Consensus and PR methods
Say we have a consensus method M that works by choosing the council C
that minimizes the maximum penalty p(C, v) for the voter that maximizes
this penalty. That is, the method finds C according to
C = arg min max p(c, v)
c v
where ties are broken in a leximax fashion (i.e. considering next to
max, then next to next to max and so on). Furthermore let the penalty
"nonnegative" in the sense that any voter with a real preference has at
least as great a penalty as a voter with no preference (the zero voter,
as it were).
Now let the modified consensus method M' be one that has the same
optimization objective, but the method is permitted to remove a Droop
quota of votes to help minimize the penalty.
So M says "what council displeases the most displeased voter the
least?", while M' says "what council displeases the most displeased
voter the least, if we can discard a Droop quota of voters from
consideration?"
Then, are there any properties for p that makes M' satisfy Droop
proportionality? Can we in general turn consensus methods of this form
into PR methods by adding a "you can discard a Droop quota" rule?
If we can, then we easily get a multiwinner version of Bucklin/MJ by
doing the following:
Let g(c, v) be the grade voter v gives to the least preferred candidate
in c.
Let the consensus method M be
C = arg max min g(c, v)
c v
Let M' permit the method to remove a Droop quota, i.e. if |V| is the
number of voters, and V is the set of voters itself:
C' = arg max c:
max x subset of V so that |x| = |V|/(seats+1):
min v in V \ x:
g(c, v)
For a single-winner election, M' is (up to tiebreaker) just MJ, because
for each potential winner c, the removal step will remove the voters who
grade c the worst, and the Droop quota for single-winner is a majority.
Then the voter grading the c the worst after half of the voters have
been removed is just the median voter.
Some thoughts about two-winner remove-voter minimax Approval follow.
Reasoning about what voter removal actually does can get kinda hairy,
thus I may very well be wrong:
In minimax Approval, p(c, v) is the Hamming distance between c and voter
v's ballot, i.e. the number of candidates in c but not approved by v
plus the number of candidates approved by v not in c.
Say we have an analogous Droop criterion for Approval: if more than k
Droop quotas approve of a set of j candidates (and nobody else), then at
least min(k, j) of these must be elected.
For two winners, there are these possibilities:
1. no Droop constraints
2. k = 2, j >= 2
3. k = 2, j = 1
4. k = 1, j >= 1
5. k = 1, j = 1
-
is no problem, because we can elect anyone we want without running
afoul of the Approval DPC. -
Since there can only be three Droop quotas in total, when we're
considering A = {C_1, C_2} with C_1 and C_2 in the set of j candidates
(call it J), we can eliminate all but the J-voters and the maximum
penalty is j-2.
In contrast, for some B = {C_x, C_y} not a subset of j, the best it can
do is eliminate a Droop quota of the J-voters. In the best case (for B),
everybody but the J-voters approve of B alone. But there still remains a
Droop quota (plus one voter) of the J-voters, and each of them gives
penalty j. So A is preferred to B.
If B = {C_1, C_x}, then even if everybody but the J-voters approve of B
alone, the J-voters give penalty j-1. So A is still preferred to B. -
Same as in 2, but let A = {C_1, C_x}, J = {C_1}. With A, we eliminate
so that only the J-voters are left, and then max penalty is 1 (for C_x).
Furthermore, every remaining voter gives penalty 1. Let B = {C_x, C_y}.
In the best case for B, a Droop quota of J-voters are eliminated and we
have a Droop quota plus one left. These all give penalty 2, which is
worse than penalty 1. So A is preferred to B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. In the best case for B here,
two Droop quotas minus a voter approve only of B, and the remaining
Droop quota plus one voter approves of J = {C_1}. Eliminating all but
one of the J-voters gives a max penalty of 3 from that one J-voter: one
point for not having C_1, and two points for having C_x and C_y. A
eliminates one of the two B-approving Droop quotas and gets a penalty of
1 from every remaining voter, which is better.
Note that I assume that C_x is approved by the B-voters. If that were
not the case, then {C_x, C_y} would already be beaten by some {C_z, C_y}
where C_z is. Note also that I don't consider the case where the
B-voters also approve of a whole load of other candidates, with the idea
of raising the penalty under A. The problem is that because only two
candidates can be elected, this would also raise their penalty under B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. The best case for B has worst
penalty j+2, since after a Droop quota of J-voters have been eliminated,
there remains a single voter who only approves of J. After eliminating
some of the B-voters, A gets penalty j from the J-voters (j-1 for the
members of J not part of {C_1, C_x} and one more for C_x which is not
approved by them), and one penalty point from the B-voters.
Here it'd seem that adding loads of candidates to the B-voters would
make things hard. Can it be salvaged?
Suppose there are J-voters and C-voters. B is a subset of C.
When considering outcome B, before excluding a Droop quota, every
J-voter gives a penalty of j+2 and every C-voter gives a penalty of c-2
where c=|C|.
Under outcome A, before excluding, every J-voter gives j, and every
B-voter gives c (-1 for having C_x, +1 for having C_1).
If j+2 > c, then we're in the domain above, and no problem.
If c > j+2, then the excluded candidates under both A and B are C-voters.
So under B we have a Droop quota of C-voters with penalty c-2, and a
Droop quota plus one of J-voters at j+2.
Under A we have a Droop quota of C-voters with penalty c, and a Droop
quota plus one of J-voters at j.
So unless I made a mistake, Hamming distance is not good enough. But I
might have made a mistake, because it seems strange that even in
ordinary minimax Approval, a coalition can increase its power by
approving a lot of clones. E.g. suppose in ordinary minimax Approval
that there are two coalitions of almost equal size:
n+1: A B
n: C1 C2 C3 ... Cq
{A, B} gets worst penalty q+2 (there are n of these and n+1 zeroes)
{A, C1} gets worst penalty q (n voters like C1 but not A)
{C1, C2} gets worst penalty q-2 (n voters give this penalty, and then
n+1 give penalty 4).
... does that mean an arbitrarily small minority can force its own
council to win by just approving enough clones that they set the worst
penalty in every outcome? That feels rather wrong.
So, the academic world has no consensus or standard model of election
method?
On 03/03/2018 19:57, Kristofer Munsterhjelm wrote:
Say we have a consensus method M that works by choosing the council C
that minimizes the maximum penalty p(C, v) for the voter that
maximizes this penalty. That is, the method finds C according to
C = arg min max p(c, v)
c v
where ties are broken in a leximax fashion (i.e. considering next to
max, then next to next to max and so on). Furthermore let the penalty
"nonnegative" in the sense that any voter with a real preference has
at least as great a penalty as a voter with no preference (the zero
voter, as it were).
Now let the modified consensus method M' be one that has the same
optimization objective, but the method is permitted to remove a Droop
quota of votes to help minimize the penalty.
So M says "what council displeases the most displeased voter the
least?", while M' says "what council displeases the most displeased
voter the least, if we can discard a Droop quota of voters from
consideration?"
Then, are there any properties for p that makes M' satisfy Droop
proportionality? Can we in general turn consensus methods of this form
into PR methods by adding a "you can discard a Droop quota" rule?
If we can, then we easily get a multiwinner version of Bucklin/MJ by
doing the following:
Let g(c, v) be the grade voter v gives to the least preferred
candidate in c.
Let the consensus method M be
C = arg max min g(c, v)
c v
Let M' permit the method to remove a Droop quota, i.e. if |V| is the
number of voters, and V is the set of voters itself:
C' = arg max c:
max x subset of V so that |x| = |V|/(seats+1):
min v in V \ x:
g(c, v)
For a single-winner election, M' is (up to tiebreaker) just MJ,
because for each potential winner c, the removal step will remove the
voters who grade c the worst, and the Droop quota for single-winner is
a majority. Then the voter grading the c the worst after half of the
voters have been removed is just the median voter.
Some thoughts about two-winner remove-voter minimax Approval follow.
Reasoning about what voter removal actually does can get kinda hairy,
thus I may very well be wrong:
In minimax Approval, p(c, v) is the Hamming distance between c and
voter v's ballot, i.e. the number of candidates in c but not approved
by v plus the number of candidates approved by v not in c.
Say we have an analogous Droop criterion for Approval: if more than k
Droop quotas approve of a set of j candidates (and nobody else), then
at least min(k, j) of these must be elected.
For two winners, there are these possibilities:
1. no Droop constraints
2. k = 2, j >= 2
3. k = 2, j = 1
4. k = 1, j >= 1
5. k = 1, j = 1
-
is no problem, because we can elect anyone we want without running
afoul of the Approval DPC. -
Since there can only be three Droop quotas in total, when we're
considering A = {C_1, C_2} with C_1 and C_2 in the set of j candidates
(call it J), we can eliminate all but the J-voters and the maximum
penalty is j-2.
In contrast, for some B = {C_x, C_y} not a subset of j, the best it
can do is eliminate a Droop quota of the J-voters. In the best case
(for B), everybody but the J-voters approve of B alone. But there
still remains a Droop quota (plus one voter) of the J-voters, and each
of them gives penalty j. So A is preferred to B.
If B = {C_1, C_x}, then even if everybody but the J-voters approve of
B alone, the J-voters give penalty j-1. So A is still preferred to B. -
Same as in 2, but let A = {C_1, C_x}, J = {C_1}. With A, we
eliminate so that only the J-voters are left, and then max penalty is
1 (for C_x). Furthermore, every remaining voter gives penalty 1. Let B
= {C_x, C_y}. In the best case for B, a Droop quota of J-voters are
eliminated and we have a Droop quota plus one left. These all give
penalty 2, which is worse than penalty 1. So A is preferred to B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. In the best case for B here,
two Droop quotas minus a voter approve only of B, and the remaining
Droop quota plus one voter approves of J = {C_1}. Eliminating all but
one of the J-voters gives a max penalty of 3 from that one J-voter:
one point for not having C_1, and two points for having C_x and C_y. A
eliminates one of the two B-approving Droop quotas and gets a penalty
of 1 from every remaining voter, which is better.
Note that I assume that C_x is approved by the B-voters. If that were
not the case, then {C_x, C_y} would already be beaten by some {C_z,
C_y} where C_z is. Note also that I don't consider the case where the
B-voters also approve of a whole load of other candidates, with the
idea of raising the penalty under A. The problem is that because only
two candidates can be elected, this would also raise their penalty
under B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. The best case for B has
worst penalty j+2, since after a Droop quota of J-voters have been
eliminated, there remains a single voter who only approves of J. After
eliminating some of the B-voters, A gets penalty j from the J-voters
(j-1 for the members of J not part of {C_1, C_x} and one more for C_x
which is not approved by them), and one penalty point from the B-voters.
Here it'd seem that adding loads of candidates to the B-voters would
make things hard. Can it be salvaged?
Suppose there are J-voters and C-voters. B is a subset of C.
When considering outcome B, before excluding a Droop quota, every
J-voter gives a penalty of j+2 and every C-voter gives a penalty of
c-2 where c=|C|.
Under outcome A, before excluding, every J-voter gives j, and every
B-voter gives c (-1 for having C_x, +1 for having C_1).
If j+2 > c, then we're in the domain above, and no problem.
If c > j+2, then the excluded candidates under both A and B are C-voters.
So under B we have a Droop quota of C-voters with penalty c-2, and a
Droop quota plus one of J-voters at j+2.
Under A we have a Droop quota of C-voters with penalty c, and a Droop
quota plus one of J-voters at j.
So unless I made a mistake, Hamming distance is not good enough. But I
might have made a mistake, because it seems strange that even in
ordinary minimax Approval, a coalition can increase its power by
approving a lot of clones. E.g. suppose in ordinary minimax Approval
that there are two coalitions of almost equal size:
n+1: A B
n: C1 C2 C3 ... Cq
{A, B} gets worst penalty q+2 (there are n of these and n+1 zeroes)
{A, C1} gets worst penalty q (n voters like C1 but not A)
{C1, C2} gets worst penalty q-2 (n voters give this penalty, and then
n+1 give penalty 4).
... does that mean an arbitrarily small minority can force its own
council to win by just approving enough clones that they set the worst
penalty in every outcome? That feels rather wrong.
Election-Methods mailing list - see http://electorama.com/em for list
info
--
Richard Lung.
http://www.voting.ukscientists.com
Democracy Science series 3 free e-books in pdf:
https://plus.google.com/106191200795605365085
E-books in epub format:
https://www.smashwords.com/profile/view/democracyscience
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus
Consensus in academia? Maybe that cigarettes cause cancer. Maybe.
I jest.
This paper may be helpful:
http://personal.lse.ac.uk/hix/Working_Papers/Carey-Hix-AJPS2011.pdf
On Wed, Mar 7, 2018 at 2:14 PM, Richard Lung voting@ukscientists.com
wrote:
So, the academic world has no consensus or standard model of election
method?
On 03/03/2018 19:57, Kristofer Munsterhjelm wrote:
Say we have a consensus method M that works by choosing the council C
that minimizes the maximum penalty p(C, v) for the voter that maximizes
this penalty. That is, the method finds C according to
C = arg min max p(c, v)
c v
where ties are broken in a leximax fashion (i.e. considering next to max,
then next to next to max and so on). Furthermore let the penalty
"nonnegative" in the sense that any voter with a real preference has at
least as great a penalty as a voter with no preference (the zero voter, as
it were).
Now let the modified consensus method M' be one that has the same
optimization objective, but the method is permitted to remove a Droop quota
of votes to help minimize the penalty.
So M says "what council displeases the most displeased voter the least?",
while M' says "what council displeases the most displeased voter the least,
if we can discard a Droop quota of voters from consideration?"
Then, are there any properties for p that makes M' satisfy Droop
proportionality? Can we in general turn consensus methods of this form into
PR methods by adding a "you can discard a Droop quota" rule?
If we can, then we easily get a multiwinner version of Bucklin/MJ by
doing the following:
Let g(c, v) be the grade voter v gives to the least preferred candidate
in c.
Let the consensus method M be
C = arg max min g(c, v)
c v
Let M' permit the method to remove a Droop quota, i.e. if |V| is the
number of voters, and V is the set of voters itself:
C' = arg max c:
max x subset of V so that |x| = |V|/(seats+1):
min v in V \ x:
g(c, v)
For a single-winner election, M' is (up to tiebreaker) just MJ, because
for each potential winner c, the removal step will remove the voters who
grade c the worst, and the Droop quota for single-winner is a majority.
Then the voter grading the c the worst after half of the voters have been
removed is just the median voter.
Some thoughts about two-winner remove-voter minimax Approval follow.
Reasoning about what voter removal actually does can get kinda hairy, thus
I may very well be wrong:
In minimax Approval, p(c, v) is the Hamming distance between c and voter
v's ballot, i.e. the number of candidates in c but not approved by v plus
the number of candidates approved by v not in c.
Say we have an analogous Droop criterion for Approval: if more than k
Droop quotas approve of a set of j candidates (and nobody else), then at
least min(k, j) of these must be elected.
For two winners, there are these possibilities:
1. no Droop constraints
2. k = 2, j >= 2
3. k = 2, j = 1
4. k = 1, j >= 1
5. k = 1, j = 1
-
is no problem, because we can elect anyone we want without running
afoul of the Approval DPC. -
Since there can only be three Droop quotas in total, when we're
considering A = {C_1, C_2} with C_1 and C_2 in the set of j candidates
(call it J), we can eliminate all but the J-voters and the maximum penalty
is j-2.
In contrast, for some B = {C_x, C_y} not a subset of j, the best it can
do is eliminate a Droop quota of the J-voters. In the best case (for B),
everybody but the J-voters approve of B alone. But there still remains a
Droop quota (plus one voter) of the J-voters, and each of them gives
penalty j. So A is preferred to B.
If B = {C_1, C_x}, then even if everybody but the J-voters approve of B
alone, the J-voters give penalty j-1. So A is still preferred to B. -
Same as in 2, but let A = {C_1, C_x}, J = {C_1}. With A, we eliminate
so that only the J-voters are left, and then max penalty is 1 (for C_x).
Furthermore, every remaining voter gives penalty 1. Let B = {C_x, C_y}. In
the best case for B, a Droop quota of J-voters are eliminated and we have a
Droop quota plus one left. These all give penalty 2, which is worse than
penalty 1. So A is preferred to B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. In the best case for B here,
two Droop quotas minus a voter approve only of B, and the remaining Droop
quota plus one voter approves of J = {C_1}. Eliminating all but one of the
J-voters gives a max penalty of 3 from that one J-voter: one point for not
having C_1, and two points for having C_x and C_y. A eliminates one of the
two B-approving Droop quotas and gets a penalty of 1 from every remaining
voter, which is better.
Note that I assume that C_x is approved by the B-voters. If that were not
the case, then {C_x, C_y} would already be beaten by some {C_z, C_y} where
C_z is. Note also that I don't consider the case where the B-voters also
approve of a whole load of other candidates, with the idea of raising the
penalty under A. The problem is that because only two candidates can be
elected, this would also raise their penalty under B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. The best case for B has worst
penalty j+2, since after a Droop quota of J-voters have been eliminated,
there remains a single voter who only approves of J. After eliminating some
of the B-voters, A gets penalty j from the J-voters (j-1 for the members of
J not part of {C_1, C_x} and one more for C_x which is not approved by
them), and one penalty point from the B-voters.
Here it'd seem that adding loads of candidates to the B-voters would make
things hard. Can it be salvaged?
Suppose there are J-voters and C-voters. B is a subset of C.
When considering outcome B, before excluding a Droop quota, every J-voter
gives a penalty of j+2 and every C-voter gives a penalty of c-2 where c=|C|.
Under outcome A, before excluding, every J-voter gives j, and every
B-voter gives c (-1 for having C_x, +1 for having C_1).
If j+2 > c, then we're in the domain above, and no problem.
If c > j+2, then the excluded candidates under both A and B are C-voters.
So under B we have a Droop quota of C-voters with penalty c-2, and a
Droop quota plus one of J-voters at j+2.
Under A we have a Droop quota of C-voters with penalty c, and a Droop
quota plus one of J-voters at j.
So unless I made a mistake, Hamming distance is not good enough. But I
might have made a mistake, because it seems strange that even in ordinary
minimax Approval, a coalition can increase its power by approving a lot of
clones. E.g. suppose in ordinary minimax Approval that there are two
coalitions of almost equal size:
n+1: A B
n: C1 C2 C3 ... Cq
{A, B} gets worst penalty q+2 (there are n of these and n+1 zeroes)
{A, C1} gets worst penalty q (n voters like C1 but not A)
{C1, C2} gets worst penalty q-2 (n voters give this penalty, and then n+1
give penalty 4).
... does that mean an arbitrarily small minority can force its own
council to win by just approving enough clones that they set the worst
penalty in every outcome? That feels rather wrong.
Election-Methods mailing list - see http://electorama.com/em for list
info
--
Richard Lung.
http://www.voting.ukscientists.com
Democracy Science series 3 free e-books in pdf:
https://plus.google.com/106191200795605365085
E-books in epub format:
https://www.smashwords.com/profile/view/democracyscience
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus
Election-Methods mailing list - see http://electorama.com/em for list info
--
Jack Santucci, Ph.D.
Independent scholar
http://www.jacksantucci.com
If my old memory serves me tolerably well, isn't this paper something
like an article entitled The Best of Both Worlds, where the authors did
a survey of a tendency for European electoral systems, over the decades,
to have decreased their average magnitude. I forget the details, just
about everything actually. But it may have gone something like: the
constiuencies shrank and the thresholds got higher.
It was an informative statistical survey.
But I think it went awry on what academics are fond of calling
"normative" considerations. Or on the stricture of David Hume, that what
is, is not necessarily right.
I would have put to the authors, as a critic. That was this trend, they
so diligently exposed, but the moving to a "sweet spot" for political
incumbents, with precious little to do with democracy and effective
elections for the voters?
from
Richard Lung.
On 07/03/2018 19:19, Jack Santucci wrote:
Consensus in academia? Maybe that cigarettes cause cancer. Maybe.
I jest.
This paper may be helpful:
http://personal.lse.ac.uk/hix/Working_Papers/Carey-Hix-AJPS2011.pdf
On Wed, Mar 7, 2018 at 2:14 PM, Richard Lung <voting@ukscientists.com
mailto:voting@ukscientists.com> wrote:
So, the academic world has no consensus or standard model of
election method?
On 03/03/2018 19:57, Kristofer Munsterhjelm wrote:
Say we have a consensus method M that works by choosing the
council C that minimizes the maximum penalty p(C, v) for the
voter that maximizes this penalty. That is, the method finds C
according to
C = arg min max p(c, v)
c v
where ties are broken in a leximax fashion (i.e. considering
next to max, then next to next to max and so on). Furthermore
let the penalty "nonnegative" in the sense that any voter with
a real preference has at least as great a penalty as a voter
with no preference (the zero voter, as it were).
Now let the modified consensus method M' be one that has the
same optimization objective, but the method is permitted to
remove a Droop quota of votes to help minimize the penalty.
So M says "what council displeases the most displeased voter
the least?", while M' says "what council displeases the most
displeased voter the least, if we can discard a Droop quota of
voters from consideration?"
Then, are there any properties for p that makes M' satisfy
Droop proportionality? Can we in general turn consensus
methods of this form into PR methods by adding a "you can
discard a Droop quota" rule?
If we can, then we easily get a multiwinner version of
Bucklin/MJ by doing the following:
Let g(c, v) be the grade voter v gives to the least preferred
candidate in c.
Let the consensus method M be
C = arg max min g(c, v)
c v
Let M' permit the method to remove a Droop quota, i.e. if |V|
is the number of voters, and V is the set of voters itself:
C' = arg max c:
max x subset of V so that |x| = |V|/(seats+1):
min v in V \ x:
g(c, v)
For a single-winner election, M' is (up to tiebreaker) just
MJ, because for each potential winner c, the removal step will
remove the voters who grade c the worst, and the Droop quota
for single-winner is a majority. Then the voter grading the c
the worst after half of the voters have been removed is just
the median voter.
Some thoughts about two-winner remove-voter minimax Approval
follow. Reasoning about what voter removal actually does can
get kinda hairy, thus I may very well be wrong:
In minimax Approval, p(c, v) is the Hamming distance between c
and voter v's ballot, i.e. the number of candidates in c but
not approved by v plus the number of candidates approved by v
not in c.
Say we have an analogous Droop criterion for Approval: if more
than k Droop quotas approve of a set of j candidates (and
nobody else), then at least min(k, j) of these must be elected.
For two winners, there are these possibilities:
1. no Droop constraints
2. k = 2, j >= 2
3. k = 2, j = 1
4. k = 1, j >= 1
5. k = 1, j = 1
1. is no problem, because we can elect anyone we want without
running afoul of the Approval DPC.
2. Since there can only be three Droop quotas in total, when
we're considering A = {C_1, C_2} with C_1 and C_2 in the set
of j candidates (call it J), we can eliminate all but the
J-voters and the maximum penalty is j-2.
In contrast, for some B = {C_x, C_y} not a subset of j, the
best it can do is eliminate a Droop quota of the J-voters. In
the best case (for B), everybody but the J-voters approve of B
alone. But there still remains a Droop quota (plus one voter)
of the J-voters, and each of them gives penalty j. So A is
preferred to B.
If B = {C_1, C_x}, then even if everybody but the J-voters
approve of B alone, the J-voters give penalty j-1. So A is
still preferred to B.
3. Same as in 2, but let A = {C_1, C_x}, J = {C_1}. With A, we
eliminate so that only the J-voters are left, and then max
penalty is 1 (for C_x). Furthermore, every remaining voter
gives penalty 1. Let B = {C_x, C_y}. In the best case for B, a
Droop quota of J-voters are eliminated and we have a Droop
quota plus one left. These all give penalty 2, which is worse
than penalty 1. So A is preferred to B.
5. Let A = {C_1, C_x} and B = {C_x, C_y}. In the best case for
B here, two Droop quotas minus a voter approve only of B, and
the remaining Droop quota plus one voter approves of J =
{C_1}. Eliminating all but one of the J-voters gives a max
penalty of 3 from that one J-voter: one point for not having
C_1, and two points for having C_x and C_y. A eliminates one
of the two B-approving Droop quotas and gets a penalty of 1
from every remaining voter, which is better.
Note that I assume that C_x is approved by the B-voters. If
that were not the case, then {C_x, C_y} would already be
beaten by some {C_z, C_y} where C_z is. Note also that I don't
consider the case where the B-voters also approve of a whole
load of other candidates, with the idea of raising the penalty
under A. The problem is that because only two candidates can
be elected, this would also raise their penalty under B.
4. Let A = {C_1, C_x} and B = {C_x, C_y}. The best case for B
has worst penalty j+2, since after a Droop quota of J-voters
have been eliminated, there remains a single voter who only
approves of J. After eliminating some of the B-voters, A gets
penalty j from the J-voters (j-1 for the members of J not part
of {C_1, C_x} and one more for C_x which is not approved by
them), and one penalty point from the B-voters.
Here it'd seem that adding loads of candidates to the B-voters
would make things hard. Can it be salvaged?
Suppose there are J-voters and C-voters. B is a subset of C.
When considering outcome B, before excluding a Droop quota,
every J-voter gives a penalty of j+2 and every C-voter gives a
penalty of c-2 where c=|C|.
Under outcome A, before excluding, every J-voter gives j, and
every B-voter gives c (-1 for having C_x, +1 for having C_1).
If j+2 > c, then we're in the domain above, and no problem.
If c > j+2, then the excluded candidates under both A and B
are C-voters.
So under B we have a Droop quota of C-voters with penalty c-2,
and a Droop quota plus one of J-voters at j+2.
Under A we have a Droop quota of C-voters with penalty c, and
a Droop quota plus one of J-voters at j.
So unless I made a mistake, Hamming distance is not good
enough. But I might have made a mistake, because it seems
strange that even in ordinary minimax Approval, a coalition
can increase its power by approving a lot of clones. E.g.
suppose in ordinary minimax Approval that there are two
coalitions of almost equal size:
n+1: A B
n: C1 C2 C3 ... Cq
{A, B} gets worst penalty q+2 (there are n of these and n+1
zeroes)
{A, C1} gets worst penalty q (n voters like C1 but not A)
{C1, C2} gets worst penalty q-2 (n voters give this penalty,
and then n+1 give penalty 4).
... does that mean an arbitrarily small minority can force its
own council to win by just approving enough clones that they
set the worst penalty in every outcome? That feels rather wrong.
----
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Political scientists like their parties to be few and disciplined. This is said to promote accountability.
Sent from my iPhone
On Mar 7, 2018, at 20:11, Richard Lung voting@ukscientists.com wrote:
If my old memory serves me tolerably well, isn't this paper something like an article entitled The Best of Both Worlds, where the authors did a survey of a tendency for European electoral systems, over the decades, to have decreased their average magnitude. I forget the details, just about everything actually. But it may have gone something like: the constiuencies shrank and the thresholds got higher.
It was an informative statistical survey.
But I think it went awry on what academics are fond of calling "normative" considerations. Or on the stricture of David Hume, that what is, is not necessarily right.
I would have put to the authors, as a critic. That was this trend, they so diligently exposed, but the moving to a "sweet spot" for political incumbents, with precious little to do with democracy and effective elections for the voters?
from
Richard Lung.
On 07/03/2018 19:19, Jack Santucci wrote:
Consensus in academia? Maybe that cigarettes cause cancer. Maybe.
I jest.
This paper may be helpful: http://personal.lse.ac.uk/hix/Working_Papers/Carey-Hix-AJPS2011.pdf
On Wed, Mar 7, 2018 at 2:14 PM, Richard Lung voting@ukscientists.com wrote:
So, the academic world has no consensus or standard model of election method?
On 03/03/2018 19:57, Kristofer Munsterhjelm wrote:
Say we have a consensus method M that works by choosing the council C that minimizes the maximum penalty p(C, v) for the voter that maximizes this penalty. That is, the method finds C according to
C = arg min max p(c, v)
c v
where ties are broken in a leximax fashion (i.e. considering next to max, then next to next to max and so on). Furthermore let the penalty "nonnegative" in the sense that any voter with a real preference has at least as great a penalty as a voter with no preference (the zero voter, as it were).
Now let the modified consensus method M' be one that has the same optimization objective, but the method is permitted to remove a Droop quota of votes to help minimize the penalty.
So M says "what council displeases the most displeased voter the least?", while M' says "what council displeases the most displeased voter the least, if we can discard a Droop quota of voters from consideration?"
Then, are there any properties for p that makes M' satisfy Droop proportionality? Can we in general turn consensus methods of this form into PR methods by adding a "you can discard a Droop quota" rule?
If we can, then we easily get a multiwinner version of Bucklin/MJ by doing the following:
Let g(c, v) be the grade voter v gives to the least preferred candidate in c.
Let the consensus method M be
C = arg max min g(c, v)
c v
Let M' permit the method to remove a Droop quota, i.e. if |V| is the number of voters, and V is the set of voters itself:
C' = arg max c:
max x subset of V so that |x| = |V|/(seats+1):
min v in V \ x:
g(c, v)
For a single-winner election, M' is (up to tiebreaker) just MJ, because for each potential winner c, the removal step will remove the voters who grade c the worst, and the Droop quota for single-winner is a majority. Then the voter grading the c the worst after half of the voters have been removed is just the median voter.
Some thoughts about two-winner remove-voter minimax Approval follow. Reasoning about what voter removal actually does can get kinda hairy, thus I may very well be wrong:
In minimax Approval, p(c, v) is the Hamming distance between c and voter v's ballot, i.e. the number of candidates in c but not approved by v plus the number of candidates approved by v not in c.
Say we have an analogous Droop criterion for Approval: if more than k Droop quotas approve of a set of j candidates (and nobody else), then at least min(k, j) of these must be elected.
For two winners, there are these possibilities:
1. no Droop constraints
2. k = 2, j >= 2
3. k = 2, j = 1
4. k = 1, j >= 1
5. k = 1, j = 1
-
is no problem, because we can elect anyone we want without running afoul of the Approval DPC.
-
Since there can only be three Droop quotas in total, when we're considering A = {C_1, C_2} with C_1 and C_2 in the set of j candidates (call it J), we can eliminate all but the J-voters and the maximum penalty is j-2.
In contrast, for some B = {C_x, C_y} not a subset of j, the best it can do is eliminate a Droop quota of the J-voters. In the best case (for B), everybody but the J-voters approve of B alone. But there still remains a Droop quota (plus one voter) of the J-voters, and each of them gives penalty j. So A is preferred to B.
If B = {C_1, C_x}, then even if everybody but the J-voters approve of B alone, the J-voters give penalty j-1. So A is still preferred to B. -
Same as in 2, but let A = {C_1, C_x}, J = {C_1}. With A, we eliminate so that only the J-voters are left, and then max penalty is 1 (for C_x). Furthermore, every remaining voter gives penalty 1. Let B = {C_x, C_y}. In the best case for B, a Droop quota of J-voters are eliminated and we have a Droop quota plus one left. These all give penalty 2, which is worse than penalty 1. So A is preferred to B.
-
Let A = {C_1, C_x} and B = {C_x, C_y}. In the best case for B here, two Droop quotas minus a voter approve only of B, and the remaining Droop quota plus one voter approves of J = {C_1}. Eliminating all but one of the J-voters gives a max penalty of 3 from that one J-voter: one point for not having C_1, and two points for having C_x and C_y. A eliminates one of the two B-approving Droop quotas and gets a penalty of 1 from every remaining voter, which is better.
Note that I assume that C_x is approved by the B-voters. If that were not the case, then {C_x, C_y} would already be beaten by some {C_z, C_y} where C_z is. Note also that I don't consider the case where the B-voters also approve of a whole load of other candidates, with the idea of raising the penalty under A. The problem is that because only two candidates can be elected, this would also raise their penalty under B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. The best case for B has worst penalty j+2, since after a Droop quota of J-voters have been eliminated, there remains a single voter who only approves of J. After eliminating some of the B-voters, A gets penalty j from the J-voters (j-1 for the members of J not part of {C_1, C_x} and one more for C_x which is not approved by them), and one penalty point from the B-voters.
Here it'd seem that adding loads of candidates to the B-voters would make things hard. Can it be salvaged?
Suppose there are J-voters and C-voters. B is a subset of C.
When considering outcome B, before excluding a Droop quota, every J-voter gives a penalty of j+2 and every C-voter gives a penalty of c-2 where c=|C|.
Under outcome A, before excluding, every J-voter gives j, and every B-voter gives c (-1 for having C_x, +1 for having C_1).
If j+2 > c, then we're in the domain above, and no problem.
If c > j+2, then the excluded candidates under both A and B are C-voters.
So under B we have a Droop quota of C-voters with penalty c-2, and a Droop quota plus one of J-voters at j+2.
Under A we have a Droop quota of C-voters with penalty c, and a Droop quota plus one of J-voters at j.
So unless I made a mistake, Hamming distance is not good enough. But I might have made a mistake, because it seems strange that even in ordinary minimax Approval, a coalition can increase its power by approving a lot of clones. E.g. suppose in ordinary minimax Approval that there are two coalitions of almost equal size:
n+1: A B
n: C1 C2 C3 ... Cq
{A, B} gets worst penalty q+2 (there are n of these and n+1 zeroes)
{A, C1} gets worst penalty q (n voters like C1 but not A)
{C1, C2} gets worst penalty q-2 (n voters give this penalty, and then n+1 give penalty 4).
... does that mean an arbitrarily small minority can force its own council to win by just approving enough clones that they set the worst penalty in every outcome? That feels rather wrong.
Election-Methods mailing list - see http://electorama.com/em for list info
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---------------------------- Original Message ----------------------------
Subject: Re: [EM] Consensus and PR methods
From: "Jack Santucci" jms346@georgetown.edu
Date: Wed, March 7, 2018 9:44 pm
To: "Richard Lung" voting@ukscientists.com
Cc: "EM" election-methods@lists.electorama.com
Political scientists like their parties to be few and disciplined. This is said to promote accountability.
"few" makes sense if it is at least 3 parties that are viable and get people elected.� then it wouldn't always fall into that "If you ain't fer us, you agin' us!" syndrome of the two-party polarization.
and i certainly don't want it to be as few as 1 or
0.� that would promote no accountability.� if you toss in libertarians and communitarians and maybe some wing on the left or right (like Greenies or Vermont Progs or John Birch), "few" might mean 6 or 8.
i like that we have parties (i just wish more than the Dems and GOP)
and i think they serve a useful purpose in accomplishing sometimes difficult political goals.� they do this by consolidation and coalition, and with combined numbers, they can demonstrate political will and popular authority.� this is why i am an unabashed advocate for RCV decided by
Condorcet.� It can demonstrate genuine mandate when such might be obscured by a polarized 2-party environment.
r b-j� � � � � � � � � � � � �rbj@audioimagination.com
"Imagination is more important than knowledge."
�
�
�
�
Few, but more than 2. Harmful equilibria are stable with 2 parties; never
stable for long with 3 or more.
And those who don't like any of the viable options (whether there are 2 or
3 or 10) should be able to vote in a way that reflects that, without
necessarily having their votes ignored. That means any good voting method
should allow cross-party voting and/or transfers somehow.
2018-03-07 21:44 GMT-05:00 Jack Santucci jms346@georgetown.edu:
Political scientists like their parties to be few and disciplined. This is
said to promote accountability.
Sent from my iPhone
On Mar 7, 2018, at 20:11, Richard Lung voting@ukscientists.com wrote:
If my old memory serves me tolerably well, isn't this paper something like
an article entitled The Best of Both Worlds, where the authors did a survey
of a tendency for European electoral systems, over the decades, to have
decreased their average magnitude. I forget the details, just about
everything actually. But it may have gone something like: the constiuencies
shrank and the thresholds got higher.
It was an informative statistical survey.
But I think it went awry on what academics are fond of calling "normative"
considerations. Or on the stricture of David Hume, that what is, is not
necessarily right.
I would have put to the authors, as a critic. That was this trend, they so
diligently exposed, but the moving to a "sweet spot" for political
incumbents, with precious little to do with democracy and effective
elections for the voters?
from
Richard Lung.
On 07/03/2018 19:19, Jack Santucci wrote:
Consensus in academia? Maybe that cigarettes cause cancer. Maybe.
I jest.
This paper may be helpful: http://personal.lse.ac.uk/hix/Working_Papers/
Carey-Hix-AJPS2011.pdf
On Wed, Mar 7, 2018 at 2:14 PM, Richard Lung voting@ukscientists.com
wrote:
So, the academic world has no consensus or standard model of election
method?
On 03/03/2018 19:57, Kristofer Munsterhjelm wrote:
Say we have a consensus method M that works by choosing the council C
that minimizes the maximum penalty p(C, v) for the voter that maximizes
this penalty. That is, the method finds C according to
C = arg min max p(c, v)
c v
where ties are broken in a leximax fashion (i.e. considering next to
max, then next to next to max and so on). Furthermore let the penalty
"nonnegative" in the sense that any voter with a real preference has at
least as great a penalty as a voter with no preference (the zero voter, as
it were).
Now let the modified consensus method M' be one that has the same
optimization objective, but the method is permitted to remove a Droop quota
of votes to help minimize the penalty.
So M says "what council displeases the most displeased voter the
least?", while M' says "what council displeases the most displeased voter
the least, if we can discard a Droop quota of voters from consideration?"
Then, are there any properties for p that makes M' satisfy Droop
proportionality? Can we in general turn consensus methods of this form into
PR methods by adding a "you can discard a Droop quota" rule?
If we can, then we easily get a multiwinner version of Bucklin/MJ by
doing the following:
Let g(c, v) be the grade voter v gives to the least preferred candidate
in c.
Let the consensus method M be
C = arg max min g(c, v)
c v
Let M' permit the method to remove a Droop quota, i.e. if |V| is the
number of voters, and V is the set of voters itself:
C' = arg max c:
max x subset of V so that |x| = |V|/(seats+1):
min v in V \ x:
g(c, v)
For a single-winner election, M' is (up to tiebreaker) just MJ, because
for each potential winner c, the removal step will remove the voters who
grade c the worst, and the Droop quota for single-winner is a majority.
Then the voter grading the c the worst after half of the voters have been
removed is just the median voter.
Some thoughts about two-winner remove-voter minimax Approval follow.
Reasoning about what voter removal actually does can get kinda hairy, thus
I may very well be wrong:
In minimax Approval, p(c, v) is the Hamming distance between c and voter
v's ballot, i.e. the number of candidates in c but not approved by v plus
the number of candidates approved by v not in c.
Say we have an analogous Droop criterion for Approval: if more than k
Droop quotas approve of a set of j candidates (and nobody else), then at
least min(k, j) of these must be elected.
For two winners, there are these possibilities:
1. no Droop constraints
2. k = 2, j >= 2
3. k = 2, j = 1
4. k = 1, j >= 1
5. k = 1, j = 1
-
is no problem, because we can elect anyone we want without running
afoul of the Approval DPC. -
Since there can only be three Droop quotas in total, when we're
considering A = {C_1, C_2} with C_1 and C_2 in the set of j candidates
(call it J), we can eliminate all but the J-voters and the maximum penalty
is j-2.
In contrast, for some B = {C_x, C_y} not a subset of j, the best it can
do is eliminate a Droop quota of the J-voters. In the best case (for B),
everybody but the J-voters approve of B alone. But there still remains a
Droop quota (plus one voter) of the J-voters, and each of them gives
penalty j. So A is preferred to B.
If B = {C_1, C_x}, then even if everybody but the J-voters approve of B
alone, the J-voters give penalty j-1. So A is still preferred to B. -
Same as in 2, but let A = {C_1, C_x}, J = {C_1}. With A, we eliminate
so that only the J-voters are left, and then max penalty is 1 (for C_x).
Furthermore, every remaining voter gives penalty 1. Let B = {C_x, C_y}. In
the best case for B, a Droop quota of J-voters are eliminated and we have a
Droop quota plus one left. These all give penalty 2, which is worse than
penalty 1. So A is preferred to B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. In the best case for B here,
two Droop quotas minus a voter approve only of B, and the remaining Droop
quota plus one voter approves of J = {C_1}. Eliminating all but one of the
J-voters gives a max penalty of 3 from that one J-voter: one point for not
having C_1, and two points for having C_x and C_y. A eliminates one of the
two B-approving Droop quotas and gets a penalty of 1 from every remaining
voter, which is better.
Note that I assume that C_x is approved by the B-voters. If that were
not the case, then {C_x, C_y} would already be beaten by some {C_z, C_y}
where C_z is. Note also that I don't consider the case where the B-voters
also approve of a whole load of other candidates, with the idea of raising
the penalty under A. The problem is that because only two candidates can be
elected, this would also raise their penalty under B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. The best case for B has worst
penalty j+2, since after a Droop quota of J-voters have been eliminated,
there remains a single voter who only approves of J. After eliminating some
of the B-voters, A gets penalty j from the J-voters (j-1 for the members of
J not part of {C_1, C_x} and one more for C_x which is not approved by
them), and one penalty point from the B-voters.
Here it'd seem that adding loads of candidates to the B-voters would
make things hard. Can it be salvaged?
Suppose there are J-voters and C-voters. B is a subset of C.
When considering outcome B, before excluding a Droop quota, every
J-voter gives a penalty of j+2 and every C-voter gives a penalty of c-2
where c=|C|.
Under outcome A, before excluding, every J-voter gives j, and every
B-voter gives c (-1 for having C_x, +1 for having C_1).
If j+2 > c, then we're in the domain above, and no problem.
If c > j+2, then the excluded candidates under both A and B are C-voters.
So under B we have a Droop quota of C-voters with penalty c-2, and a
Droop quota plus one of J-voters at j+2.
Under A we have a Droop quota of C-voters with penalty c, and a Droop
quota plus one of J-voters at j.
So unless I made a mistake, Hamming distance is not good enough. But I
might have made a mistake, because it seems strange that even in ordinary
minimax Approval, a coalition can increase its power by approving a lot of
clones. E.g. suppose in ordinary minimax Approval that there are two
coalitions of almost equal size:
n+1: A B
n: C1 C2 C3 ... Cq
{A, B} gets worst penalty q+2 (there are n of these and n+1 zeroes)
{A, C1} gets worst penalty q (n voters like C1 but not A)
{C1, C2} gets worst penalty q-2 (n voters give this penalty, and then
n+1 give penalty 4).
... does that mean an arbitrarily small minority can force its own
council to win by just approving enough clones that they set the worst
penalty in every outcome? That feels rather wrong.
Election-Methods mailing list - see http://electorama.com/em for list
info
--
Richard Lung.
http://www.voting.ukscientists.com
Democracy Science series 3 free e-books in pdf:
https://plus.google.com/106191200795605365085
E-books in epub format:
https://www.smashwords.com/profile/view/democracyscience
This email has been checked for viruses by Avast antivirus software.
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--
Jack Santucci, Ph.D.
Independent scholar
http://www.jacksantucci.com
--
Richard Lung.http://www.voting.ukscientists.com
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E-books in epub format:https://www.smashwords.com/profile/view/democracyscience
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Exactly the conclusion I reached when I designed the crurch option for SPPA...!
Harmful equilibria are stable with 2 parties; never stable for long with 3 or more.
And SPPA allows transfer between political parties by rallying.
Envoyé de mon iPhone
Le 8 mars 2018 à 11:46, Jameson Quinn jameson.quinn@gmail.com a écrit :
Few, but more than 2. Harmful equilibria are stable with 2 parties; never stable for long with 3 or more.
And those who don't like any of the viable options (whether there are 2 or 3 or 10) should be able to vote in a way that reflects that, without necessarily having their votes ignored. That means any good voting method should allow cross-party voting and/or transfers somehow.
2018-03-07 21:44 GMT-05:00 Jack Santucci jms346@georgetown.edu:
Political scientists like their parties to be few and disciplined. This is said to promote accountability.
Sent from my iPhone
On Mar 7, 2018, at 20:11, Richard Lung voting@ukscientists.com wrote:
If my old memory serves me tolerably well, isn't this paper something like an article entitled The Best of Both Worlds, where the authors did a survey of a tendency for European electoral systems, over the decades, to have decreased their average magnitude. I forget the details, just about everything actually. But it may have gone something like: the constiuencies shrank and the thresholds got higher.
It was an informative statistical survey.
But I think it went awry on what academics are fond of calling "normative" considerations. Or on the stricture of David Hume, that what is, is not necessarily right.
I would have put to the authors, as a critic. That was this trend, they so diligently exposed, but the moving to a "sweet spot" for political incumbents, with precious little to do with democracy and effective elections for the voters?
from
Richard Lung.
On 07/03/2018 19:19, Jack Santucci wrote:
Consensus in academia? Maybe that cigarettes cause cancer. Maybe.
I jest.
This paper may be helpful: http://personal.lse.ac.uk/hix/Working_Papers/Carey-Hix-AJPS2011.pdf
On Wed, Mar 7, 2018 at 2:14 PM, Richard Lung voting@ukscientists.com wrote:
So, the academic world has no consensus or standard model of election method?
On 03/03/2018 19:57, Kristofer Munsterhjelm wrote:
Say we have a consensus method M that works by choosing the council C that minimizes the maximum penalty p(C, v) for the voter that maximizes this penalty. That is, the method finds C according to
C = arg min max p(c, v)
c v
where ties are broken in a leximax fashion (i.e. considering next to max, then next to next to max and so on). Furthermore let the penalty "nonnegative" in the sense that any voter with a real preference has at least as great a penalty as a voter with no preference (the zero voter, as it were).
Now let the modified consensus method M' be one that has the same optimization objective, but the method is permitted to remove a Droop quota of votes to help minimize the penalty.
So M says "what council displeases the most displeased voter the least?", while M' says "what council displeases the most displeased voter the least, if we can discard a Droop quota of voters from consideration?"
Then, are there any properties for p that makes M' satisfy Droop proportionality? Can we in general turn consensus methods of this form into PR methods by adding a "you can discard a Droop quota" rule?
If we can, then we easily get a multiwinner version of Bucklin/MJ by doing the following:
Let g(c, v) be the grade voter v gives to the least preferred candidate in c.
Let the consensus method M be
C = arg max min g(c, v)
c v
Let M' permit the method to remove a Droop quota, i.e. if |V| is the number of voters, and V is the set of voters itself:
C' = arg max c:
max x subset of V so that |x| = |V|/(seats+1):
min v in V \ x:
g(c, v)
For a single-winner election, M' is (up to tiebreaker) just MJ, because for each potential winner c, the removal step will remove the voters who grade c the worst, and the Droop quota for single-winner is a majority. Then the voter grading the c the worst after half of the voters have been removed is just the median voter.
Some thoughts about two-winner remove-voter minimax Approval follow. Reasoning about what voter removal actually does can get kinda hairy, thus I may very well be wrong:
In minimax Approval, p(c, v) is the Hamming distance between c and voter v's ballot, i.e. the number of candidates in c but not approved by v plus the number of candidates approved by v not in c.
Say we have an analogous Droop criterion for Approval: if more than k Droop quotas approve of a set of j candidates (and nobody else), then at least min(k, j) of these must be elected.
For two winners, there are these possibilities:
1. no Droop constraints
2. k = 2, j >= 2
3. k = 2, j = 1
4. k = 1, j >= 1
5. k = 1, j = 1
-
is no problem, because we can elect anyone we want without running afoul of the Approval DPC.
-
Since there can only be three Droop quotas in total, when we're considering A = {C_1, C_2} with C_1 and C_2 in the set of j candidates (call it J), we can eliminate all but the J-voters and the maximum penalty is j-2.
In contrast, for some B = {C_x, C_y} not a subset of j, the best it can do is eliminate a Droop quota of the J-voters. In the best case (for B), everybody but the J-voters approve of B alone. But there still remains a Droop quota (plus one voter) of the J-voters, and each of them gives penalty j. So A is preferred to B.
If B = {C_1, C_x}, then even if everybody but the J-voters approve of B alone, the J-voters give penalty j-1. So A is still preferred to B. -
Same as in 2, but let A = {C_1, C_x}, J = {C_1}. With A, we eliminate so that only the J-voters are left, and then max penalty is 1 (for C_x). Furthermore, every remaining voter gives penalty 1. Let B = {C_x, C_y}. In the best case for B, a Droop quota of J-voters are eliminated and we have a Droop quota plus one left. These all give penalty 2, which is worse than penalty 1. So A is preferred to B.
-
Let A = {C_1, C_x} and B = {C_x, C_y}. In the best case for B here, two Droop quotas minus a voter approve only of B, and the remaining Droop quota plus one voter approves of J = {C_1}. Eliminating all but one of the J-voters gives a max penalty of 3 from that one J-voter: one point for not having C_1, and two points for having C_x and C_y. A eliminates one of the two B-approving Droop quotas and gets a penalty of 1 from every remaining voter, which is better.
Note that I assume that C_x is approved by the B-voters. If that were not the case, then {C_x, C_y} would already be beaten by some {C_z, C_y} where C_z is. Note also that I don't consider the case where the B-voters also approve of a whole load of other candidates, with the idea of raising the penalty under A. The problem is that because only two candidates can be elected, this would also raise their penalty under B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. The best case for B has worst penalty j+2, since after a Droop quota of J-voters have been eliminated, there remains a single voter who only approves of J. After eliminating some of the B-voters, A gets penalty j from the J-voters (j-1 for the members of J not part of {C_1, C_x} and one more for C_x which is not approved by them), and one penalty point from the B-voters.
Here it'd seem that adding loads of candidates to the B-voters would make things hard. Can it be salvaged?
Suppose there are J-voters and C-voters. B is a subset of C.
When considering outcome B, before excluding a Droop quota, every J-voter gives a penalty of j+2 and every C-voter gives a penalty of c-2 where c=|C|.
Under outcome A, before excluding, every J-voter gives j, and every B-voter gives c (-1 for having C_x, +1 for having C_1).
If j+2 > c, then we're in the domain above, and no problem.
If c > j+2, then the excluded candidates under both A and B are C-voters.
So under B we have a Droop quota of C-voters with penalty c-2, and a Droop quota plus one of J-voters at j+2.
Under A we have a Droop quota of C-voters with penalty c, and a Droop quota plus one of J-voters at j.
So unless I made a mistake, Hamming distance is not good enough. But I might have made a mistake, because it seems strange that even in ordinary minimax Approval, a coalition can increase its power by approving a lot of clones. E.g. suppose in ordinary minimax Approval that there are two coalitions of almost equal size:
n+1: A B
n: C1 C2 C3 ... Cq
{A, B} gets worst penalty q+2 (there are n of these and n+1 zeroes)
{A, C1} gets worst penalty q (n voters like C1 but not A)
{C1, C2} gets worst penalty q-2 (n voters give this penalty, and then n+1 give penalty 4).
... does that mean an arbitrarily small minority can force its own council to win by just approving enough clones that they set the worst penalty in every outcome? That feels rather wrong.
Election-Methods mailing list - see http://electorama.com/em for list info
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--
Richard Lung.
http://www.voting.ukscientists.com
Democracy Science series 3 free e-books in pdf:
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Sorry: "crutch option"
Envoyé de mon iPhone
Le 8 mars 2018 à 14:14, Stéphane Rouillon stephane.rouillon@sympatico.ca a écrit :
Exactly the conclusion I reached when I designed the crurch option for SPPA...!
Harmful equilibria are stable with 2 parties; never stable for long with 3 or more.
And SPPA allows transfer between political parties by rallying.
Envoyé de mon iPhone
Le 8 mars 2018 à 11:46, Jameson Quinn jameson.quinn@gmail.com a écrit :
Few, but more than 2. Harmful equilibria are stable with 2 parties; never stable for long with 3 or more.
And those who don't like any of the viable options (whether there are 2 or 3 or 10) should be able to vote in a way that reflects that, without necessarily having their votes ignored. That means any good voting method should allow cross-party voting and/or transfers somehow.
2018-03-07 21:44 GMT-05:00 Jack Santucci jms346@georgetown.edu:
Political scientists like their parties to be few and disciplined. This is said to promote accountability.
Sent from my iPhone
On Mar 7, 2018, at 20:11, Richard Lung voting@ukscientists.com wrote:
If my old memory serves me tolerably well, isn't this paper something like an article entitled The Best of Both Worlds, where the authors did a survey of a tendency for European electoral systems, over the decades, to have decreased their average magnitude. I forget the details, just about everything actually. But it may have gone something like: the constiuencies shrank and the thresholds got higher.
It was an informative statistical survey.
But I think it went awry on what academics are fond of calling "normative" considerations. Or on the stricture of David Hume, that what is, is not necessarily right.
I would have put to the authors, as a critic. That was this trend, they so diligently exposed, but the moving to a "sweet spot" for political incumbents, with precious little to do with democracy and effective elections for the voters?
from
Richard Lung.
On 07/03/2018 19:19, Jack Santucci wrote:
Consensus in academia? Maybe that cigarettes cause cancer. Maybe.
I jest.
This paper may be helpful: http://personal.lse.ac.uk/hix/Working_Papers/Carey-Hix-AJPS2011.pdf
On Wed, Mar 7, 2018 at 2:14 PM, Richard Lung voting@ukscientists.com wrote:
So, the academic world has no consensus or standard model of election method?
On 03/03/2018 19:57, Kristofer Munsterhjelm wrote:
Say we have a consensus method M that works by choosing the council C that minimizes the maximum penalty p(C, v) for the voter that maximizes this penalty. That is, the method finds C according to
C = arg min max p(c, v)
c v
where ties are broken in a leximax fashion (i.e. considering next to max, then next to next to max and so on). Furthermore let the penalty "nonnegative" in the sense that any voter with a real preference has at least as great a penalty as a voter with no preference (the zero voter, as it were).
Now let the modified consensus method M' be one that has the same optimization objective, but the method is permitted to remove a Droop quota of votes to help minimize the penalty.
So M says "what council displeases the most displeased voter the least?", while M' says "what council displeases the most displeased voter the least, if we can discard a Droop quota of voters from consideration?"
Then, are there any properties for p that makes M' satisfy Droop proportionality? Can we in general turn consensus methods of this form into PR methods by adding a "you can discard a Droop quota" rule?
If we can, then we easily get a multiwinner version of Bucklin/MJ by doing the following:
Let g(c, v) be the grade voter v gives to the least preferred candidate in c.
Let the consensus method M be
C = arg max min g(c, v)
c v
Let M' permit the method to remove a Droop quota, i.e. if |V| is the number of voters, and V is the set of voters itself:
C' = arg max c:
max x subset of V so that |x| = |V|/(seats+1):
min v in V \ x:
g(c, v)
For a single-winner election, M' is (up to tiebreaker) just MJ, because for each potential winner c, the removal step will remove the voters who grade c the worst, and the Droop quota for single-winner is a majority. Then the voter grading the c the worst after half of the voters have been removed is just the median voter.
Some thoughts about two-winner remove-voter minimax Approval follow. Reasoning about what voter removal actually does can get kinda hairy, thus I may very well be wrong:
In minimax Approval, p(c, v) is the Hamming distance between c and voter v's ballot, i.e. the number of candidates in c but not approved by v plus the number of candidates approved by v not in c.
Say we have an analogous Droop criterion for Approval: if more than k Droop quotas approve of a set of j candidates (and nobody else), then at least min(k, j) of these must be elected.
For two winners, there are these possibilities:
1. no Droop constraints
2. k = 2, j >= 2
3. k = 2, j = 1
4. k = 1, j >= 1
5. k = 1, j = 1
-
is no problem, because we can elect anyone we want without running afoul of the Approval DPC.
-
Since there can only be three Droop quotas in total, when we're considering A = {C_1, C_2} with C_1 and C_2 in the set of j candidates (call it J), we can eliminate all but the J-voters and the maximum penalty is j-2.
In contrast, for some B = {C_x, C_y} not a subset of j, the best it can do is eliminate a Droop quota of the J-voters. In the best case (for B), everybody but the J-voters approve of B alone. But there still remains a Droop quota (plus one voter) of the J-voters, and each of them gives penalty j. So A is preferred to B.
If B = {C_1, C_x}, then even if everybody but the J-voters approve of B alone, the J-voters give penalty j-1. So A is still preferred to B. -
Same as in 2, but let A = {C_1, C_x}, J = {C_1}. With A, we eliminate so that only the J-voters are left, and then max penalty is 1 (for C_x). Furthermore, every remaining voter gives penalty 1. Let B = {C_x, C_y}. In the best case for B, a Droop quota of J-voters are eliminated and we have a Droop quota plus one left. These all give penalty 2, which is worse than penalty 1. So A is preferred to B.
-
Let A = {C_1, C_x} and B = {C_x, C_y}. In the best case for B here, two Droop quotas minus a voter approve only of B, and the remaining Droop quota plus one voter approves of J = {C_1}. Eliminating all but one of the J-voters gives a max penalty of 3 from that one J-voter: one point for not having C_1, and two points for having C_x and C_y. A eliminates one of the two B-approving Droop quotas and gets a penalty of 1 from every remaining voter, which is better.
Note that I assume that C_x is approved by the B-voters. If that were not the case, then {C_x, C_y} would already be beaten by some {C_z, C_y} where C_z is. Note also that I don't consider the case where the B-voters also approve of a whole load of other candidates, with the idea of raising the penalty under A. The problem is that because only two candidates can be elected, this would also raise their penalty under B. -
Let A = {C_1, C_x} and B = {C_x, C_y}. The best case for B has worst penalty j+2, since after a Droop quota of J-voters have been eliminated, there remains a single voter who only approves of J. After eliminating some of the B-voters, A gets penalty j from the J-voters (j-1 for the members of J not part of {C_1, C_x} and one more for C_x which is not approved by them), and one penalty point from the B-voters.
Here it'd seem that adding loads of candidates to the B-voters would make things hard. Can it be salvaged?
Suppose there are J-voters and C-voters. B is a subset of C.
When considering outcome B, before excluding a Droop quota, every J-voter gives a penalty of j+2 and every C-voter gives a penalty of c-2 where c=|C|.
Under outcome A, before excluding, every J-voter gives j, and every B-voter gives c (-1 for having C_x, +1 for having C_1).
If j+2 > c, then we're in the domain above, and no problem.
If c > j+2, then the excluded candidates under both A and B are C-voters.
So under B we have a Droop quota of C-voters with penalty c-2, and a Droop quota plus one of J-voters at j+2.
Under A we have a Droop quota of C-voters with penalty c, and a Droop quota plus one of J-voters at j.
So unless I made a mistake, Hamming distance is not good enough. But I might have made a mistake, because it seems strange that even in ordinary minimax Approval, a coalition can increase its power by approving a lot of clones. E.g. suppose in ordinary minimax Approval that there are two coalitions of almost equal size:
n+1: A B
n: C1 C2 C3 ... Cq
{A, B} gets worst penalty q+2 (there are n of these and n+1 zeroes)
{A, C1} gets worst penalty q (n voters like C1 but not A)
{C1, C2} gets worst penalty q-2 (n voters give this penalty, and then n+1 give penalty 4).
... does that mean an arbitrarily small minority can force its own council to win by just approving enough clones that they set the worst penalty in every outcome? That feels rather wrong.
Election-Methods mailing list - see http://electorama.com/em for list info
--
Richard Lung.
http://www.voting.ukscientists.com
Democracy Science series 3 free e-books in pdf:
https://plus.google.com/106191200795605365085
E-books in epub format:
https://www.smashwords.com/profile/view/democracyscience
This email has been checked for viruses by Avast antivirus software.
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Independent scholar
http://www.jacksantucci.com
--
Richard Lung.
http://www.voting.ukscientists.com
Democracy Science series 3 free e-books in pdf:
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E-books in epub format:
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Hi Jack and Robert,
Much more inline below
Date: Wed, March 7, 2018 9:44 pm, From: "Jack Santucci" jms346@georgetown.edu
Political scientists like their parties to be few and disciplined. This is
said to promote accountability.
Yes, this. It's very easy to discount the value of a party
infrastructure, but political parties at their best provide valuable
vetting and coordination. In a capitalist society, we are frequently
confronted by more choices than we know what to do with. Party labels
provide branding marks that are as useful when voting as branded food
items are at the grocery store.
(and thank you for the Carey and Hix reference; that's going on my reading list)
On Wed, Mar 7, 2018 at 10:58 PM, robert bristow-johnson
rbj@audioimagination.com wrote:
"few" makes sense if it is at least 3 parties that are viable and get people
elected. then it wouldn't always fall into that "If you ain't fer us, you
agin' us!" syndrome of the two-party polarization. [...] and i certainly
don't want it to be as few as 1 or 0. that would promote no
accountability. if you toss in libertarians and communitarians and maybe
some wing on the left or right (like Greenies or Vermont Progs or John
Birch), "few" might mean 6 or 8.
i like that we have parties (i just wish more than the Dems and GOP) and i
think they serve a useful purpose in accomplishing sometimes difficult
political goals. they do this by consolidation and coalition, and with
combined numbers, they can demonstrate political will and popular authority.
Agreed! I think there is a big opportunity right now for the
Democrats to stake out a claim as the hub in a hub-and-spoke model of
party-based partisanship. The Clinton New Democrats could be the
"let's not do anything too crazy" party at the hub of the system,
built around consensus-based leadership, with room for left-leaning
coalitions to split off who want a more concerted push for specific
policies. Prudence and respect for governing expertise were taken for
granted prior to 2016, but we've all come to realize that it was a
norm waiting to be broken. Having one honest broker party that
considers respect for governing and democratic ideals the topmost
priorities on their agenda would provide an anchor of administrative
continuity that we're lacking right now. A good system would let a
this is why i am an unabashed advocate for RCV decided by Condorcet. It can
demonstrate genuine mandate when such might be obscured by a polarized
2-party environment.
I've warmed up to Approval voting and simplified Score Voting (as
Kenneth Arrow seemed to like, with "three or four classes" or scores
to keep the strategic voting to a minimum)[1]
[1]: https://electology.org/podcasts/2012-10-06_kenneth_arrow
Being able to tell people "the candidate with the highest approval
rating wins" is powerful in its simplicity. Plus, Brams' and
Fishburn's assertion that Approval almost always picks the Condorcet
winner2 sounds compelling and seems intuitive
Still, methods that are guaranteed to conform to the Condorcet
criterion are still very appealing to me. I'm thinking of revisiting
the "Copeland Majority" method I proposed back in 2005:
http://election-methods.5485.n7.nabble.com/Copeland-s-criteria-td18953.html#a18954
Rather than starting a new effort, an interesting possibility is to
amend many other methods which are more definitive (e.g. Score/Range,
IRV) but have boundary conditions that make people nervous. A
"Copeland Majority" winner (i.e. a Condorcet winner) could be allowed
to challenge the IRV or Score winner, which would result in a special
election.
So, for example, one could amend the law in a place that already has
IRV to avert the situation that happened in Burlington in 2009.3
In that election, if a "Copeland Majority Challenge" was possible,
Andy Montroll could have issued a challenge to Bob Kiss, and they
could have had a head-to-head special election. That would have been
better than what did happen, which was that they held an election to
eliminate IRV, and switch to top-two jungle primary4. Note that
Kurt Wright would not have been eligible to challenge
Here in San Francisco, we're using IRV now (as is Oakland, Berkeley,
and San Leandro) I'm not aware of anything as bad as the Burlington
2009 election happening here, though my understanding is that Jean
Quan's election in Oakland in 2011 may have had similar issues (and,
unlike Burlington, it may be that Oakland doesn't publish the
information necessary to analyze it).
There are many things I don't like about Copeland (either in it's
classic form or in the modification I proposed). But the thing I like
a lot about it is that it's simple to explain to anyone who follows
sports[5]
[5]: https://en.wikipedia.org/wiki/2017%E2%80%9318_NBA_season#By_conference
Would it be possible to come up with an addition like a "Copeland
Majority Challenge" to make Range and/or IRV more appealing to
Condorcet advocates? (besides myself)?
Rob