Re: [EM] Strong Reverse Symmetry Clone Proof Borda
To give credit where credit is due, I want to thank Richard Lung for
getting us thinking along the inclusion/exclusion axis corresponding to
near-first/near-last ballot ranks ... That contributed to my confidence in
the lottery/anti-lottery idea as a means of de-cloning Borda, Copeland. and
Kemeny-Young, which has panned out so beautifully.
Thanks, Richard!!!
El vie., 14 de ene. de 2022 8:18 a. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Two equivalent ways of decloning Borda ... both relying on a combination
of the benchmark and anti-benchmark lotteries BML and ABML.
First method:
Ballot B contributes to candidate X's score the sum of BML(j) [summed over
candidates j ranked below X] minus the sum of ABML(k) [summed over
candidates k ranked above X].
Think of it this way ... the importance of candidate X to the voter of
ballot B is the estimated strength of the candidates she likes less than X,
minus the estimated weakness of the candidates she likes better:
[If the ones you like better are weak and the ones you dislike in
comparison are strong, you should support X.]
[Also this is a natural DSV kind of way to specify an approval cutoff]
Method 2.
The score for X is the difference in the BML expectation of X's pairwise
matrix row and the ABML expectation of X's pairwise matrix column.
The former is a source of pride for X ...the weighted average pairwise
scores of X over the candidates, where the weights are estimates of the
strengths of those candidates.
The latter expectation is a source of chagrin for X ... the weighted
average of the scores against X by the other candidates, where the weights
are estimates of the weakness of those candidates... if you are soundly
trounced by a weak candidate, you should feel chagrin, if not shame!
This gives another idea for a kind of MinMax/MaxMin method:
Let M(X) be the Max value of the product of ABML(k)*P(k,X), where P(k, X)
is the percentage of ballots on which k is ranked ahead of X.
Similarly, let m(X) be the min value of P(X, j)*BML(j), where P(X,j) is
the percentage of ballots ranking X above j.
Elect argmin(M(X)-m(X)).
[Maybe this is another key to fpC-fpA]
Note the strong symmetry of this method, also.
All in all, I'm pretty excited about the usefulness of the anti-benchmark
lottery in conjunction with the benchmark lottery.
If BML(X) is an estimate of X's viability, then ABML(X) is an estimate of
X's weakness.
Of course, the BML is not the only clone-free lottery ...Jobst's MaxParC,
the Martin Harper Lottery, the Random Approval Lottery, the Nash Lottery
and its "Ultimate Lottery" generalization ...all of these and others have
their corresponding anti-lottery counterparts ... just reverse the ballots
to get them.
Why didn't we think of this before? It was staring us right in the face!
What other obvious ideas are still invisible to us?
-Forest
El vie., 14 de ene. de 2022 6:24 a. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Kristofer,
I have a feeling you might like this method:
For simplicity assume complete rankings.
For each candidate X, let fp(X) and lp(X), respectively, be the total
number of first place votes and last place votes, respectively, of the
candidates that are pairwise defeated by X, and pairwise defeat X,
respectively.
Elect argmax(fp(X)-lp(X))
What do you think?
-Forest
El vie., 14 de ene. de 2022 2:11 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
On 14.01.2022 11:01, Kristofer Munsterhjelm wrote:
On 14.01.2022 04:05, Forest Simmons wrote:
I've always admired the basic idea of Kemeny-Young, namely a cost
metric
for conversion of one ranking into another ... simply the number of
transpositions (elementary swaps) required.
The deal-breaker draw-back of the Kemeny cost metric is that it is
clone
dependent ... if it were independent of clones the " cost" of
transposing AB to BA would be the same as the net cost of transposing
the rank order of respective clones of A and B:
the order a1a2a3b1b2b3
to the order b3b2b1a3a2a1
Of course, I have to restate here that Ranked Pairs is such a method:
the metric is just leximax. (ordering a is better than ordering b if
the
greatest pairwise victory consistent with a is higher than the greatest
pairwise victory consistent with b, with tiebreaks for second greatest,
third greatest, etc.) And it's cloneproof.
Here's a thought: let parameterized p-Kemeny be: take the lp norm of the
vector of all pairwise victories consistent with the output ordering.
Break ties by gradually lowering p, i.e. l(p-epsilon), further ties by
l(p-2 epsilon), etc.
Then Kemeny is l_1 and RP is the limit as p approaches infinity, l_inf.
I think? Are any other values of p interesting?
-km
It's gratifying to be praised for helping something you don't yourself
comprehend. Forest has that good nature.
It may be worth mentioning that the binomial stv use of an election
count and an exclusion count may both benefit a candidate. If the
candidate is popular enough to gain an election quota, but not too
unpopular to gain an exclusion quota, then a poor exclusion count can
consolidate a good election count, for the same candidate..
Unlike classical Borda count, but in iine with the Gregory count, later
preferences do not count against former preferences. When there is a
conflict with a candidate both popular (in the election count) and
unpopular (in the exclusion count) it is a reflection of real
differences in opinion between different factions of the electorate.
Unlike traditional stv, the Gregory count, via an extension of Meek
method keep-value counting to all candidates, not just elected
candidates, is repeated on the voters preferences, in reverse, as an
unpreference vote. This necessitates the counting of abstentions, to
ascertain the balance of liking and disliking of the candidates, by the
voters.
This in turn means all the preferences are counted, which amounts to a
complete dimension of choice counted -- not merely an uncertain
fraction of a dimension counted.
Real elections with binomial stv should show how the entropy of the
voters usual exponential decline in preferences -- or unpreferences --
works or fails to work for decisive results, in practise.
Sorry, if this explanation was too cryptic.
Regards,
Richard Lung.
To give credit where credit is due, I want to thank Richard Lung for
getting us thinking along the inclusion/exclusion axis corresponding to
near-first/near-last ballot ranks ... That contributed to my confidence
in the lottery/anti-lottery idea as a means of de-cloning Borda,
Copeland. and Kemeny-Young, which has panned out so beautifully.
Thanks, Richard!!!
El vie., 14 de ene. de 2022 8:18 a. m., Forest Simmons
forest.simmons21@gmail.com escribió:
Two equivalent ways of decloning Borda ... both relying on a
combination of the benchmark and anti-benchmark lotteries BML and ABML.
First method:
Ballot B contributes to candidate X's score the sum of BML(j)
[summed over candidates j ranked below X] minus the sum of ABML(k)
[summed over candidates k ranked above X].
Think of it this way ... the importance of candidate X to the voter
of ballot B is the estimated strength of the candidates she likes
less than X, minus the estimated weakness of the candidates she
likes better:
[If the ones you like better are weak and the ones you dislike in
comparison are strong, you should support X.]
[Also this is a natural DSV kind of way to specify an approval cutoff]
Method 2.
The score for X is the difference in the BML expectation of X's
pairwise matrix row and the ABML expectation of X's pairwise matrix
column.
The former is a source of pride for X ...the weighted average
pairwise scores of X over the candidates, where the weights are
estimates of the strengths of those candidates.
The latter expectation is a source of chagrin for X ... the weighted
average of the scores against X by the other candidates, where the
weights are estimates of the weakness of those candidates... if you
are soundly trounced by a weak candidate, you should feel chagrin,
if not shame!
This gives another idea for a kind of MinMax/MaxMin method:
Let M(X) be the Max value of the product of ABML(k)*P(k,X), where
P(k, X) is the percentage of ballots on which k is ranked ahead of X.
Similarly, let m(X) be the min value of P(X, j)*BML(j), where P(X,j)
is the percentage of ballots ranking X above j.
Elect argmin(M(X)-m(X)).
[Maybe this is another key to fpC-fpA]
Note the strong symmetry of this method, also.
All in all, I'm pretty excited about the usefulness of the
anti-benchmark lottery in conjunction with the benchmark lottery.
If BML(X) is an estimate of X's viability, then ABML(X) is an
estimate of X's weakness.
Of course, the BML is not the only clone-free lottery ...Jobst's
MaxParC, the Martin Harper Lottery, the Random Approval Lottery, the
Nash Lottery and its "Ultimate Lottery" generalization ...all of
these and others have their corresponding anti-lottery counterparts
... just reverse the ballots to get them.
Why didn't we think of this before? It was staring us right in the
face!
What other obvious ideas are still invisible to us?
-Forest
El vie., 14 de ene. de 2022 6:24 a. m., Forest Simmons
<forest.simmons21@gmail.com> escribió:
Kristofer,
I have a feeling you might like this method:
For simplicity assume complete rankings.
For each candidate X, let fp(X) and lp(X), respectively, be the
total number of first place votes and last place votes,
respectively, of the candidates that are pairwise defeated by X,
and pairwise defeat X, respectively.
Elect argmax(fp(X)-lp(X))
What do you think?
-Forest
El vie., 14 de ene. de 2022 2:11 a. m., Kristofer Munsterhjelm
<km_elmet@t-online.de> escribió:
On 14.01.2022 11:01, Kristofer Munsterhjelm wrote:
On 14.01.2022 04:05, Forest Simmons wrote:
I've always admired the basic idea of Kemeny-Young,
namely a cost metric
for conversion of one ranking into another ... simply
the number of
transpositions (elementary swaps) required.
The deal-breaker draw-back of the Kemeny cost metric is
that it is clone
dependent ... if it were independent of clones the "
cost" of
transposing AB to BA would be the same as the net cost
of transposing
the rank order of respective clones of A and B:
the order a1a2a3b1b2b3
to the order b3b2b1a3a2a1
Of course, I have to restate here that Ranked Pairs is
such a method:
the metric is just leximax. (ordering a is better than
ordering b if the
greatest pairwise victory consistent with a is higher
than the greatest
pairwise victory consistent with b, with tiebreaks for
second greatest,
third greatest, etc.) And it's cloneproof.
Here's a thought: let parameterized p-Kemeny be: take the lp
norm of the
vector of all pairwise victories consistent with the output
ordering.
Break ties by gradually lowering p, i.e. l(p-epsilon),
further ties by
l(p-2 epsilon), etc.
Then Kemeny is l_1 and RP is the limit as p approaches
infinity, l_inf.
I think? Are any other values of p interesting?
-km