Ideas for Schulze tie-breaking
Like minmax, Schulze sometimes has a lot of ties, particularly for
elections with few voters. But perhaps this way could work to break ties
in a consistent way:
Let p[A, B] be the strength of the strongest beatpath going from A to B.
Now do ext-Minmax on the p matrix. This is minmax (i.e. A's score is A's
weakest pairwise contest vs B), but ties are broken by looking at second
weakest pairwise contests, third weakest, etc.
Since there always exists a candidate A who is the "Condorcet winner"
according to p, and this is the same as the Schulze winner, and both
Minmax and ext-Minmax pass Condorcet, this method should agree with
Schulze when there are no ties.
But do we lose any criteria from this? I don't know; it's just a thought
that occurred to me as a way to make ext-Minmax more Schulze-like.
Possibly there exist situations where there's a tie according to
Schulze, and this breaks the tie in a way that (say) depends on clones;
but I'm not sure how to construct such an example.
Another possible drawback is that the social order may suffer, because
minmax doesn't pass Condorcet loser. So it might be that the social
ranking is A>B>C>D by Schulze, and this is strict (A has 3 strong
beatpaths, B has 2, C has 1, D has none), but D's weakest beatpath is
stronger than C's; and then the ext-Minmax modification returns a
different social order even though there are no ties at any point.
This might point to the idea being not entirely defensible. One could,
of course, only break the ties that exist by ext-Minmax, but that feels
rather like a hack.
Perhaps it would be doable to augment Floyd-Warshall to maximize the
leximin of the beatpath instead of just its minimum - among the paths
with the strongest weakest link, find the one with the strongest
second-weakest link, etc. Then comparing the full vectors of defeats
along the beatpath thus recovered could resolve ties in a way more
consistent with the Schulze method itself.
I'm still only working by intuition, though; I haven't rigorously
checked any of these ideas.
-km
Define the total resistance of the losers in a beatpath as the sum of the
losing votes. Unfortunately, minimization of this total beatpath resistance
is a clone dependent method.
However, the method can be decloned.
Let L be the vector of losing votes along the beatpath. Let F be the vector
of corresponding favorite votes, i.e. F_k is the number of ballots on which
candidate k is designated as favorite.
The decloned resistance of the beatpath is the dot product of L and F.
Minimizing this dot product over beatpaths is a clone free method that
could be called de-cloned Min Resistance BeatPath (dcMinRBP).
Use this method to break ties.
Better yet, use dcMinRBP as the main method, and break ties with Schulze.
El dom., 27 de feb. de 2022 6:05 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
Like minmax, Schulze sometimes has a lot of ties, particularly for
elections with few voters. But perhaps this way could work to break ties
in a consistent way:
Let p[A, B] be the strength of the strongest beatpath going from A to B.
Now do ext-Minmax on the p matrix. This is minmax (i.e. A's score is A's
weakest pairwise contest vs B), but ties are broken by looking at second
weakest pairwise contests, third weakest, etc.
Since there always exists a candidate A who is the "Condorcet winner"
according to p, and this is the same as the Schulze winner, and both
Minmax and ext-Minmax pass Condorcet, this method should agree with
Schulze when there are no ties.
But do we lose any criteria from this? I don't know; it's just a thought
that occurred to me as a way to make ext-Minmax more Schulze-like.
Possibly there exist situations where there's a tie according to
Schulze, and this breaks the tie in a way that (say) depends on clones;
but I'm not sure how to construct such an example.
Another possible drawback is that the social order may suffer, because
minmax doesn't pass Condorcet loser. So it might be that the social
ranking is A>B>C>D by Schulze, and this is strict (A has 3 strong
beatpaths, B has 2, C has 1, D has none), but D's weakest beatpath is
stronger than C's; and then the ext-Minmax modification returns a
different social order even though there are no ties at any point.
This might point to the idea being not entirely defensible. One could,
of course, only break the ties that exist by ext-Minmax, but that feels
rather like a hack.
Perhaps it would be doable to augment Floyd-Warshall to maximize the
leximin of the beatpath instead of just its minimum - among the paths
with the strongest weakest link, find the one with the strongest
second-weakest link, etc. Then comparing the full vectors of defeats
along the beatpath thus recovered could resolve ties in a way more
consistent with the Schulze method itself.
I'm still only working by intuition, though; I haven't rigorously
checked any of these ideas.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
I wrote this in a hurry ... the basic method should be called
MinMaxLeastTotalResistanceBeatPath.
The total resistance TR of a beatpath BP is the sum of its losing votes.
For each pair of candidates (j,k) let TLR(j,k) be the total resistance of
the beatpath of least resistance from j to k. Elect the candidate j that
minimizes the maximum value of TLR(j,k) as k varies over the other
candidates.
To get the de-cloned version, weight each losing vote by the number of
ballots on which the pairwise loser is designated favorite.
Any electrical engineer should be able to design an "electo-meter" network
that has, for each pair of candidates (j,k), a diode from node j to node k
that allows current to move from j to k with resistance jointly
proportional to the number of ballots on which k is ranked ahead of j and
the number of ballots on which k is designated favorite.
Elect the candidate j that minimizes the maximum network resistance from
node j to node k as k varies over the other nodes. [Put the source probe of
a directional ohm meter at j while varying the sink probe over the other
nodes (k).]
One of ypu engineers correct me if my EE terminology isn't quite right.
-Forest
El mar., 1 de mar. de 2022 9:25 a. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Define the total resistance of the losers in a beatpath as the sum of the
losing votes. Unfortunately, minimization of this total beatpath resistance
is a clone dependent method.
However, the method can be decloned.
Let L be the vector of losing votes along the beatpath. Let F be the
vector of corresponding favorite votes, i.e. F_k is the number of ballots
on which candidate k is designated as favorite.
The decloned resistance of the beatpath is the dot product of L and F.
Minimizing this dot product over beatpaths is a clone free method that
could be called de-cloned Min Resistance BeatPath (dcMinRBP).
Use this method to break ties.
Better yet, use dcMinRBP as the main method, and break ties with Schulze.
El dom., 27 de feb. de 2022 6:05 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
Like minmax, Schulze sometimes has a lot of ties, particularly for
elections with few voters. But perhaps this way could work to break ties
in a consistent way:
Let p[A, B] be the strength of the strongest beatpath going from A to B.
Now do ext-Minmax on the p matrix. This is minmax (i.e. A's score is A's
weakest pairwise contest vs B), but ties are broken by looking at second
weakest pairwise contests, third weakest, etc.
Since there always exists a candidate A who is the "Condorcet winner"
according to p, and this is the same as the Schulze winner, and both
Minmax and ext-Minmax pass Condorcet, this method should agree with
Schulze when there are no ties.
But do we lose any criteria from this? I don't know; it's just a thought
that occurred to me as a way to make ext-Minmax more Schulze-like.
Possibly there exist situations where there's a tie according to
Schulze, and this breaks the tie in a way that (say) depends on clones;
but I'm not sure how to construct such an example.
Another possible drawback is that the social order may suffer, because
minmax doesn't pass Condorcet loser. So it might be that the social
ranking is A>B>C>D by Schulze, and this is strict (A has 3 strong
beatpaths, B has 2, C has 1, D has none), but D's weakest beatpath is
stronger than C's; and then the ext-Minmax modification returns a
different social order even though there are no ties at any point.
This might point to the idea being not entirely defensible. One could,
of course, only break the ties that exist by ext-Minmax, but that feels
rather like a hack.
Perhaps it would be doable to augment Floyd-Warshall to maximize the
leximin of the beatpath instead of just its minimum - among the paths
with the strongest weakest link, find the one with the strongest
second-weakest link, etc. Then comparing the full vectors of defeats
along the beatpath thus recovered could resolve ties in a way more
consistent with the Schulze method itself.
I'm still only working by intuition, though; I haven't rigorously
checked any of these ideas.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info
Here is a highly flawed first attempt at decloning the "electometer"
network:
Unlike the straightforward case of decloning the beatpath resistance
method, the electrical network resistance model allows for parallel paths
of current, which introduces another source of clone dependence.
To compensate for that we need to include another factor in the resistance
diodes.
Recall that every pair of candidates (j, k) is connected by two diodes (or
possibly one equivalent composite circuit element, but let's go with two
dodes for clarity):
The diode for current from node j to k has a rectifier that enforces that
direction in series with a resistor whose resistance is jointly
proportional with three factors ... (1) the number of ballots on which
candidate k is designated favorite, (2) the number of of ballots on which
candidate k is ranked ahead of candidate j, and (3) the number of ballots
on which j is designated as "anti-favorite."
It is this last factor of resistance that allows for candidate j to be
cloned in the context of current flowing out of j into k, just as the first
factor allows for k to be cloned in the context of current flowing into k
from j.
If j and k, respectively, are replaced by clone sets of cardinality #j and
#k, respectively, then the directional diode from j to k will be replaced
by #j*#k diodes whose combined resistance will be the same as the single
resistance described above for the diode from j to k.
Unfortunately, this last claim of combined resistance cannot be correct
because only in series (as opposed to parallel) is resistance additive.
Does anybody see how to rescue this approach?
Perhaps through conductance, which is additive in parallel ...
-Forest
El mar., 1 de mar. de 2022 1:38 p. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
I wrote this in a hurry ... the basic method should be called
MinMaxLeastTotalResistanceBeatPath.
The total resistance TR of a beatpath BP is the sum of its losing votes.
For each pair of candidates (j,k) let TLR(j,k) be the total resistance of
the beatpath of least resistance from j to k. Elect the candidate j that
minimizes the maximum value of TLR(j,k) as k varies over the other
candidates.
To get the de-cloned version, weight each losing vote by the number of
ballots on which the pairwise loser is designated favorite.
Any electrical engineer should be able to design an "electo-meter" network
that has, for each pair of candidates (j,k), a diode from node j to node k
that allows current to move from j to k with resistance jointly
proportional to the number of ballots on which k is ranked ahead of j and
the number of ballots on which k is designated favorite.
Elect the candidate j that minimizes the maximum network resistance from
node j to node k as k varies over the other nodes. [Put the source probe of
a directional ohm meter at j while varying the sink probe over the other
nodes (k).]
One of ypu engineers correct me if my EE terminology isn't quite right.
-Forest
El mar., 1 de mar. de 2022 9:25 a. m., Forest Simmons <
forest.simmons21@gmail.com> escribió:
Define the total resistance of the losers in a beatpath as the sum of the
losing votes. Unfortunately, minimization of this total beatpath resistance
is a clone dependent method.
However, the method can be decloned.
Let L be the vector of losing votes along the beatpath. Let F be the
vector of corresponding favorite votes, i.e. F_k is the number of ballots
on which candidate k is designated as favorite.
The decloned resistance of the beatpath is the dot product of L and F.
Minimizing this dot product over beatpaths is a clone free method that
could be called de-cloned Min Resistance BeatPath (dcMinRBP).
Use this method to break ties.
Better yet, use dcMinRBP as the main method, and break ties with Schulze.
El dom., 27 de feb. de 2022 6:05 a. m., Kristofer Munsterhjelm <
km_elmet@t-online.de> escribió:
Like minmax, Schulze sometimes has a lot of ties, particularly for
elections with few voters. But perhaps this way could work to break ties
in a consistent way:
Let p[A, B] be the strength of the strongest beatpath going from A to B.
Now do ext-Minmax on the p matrix. This is minmax (i.e. A's score is A's
weakest pairwise contest vs B), but ties are broken by looking at second
weakest pairwise contests, third weakest, etc.
Since there always exists a candidate A who is the "Condorcet winner"
according to p, and this is the same as the Schulze winner, and both
Minmax and ext-Minmax pass Condorcet, this method should agree with
Schulze when there are no ties.
But do we lose any criteria from this? I don't know; it's just a thought
that occurred to me as a way to make ext-Minmax more Schulze-like.
Possibly there exist situations where there's a tie according to
Schulze, and this breaks the tie in a way that (say) depends on clones;
but I'm not sure how to construct such an example.
Another possible drawback is that the social order may suffer, because
minmax doesn't pass Condorcet loser. So it might be that the social
ranking is A>B>C>D by Schulze, and this is strict (A has 3 strong
beatpaths, B has 2, C has 1, D has none), but D's weakest beatpath is
stronger than C's; and then the ext-Minmax modification returns a
different social order even though there are no ties at any point.
This might point to the idea being not entirely defensible. One could,
of course, only break the ties that exist by ext-Minmax, but that feels
rather like a hack.
Perhaps it would be doable to augment Floyd-Warshall to maximize the
leximin of the beatpath instead of just its minimum - among the paths
with the strongest weakest link, find the one with the strongest
second-weakest link, etc. Then comparing the full vectors of defeats
along the beatpath thus recovered could resolve ties in a way more
consistent with the Schulze method itself.
I'm still only working by intuition, though; I haven't rigorously
checked any of these ideas.
-km
Election-Methods mailing list - see https://electorama.com/em for list
info