial resistant ... no matter how I define defeat strength.
My current thinking (based on geometrically motivated ballot sets) is that
non trivial top cycles are more likely to be artificial than sincere, and
the simplest, most likely mechanism (for creating promising cycles) is
unilateral burial or strategic truncation of the sincere CW by some faction.
And it seems to me if this guess is more or less accurate, our top
Condorcet design priority should be to "not reward the buriers" ... as
opposed to detecting a sincere signal in a noisy ballot set ... i.e. one
riddled by mistaken but sincere judgments ... the kind of ballot set where
weakest link cycle breaking would be indicated.
Forest
Also B is Ranked on fewer ballots (40) than A is ranked Top (45) so
Plurality requires B to lose as well.
The Plurality criterion was coined by Douglas Woodall, who only discussed
ballots with strict ranking from
the top with truncation allowed. So it says that that B isn't allowed to
win if B is voted above bottom on fewer
ballots than A is voted alone above all others.
So it is generally accepted that Winning Votes meets the ("normal",
original) Plurality criterion.
Chris Benham
On Thu, May 4, 2023, 1:25 AM C.Benham cbenham@adam.com.au wrote:
My favourite method that meets both Condorcet and Chicken Dilemma is 'Min
Losing-Votes (equal-ranking whole) Sorted Margins Elimination':
*Voters rank from the top whatever number of candidates they like.
Equal-ranking and truncation are allowed.
For the purpose of determining candidates' pairwise scores:
a ballot that votes both X and Y above no other (remaining) candidates
contributes nothing to X's pairwise score versus Y and vice versa,
a ballot that ranks X and Y equal and above at least one (remaining)
candidate contributes a whole vote to X's pairwise score versus Y and vice
versa,
a ballot that ranks X above Y contributes a whole vote to X's pairwise
score versus Y and nothing to Y's pairwise score
versus X.
Give each candidate X a score equal to X's smallest losing pairwise
score.
Initially order the candidates from highest-scored to lowest scored. If
any adjacent pair is out-of-order pairwise, then swap
the out-of-order pair with the smallest score-difference. If there is a
tie for that then swap the tied pair that is lowest in
the order. Repeat until no adjacent pair is pairwise out-of-order, and
then eliminate the lowest-ordered candidate.
Repeat (disregarding any pairwise scores with eliminated candidates)
until one candidate remains. *
Some examples:
46 A>B
44 B>C (sincere is B or B>A)
05 C>A
05 C>B
A>B 51-49, B>C 90-10, C>A 54-46.
MinLV(erw) scores: B49 > A46 > C10.
Both adjacent pairs (B>A and A>C) are pairwise out of order. The B>A
score-difference is the smallest of the two
(3 versus 36) so we first swap that order to give
A49 > B51 > C10
Now neither pair of adjacent candidates is pairwise out of order so C is
eliminated and A wins.
Winning Votes, Margins, MMPO elect the Burier's candidate.
25 A>B
26 B>C
23 C>A
26 C
C>A 75-25, A>B 48-26, B>C 51-49.
MinLV(erw) scores: C49 > B26 > A25.
Both adjacent pairs (C>B and B>A) are pairwise out-of-order. The B-A
score difference is by
far the smallest, so we swap the B>A order to give
C > A > B. That order is final and C wins. C is the most top ranked
and the most above-bottom ranked
candidate. WV, MMPO, IRV, Benham elect B.
35 A
10 A=B
30 B>C
25 C
C>A 55-45, A>B 45-40 (note 10A=B effect), B>C 40-25.
I noticed that A has more losing votes (45) than B has wining votes (40).
It seems to me that this fact (by itself) should disqualify B.
Also B is Ranked on feet ballots (40) than A is ranked Top (45) so
Plurality requires B to los as well.
So how about this as a tournament versión of Plurality:
If B's maxPairwiseSuppoft is less than A's minPS, then B should not win.
Anybody ever proposed this Criterion before?
MinLV(erw) scores: A45 > B40 > C25. Neither adjacent pair is pairwise
out-of-order so the order is final
and A wins.
A both pairwise-beats and positionally dominates B, but WV, Margins, MMPO
all elect B.
Chris Benham
Hi Chris/Forest,
Le lundi 5 juin 2023 à 01:03:25 UTC−5, C.Benham cbenham@adam.com.au a écrit :
On 5/06/2023 9:16 am, Forest Simmons wrote:
35 A
10 A=B
30 B>C
25 C
C>A 55-45, A>B 45-40 (note 10A=B effect), B>C 40-25.
I noticed that A has more losing votes (45) than B has wining votes (40).
It seems to me that this fact (by itself) should disqualify B.
So how about this as a tournament versión of Plurality:
If B's maxPairwiseSuppoft is less than A's minPS, then B should not win.
Anybody ever proposed this Criterion before?
I and/or Kevin Venzke may have mentioned it in passing. The example is originally from Kevin.
There was a concept of "pairwise Plurality" under which the problem is that B's maxPS is
lower than the max opposition to B. So nothing to do with A. Similarly you could have
"pairwise minimal defense" which says if you have a majority loss but no majority win, then
you lose. It's considerably harder to meet than normal MD.
B's maxPS is not actually less than A's minPS. The 45-40 above is based on MinLV's adjusted
matrix.
I think that minPS should always be understood to be zero, so that there isn't an inherent
Clone-Winner issue with it.
I think that max losing votes comes across as inherently non-monotone, since if a candidate
X has two losing votes scores of 30 and 40, but then the latter is increased to 42 votes and
the contest becomes a win, their max losing votes will drop from 40 to 30. It's then not
clear how to guarantee that the decrease of this score is compensated for, in its effect, by
the gain of a pairwise win.
I agree that is a good criterion and should be a strong standard. I quite a while ago rejected the idea
that the best Condorcet methods were those that focused purely on "defeat strengths" with a view to
simply "break the cycle at its weakest link".
Also B is Ranked on fewer ballots (40) than A is ranked Top (45) so Plurality requires B to lose as well.
The Plurality criterion was coined by Douglas Woodall, who only discussed ballots with strict ranking from
the top with truncation allowed. So it says that that B isn't allowed to win if B is voted above bottom on fewer
ballots than A is voted alone above all others.
So it generally accepted that Winning Votes meets the ("normal", original) Plurality criterion.
Yes, it was noticed as an aberration that methods accepted to satisfy Plurality would elect B.
I've always implemented Plurality as checking against strict first preferences only.
When you use WV, in fact MinMax still doesn't satisfy Plurality:
39: C
29: A>B
20: D>A>B
12: B>D>C
D wins, agreeing with MMPO but basically nothing else.
(Margins would elect A, which is allowed by Plurality, but is still a very unusual choice.
To compare, generally MD methods elect B, and LNHarm methods pick C.)
Kevin
votingmethods.net
On 9/06/2023 9:31 pm, Kevin Venzke wrote:
I think that max losing votes comes across as inherently non-monotone, since if a candidate
X has two losing votes scores of 30 and 40, but then the latter is increased to 42 votes and
the contest becomes a win, their max losing votes will drop from 40 to 30. It's then not
clear how to guarantee that the decrease of this score is compensated for, in its effect, by
the gain of a pairwise win.
Kevin,
Are talking about the MinLV(erw) Sorted Margins Elimination method, and
"max" instead of min is a typo?
If so, in the example I assume you mean the 30 (not the 40) is increased
enough for it to stop being a loss.
It is very very rare for the Smith set to contain more than three
members. The members of the Smith set at some point
will be the only candidates left after the rest have been eliminated.
Then it should be clear that X "gaining a pairwise win"
simply makes X the CW.
Chris
On 9/06/2023 9:31 pm, Kevin Venzke wrote:
Hi Chris/Forest,
Le lundi 5 juin 2023 à 01:03:25 UTC−5, C.Benham cbenham@adam.com.au a écrit :
On 5/06/2023 9:16 am, Forest Simmons wrote:
35 A
10 A=B
30 B>C
25 C
C>A 55-45, A>B 45-40 (note 10A=B effect), B>C 40-25.
I noticed that A has more losing votes (45) than B has wining votes (40).
It seems to me that this fact (by itself) should disqualify B.
So how about this as a tournament versión of Plurality:
If B's maxPairwiseSuppoft is less than A's minPS, then B should not win.
Anybody ever proposed this Criterion before?
I and/or Kevin Venzke may have mentioned it in passing. The example is originally from Kevin.
There was a concept of "pairwise Plurality" under which the problem is that B's maxPS is
lower than the max opposition to B. So nothing to do with A. Similarly you could have
"pairwise minimal defense" which says if you have a majority loss but no majority win, then
you lose. It's considerably harder to meet than normal MD.
B's maxPS is not actually less than A's minPS. The 45-40 above is based on MinLV's adjusted
matrix.
I think that minPS should always be understood to be zero, so that there isn't an inherent
Clone-Winner issue with it.
I think that max losing votes comes across as inherently non-monotone, since if a candidate
X has two losing votes scores of 30 and 40, but then the latter is increased to 42 votes and
the contest becomes a win, their max losing votes will drop from 40 to 30. It's then not
clear how to guarantee that the decrease of this score is compensated for, in its effect, by
the gain of a pairwise win.
I agree that is a good criterion and should be a strong standard. I quite a while ago rejected the idea
that the best Condorcet methods were those that focused purely on "defeat strengths" with a view to
simply "break the cycle at its weakest link".
Also B is Ranked on fewer ballots (40) than A is ranked Top (45) so Plurality requires B to lose as well.
The Plurality criterion was coined by Douglas Woodall, who only discussed ballots with strict ranking from
the top with truncation allowed. So it says that that B isn't allowed to win if B is voted above bottom on fewer
ballots than A is voted alone above all others.
So it generally accepted that Winning Votes meets the ("normal", original) Plurality criterion.
Yes, it was noticed as an aberration that methods accepted to satisfy Plurality would elect B.
I've always implemented Plurality as checking against strict first preferences only.
When you use WV, in fact MinMax still doesn't satisfy Plurality:
39: C
29: A>B
20: D>A>B
12: B>D>C
D wins, agreeing with MMPO but basically nothing else.
(Margins would elect A, which is allowed by Plurality, but is still a very unusual choice.
To compare, generally MD methods elect B, and LNHarm methods pick C.)
Kevin
votingmethods.net
Hi Chris, no, I was referring to Forest's observation about A's (presumably maximum) losing votes. No specific method.
Kevin
(end)
Le dimanche 18 juin 2023 à 09:39:48 UTC−5, C.Benham cbenham@adam.com.au a écrit :
On 9/06/2023 9:31 pm, Kevin Venzke wrote:
I think that max losing votes comes across as inherently non-monotone, since if a candidate
X has two losing votes scores of 30 and 40, but then the latter is increased to 42 votes and
the contest becomes a win, their max losing votes will drop from 40 to 30. It's then not
clear how to guarantee that the decrease of this score is compensated for, in its effect, by
the gain of a pairwise win.
Kevin,
Are talking about the MinLV(erw) Sorted Margins Elimination method, and
"max" instead of min is a typo?
If so, in the example I assume you mean the 30 (not the 40) is increased
enough for it to stop being a loss.
It is very very rare for the Smith set to contain more than three
members. The members of the Smith set at some point
will be the only candidates left after the rest have been eliminated.
Then it should be clear that X "gaining a pairwise win"
simply makes X the CW.
Chris
On 9/06/2023 9:31 pm, Kevin Venzke wrote:
Hi Chris/Forest,
Le lundi 5 juin 2023 à 01:03:25 UTC−5, C.Benham cbenham@adam.com.au a écrit :
On 5/06/2023 9:16 am, Forest Simmons wrote:
35 A
10 A=B
30 B>C
25 C
C>A 55-45, A>B 45-40 (note 10A=B effect), B>C 40-25.
I noticed that A has more losing votes (45) than B has wining votes (40).