[EM] smith/schwartz/landau

[Curt]

Hi, as a side project I’m writing a scala script that determines the smith set and schwartz set from ranked ballots. (I’m not currently interested in implementing any tiebreakers/completion algorithms. Just condorcet winner, smith set, schwartz set.)

I’m uncertain about my implementations for smith and schwartz. I followed some pseudocode algorithms I found online but I’m not sure they’re quite right. The schwartz algorithm I found looks more like how landau is described. Part of my challenge is also that I’m trying to do them functional-style without mutable variables like loops and counters.

Could someone post an example of a simple collection of ballots where the smith set, schwartz set, and landau sets are all different?

Also if we have a collection of ballot example somewhere that show odd examples of condorcet winner, smith set, schwartz set, etc, and what the count should yield, I’d love to collect them because they’d make a good test suite.

Thanks,
Curt

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Election-Methods mailing

[Kristofer Munsterhjelm]

Post by Curt
Hi, as a side project I’m writing a scala script that determines the smith set and schwartz set from ranked ballots. (I’m not currently interested in implementing any tiebreakers/completion algorithms. Just condorcet winner, smith set, schwartz set.)
I’m uncertain about my implementations for smith and schwartz. I followed some pseudocode algorithms I found online but I’m not sure they’re quite right. The schwartz algorithm I found looks more like how landau is described. Part of my challenge is also that I’m trying to do them functional-style without mutable variables like loops and counters.
Could someone post an example of a simple collection of ballots where the smith set, schwartz set, and landau sets are all different?
Also if we have a collection of ballot example somewhere that show odd examples of condorcet winner, smith set, schwartz set, etc, and what the count should yield, I’d love to collect them because they’d make a good test suite.

A partial answer: The Wikipedia article on the Schwartz set,
https://en.wikipedia.org/wiki/Schwartz_set, gives an example where the
Schwartz set is a single candidate but the Smith set consists of every
candidate.

Sets like Schwartz and Smith are usually maximal elements of a partially
ordered set.
https://wiki.electorama.com/wiki/Maximal_elements_algorithms gives more
information about this, as well as how to use Floyd-Warshall or
Kosaraju's algorithms to find maximal elements.

In graph theory terms, we're interested in the smallest set so that
there's a cycle from any candidate in that set to any other candidate in
that set, according to a relation (X beats Y pairwise for the Schwartz
set, X beats or ties Y pairwise for the Smith set). Any strongly
connected components algorithm should work, if given a directed graph
where there's an edge from X to Y iff X is ranked ahead of Y according
to the relation being used.

https://stackoverflow.com/a/15905986 links to a paper detailing
(somewhat complex) Haskell code for Kosaraju's algorithm, and the
accepted answer has Scala code for Tarjan's algorithm (which also finds
strongly connected components). I imagine that Floyd-Warshall could be
implemented relatively simply in a functional style by just using the
dynamic programming equation that drives it, and memoization.
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[Curt]

The wiki page https://wiki.electorama.com/wiki/Maximal_elements_algorithms <https://wiki.electorama.com/wiki/Maximal_elements_algorithms> confused me because it says (as you paraphrased) “X beats Y pairwise for the Schwartz set, X beat or ties Y pairwise for the Smith Set”.

Isn’t that backwards? I thought the Smith set was the smallest set of candidates who all *defeat* candidates outside of the set (as wikipedia says), while Schwartz set was the one that allowed ties.

Sets like Schwartz and Smith are usually maximal elements of a partially ordered set. https://wiki.electorama.com/wiki/Maximal_elements_algorithms gives more information about this, as well as how to use Floyd-Warshall or Kosaraju's algorithms to find maximal elements.
In graph theory terms, we're interested in the smallest set so that there's a cycle from any candidate in that set to any other candidate in that set, according to a relation (X beats Y pairwise for the Schwartz set, X beats or ties Y pairwise for the Smith set). Any strongly connected components algorithm should work, if given a directed graph where there's an edge from X to Y iff X is ranked ahead of Y according to the relation being used.

[Kristofer Munsterhjelm]

Post by Curt
The wiki page
https://wiki.electorama.com/wiki/Maximal_elements_algorithms confused me
because it says (as you paraphrased) “X beats Y pairwise for the
Schwartz set, X beat or ties Y pairwise for the Smith Set”.
Isn’t that backwards? I thought the Smith set was the smallest set of
candidates who all *defeat* candidates outside of the set (as wikipedia
says), while Schwartz set was the one that allowed ties.

Suppose the relation is beats-or-ties. Then that implies that for all
pairs of candidates X, Y inside the Smith set, there's a path from X to
Y where X beats or ties A who beats or ties B who beats or ties... who
beats or ties Y.

Then all candidates in the Smith set must beat every candidate outside.
Suppose that this were not true; that X is in the Smith set and Z is
not, but X doesn't beat Z. Then there are two possibilities. Either X
ties Z, or Z beats X.

If X ties Z, then Z is in the Smith set because there's a path from Z
(through X) to everyone else in the Smith set. If Z beats X, then
there's similarly also a path from Z to everyone else in the Smith set.
Thus Z must be in the Smith set as well.

For the Schwartz set, there's a beat (no tie) path from any X to any Y
in the Schwartz set. Because you can no longer bridge a gap using ties,
the Schwartz set is a subset of the Smith set. So the Schwartz set is
never larger than the Smith set.

Since there's a path from any X in the Schwartz set to any Y also in it
using only the beats relation, every member in the Schwartz set must
beat or tie everyone outside.

So the confusion is, I think, about whether you're on the outside
looking in or on the inside looking at others on the inside. Every Smith
set member beats everyone outside, but the Smith set consists of
candidates with beat-or-tie paths to each other. Every Schwartz set
member beats or ties everyone outside, but the Schwartz set consists of
candidates with beat-relation paths to each other.
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[Kevin Venzke]

I think Schwartz members don't actually have to have beatpaths to each other though. Suppose there are five candidates. A>B>C>A cycle. E is the Condorcet loser. All of D's contests are ties except for D>E. Then the Schwartz set is {A,B,C,D}.
Kevin

De : Kristofer Munsterhjelm <***@t-online.de>
À : Curt <***@museworld.com>; election-***@lists.electorama.com
Envoyé le : Vendredi 9 mars 2018 6h22
Objet : Re: [EM] smith/schwartz/landau

Post by Curt
The wiki page
https://wiki.electorama.com/wiki/Maximal_elements_algorithms confused me
because it says (as you paraphrased) “X beats Y pairwise for the
Schwartz set, X beat or ties Y pairwise for the Smith Set”.
Isn’t that backwards? I thought the Smith set was the smallest set of
candidates who all *defeat* candidates outside of the set (as wikipedia
says), while Schwartz set was the one that allowed ties.

Suppose the relation is beats-or-ties. Then that implies that for all
pairs of candidates X, Y inside the Smith set, there's a path from X to
Y where X beats or ties A who beats or ties B who beats or ties... who
beats or ties Y.

Then all candidates in the Smith set must beat every candidate outside.
Suppose that this were not true; that X is in the Smith set and Z is
not, but X doesn't beat Z. Then there are two possibilities. Either X
ties Z, or Z beats X.

If X ties Z, then Z is in the Smith set because there's a path from Z
(through X) to everyone else in the Smith set. If Z beats X, then
there's similarly also a path from Z to everyone else in the Smith set.
Thus Z must be in the Smith set as well.

For the Schwartz set, there's a beat (no tie) path from any X to any Y
in the Schwartz set. Because you can no longer bridge a gap using ties,
the Schwartz set is a subset of the Smith set. So the Schwartz set is
never larger than the Smith set.

Since there's a path from any X in the Schwartz set to any Y also in it
using only the beats relation, every member in the Schwartz set must
beat or tie everyone outside.

So the confusion is, I think, about whether you're on the outside
looking in or on the inside looking at others on the inside. Every Smith
set member beats everyone outside, but the Smith set consists of
candidates with beat-or-tie paths to each other. Every Schwartz set
member beats or ties everyone outside, but the Schwartz set consists of
candidates with beat-relation paths to each other.
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[Kristofer Munsterhjelm]

Post by Kevin Venzke
I think Schwartz members don't actually have to have beatpaths to each
other though. Suppose there are five candidates. A>B>C>A cycle. E is the
Condorcet loser. All of D's contests are ties except for D>E. Then the
Schwartz set is {A,B,C,D}.

Ah, right, for the Schwartz set there may be disjoint smallest sets of
candidates having beatpaths to everybody else inside that set. The
smaller sets are called Schwartz set components, and the Schwartz set
itself is the union of all those disjoint sets.

I think that strictly speaking holds for Smith, too, but the use of ties
to bridge gaps means there won't be more than one such smallest set
(unless you use a pairwise matrix setup that allows for, say, pairwise
contests that are entirely unknown, so you neither have A beats B, A
ties B, or B beats A).

Post by Kevin Venzke
I like to define the Schwartz set as every candidate X who has a
beatpath to every other candidate Y who has a beatpath to X.

That works for both Schwartz and Smith with respectively the beats
relation and the beats-or-ties relation. That is,

X is in the set if
for every other Y where Y has a path to X using the relation,
X also has a path to Y using the relation.

D above is in the Schwartz set because nobody has a path of pairwise
defeats to D, and D trivially has a path of pairwise defeats to every
candidate in the empty set.
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[Curt]

Thanks to Kristofer for explaining my “beats” vs “beats or ties” confusion.

For anyone interested, here is the software package of me using scala to compute Smith and Schwartz sets. It’s not super-advanced, but it at least avoids mutable variables. In the future I may try to use more expressive FP concepts, and pull in one of the faster Schwartz algorithms. I don’t entirely understand the graph algorithms yet.

https://github.com/tunesmith/condorcet-counter <https://github.com/tunesmith/condorcet-counter>

I opined a bit in the README but that’s not really the point of the project. I just wanted an easy way to identify Smith and Schwartz sets for myself.

Curt

Post by Kevin Venzke
I think Schwartz members don't actually have to have beatpaths to each other though. Suppose there are five candidates. A>B>C>A cycle. E is the Condorcet loser. All of D's contests are ties except for D>E. Then the Schwartz set is {A,B,C,D}.

Ah, right, for the Schwartz set there may be disjoint smallest sets of candidates having beatpaths to everybody else inside that set. The smaller sets are called Schwartz set components, and the Schwartz set itself is the union of all those disjoint sets.
I think that strictly speaking holds for Smith, too, but the use of ties to bridge gaps means there won't be more than one such smallest set (unless you use a pairwise matrix setup that allows for, say, pairwise contests that are entirely unknown, so you neither have A beats B, A ties B, or B beats A).

Post by Kevin Venzke
I like to define the Schwartz set as every candidate X who has a
beatpath to every other candidate Y who has a beatpath to X.

That works for both Schwartz and Smith with respectively the beats relation and the beats-or-ties relation. That is,
X is in the set if
for every other Y where Y has a path to X using the relation,
X also has a path to Y using the relation.
D above is in the Schwartz set because nobody has a path of pairwise defeats to D, and D trivially has a path of pairwise defeats to every candidate in the empty set.

[robert bristow-johnson]

---------------------------- Original Message ----------------------------

Subject: Re: [EM] smith/schwartz/landau

From: "Curt" <***@museworld.com>

Date: Fri, March 23, 2018 4:33 pm

To: "election-***@lists.electorama.com" <election-***@lists.electorama.com>

--------------------------------------------------------------------------

Thanks to Kristofer for explaining my &ldquo;beats&rdquo; vs &ldquo;beats or ties&rdquo; confusion.
For anyone interested, here is the software package of me using scala to compute Smith and Schwartz sets. It&rsquo;s not super-advanced, but it at least avoids mutable variables. In the future I may try to use more expressive FP concepts, and pull in one of the faster Schwartz algorithms. I

don&rsquo;t entirely understand the graph algorithms yet.

https://github.com/tunesmith/condorcet-counter <https://github.com/tunesmith/condorcet-counter>
I opined a bit in the README but that&rsquo;s not really the point of the project. I just wanted an easy way to identify Smith and Schwartz sets for myself.

 
to wit: "It's this author's view that a
method should only be called a Condorcet if it is limited to identifying the Smith Set,"
seems to me that your view is that the current definition of "Condorcet-compliant-method" should be changed.  so is Tideman Ranked-Pairs or Schulze Beat-Path methods not
"Condorcet methods"?

--



r b-j                         ***@audioimagination.com



"Imagination is more important than knowledge."

 
 
 
 

[Curt]

Post by robert bristow-johnson
---------------------------- Original Message ----------------------------
--------------------------------------------------------------------------

Post by Curt
https://github.com/tunesmith/condorcet-counter <https://github.com/tunesmith/condorcet-counter>
I opined a bit in the README but that’s not really the point of the project. I just wanted an easy way to identify Smith and Schwartz sets for myself.

to wit: "It's this author's view that a method should only be called a Condorcet if it is limited to identifying the Smith Set,"
seems to me that your view is that the current definition of "Condorcet-compliant-method" should be changed. so is Tideman Ranked-Pairs or Schulze Beat-Path methods not "Condorcet methods"?

Yes, that is my view. They should not be called “Condorcet Methods”, because they are not guaranteed to select the candidate(s) that would defeat all other candidates.

There is a clear difference between a “Condorcet Method” that finds a Condorcet Winner or Smith Set, and a “tiebreaking" method that tries to pick a single winner from a multi-candidate Smith Set. The latter fails criteria that the former does not. I don’t know what we should call these tiebreaking algorithms - perhaps they are not quite “tiebreaking” methods since a cycle is not exactly a tie. But it isn’t appropriate to call them “Completion” methods either as that implies something that it isn’t.

The actual objection I have is that when both are described as “Condorcet Methods”, then it’s too easy in the literature (and the blogs, and the wikipedia articles, particularly from Condorcet detractors) to paint with a broad brush and argue that all Condorcet Methods are flawed in some manner, same as how all other methods are “unfair” in some way, which ultimately does a disservice to the Condorcet Method. An election that has a Condorcet Winner is not unfair in those ways, compared to something like IRV or Plurality or Top-Two.

It might worth a survey of what a Smith Set actually *means*. I believe it signifies something valuable about the electorate, beyond just an indication that the election is “incomplete” and that we should apply some algorithm to divine a single winner from it.

(Regarding the typo in my README - I meant “should only be called a Condorcet *Method* if it is
”)

Curt

[robert bristow-johnson]

---------------------------- Original Message ----------------------------

Subject: Re: [EM] smith/schwartz/landau

From: "Curt" <***@museworld.com>

Date: Sat, March 24, 2018 12:20 am

To: "election-***@lists.electorama.com" <election-***@lists.electorama.com>

--------------------------------------------------------------------------

Post by robert bristow-johnson
---------------------------- Original Message ----------------------------
--------------------------------------------------------------------------

Post by Curt
https://github.com/tunesmith/condorcet-counter <https://github.com/tunesmith/condorcet-counter>
I opined a bit in the README but that&rsquo;s not really the point of the project. I just wanted an easy way to identify Smith and Schwartz sets for myself.

to wit: "It's this author's view that a method should only be called a Condorcet if it is limited to identifying the Smith Set,"
seems to me that your view is that the current definition of "Condorcet-compliant-method" should be changed. so is Tideman Ranked-Pairs or Schulze Beat-Path methods not "Condorcet methods"?

Yes, that is my view. They should not be called &ldquo;Condorcet Methods&rdquo;, because they are not guaranteed to select the candidate(s) that would defeat all other candidates.
There is a clear difference between a &ldquo;Condorcet Method&rdquo; that finds a Condorcet Winner or Smith Set, and a &ldquo;tiebreaking" method that tries to pick a single winner from a multi-candidate Smith Set. The latter fails criteria that the former does not. I don&rsquo;t know what

we should call these tiebreaking algorithms - perhaps they are not quite &ldquo;tiebreaking&rdquo; methods since a cycle is not exactly a tie. But it isn&rsquo;t appropriate to call them &ldquo;Completion&rdquo; methods either as that implies something that it isn&rsquo;t.

The actual objection I have is that when both are described as &ldquo;Condorcet Methods&rdquo;, then it&rsquo;s too easy in the literature (and the blogs, and the wikipedia articles, particularly from Condorcet detractors) to paint with a broad brush and argue that all Condorcet Methods are

flawed in some manner, same as how all other methods are &ldquo;unfair&rdquo; in some way, which ultimately does a disservice to the Condorcet Method. An election that has a Condorcet Winner is not unfair in those ways, compared to something like IRV or Plurality or Top-Two.

It might worth a survey of what a Smith Set actually *means*. I believe it signifies something valuable about the electorate, beyond just an indication that the election is &ldquo;incomplete&rdquo; and that we should apply some algorithm to divine a single winner from it.

well, arguing
about the semantics is one thing, arguing about theory or ideals is another, and arguing about practice is yet another.


so whether you call it "Condorcet-compliant" or call it a turnip, there are ranked-ballot tabulation procedures that are 1. Decisive (they will elect someone) and 2. If a Condorcet Winner exists, the procedure will elect the CW.  we gotta call that something and
"Condorcet-compliant method" is more descriptive than "turnip".  And it serves a purpose.  It separates both practice and "theory and ideals" of these aforementioned procedures from other ranked-ballot systems such as IRV, Borda, or Bucklin.


both Ranked-Pairs and Schulze are fundamentally **not** defined as some post algorithm to divine a single winner from a Smith set that is larger than 1.  They are well-defined procedures, in their own right, that **happen** to elect the CW if a CW exists.  IRV *may* elect the CW if one
exists, but of course we know that hasn't always been the case in practice.  They are not "tiebreaking" a Smith set.  they are not a procedure to be applied **after** it is discovered no single CW exists.  they are procedures that will elect a candidate by the same rules
whether a CW exists or not.  But the candidate elected will be the CW if one exists. So the only practical difference between these turnips is what their outcome might be if there was a Smith set greater than 3 and some weird voting alignment.


So we need a term to draw the line between Ranked-Pairs or Schulze or BTR-IRV on one side and Bucklin, Borda, or IRV on the other.  What semantic would you suggest?



the salient difference is simply what happened in my town 9 years ago: A candidate for mayor was elected to office when the voters in the city unambiguously marked their ballots that they preferred a different **specific** candidate.  that is the problem with **any** non-turnip method. 
it's the converse of who a CW is.  if everyone's vote carries the same weight (this is the "one-person-one-vote" principle), if Candidate A is preferred by more voters than Candidate B, what possible reason in the world should the less-preferred candidate be tapped to serve than the
more-preferred candidate?  especially when **no** other candidate is preferred over the more-preferred candidate?


When at all possible (because it isn't always, at least in theory), if more voters mark their ballots preferring Candidate A over Candidate B than the number of voters marking their ballots to the contrary, then Candidate B is not elected.  What semantic should be used for the previous
sentence?

--



r b-j                         ***@audioimagination.com



"Imagination is more important than knowledge."

 
 
 
 

[Curt]

Yes, I’ve used the same kind of argument. If, in a two-candidate democratic election, A has more votes than B, should A be the winner? I would argue yes.

If, in a 99-voter democratic election, A has 50 votes and B has 49, should A be the winner? I would argue yes.

If, in a 99-voter democratic election, A has 50 unenthusiastic votes and B has 49 wildly enthusiastic votes, should A be the winner? I would argue yes. There are others that argue no, that B has more social utility. I would say this is a difference of opinion that rests not on logic or voting criteria, but personal values. The two camps can respectfully disagree with each other. Call it the “Majority” versus “Utility” disagreement. I also think there are election types (private organizations, clubs, whatever) where the “Utility” direction might be more appropriate than the “Majority” direction. That’s fine.

But, for those elections where we believe that A should be the winner in that scenario -- the “Majority” believers -- that is what leads us to the Condorcet camp, as opposed to Borda, score, range, etc.

And expanding to multiple candidates, if Candidate A would beat all other candidates head to head, then A should be the winner. A is the Condorcet Winner, just the same as if A is the Condorcet Winner if he has more votes in a two-candidate election.

(In your final paragraphs, I am not sure if you are talking about a candidate other than the Condorcet Winner, or, a candidate from a multi-candidate Smith Set that would (in the case of a cycle) by definition have another candidate that is preferred over it.)

But yes, I definitely agree that there should be a bright line between methods that
A: “elect a Condorcet Winner if one exists”
and methods that might
B: “elect a winner other than the Condorcet Winner”.

For us “Majority” believers, we are in violent agreement that group A is superior to group B.

But I also believe that there should be a bright line between methods that
C: identify a “candidate or candidates that would defeat all other candidates head to head”
and methods that might
D: “elect a single winner that is not a Condorcet Winner if a CW does not exist”.

Group C stops with the identification of the Condorcet Winner, or the Smith Set if the CW does not exist. (Or, Group C might stop with the identification of the Weak Condorcet Winner, or the Schwartz Set if the WCW does not exist, *if* beats-or-ties is deemed allowable.)

Group D contains ranked-pairs, beatpaths, etc.

The reason I believe in the distinction is because D fails criteria that C does not. And if C and D are conflated, it does a disservice to C. When in large elections with a limited number of candidates, a CW is much more probable than a cycle. It does Condorcet proponents no favors to have Condorcet Methods described as “flawed” in the way group D is.

Group D is “decisive” where Group C is not. In these cases I would argue decisiveness is overvalued.

What do you believe the Smith Set signifies? Is it meaningless to you other than something from which a winner should be algorithmically selected?

Thank you for the exchange,
Curt

Post by robert bristow-johnson
---------------------------- Original Message ----------------------------
Subject: Re: [EM] smith/schwartz/landau
Date: Sat, March 24, 2018 12:20 am
--------------------------------------------------------------------------

Post by Curt

Post by robert bristow-johnson
---------------------------- Original Message ----------------------------
--------------------------------------------------------------------------

Post by Curt
https://github.com/tunesmith/condorcet-counter <https://github.com/tunesmith/condorcet-counter>
I opined a bit in the README but that’s not really the point of the project. I just wanted an easy way to identify Smith and Schwartz sets for myself.

to wit: "It's this author's view that a method should only be called a Condorcet if it is limited to identifying the Smith Set,"
seems to me that your view is that the current definition of "Condorcet-compliant-method" should be changed. so is Tideman Ranked-Pairs or Schulze Beat-Path methods not "Condorcet methods"?

Yes, that is my view. They should not be called “Condorcet Methods”, because they are not guaranteed to select the candidate(s) that would defeat all other candidates.
There is a clear difference between a “Condorcet Method” that finds a Condorcet Winner or Smith Set, and a “tiebreaking" method that tries to pick a single winner from a multi-candidate Smith Set. The latter fails criteria that the former does not. I don’t know what we should call these tiebreaking algorithms - perhaps they are not quite “tiebreaking” methods since a cycle is not exactly a tie. But it isn’t appropriate to call them “Completion” methods either as that implies something that it isn’t.
The actual objection I have is that when both are described as “Condorcet Methods”, then it’s too easy in the literature (and the blogs, and the wikipedia articles, particularly from Condorcet detractors) to paint with a broad brush and argue that all Condorcet Methods are flawed in some manner, same as how all other methods are “unfair” in some way, which ultimately does a disservice to the Condorcet Method. An election that has a Condorcet Winner is not unfair in those ways, compared to something like IRV or Plurality or Top-Two.
It might worth a survey of what a Smith Set actually *means*. I believe it signifies something valuable about the electorate, beyond just an indication that the election is “incomplete” and that we should apply some algorithm to divine a single winner from it.

well, arguing about the semantics is one thing, arguing about theory or ideals is another, and arguing about practice is yet another.
so whether you call it "Condorcet-compliant" or call it a turnip, there are ranked-ballot tabulation procedures that are 1. Decisive (they will elect someone) and 2. If a Condorcet Winner exists, the procedure will elect the CW. we gotta call that something and "Condorcet-compliant method" is more descriptive than "turnip". And it serves a purpose. It separates both practice and "theory and ideals" of these aforementioned procedures from other ranked-ballot systems such as IRV, Borda, or Bucklin.
both Ranked-Pairs and Schulze are fundamentally **not** defined as some post algorithm to divine a single winner from a Smith set that is larger than 1. They are well-defined procedures, in their own right, that **happen** to elect the CW if a CW exists. IRV *may* elect the CW if one exists, but of course we know that hasn't always been the case in practice. They are not "tiebreaking" a Smith set. they are not a procedure to be applied **after** it is discovered no single CW exists. they are procedures that will elect a candidate by the same rules whether a CW exists or not. But the candidate elected will be the CW if one exists. So the only practical difference between these turnips is what their outcome might be if there was a Smith set greater than 3 and some weird voting alignment.
So we need a term to draw the line between Ranked-Pairs or Schulze or BTR-IRV on one side and Bucklin, Borda, or IRV on the other. What semantic would you suggest?
the salient difference is simply what happened in my town 9 years ago: A candidate for mayor was elected to office when the voters in the city unambiguously marked their ballots that they preferred a different **specific** candidate. that is the problem with **any** non-turnip method. it's the converse of who a CW is. if everyone's vote carries the same weight (this is the "one-person-one-vote" principle), if Candidate A is preferred by more voters than Candidate B, what possible reason in the world should the less-preferred candidate be tapped to serve than the more-preferred candidate? especially when **no** other candidate is preferred over the more-preferred candidate?
When at all possible (because it isn't always, at least in theory), if more voters mark their ballots preferring Candidate A over Candidate B than the number of voters marking their ballots to the contrary, then Candidate B is not elected. What semantic should be used for the previous sentence?
--
"Imagination is more important than knowledge."
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[robert bristow-johnson]

---------------------------- Original Message ----------------------------

Subject: Re: [EM] smith/schwartz/landau

From: "Curt" <***@museworld.com>

Date: Sat, March 24, 2018 11:30 pm

To: "election-***@lists.electorama.com" <election-***@lists.electorama.com>

--------------------------------------------------------------------------

Yes, I&rsquo;ve used the same kind of argument. If, in a two-candidate democratic election, A has more votes than B, should A be the winner? I would argue yes.
If, in a 99-voter democratic election, A has 50 votes and B has 49, should A be the winner? I would argue yes.
If, in a 99-voter democratic election, A has 50 unenthusiastic votes and B has 49 wildly enthusiastic votes, should A be the winner? I would argue yes. There are others that argue no, that B has more social utility. I would say this is a difference of opinion that rests not on logic or voting

criteria, but personal values. The two camps can respectfully disagree with each other. Call it the &ldquo;Majority&rdquo; versus &ldquo;Utility&rdquo; disagreement. I also think there are election types (private organizations, clubs, whatever) where the &ldquo;Utility&rdquo; direction might be more
appropriate than the &ldquo;Majority&rdquo; direction. That&rsquo;s fine.

But, for those elections where we believe that A should be the winner in that scenario -- the &ldquo;Majority&rdquo; believers -- that is what leads us to the Condorcet camp, as opposed to Borda, score, range, etc.
And expanding to multiple candidates, if Candidate A would beat all other candidates head to head, then A should be the winner. A is the Condorcet Winner, just the same as if A is the Condorcet Winner if he has more votes in a two-candidate election.
(In your final paragraphs, I am not sure if you are talking about a candidate other than the Condorcet Winner, or, a candidate from a multi-candidate Smith Set that would (in the case of a cycle) by definition have another candidate that is preferred over it.)
But yes, I definitely agree that there should be a bright line between methods that
A: &ldquo;elect a Condorcet Winner if one exists&rdquo;
and methods that might
B: &ldquo;elect a winner other than the Condorcet Winner&rdquo;.
For us &ldquo;Majority&rdquo; believers, we are in violent agreement that group A is superior to group B.
But I also believe that there should be a bright line between methods that
C: identify a &ldquo;candidate or candidates that would defeat all other candidates head to head&rdquo;
and methods that might
D: &ldquo;elect a single winner that is not a Condorcet Winner if a CW does not exist&rdquo;.
Group C stops with the identification of the Condorcet Winner, or the Smith Set if the CW does not exist. (Or, Group C might stop with the identification of the Weak Condorcet Winner, or the Schwartz Set if the WCW does not exist, *if* beats-or-ties is deemed allowable.)
Group D contains ranked-pairs, beatpaths, etc.
The reason I believe in the distinction is because D fails criteria that C does not. And if C and D are conflated, it does a disservice to C. When in large elections with a limited number of candidates, a CW is much more probable than a cycle. It does Condorcet proponents no favors to have

Condorcet Methods described as &ldquo;flawed&rdquo; in the way group D is.

Group D is &ldquo;decisive&rdquo; where Group C is not. In these cases I would argue decisiveness is overvalued.

well, organizations and governments have to move on.  they *need* answers and elections are held to provide answers.



for a single-seat (usually executive) office, what would you suggest?  a runoff?

What do you believe the Smith Set signifies? Is it meaningless to you other than something from which a winner should be algorithmically selected?

 
it's not meaningless.  it just need not be a concept coded in election law.  remember (i am not sure you got this point), Ranked-Pairs and Schulze do **not** select a winner from the Smith set.  RP and Schulze select a winner from the entire field of candidates using a
consistent rule and, it turns out, that the winner selected by RP or Schulze **is** the CW if a CW exists.  that's in Group A.


"Group A" is a crappy term.  as bad as "turnip".  so here's a quite abridged taxonomy:



Mark-Only-One ballot:

   1. FPTP  (plurality.  decisive)

   2. FPTP runoff if no majority



Mark as many as you want:

   3. Approval
Mark ballot scoring candidates:

   4.  Score voting or Range voting



Ranked-choice ballot:

   5. Bucklin

   6. Borda

   7. IRV

   8. Ranked Pairs (let's say based on margins)

   9. Schulze (margins)

 10. Min/Max

 11. IRV-BTR



then there are all of these esoteric methods that are promoted by some people in this group.
what should be the term that differentiates 8, 9, 10, 11 from 5, 6, 7?
--



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"Imagination is more important than knowledge."

 
 
 
 

[Curt]

Post by robert bristow-johnson
---------------------------- Original Message ----------------------------
--------------------------------------------------------------------------

Post by Curt
A: “elect a Condorcet Winner if one exists”
B: “elect a winner other than the Condorcet Winner”.
[within A: ]
C: identify a “candidate or candidates that would defeat all other candidates head to head”
D: “elect a single winner that is not a Condorcet Winner if a CW does not exist”.
[
]
Group D is “decisive” where Group C is not. In these cases I would argue decisiveness is overvalued.

well, organizations and governments have to move on. they *need* answers and elections are held to provide answers.
for a single-seat (usually executive) office, what would you suggest? a runoff?

Appropriate remedies might be a runoff after another period of consideration, or a power-sharing agreement. But this ties into what it means to have a multi-candidate Smith Set.

Post by robert bristow-johnson

Post by Curt
What do you believe the Smith Set signifies? Is it meaningless to you other than something from which a winner should be algorithmically selected?

it's not meaningless. it just need not be a concept coded in election law.

But I'm honestly curious what you believe the multi-candidate Smith Set signifies or means.

Post by robert bristow-johnson
remember (i am not sure you got this point), Ranked-Pairs and Schulze do **not** select a winner from the Smith set.

Is this a semantic argument, or are you saying that ranked-pairs and schulze can elect single winners that are not in the Smith Set? Can you produce an example ballot set? I’m currently under the impression that is incorrect, as Wikipedia says both are Smith-compliant. I would love to be corrected if Wikipedia is wrong. I would not regard a method as Condorcet-compliant if it is not Smith-compliant.

At any rate, I am not here to lobby for a particular term for the methods in Group D. Only to argue that a distinction between groups C and D should exist, and without it, we do a disservice to Condorcet Methods by allowing them to be describe as “flawed” in ways that only group D is. I am open to suggestions.

[robert bristow-johnson]

---------------------------- Original Message ----------------------------

Subject: Re: [EM] smith/schwartz/landau

From: "Curt" <***@museworld.com>

Date: Sun, March 25, 2018 2:12 am

To: "election-***@lists.electorama.com" <election-***@lists.electorama.com>

--------------------------------------------------------------------------

Post by Curt

Post by robert bristow-johnson
---------------------------- Original Message ----------------------------
--------------------------------------------------------------------------

A: &ldquo;elect a Condorcet Winner if one exists&rdquo;
B: &ldquo;elect a winner other than the Condorcet Winner&rdquo;.
[within A: ]
C: identify a &ldquo;candidate or candidates that would defeat all other candidates head to head&rdquo;
D: &ldquo;elect a single winner that is not a Condorcet Winner if a CW does not exist&rdquo;.
[&hellip;]
Group D is &ldquo;decisive&rdquo; where Group C is not. In these cases I would argue decisiveness is overvalued.

well, organizations and governments have to move on. they *need* answers and elections are held to provide answers.
for a single-seat (usually executive) office, what would you suggest? a runoff?

Appropriate remedies might be a runoff after another period of consideration, or a power-sharing agreement. But this ties into what it means to have a multi-candidate Smith Set.

Post by robert bristow-johnson

What do you believe the Smith Set signifies? Is it meaningless to you other than something from which a winner should be algorithmically selected?

it's not meaningless. it just need not be a concept coded in election law.

But I'm honestly curious what you believe the multi-candidate Smith Set signifies or means.

well, what i think about the Smith Set, whether it be single or multi-candidate set in a single-winner race, is that there is no compelling reason that any candidate in the complement of the Smith
set should be elected.  because the electorate is clear that there are other, more preferred, candidates in *any* scenario.  but it doesn't have to be encoded into law.

Post by Curt

Post by robert bristow-johnson
remember (i am not sure you got this point), Ranked-Pairs and Schulze do **not** select a winner from the Smith set.

Is this a semantic argument, or are you saying that ranked-pairs and schulze can elect single winners that are not in the Smith Set? Can you produce an example ballot set?

we're having a semantic difference.  RP and Schulze elect a candidate drawing from the entire pool of
candidates.  they do not identify a Smith set and then proceed to select from the Smith set.


so it's just an election procedure say like IRV is or Bucklin is.  same ballots, different way of looking at the ballots. and **if** a CW exists, the RP winner and Schulze winner will be the CW. (and it turns out that RP and Schulze will elect the same candidate in the case of a Smith set of 3,
and i am convinced a bigger Smith set is highly unlikely to occur in a real governmental election.  RP is simpler to explain to legislators and the public than is Schulze, in my opinion.  so it really just matters whether or not we consider the RP winner to be the best indication of the
least disapproved candidate.)

Post by Curt
I&rsquo;m currently under the impression that is incorrect, as Wikipedia says both are Smith-compliant. I would love to be corrected if Wikipedia is wrong. I would not regard a method as Condorcet-compliant if it is not Smith-compliant.

i am not taking an issue with that.  i
understand why, in RP at least, why the winner is in the Smith set.  but it's not the same as a method that acts in one way to get the CW, and then, if no CW is forthcoming, does a secondary algorithm to pick the winner.  RP and Schulze do what they do on the entire pool of
candidates.

Post by Curt
At any rate, I am not here to lobby for a particular term for the methods in Group D. Only to argue that a distinction between groups C and D should exist, and without it, we do a disservice to Condorcet Methods by allowing them to be describe as &ldquo;flawed&rdquo; in ways

that only group D is. I am open to suggestions.
i just wanna know what we wanna call a ranked-ballot election method that elects the Condocet Winner when such exists.  what should such a class of methods be called?
--



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"Imagination is more important than knowledge."

 
 
 
 

[robert bristow-johnson]

---------------------------- Original Message ----------------------------

Subject: Re: [EM] smith/schwartz/landau

From: "Curt" <***@museworld.com>

Date: Sat, March 24, 2018 11:30 pm

To: "election-***@lists.electorama.com" <election-***@lists.electorama.com>

--------------------------------------------------------------------------

Post by Curt
(In your final paragraphs, I am not sure if you are talking about a candidate other than the Condorcet Winner, or, a candidate from a multi-candidate Smith Set that would (in the case of a cycle) by definition have another candidate that is preferred over it.)

i was referring to the 2009 mayoral race in Burlington Vermont.  IRV was used, but the ballot data was publically available and we could re-tabulate the ballots according to other rules and we discovered that a CW existed and was not the IRV winner (which meant that the CW came in
"third" by IRV reckoning).


there was no cycle.  the Smith set was of size 1.



the CW was the more centrist Democrat

the IRV winner (and election winner) was Prog

the plurality winner (counting first-choice votes) was GOP



so three different candidates all having some claim of legitimacy for office.  but the law was IRV and the Prog won even though the CW beat him by about 5% in a head-to-head.
the next year, the city voters voted to repeal IRV and we have reverted back to Plurality with a runoff if 40%
cannot be obtained.

--



r b-j                         ***@audioimagination.com



"Imagination is more important than knowledge."

 
 
 
 

[Juho Laatu]

Post by Curt
What do you believe the Smith Set signifies? Is it meaningless to you other than something from which a winner should be algorithmically selected?

To me Smith Set is a criterion that in some sense and at first sight looks natural, but on second thought does not cover all possible scenarios well. I mean that if there is a group of candidates that are a unified group, and they beat all others, then yes, one of them should at least in most cases win. This is related to clones. If all the Smith Set candidates can be considered to be clones, then nominating only one of them would probably lead to electing that candidate as a Condorcet Winner.

What are the problems then? One problem is that those Condorcet methods that are based on rankings only, and possibly a pairwise matrix only, can carry only limited information on what the preferences of the voters are. There can be multiple explanations to what the voter preferences might have been. There are scenarios where electing from the Smith Set may not be natural.

Another problem of the Smith Set is that it may look more natural than it is, when one draws the end results (in paper or in one's mind) so that all the Smith Set candidates are at top, and all others below. This drawing technique to some extent hides the defeats within the Smith Set from the eye.

The best I can do to demonstrate these problems is to give you one particular (old) example scenario where selecting the winner outside of the Smith Set seems quite natural. In some extreme situations Smith Set may thus not be the right choice.

17: A > B > d > C
16: A > d > B > C
17: B > C > d > A
16: B > d > C > A
17: C > A > d > B
16: C > d > A > B

This example is a classic strong cycle of A, B and C, with one more candidate (d) added. Candidates A, B and C are not clones since they are not next to each others in the ballots. Candidate d is not in the Smith Set, but is very close to being a Condorcet Winner (2 votes short). Candidates A, B and C are very far from being Condorcet Winners.

BR, Juho

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[Curt]

Hi Juho, thank you for this example ballot set. I have added it to the codebase as a test case with some documentation. (Both Smith and Schwartz should identify A, B, and C.)

A, B, and C all defeat each other 66:33. And they each defeat d 50:49. I understand the urge to award d the win, given those numbers. But I believe that that urge ascribes “intensity of preference” to A, B, and C - when for Condorcet, which is purely ordinal, we have no idea. If I imagine those voters as voting stoically, dispassionately, poker-face, then I have no idea whether they passionately prefer A to B, or if, for instance, they are completely torn but have some consistent but trivial reason to pick one over the other. So in the absence of intensity-of-preference data, we really don’t have enough data to conclude that d should be the winner - for all we know, there actually is more passion in those 50:49 splits.

So I am having trouble seeing it as a flaw with the Smith Set concept itself. I do agree that it points to some sort of flaw, but I think the proper identification of the flaw’s home requires zooming out and looking at the framework.

Each time we step close to determining an election winner, we make a choice to lose fidelity in some fashion.

1. We have a need to collectively decide something, and so we voice that need by identifying a question to answer. But in identifying or voicing that question, we risk losing some essential part of the real question. In other words, we risk a failure of specification. But, we need to move forward, so we accept that risk, and move forward with the question as asked.

2. In asking the question, we identify options for a solution - the candidates in an election. But in doing so, we risk omitting some of the proper solution spectrum. The collected candidates may still be insufficient in some way. But, we have to draw the line somewhere, so voters are restricted to choosing between a potentially imperfect slate of candidates.

3. We set a time to choose. But in doing so, we risk the deadline being too soon for some voters, in that they might not be finished with their decision process. So we lose some fidelity there in measuring voter preference exactly.

4. We want to protect against bullying, intimidation, and some people being convinced to make their vote count “less” than someone else’s, and so we decide to protect the ideal of “one person, one vote” where every person’s vote counts the same. But there we lose some fidelity in measuring intensity of preference among voters.

So I’m inclined to think that the example below is more about the costs associated with something like #4 above. In fact, the scenario seems very similar to me, to the one I sketched out earlier - if, in a two-candidate election, A defeats B 50:49, but A’s support is lukewarm and B’s is passionate, should B win? I think you can defend that if you value “social utility” and “intensity of preference” more than “one-person-one-vote” or "majority", but I wouldn’t count that as a flaw with the Condorcet method in particular, since it’s a majority-type of voting method.

Regards,
Curt

Post by Juho Laatu

Post by Curt
What do you believe the Smith Set signifies? Is it meaningless to you other than something from which a winner should be algorithmically selected?

To me Smith Set is a criterion that in some sense and at first sight looks natural, but on second thought does not cover all possible scenarios well. I mean that if there is a group of candidates that are a unified group, and they beat all others, then yes, one of them should at least in most cases win. This is related to clones. If all the Smith Set candidates can be considered to be clones, then nominating only one of them would probably lead to electing that candidate as a Condorcet Winner.
What are the problems then? One problem is that those Condorcet methods that are based on rankings only, and possibly a pairwise matrix only, can carry only limited information on what the preferences of the voters are. There can be multiple explanations to what the voter preferences might have been. There are scenarios where electing from the Smith Set may not be natural.
Another problem of the Smith Set is that it may look more natural than it is, when one draws the end results (in paper or in one's mind) so that all the Smith Set candidates are at top, and all others below. This drawing technique to some extent hides the defeats within the Smith Set from the eye.
The best I can do to demonstrate these problems is to give you one particular (old) example scenario where selecting the winner outside of the Smith Set seems quite natural. In some extreme situations Smith Set may thus not be the right choice.
17: A > B > d > C
16: A > d > B > C
17: B > C > d > A
16: B > d > C > A
17: C > A > d > B
16: C > d > A > B
This example is a classic strong cycle of A, B and C, with one more candidate (d) added. Candidates A, B and C are not clones since they are not next to each others in the ballots. Candidate d is not in the Smith Set, but is very close to being a Condorcet Winner (2 votes short). Candidates A, B and C are very far from being Condorcet Winners.
BR, Juho
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[Juho Laatu]

Post by Curt
Hi Juho, thank you for this example ballot set. I have added it to the codebase as a test case with some documentation. (Both Smith and Schwartz should identify A, B, and C.)
A, B, and C all defeat each other 66:33. And they each defeat d 50:49. I understand the urge to award d the win, given those numbers. But I believe that that urge ascribes “intensity of preference” to A, B, and C - when for Condorcet, which is purely ordinal, we have no idea. If I imagine those voters as voting stoically, dispassionately, poker-face, then I have no idea whether they passionately prefer A to B, or if, for instance, they are completely torn but have some consistent but trivial reason to pick one over the other. So in the absence of intensity-of-preference data, we really don’t have enough data to conclude that d should be the winner - for all we know, there actually is more passion in those 50:49 splits.

We don't know the strengths of preferences (strength), but we know the number of voters (count) on each side in every pairwise preference. A, B and C are preferred over d weakly (count). A is preferred over B strongly (count). In the following examples preferences vary a lot if we study the strengths of preferences (strength).

In the following example d is weak (strength). (only ">>>" added to the original example)

17: A > B >>> d > C
16: A >>> d > B > C
17: B > C >>> d > A
16: B >>> d > C > A
17: C > A >>> d > B
16: C >>> d > A > B

In the following example d is strong (strength). (only ">>>" added)

17: A > B > d >>> C
16: A > d >>> B > C
17: B > C > d >>> A
16: B > d >>> C > A
17: C > A > d >>> B
16: C > d >>> A > B

Votes in the original example are almost symmetric in the sense that A, B and C are about as often below and above d, and the votes are about symmetric whether you read them from left to right or from right to left. This makes the ABC group and d about equal when thinking in terms of having them listed "on the right" or "on the left". The Smith Set candidates are in the ballots only marginally more "on the left" than candidate d. Therefore also their preference strengths (strength) are with good probability (under some randomness assumptions) about the same.

In summary, I don't see how to make any conclusions on the relative preferences (strength) of the ABC group and d. Just like Condorcet Winners could be popular or not (strength), also Smith Set members and non Smith Set members could be popular or not (strength). The example shows a situation where preferences can be whatever or about the same (strength), or in favour of d (count). There are three candidates that beat d, but that should not carry much weight since number of pairwise victories is known to be a poor criterion (because of clone problems).

Post by Curt
So I am having trouble seeing it as a flaw with the Smith Set concept itself. I do agree that it points to some sort of flaw, but I think the proper identification of the flaw’s home requires zooming out and looking at the framework.

I think the interesting question is if electing from the Smith Set makes sense as often as electing a Condorcet Winner. Many people on this list think that in many elections it would make sense to always elect the CW (knowing that the strengths of preferences (strength) are not well known). People who think that way should ask themselves if they require the winner to come from the SS in the given example, of if d would be a better choice. If d is ok (again assuming no knowledge of the strengths of preferences (strength)), then SS should not be seen as a requirement (although 99.9% of the elections would still elect from the SS, since the situation given in the example is very rare).

To me that example tells that sometimes candidates outside the Smith Set can indeed be almost ideal (2 votes short of being a Condorcet Winner), and candidates in the Smith Set can sometimes be quite poor (creating a strong unified opposition with interest to change winner A to C). This kind of ballot sets are however very unusual. Practical Condorcet methods should therefore elect almost always from the Smith Set. But as far as I'm concerned, not necessarily as a requirement, because of special cases like this.

Post by Curt
Each time we step close to determining an election winner, we make a choice to lose fidelity in some fashion.
1. We have a need to collectively decide something, and so we voice that need by identifying a question to answer. But in identifying or voicing that question, we risk losing some essential part of the real question. In other words, we risk a failure of specification. But, we need to move forward, so we accept that risk, and move forward with the question as asked.
2. In asking the question, we identify options for a solution - the candidates in an election. But in doing so, we risk omitting some of the proper solution spectrum. The collected candidates may still be insufficient in some way. But, we have to draw the line somewhere, so voters are restricted to choosing between a potentially imperfect slate of candidates.
3. We set a time to choose. But in doing so, we risk the deadline being too soon for some voters, in that they might not be finished with their decision process. So we lose some fidelity there in measuring voter preference exactly.
4. We want to protect against bullying, intimidation, and some people being convinced to make their vote count “less” than someone else’s, and so we decide to protect the ideal of “one person, one vote” where every person’s vote counts the same. But there we lose some fidelity in measuring intensity of preference among voters.
So I’m inclined to think that the example below is more about the costs associated with something like #4 above.

I don't see any strong connection to strength of preferences (strength). The example is intended to follow the "one person, one vote" and "majority" tradition of explaining how to design pure ranked (Condorcet) methods. To me the key question thus is, when measuring only pairwise preferences, are there situations where the best winner might come outside of the Smith Set. The given example is the most obvious and most extreme to me. And the key point there is that d is two votes short of being a Condorcet Winner. Let's assume that we all would like d to win if it was a CW. A, B and C on the other hand are far from being Condorcet winners. This can be seen as a question of having one major defeat (A, B, C) to some other candidate, or having three marginal defeats (d) to other candidates.

That's just my two cents. I'm just trying to encourage people to think if Smith Set should be seen as a requirement, or just as a the most common outcome when there is a top cycle.

BB, Juho


P.S. I note once more that the Smith Set should not be visualised as a group _above_ d (although that is the way people always draw it). Cyclic preferences do not have any obvious geometric presentation on a 2D paper. An alternative drawing approach would be to draw all the candidates as far from the winner position as they have distance to being a Condorcet Winner. That's how I tend to see the given example. (There are however few alternative ways to count the distance of each candidate to becoming a Condorcet Winner.)

Post by Curt
In fact, the scenario seems very similar to me, to the one I sketched out earlier - if, in a two-candidate election, A defeats B 50:49, but A’s support is lukewarm and B’s is passionate, should B win? I think you can defend that if you value “social utility” and “intensity of preference” more than “one-person-one-vote” or "majority", but I wouldn’t count that as a flaw with the Condorcet method in particular, since it’s a majority-type of voting method.
Regards,
Curt

Post by Juho Laatu

Post by Curt
What do you believe the Smith Set signifies? Is it meaningless to you other than something from which a winner should be algorithmically selected?

To me Smith Set is a criterion that in some sense and at first sight looks natural, but on second thought does not cover all possible scenarios well. I mean that if there is a group of candidates that are a unified group, and they beat all others, then yes, one of them should at least in most cases win. This is related to clones. If all the Smith Set candidates can be considered to be clones, then nominating only one of them would probably lead to electing that candidate as a Condorcet Winner.
What are the problems then? One problem is that those Condorcet methods that are based on rankings only, and possibly a pairwise matrix only, can carry only limited information on what the preferences of the voters are. There can be multiple explanations to what the voter preferences might have been. There are scenarios where electing from the Smith Set may not be natural.
Another problem of the Smith Set is that it may look more natural than it is, when one draws the end results (in paper or in one's mind) so that all the Smith Set candidates are at top, and all others below. This drawing technique to some extent hides the defeats within the Smith Set from the eye.
The best I can do to demonstrate these problems is to give you one particular (old) example scenario where selecting the winner outside of the Smith Set seems quite natural. In some extreme situations Smith Set may thus not be the right choice.
17: A > B > d > C
16: A > d > B > C
17: B > C > d > A
16: B > d > C > A
17: C > A > d > B
16: C > d > A > B
This example is a classic strong cycle of A, B and C, with one more candidate (d) added. Candidates A, B and C are not clones since they are not next to each others in the ballots. Candidate d is not in the Smith Set, but is very close to being a Condorcet Winner (2 votes short). Candidates A, B and C are very far from being Condorcet Winners.
BR, Juho
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[Juho Laatu]

Here's yet one more example that explains my line of thought.

This is an example of a weak (strength) Condorcet Winner. This is the same as the earlier example of a weak (strength) d, but numbers "16" and "17" have been swapped. The point is to demonstrate that Condorcet Winners can be weak or strong (strength) in the same way as candidates outside the Smith Set can be, and the difference is sometimes not big (swap of "16" and "17"). You can also change the location of the ">>>" marks if you want to make d strong (strength) as a Condorcet Winner or as a non Smith Set member.

16: A > B >>> d > C
17: A >>> d > B > C
16: B > C >>> d > A
17: B >>> d > C > A
16: C > A >>> d > B
17: C >>> d > A > B

Also this example should be evaluated assuming that the pairwise preferences are exactly known (count), the individual ballots are possibly known (count), and the strengths of preferences (strength) are totally unknown (i.e. forget the ">>>" marks). They (strengths) could be anything, but some distributions are more common than some others (for statistical reasons, although we know nothing bout the voters).

BR, Juho

Post by Juho Laatu

Post by Curt
Hi Juho, thank you for this example ballot set. I have added it to the codebase as a test case with some documentation. (Both Smith and Schwartz should identify A, B, and C.)
A, B, and C all defeat each other 66:33. And they each defeat d 50:49. I understand the urge to award d the win, given those numbers. But I believe that that urge ascribes “intensity of preference” to A, B, and C - when for Condorcet, which is purely ordinal, we have no idea. If I imagine those voters as voting stoically, dispassionately, poker-face, then I have no idea whether they passionately prefer A to B, or if, for instance, they are completely torn but have some consistent but trivial reason to pick one over the other. So in the absence of intensity-of-preference data, we really don’t have enough data to conclude that d should be the winner - for all we know, there actually is more passion in those 50:49 splits.

We don't know the strengths of preferences (strength), but we know the number of voters (count) on each side in every pairwise preference. A, B and C are preferred over d weakly (count). A is preferred over B strongly (count). In the following examples preferences vary a lot if we study the strengths of preferences (strength).
In the following example d is weak (strength). (only ">>>" added to the original example)
17: A > B >>> d > C
16: A >>> d > B > C
17: B > C >>> d > A
16: B >>> d > C > A
17: C > A >>> d > B
16: C >>> d > A > B
In the following example d is strong (strength). (only ">>>" added)
17: A > B > d >>> C
16: A > d >>> B > C
17: B > C > d >>> A
16: B > d >>> C > A
17: C > A > d >>> B
16: C > d >>> A > B
Votes in the original example are almost symmetric in the sense that A, B and C are about as often below and above d, and the votes are about symmetric whether you read them from left to right or from right to left. This makes the ABC group and d about equal when thinking in terms of having them listed "on the right" or "on the left". The Smith Set candidates are in the ballots only marginally more "on the left" than candidate d. Therefore also their preference strengths (strength) are with good probability (under some randomness assumptions) about the same.
In summary, I don't see how to make any conclusions on the relative preferences (strength) of the ABC group and d. Just like Condorcet Winners could be popular or not (strength), also Smith Set members and non Smith Set members could be popular or not (strength). The example shows a situation where preferences can be whatever or about the same (strength), or in favour of d (count). There are three candidates that beat d, but that should not carry much weight since number of pairwise victories is known to be a poor criterion (because of clone problems).

Post by Curt
So I am having trouble seeing it as a flaw with the Smith Set concept itself. I do agree that it points to some sort of flaw, but I think the proper identification of the flaw’s home requires zooming out and looking at the framework.

I think the interesting question is if electing from the Smith Set makes sense as often as electing a Condorcet Winner. Many people on this list think that in many elections it would make sense to always elect the CW (knowing that the strengths of preferences (strength) are not well known). People who think that way should ask themselves if they require the winner to come from the SS in the given example, of if d would be a better choice. If d is ok (again assuming no knowledge of the strengths of preferences (strength)), then SS should not be seen as a requirement (although 99.9% of the elections would still elect from the SS, since the situation given in the example is very rare).
To me that example tells that sometimes candidates outside the Smith Set can indeed be almost ideal (2 votes short of being a Condorcet Winner), and candidates in the Smith Set can sometimes be quite poor (creating a strong unified opposition with interest to change winner A to C). This kind of ballot sets are however very unusual. Practical Condorcet methods should therefore elect almost always from the Smith Set. But as far as I'm concerned, not necessarily as a requirement, because of special cases like this.

Post by Curt
Each time we step close to determining an election winner, we make a choice to lose fidelity in some fashion.
1. We have a need to collectively decide something, and so we voice that need by identifying a question to answer. But in identifying or voicing that question, we risk losing some essential part of the real question. In other words, we risk a failure of specification. But, we need to move forward, so we accept that risk, and move forward with the question as asked.
2. In asking the question, we identify options for a solution - the candidates in an election. But in doing so, we risk omitting some of the proper solution spectrum. The collected candidates may still be insufficient in some way. But, we have to draw the line somewhere, so voters are restricted to choosing between a potentially imperfect slate of candidates.
3. We set a time to choose. But in doing so, we risk the deadline being too soon for some voters, in that they might not be finished with their decision process. So we lose some fidelity there in measuring voter preference exactly.
4. We want to protect against bullying, intimidation, and some people being convinced to make their vote count “less” than someone else’s, and so we decide to protect the ideal of “one person, one vote” where every person’s vote counts the same. But there we lose some fidelity in measuring intensity of preference among voters.
So I’m inclined to think that the example below is more about the costs associated with something like #4 above.

I don't see any strong connection to strength of preferences (strength). The example is intended to follow the "one person, one vote" and "majority" tradition of explaining how to design pure ranked (Condorcet) methods. To me the key question thus is, when measuring only pairwise preferences, are there situations where the best winner might come outside of the Smith Set. The given example is the most obvious and most extreme to me. And the key point there is that d is two votes short of being a Condorcet Winner. Let's assume that we all would like d to win if it was a CW. A, B and C on the other hand are far from being Condorcet winners. This can be seen as a question of having one major defeat (A, B, C) to some other candidate, or having three marginal defeats (d) to other candidates.
That's just my two cents. I'm just trying to encourage people to think if Smith Set should be seen as a requirement, or just as a the most common outcome when there is a top cycle.
BB, Juho
P.S. I note once more that the Smith Set should not be visualised as a group _above_ d (although that is the way people always draw it). Cyclic preferences do not have any obvious geometric presentation on a 2D paper. An alternative drawing approach would be to draw all the candidates as far from the winner position as they have distance to being a Condorcet Winner. That's how I tend to see the given example. (There are however few alternative ways to count the distance of each candidate to becoming a Condorcet Winner.)

Post by Curt
In fact, the scenario seems very similar to me, to the one I sketched out earlier - if, in a two-candidate election, A defeats B 50:49, but A’s support is lukewarm and B’s is passionate, should B win? I think you can defend that if you value “social utility” and “intensity of preference” more than “one-person-one-vote” or "majority", but I wouldn’t count that as a flaw with the Condorcet method in particular, since it’s a majority-type of voting method.
Regards,
Curt

Post by Juho Laatu

Post by Curt
What do you believe the Smith Set signifies? Is it meaningless to you other than something from which a winner should be algorithmically selected?

To me Smith Set is a criterion that in some sense and at first sight looks natural, but on second thought does not cover all possible scenarios well. I mean that if there is a group of candidates that are a unified group, and they beat all others, then yes, one of them should at least in most cases win. This is related to clones. If all the Smith Set candidates can be considered to be clones, then nominating only one of them would probably lead to electing that candidate as a Condorcet Winner.
What are the problems then? One problem is that those Condorcet methods that are based on rankings only, and possibly a pairwise matrix only, can carry only limited information on what the preferences of the voters are. There can be multiple explanations to what the voter preferences might have been. There are scenarios where electing from the Smith Set may not be natural.
Another problem of the Smith Set is that it may look more natural than it is, when one draws the end results (in paper or in one's mind) so that all the Smith Set candidates are at top, and all others below. This drawing technique to some extent hides the defeats within the Smith Set from the eye.
The best I can do to demonstrate these problems is to give you one particular (old) example scenario where selecting the winner outside of the Smith Set seems quite natural. In some extreme situations Smith Set may thus not be the right choice.
17: A > B > d > C
16: A > d > B > C
17: B > C > d > A
16: B > d > C > A
17: C > A > d > B
16: C > d > A > B
This example is a classic strong cycle of A, B and C, with one more candidate (d) added. Candidates A, B and C are not clones since they are not next to each others in the ballots. Candidate d is not in the Smith Set, but is very close to being a Condorcet Winner (2 votes short). Candidates A, B and C are very far from being Condorcet Winners.
BR, Juho
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[Kristofer Munsterhjelm]

Post by Juho Laatu
In summary, I don't see how to make any conclusions on the relative preferences (strength) of the ABC group and d. Just like Condorcet Winners could be popular or not (strength), also Smith Set members and non Smith Set members could be popular or not (strength). The example shows a situation where preferences can be whatever or about the same (strength), or in favour of d (count). There are three candidates that beat d, but that should not carry much weight since number of pairwise victories is known to be a poor criterion (because of clone problems).

Isn't this analogous to majority vs utility? Suppose you have a ranked
election like this:

51: A>B>C
48: B>C>A
2: C>B>A

A is the majority winner, but it only takes two voters changing from
A>B>C to B>A>C to make B the majority winner, analogous to how {ABC} is
the Smith set but {d} is close to being the CW. A voting method that
passes the majority criterion will always elect A, but a more
consensus-focused method like Borda elects B.

If we knew the utilities, then we could determine whether this is a

51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0

election (A should win if we're counting utilities), or a

51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0

election (B should win). But we don't, so we can't. Almost any situation
where A is the superior candidate can be matched by a parallel situation
where B is the superior candidate. Lacking utility information, we can't
establish which is correct, and so there seem to be only two ways to get
out of the problem.

The first way is to say that we have a model of how the electorate
behaves, which lets us determine which rated election is most likely
given the ranked data. (E.g. in a very polarized electorate, it's more
likely to be the former than the latter.)

The second way is to say "we have no idea of which may be true, but we
want to pass the majority criterion, so that settles the matter for us".
In that case, the hypothetical "we" would choose a method that elects A.

If the Smith situation is analogous to majority, then it seems that the
most consistent choice with respect to majority (as opposed to utility)
is to elect from the Smith set. A method that doesn't elect from the
Smith set may be superior in a utilitarian sense, but it may also not
be. There's no way to know (short of taking the first approach, or by
proving that e.g. for a very wide range of utility models, some
non-Smith method has better utilitarian performance than some Smith method).

I suppose what I'm saying is that if we choose to have a method that
passes majority on the grounds on the grounds that the majority is
right, then the argument can be adapted pretty easily to Condorcet and
then to Smith. Any counter of the sort that electing from Smith can get
the wrong winner can be similarly "ported back" to the majority
situation by showing that the majority criterion can also get it wrong.

Of course, if there's a more indirect argument behind wanting majority,
then that can break the symmetry. For instance, one could argue that
passing majority is a sort of DSV; in strategic Range, everybody votes
Approval style and so the majority winner wins anyway, so why not ease
the burden of the voter? Or "a large majority has the backing to get
what it wants, better let it have what it wants on the ballot than risk
a riot". It's not obvious how such arguments would support Smith,
whereas it is pretty clear that they would support the majority criterion.

But then that argument has to be made, first; in the absence of such an
argument, Smith seems a reasonable extrapolation from majority (by way
of Condorcet).
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Post by Kristofer Munsterhjelm

Post by Juho Laatu
In summary, I don't see how to make any conclusions on the relative preferences (strength) of the ABC group and d. Just like Condorcet Winners could be popular or not (strength), also Smith Set members and non Smith Set members could be popular or not (strength). The example shows a situation where preferences can be whatever or about the same (strength), or in favour of d (count). There are three candidates that beat d, but that should not carry much weight since number of pairwise victories is known to be a poor criterion (because of clone problems).

Yes, my count vs strength corresponds to majority vs utility. I should have already introduced new standard terms for "better if measured in utility" and "better if measured in majority/plurality".

Post by Kristofer Munsterhjelm
51: A>B>C
48: B>C>A
2: C>B>A
A is the majority winner, but it only takes two voters changing from A>B>C to B>A>C to make B the majority winner, analogous to how {ABC} is the Smith set but {d} is close to being the CW. A voting method that passes the majority criterion will always elect A, but a more consensus-focused method like Borda elects B.

Condorcet fans tend to emphasize the majority approach. One key reason is the strategy concerns.

Post by Kristofer Munsterhjelm
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, and so there seem to be only two ways to get out of the problem.
The first way is to say that we have a model of how the electorate behaves, which lets us determine which rated election is most likely given the ranked data. (E.g. in a very polarized electorate, it's more likely to be the former than the latter.)

This approach is good for studying the results (based on various assumptions on voter preferences). Not that easy if one would have to decide the winner based on an agreed model on what the preferences of the voters are.

Post by Kristofer Munsterhjelm
The second way is to say "we have no idea of which may be true, but we want to pass the majority criterion, so that settles the matter for us". In that case, the hypothetical "we" would choose a method that elects A.

I believe many Condorcet fans think that getting honest rankings is about as much honest information as we can get. That means that one is to some extent forced to follow the majority logic.

On the other hand people believe in the principle of one person one vote, and in the right on majorities to decide. That makes the majoritarian approach of Condorcet again very natural.

Post by Kristofer Munsterhjelm
If the Smith situation is analogous to majority, then it seems that the most consistent choice with respect to majority (as opposed to utility) is to elect from the Smith set.

If the Smith Set forms some sort of clear majority, then yes, but that need not be the case. If A, B and C would beat each others by max n votes, and they all would beat d by more than n votes, then one could say that someone from the Smith set should win to respect majority.

One can also study the examples assuming that some of the candidates are clones. In my example A, B and C certainly are not clones. But there could be also ballots where A, B and C would be next to each others in all ballots (technical clones). This does not yet guarantee that those candidates are also real life clones (candidates of one party, in friendly terms, under the assumption the utilities of the voters are about the same for all three).

In the example that I gave, we should probably assume that there are four parties that compete with each others. If we would elect A, there would be a 66:33 opposition saying that we should have elected C. If we elect d, there would be only a weak (majority) opposition saying that we should have elected one of the others. Although A, B and C form a Smith Set, they probably do not form any kind of ideological set that should be given the right to claim victory. I.e. defeats within the Smith Set are just as damaging as defeats of d. Being part of the (technical) Smith Set does not give any additional privileges.

Post by Kristofer Munsterhjelm
A method that doesn't elect from the Smith set may be superior in a utilitarian sense, but it may also not be. There's no way to know (short of taking the first approach, or by proving that e.g. for a very wide range of utility models, some non-Smith method has better utilitarian performance than some Smith method).

I think the utility uncertainties and arguments have no specific role in the given example. Condorcet Winners can have low or hight utility. Same with members of Smith Set, and candidates outside of the Smith Set. My arguments to consider d to be a reasonable winner are based on the majoritarian pairwise comparisons only. Shortly, d is two votes short of being a majoritarian Condorcet Winner (and the others are far behind in this kind of measurements).

I think human intuition is a problem here. The Smith Set with its three candidates seems to be between d and the winning position (that is drawn above the Smith Set). That makes d look bad. But I claim that this image hides all the defeats within the Smith Set, and it ignores the fact that d is actually very close to being a Condorcet Winner, not far from it.

Post by Kristofer Munsterhjelm
I suppose what I'm saying is that if we choose to have a method that passes majority on the grounds on the grounds that the majority is right, then the argument can be adapted pretty easily to Condorcet and then to Smith. Any counter of the sort that electing from Smith can get the wrong winner can be similarly "ported back" to the majority situation by showing that the majority criterion can also get it wrong.

Majority criterion is not a solution to all problems, but we can take it as granted in a traditional majoritarian political system. I would not say that the fact that Smith Set members beat all the other candidates is a valid majority based argument supporting election from the Smith Set. The reason is that this statement would ignore all the defeats within the Smith Set. We must see the Condorcet election as a competition between individual candidates, and in that competition all pairwise victories and defeats do count.

I note one more problem with human intuition. Humans have a tendency to force group opinions to form a linear preference order. It looks natural that d is behind the Smith Set, and that also all members of the Smith Set should be forced to form a linear preference order. I think this would be a big mistake since we know that group preferences are not linear. The winner should be chosen based on the ballot or matrix preferences only, not based on any imagined linear preference order.

In my example electing someone from the Smith Set is a violation of one of the 66:33 majorities.

Post by Kristofer Munsterhjelm
Of course, if there's a more indirect argument behind wanting majority, then that can break the symmetry. For instance, one could argue that passing majority is a sort of DSV; in strategic Range, everybody votes Approval style and so the majority winner wins anyway, so why not ease the burden of the voter? Or "a large majority has the backing to get what it wants, better let it have what it wants on the ballot than risk a riot". It's not obvious how such arguments would support Smith, whereas it is pretty clear that they would support the majority criterion.
But then that argument has to be made, first; in the absence of such an argument, Smith seems a reasonable extrapolation from majority (by way of Condorcet).

I tried to avoid all more complex argumentation and stick to the pairwise majority comparisons (and potentially information from the ballots too).

I don't see Smith Set as an extension of majority. It would be if we would decide to see the Smith Set members as one unified block of candidates of one party, and their defeats of each others would be either minor, or we would decide that their defeats do not matter even if they are big (assuming a small difference in utility although there are many voters preferring one over the other).

BR, Juho

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[Curt]

Hi Juho, if you extrapolate Kristofer’s 51-48-2 example (quoted below) to your 66:33 example, the point is that *if* the objective is to only use ordinal rankings and not incorporate utility, there really isn’t any reason to consider a 66:33 cycle a “strong” cycle, or 50:49 a “weak” majority. This is because the concepts of “strong” and “weak” suggest intensity of preference. Without any supporting utility data, we can’t infer anything about the public’s willingness to accept “d” over any of A, B, and C. Considering “d” to be a reasonable winner appears to be based on more than pairwise comparisons. It appears to also be based on ascribing probable utility to the voters given the vote margins.

It’s true that these might be entirely sane assumptions given the numbers. 66:33 *probably* does correlate in some fashion to intensity of preference or utility. You’re probably right there. But the point is that concluding that d should probably be awarded the winner is the same thing as incorporating utility into the calculation of the vote.

For those votes based off of the majority criterion or “one person one vote”, that’s improper. It’s not a flaw with Condorcet or Smith Set; it’s just applying the argument that utility should matter - which is a different subject. In other words, advocating Condorcet is basically accepting the axiom that 50:49 should win even if 49 has more passion, and, similarly, that someone from the ABC 66:33 Smith Set should win over d since they all defeat d 50:49. You can always disagree with an axiom, but by doing so you are entering into the conversation of whether utility should matter in Condorcet voting.

(That said, even for Condorcet voting, I am curious about capturing score preferences on the ballots, and then using them only to select from a multi-candidate Smith Set.)

Curt

Post by Kristofer Munsterhjelm
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, [
]

[Juho Laatu]

Post by Curt
Hi Juho, if you extrapolate Kristofer’s 51-48-2 example (quoted below) to your 66:33 example, the point is that *if* the objective is to only use ordinal rankings and not incorporate utility, there really isn’t any reason to consider a 66:33 cycle a “strong” cycle, or 50:49 a “weak” majority. This is because the concepts of “strong” and “weak” suggest intensity of preference.

I think the difference is between measuring the number of voters vs measuring the strength of opinion of those voters. The former we can do (quite reliably). The latter the pure ranked methods do not measure (one reason being that one could not do that in a reliable way anyway). Numbers 66, 33, 50 and 49 can be picked from the pairwise matrix. Numbers 51, 48 and 2 can be derived from the ballots (but probably we are more interested in the matrix that those ballots will produce).

Post by Curt
Without any supporting utility data, we can’t infer anything about the public’s willingness to accept “d” over any of A, B, and C. Considering “d” to be a reasonable winner appears to be based on more than pairwise comparisons. It appears to also be based on ascribing probable utility to the voters given the vote margins.

The rankings / pairwise preferences are available. Utilities are not known and we can ignore them. My intention was to present d as a potential winner based on the rankings only.

Post by Curt
It’s true that these might be entirely sane assumptions given the numbers. 66:33 *probably* does correlate in some fashion to intensity of preference or utility. You’re probably right there. But the point is that concluding that d should probably be awarded the winner is the same thing as incorporating utility into the calculation of the vote.

I tried to avoid linking d to the utilities.

Post by Curt
For those votes based off of the majority criterion or “one person one vote”, that’s improper. It’s not a flaw with Condorcet or Smith Set;

I don't think Smith Set is flawed, but I question the idea that it would be a natural extension of the majority rule to require that the winner should always be picked from the Smith Set (in majority oriented elections).

Post by Curt
it’s just applying the argument that utility should matter - which is a different subject. In other words, advocating Condorcet is basically accepting the axiom that 50:49 should win even if 49 has more passion, and, similarly, that someone from the ABC 66:33 Smith Set should win over d since they all defeat d 50:49. You can always disagree with an axiom, but by doing so you are entering into the conversation of whether utility should matter in Condorcet voting.

There are some sensible Condorcet methods that can use also utility based information, but in this discussion I try to live under the basic assumption that Condorcet methods are pure ranked methods. That includes reasoning about d.

Post by Curt
(That said, even for Condorcet voting, I am curious about capturing score preferences on the ballots, and then using them only to select from a multi-candidate Smith Set.)

Maybe James Green-Armytage's cardinal-weighted pairwise method would be of interest to you.

BR, Juho

Post by Curt
Curt

Post by Kristofer Munsterhjelm
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, [
]

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[Curt]

Hi Juho, you question whether it is a “natural extension of the majority rule to require that the winner should always be picked from the Smith Set in majority oriented elections”, and it appears your argument depends on the concept of a “strong” cycle.

My view is that we cannot assume a 66:33 Smith Set is a “strong” cycle. And that the *only* way to see 66:33 as a “strong” cycle is to imbue it with utility or strength-of-preference data.

Because rankings are ordinal, whether the margin is 66:33, 98:1, or 50:49, we cannot infer anything about “strength of victory” in those margins. All we know is that the one with more votes is majority preferred over the other. We have zero conception of “how much” they are preferred by. Ordinal rankings cannot be used to simulate utility or strength of preference. 98:1 could be a whisker, and 50:49 could be a gulf.

Since ABC is preferred to d by 50:49, ABC being the winner makes sense to me because it is the winner by definition. It is axiomatic (when we are talking about Condorcet-style elections).

I respect the view that utility would be helpful data in an example such as the 66:33 example, but I also see it as intrinsically related to the view that utility data would be helpful in a two-candidate 50:49 election.

To illustrate how we cannot rely on 66:33 to indicate any information supporting d as a winner, here are two examples of the 66:33 scenario with utility data added in:

Here, any of A, B, or C are clearly preferred to d, and ABC are clearly tied in utility despite the 66:33 margins:

17: A:100, B:99, d:2, C:1
16: A:100, d:3, B:2, C:1
17: B:100, C:99, d:2, A:1
16: B:100, d:3, C:2, A:1
17: C:100, A:99, d:2, B:1
16: C:100, d:3, A:2, B:1

A: 1700 + 1600 + 17 + 16 + 1683 + 32 = 5048
B: 1684 + 32 + 1700 + 1600 + 17 + 16 = 5048
C: 17 + 16 + 1683 + 32 + 1700 + 1600 = 5048
d: 34 + 48 + 34 + 48 + 34 + 48 = 246

Here, d is clearly preferred to A, B, and C, but ABC are still tied:

17: A:4, B:3, d:2, C:1
16: A:100, d:99 B:2, C:1
17: B:4, C:3, d:2, A:1
16: B :100, d:99, C:2, A:1
17: C:4, A:3, d:2, B:1
16: C:100, d:99, A:2, B:1

A: 1600 + 16 + 32 + 68 + 17 + 51 = 1784
B: 1600 + 16 + 32 + 51 + 68 + 17 = 1784
C: 1600 + 16 + 32 + 17 + 51 + 68 = 1784
d: 4752 + 34 + 34 + 34 = 4854

Curt

Post by Juho Laatu

Post by Curt
Hi Juho, if you extrapolate Kristofer’s 51-48-2 example (quoted below) to your 66:33 example, the point is that *if* the objective is to only use ordinal rankings and not incorporate utility, there really isn’t any reason to consider a 66:33 cycle a “strong” cycle, or 50:49 a “weak” majority. This is because the concepts of “strong” and “weak” suggest intensity of preference.

I think the difference is between measuring the number of voters vs measuring the strength of opinion of those voters. The former we can do (quite reliably). The latter the pure ranked methods do not measure (one reason being that one could not do that in a reliable way anyway). Numbers 66, 33, 50 and 49 can be picked from the pairwise matrix. Numbers 51, 48 and 2 can be derived from the ballots (but probably we are more interested in the matrix that those ballots will produce).

Post by Curt
Without any supporting utility data, we can’t infer anything about the public’s willingness to accept “d” over any of A, B, and C. Considering “d” to be a reasonable winner appears to be based on more than pairwise comparisons. It appears to also be based on ascribing probable utility to the voters given the vote margins.

The rankings / pairwise preferences are available. Utilities are not known and we can ignore them. My intention was to present d as a potential winner based on the rankings only.

Post by Curt
It’s true that these might be entirely sane assumptions given the numbers. 66:33 *probably* does correlate in some fashion to intensity of preference or utility. You’re probably right there. But the point is that concluding that d should probably be awarded the winner is the same thing as incorporating utility into the calculation of the vote.

I tried to avoid linking d to the utilities.

Post by Curt
For those votes based off of the majority criterion or “one person one vote”, that’s improper. It’s not a flaw with Condorcet or Smith Set;

I don't think Smith Set is flawed, but I question the idea that it would be a natural extension of the majority rule to require that the winner should always be picked from the Smith Set (in majority oriented elections).

Post by Curt
it’s just applying the argument that utility should matter - which is a different subject. In other words, advocating Condorcet is basically accepting the axiom that 50:49 should win even if 49 has more passion, and, similarly, that someone from the ABC 66:33 Smith Set should win over d since they all defeat d 50:49. You can always disagree with an axiom, but by doing so you are entering into the conversation of whether utility should matter in Condorcet voting.

There are some sensible Condorcet methods that can use also utility based information, but in this discussion I try to live under the basic assumption that Condorcet methods are pure ranked methods. That includes reasoning about d.

Post by Curt
(That said, even for Condorcet voting, I am curious about capturing score preferences on the ballots, and then using them only to select from a multi-candidate Smith Set.)

Maybe James Green-Armytage's cardinal-weighted pairwise method would be of interest to you.
BR, Juho

Post by Curt
Curt

Post by Kristofer Munsterhjelm
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, [
]

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[Juho Laatu]

Post by Curt
Hi Juho, you question whether it is a “natural extension of the majority rule to require that the winner should always be picked from the Smith Set in majority oriented elections”, and it appears your argument depends on the concept of a “strong” cycle.

Yes. A strong cycle among all the Smith Set members makes the non Smith Set candidates more appealing.

Post by Curt
My view is that we cannot assume a 66:33 Smith Set is a “strong” cycle. And that the *only* way to see 66:33 as a “strong” cycle is to imbue it with utility or strength-of-preference data.

In terms of plurality (number of voters) It is hard to make strong cycles any stronger than that. And the 50:49 preferences are as weak as you can get with 99 voters.

If you observe the strength of comparisons from utility point of view, that could make those 66:33 preferences either stronger or weaker. But my argumentation (in favour of considering the non Smith Set members as potential winners) is based on plurality only, not on utilities.

Post by Curt
Because rankings are ordinal, whether the margin is 66:33, 98:1, or 50:49, we cannot infer anything about “strength of victory” in those margins.

Talking still about utilities here. There can be many kind of utility data, but if we assume that the voters have a finite range of integer values to give to the candidates, then the difference of sums of voter utilities in 66 vs 33 could be higher than in the sums of 50 vs 49.

(In some sense pure ordinal methods make the one-man-one-vote assumption and behave as if utilities correspond directly to the plurality difference (of some other function of the pluralities).)

Post by Curt
All we know is that the one with more votes is majority preferred over the other. We have zero conception of “how much” they are preferred by. Ordinal rankings cannot be used to simulate utility or strength of preference. 98:1 could be a whisker, and 50:49 could be a gulf.
Since ABC is preferred to d by 50:49, ABC being the winner makes sense to me because it is the winner by definition. It is axiomatic (when we are talking about Condorcet-style elections).

If ABC was one candidate, then yes. ABC is however a group of candidates that also win each others (strong preferences) in the strong cycle.

Post by Curt
I respect the view that utility would be helpful data in an example such as the 66:33 example, but I also see it as intrinsically related to the view that utility data would be helpful in a two-candidate 50:49 election.

Utility data would be useful. Sincere utility data could be used in different methods e.g. to make a Condorcet Winner lose, or to make a majority winner lose. But pure ranked Condorcet methods are popular since it is difficult to collect sincere utility opinions.

Post by Curt
17: A:100, B:99, d:2, C:1
16: A:100, d:3, B:2, C:1
17: B:100, C:99, d:2, A:1
16: B:100, d:3, C:2, A:1
17: C:100, A:99, d:2, B:1
16: C:100, d:3, A:2, B:1
A: 1700 + 1600 + 17 + 16 + 1683 + 32 = 5048
B: 1684 + 32 + 1700 + 1600 + 17 + 16 = 5048
C: 17 + 16 + 1683 + 32 + 1700 + 1600 = 5048
d: 34 + 48 + 34 + 48 + 34 + 48 = 246

This one is about the same as one of the examples that I gave.

Post by Curt
17: A:4, B:3, d:2, C:1
16: A:100, d:99 B:2, C:1
17: B:4, C:3, d:2, A:1
16: B :100, d:99, C:2, A:1
17: C:4, A:3, d:2, B:1
16: C:100, d:99, A:2, B:1
A: 1600 + 16 + 32 + 68 + 17 + 51 = 1784
B: 1600 + 16 + 32 + 51 + 68 + 17 = 1784
C: 1600 + 16 + 32 + 17 + 51 + 68 = 1784
d: 4752 + 34 + 34 + 34 = 4854

Yes, works too. Includes "weak" votes that are not normalised to the 100...1 range.

My question is still "whether it is a natural extension of the majority rule to require that the winner should always be picked from the Smith Set, when observing the number of voters only”.

BR, Juho

Post by Curt
Curt

Post by Juho Laatu

Post by Curt
Hi Juho, if you extrapolate Kristofer’s 51-48-2 example (quoted below) to your 66:33 example, the point is that *if* the objective is to only use ordinal rankings and not incorporate utility, there really isn’t any reason to consider a 66:33 cycle a “strong” cycle, or 50:49 a “weak” majority. This is because the concepts of “strong” and “weak” suggest intensity of preference.

I think the difference is between measuring the number of voters vs measuring the strength of opinion of those voters. The former we can do (quite reliably). The latter the pure ranked methods do not measure (one reason being that one could not do that in a reliable way anyway). Numbers 66, 33, 50 and 49 can be picked from the pairwise matrix. Numbers 51, 48 and 2 can be derived from the ballots (but probably we are more interested in the matrix that those ballots will produce).

Post by Curt
Without any supporting utility data, we can’t infer anything about the public’s willingness to accept “d” over any of A, B, and C. Considering “d” to be a reasonable winner appears to be based on more than pairwise comparisons. It appears to also be based on ascribing probable utility to the voters given the vote margins.

The rankings / pairwise preferences are available. Utilities are not known and we can ignore them. My intention was to present d as a potential winner based on the rankings only.

Post by Curt
It’s true that these might be entirely sane assumptions given the numbers. 66:33 *probably* does correlate in some fashion to intensity of preference or utility. You’re probably right there. But the point is that concluding that d should probably be awarded the winner is the same thing as incorporating utility into the calculation of the vote.

I tried to avoid linking d to the utilities.

Post by Curt
For those votes based off of the majority criterion or “one person one vote”, that’s improper. It’s not a flaw with Condorcet or Smith Set;

I don't think Smith Set is flawed, but I question the idea that it would be a natural extension of the majority rule to require that the winner should always be picked from the Smith Set (in majority oriented elections).

Post by Curt
it’s just applying the argument that utility should matter - which is a different subject. In other words, advocating Condorcet is basically accepting the axiom that 50:49 should win even if 49 has more passion, and, similarly, that someone from the ABC 66:33 Smith Set should win over d since they all defeat d 50:49. You can always disagree with an axiom, but by doing so you are entering into the conversation of whether utility should matter in Condorcet voting.

There are some sensible Condorcet methods that can use also utility based information, but in this discussion I try to live under the basic assumption that Condorcet methods are pure ranked methods. That includes reasoning about d.

Post by Curt
(That said, even for Condorcet voting, I am curious about capturing score preferences on the ballots, and then using them only to select from a multi-candidate Smith Set.)

Maybe James Green-Armytage's cardinal-weighted pairwise method would be of interest to you.
BR, Juho

Post by Curt
Curt

Post by Kristofer Munsterhjelm
If we knew the utilities, then we could determine whether this is a
51: A: 100, B: 1, C: 0
48: B: 100, C: 1, A: 0
2: C: 100, B: 1, A: 0
election (A should win if we're counting utilities), or a
51: A: 100, B: 99, C: 0
48: B: 100, C: 99, A: 0
2: C: 100, B: 99, A: 0
election (B should win). But we don't, so we can't. Almost any situation where A is the superior candidate can be matched by a parallel situation where B is the superior candidate. Lacking utility information, we can't establish which is correct, [
]

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[Kristofer Munsterhjelm]

Post by Juho Laatu
Yes, works too. Includes "weak" votes that are not normalised to the 100...1 range.
My question is still "whether it is a natural extension of the majority
rule to require that the winner should always be picked from the Smith
Set, when observing the number of voters only”.

I just thought of a way to see if the Smith criterion is a reasonable
extension of majority; or rather, if it's a reasonable extension of
Condorcet.

You say that if ABC were one candidate, he should win. What if we make
ABC originate as one candidate? One way of doing so would be by cloning
an initial A into ABC. Then we know (if we value clone independence)
that in any situation where A wins against d in a majority election
(that's where majority comes in), then if we clone A into ABC, the
winner should be one of ABC.

And if we can turn any ABCd election (Smith set of size three, plus a
Condorcet loser) into one where ABC are clones, then we get that any
clone independent Condorcet method must be Smith in the four candidate
case, just by extending majority, through Condorcet, into Smith.

Are there any clone-independent methods that meet Condorcet but not
Smith? The only one I can think of is the "obvious" method that
collapses clones and then runs Minmax. Perhaps also BTR-IRV? I'm not
sure. But both of these require access to the ballots themselves, which
makes me suspect that you can't tell a clone scenario from a non-clone
scenario if all you have is the pairwise matrix.

If I'm right, then for any given (3-cycle + 1) Condorcet matrix, I could
use my old tool, linear programming, to determine a clone ballot set
that would produce that matrix. It would be something like:

- Because ABC is a clone set, the only permitted ballots are xD or Dx
where x is one of ABC ACB BAC BCA CAB CBA, i.e. ABCD ACBD BACD BCAD CABD
CBAD DABC DACB DBAC DBCA DCAB DCBA.

- We want the pairwise strengths to be as determined beforehand.

- Usual constraints: no negative number of ballots, no infinities.

In your ABCd case, we have:

17: A > B > d > C
16: A > d > B > C
17: B > C > d > A
16: B > d > C > A
17: C > A > d > B
16: C > d > A > B

and the pairwise matrix is (for row, against column)

A B C d
A -- 66 33 50
B 33 -- 66 50
C 66 33 -- 50
d 49 49 49 --

Programming the LP gives the following clone solution:

33: A>B>C>d
17: B>C>A>d
16: d>B>C>A
33: d>C>A>B

that produces the same pairwise matrix as above.

If we collapse the clone set, we get

50: A>d
49: d>A

so any clone-independent Condorcet method that only knows the pairwise
matrix must elect from the ABC set here, and since it can't distinguish
the LP solution (where ABC is a clone set) from the original ballot set,
there's no way for it to elect d in your example.

I haven't actually proven that clone independence + Condorcet + only
look at the pairwise matrix implies 4-candidate Smith; that would
require more linear algebra than I feel like using today. But it's not
that implausible; and if it's true, that means that whatever makes a
Condorcet loser deserve to win, if anything, must come from information
not provided by the pairwise matrix.
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[Juho Laatu]

Post by Juho Laatu
Yes, works too. Includes "weak" votes that are not normalised to the 100...1 range.
My question is still "whether it is a natural extension of the majority rule to require that the winner should always be picked from the Smith Set, when observing the number of voters only”.

I just thought of a way to see if the Smith criterion is a reasonable extension of majority; or rather, if it's a reasonable extension of Condorcet.
You say that if ABC were one candidate, he should win. What if we make ABC originate as one candidate? One way of doing so would be by cloning an initial A into ABC. Then we know (if we value clone independence) that in any situation where A wins against d in a majority election (that's where majority comes in), then if we clone A into ABC, the winner should be one of ABC.

m: A=B=C > d
n: d > A=B=C

With these ballots, if n < m, we can say that majority extensions should not support electing d.

And if we can turn any ABCd election (Smith set of size three, plus a Condorcet loser) into one where ABC are clones, then we get that any clone independent Condorcet method must be Smith in the four candidate case, just by extending majority, through Condorcet, into Smith.

Let's say we have matrix M that can be derived either from ballot set B1 or B2, where B1 contains no technical clones and in B2 A, B and C are technical clones. Being a technical clone means that those candidates are ranked together in all the ballots (the usual clone definition). It doesn't say that those candidates would be similar of that the voters would consider them to be closely related political alternatives (political clones) (they could be as well from three fiercely competing parties).

I note that clone criterion is a problematic criterion in the sense that it does not cover only political clones (does of course often miss also them), but sometimes also competing groups. In methods that are based on the matrix only it forces also ballots B1 to be treated as if A, B and C would be clones, although they can be very far from being political or technical clones. This means that especially in matrix based methods clone criterion can be an overkill.

m: mix(A, B, C) > d
n: d > mix(A, B, C)

Lets use margins as the default measure of preference strength (without losing generality). Expression mix(A, B, C) refers to a mixture of any preference orders between A, B and C. The strongest unavoidable defeat (sud) of that mix ((or any set of candidates)) is defined as the strongest defeat of the candidate whose strongest defeat is smallest (minmax style). The interesting ABCd cases (from the perspective on my question) could be the ones where sud(A, B, C) > m:n. I note also that if sud(A, B, C) > m:n, one can ask if A, B and C are really political clones, even in the case that they are technical clones in the ballots (B2).

Are there any clone-independent methods that meet Condorcet but not Smith? The only one I can think of is the "obvious" method that collapses clones and then runs Minmax. Perhaps also BTR-IRV? I'm not sure. But both of these require access to the ballots themselves, which
makes me suspect that you can't tell a clone scenario from a non-clone scenario if all you have is the pairwise matrix.

Yes, the pairwise matrix hides lots of ballot information, including having candidates next to each others in the ballots. (M vs. B1 and B2)

- Because ABC is a clone set, the only permitted ballots are xD or Dx where x is one of ABC ACB BAC BCA CAB CBA, i.e. ABCD ACBD BACD BCAD CABD CBAD DABC DACB DBAC DBCA DCAB DCBA.
- We want the pairwise strengths to be as determined beforehand.
- Usual constraints: no negative number of ballots, no infinities.
17: A > B > d > C
16: A > d > B > C
17: B > C > d > A
16: B > d > C > A
17: C > A > d > B
16: C > d > A > B
and the pairwise matrix is (for row, against column)
A B C d
A -- 66 33 50
B 33 -- 66 50
C 66 33 -- 50
d 49 49 49 --
33: A>B>C>d
17: B>C>A>d
16: d>B>C>A
33: d>C>A>B
that produces the same pairwise matrix as above.
If we collapse the clone set, we get
50: A>d
49: d>A
so any clone-independent Condorcet method that only knows the pairwise matrix must elect from the ABC set here, and since it can't distinguish the LP solution (where ABC is a clone set) from the original ballot set, there's no way for it to elect d in your example.
I haven't actually proven that clone independence + Condorcet + only look at the pairwise matrix implies 4-candidate Smith; that would require more linear algebra than I feel like using today.

Fine so far.

But it's not that implausible; and if it's true, that means that whatever makes a Condorcet loser deserve to win, if anything, must come from information not provided by the pairwise matrix.

Note that the problems between matrix and ballot information mainly emerge from the clone criterion. One could say that pure clone independence / existence of clone candidates can not be measured from the matrix only (without doing the "overkill"). The non Smith Set arguments are more neutral (e.g. minmax style arguments) with respect to using the matrix only vs also the ballots (matrix is enough). The "overkill" is the problem that forces d not to be elected also when there are no technical clones. (Smith Set criterion is close to the clone independence criterion.)

The arguments in favour of d can be valid also when we have technical clones (B2). The sud(A, B, C) > m:n argument/criterion may work also with B2 (since the cycle among the technical clones / possible political clones would still cause lots of dissatisfaction if one of the technical clones would be elected).

BR; Juho

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[Kristofer Munsterhjelm]

Post by Juho Laatu

But it's not that implausible; and if it's true, that means that
whatever makes a Condorcet loser deserve to win, if anything, must
come from information not provided by the pairwise matrix.

Note that the problems between matrix and ballot information mainly
emerge from the clone criterion. One could say that pure clone
independence / existence of clone candidates can not be measured from
the matrix only (without doing the "overkill"). The non Smith Set
arguments are more neutral (e.g. minmax style arguments) with respect to
using the matrix only vs also the ballots (matrix is enough). The
"overkill" is the problem that forces d not to be elected also when
there are no technical clones. (Smith Set criterion is close to the
clone independence criterion.)

It's not just clones. Since the example's collapsed ballots are of the form

m: A>d
n: d>A

with m>n, majority implies that A should be elected. In the uncollapsed
example, that means that the set {A, B, C} is first on a majority of the
ballots, so any method that passes mutual majority must elect from this set.

That's perhaps a stronger example of how a generalization of majority
forces one of {A, B, C} to be elected, since the point of mutual
majority (as I see it, at least) is that a majority can get the
candidate they want to be elected, elected, without having to coordinate
beforehand to rank the candidates in the same order.

It's hard to be opposed to such a property. I imagine it's easier to say
"okay, the pairwise matrix doesn't supply enough information" and
require that the method use more than just the pairwise matrix to decide
the winner.
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[Juho Laatu]

Post by Kristofer Munsterhjelm

Post by Juho Laatu

But it's not that implausible; and if it's true, that means that whatever makes a Condorcet loser deserve to win, if anything, must come from information not provided by the pairwise matrix.

Note that the problems between matrix and ballot information mainly
emerge from the clone criterion. One could say that pure clone
independence / existence of clone candidates can not be measured from
the matrix only (without doing the "overkill"). The non Smith Set
arguments are more neutral (e.g. minmax style arguments) with respect to
using the matrix only vs also the ballots (matrix is enough). The
"overkill" is the problem that forces d not to be elected also when
there are no technical clones. (Smith Set criterion is close to the
clone independence criterion.)

It's not just clones. Since the example's collapsed ballots are of the form
m: A>d
n: d>A
with m>n, majority implies that A should be elected. In the uncollapsed example, that means that the set {A, B, C} is first on a majority of the ballots, so any method that passes mutual majority must elect from this set.
That's perhaps a stronger example of how a generalization of majority forces one of {A, B, C} to be elected, since the point of mutual majority (as I see it, at least) is that a majority can get the candidate they want to be elected, elected, without having to coordinate beforehand to rank the candidates in the same order.
It's hard to be opposed to such a property. I imagine it's easier to say "okay, the pairwise matrix doesn't supply enough information" and require that the method use more than just the pairwise matrix to decide the winner.

In the case that we have mutual majority the mutual majority candidates could nominate only one of them, and that candidate would win (if voter preferences would stay the same). But if they all run, then there may be also defeats among them, and those defeats could be considered worse than the defeats of some candidates outside that set. The alternative strategy is to rank A, B and C in the same order. If all ABC supporters implement this strategy (100%), they could break the ABC loop that other voters maybe generated.

Having the (theoretical) possibility to implement a working strategy should not be seen as a right to win without implementing that strategy. Implementing the first strategy could be difficult since voters may not understand if some of the competing candidates would withdraw. In the second strategy many voters might not follow the recommended strategy. In both cases it could be difficult to know if the strategic opportunity exists, and all three candidates to agree that all of them will benefit of this strategy. If d seems to have only minority in any case, then why let one of the other mutual majority candidates win for free (assuming that these three are from competing parties).

You already mentioned the problem that If one wants mutual majority without doing an overkill, then one must use the ballots instead of the matrix. In practice one would however probably use the matrix and a Smith Set compatible method (i.e. exactly measured mutual majority methods may not be very practical).

We know that some sincere winner or strategy related criteria will be violated in any case when preferences are circular. If one likes methods that may elect outside the Smith Set, then making a matrix based method compatible with mutual majority criterion probably violates the "outside the Smith Set target". If you want both and you can't, then you abandon one of them, or (maybe preferably) violate both of them but only lightly. (Just noting these facts.)

The mutual majority ballots could be as follows.

50: circular_mix_of_A_B_C > d > e > f
49: d > e > f > circular_mix_of_A_B_C

The idea is just to show that d could still be quite popular despite of the mutual majority. If one thinks that sometimes candidates outside the Smith Set could win, this might be one of those cases (despite of the mutual majority). Note that candidates d, e and f are not very far from having mutual majority, and among them it is clear that d is the best.

In this example the majority related key facts (of sincere opinions) are. 1) d loses marginally to A, B and C, 2) ABC is marginally better in mutual majority measurements than def, 3) A, B and C lose quite a lot, each to one of the others. Is the marginal mutual majority result more important than the strong (majority) defeats of A, B and C?

BR, Juho

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[Curt]

Post by Juho Laatu
If you observe the strength of comparisons from utility point of view, that could make those 66:33 preferences either stronger or weaker. But my argumentation (in favour of considering the non Smith Set members as potential winners) is based on plurality only, not on utilities.

I view a 66:33 Smith Cycle as having *no* additional useful information over a 60:40 Smith Cycle or a 50:49 Smith Cycle. The only way to see a 66:33 Smith Cycle as being “stronger” than the other cycles is to imagine that it corresponds to a strength of preference or depth of division in the electorate, when there is no data to support that.

Post by Juho Laatu
(In some sense pure ordinal methods make the one-man-one-vote assumption and behave as if utilities correspond directly to the plurality difference (of some other function of the pluralities).)

I would argue that they actually don’t behave that way, and should not be looked at this way. It assumes probability, from an imagined distribution that is not that predictable in reality. Electorate preference is not evenly distributed. It clusters all over the place in ways that are not predictable. There is nothing about a 66:33 cycle that makes “d” a more appropriate winner than a 60:40 or 50:49 cycle. To argue otherwise is to imagine a correspondence with utility.

Number of voters in a margin is meaningless to Condorcet methods. All that matters is whether a majority is reached. Beyond that, it is impossible to characterize that majority.

Incidentally, this does point to what I believe *is* a major downside of a Condorcet method - it’s by definition entirely unsuitable for figuring proportional representation, because that again would imbuing vote margins with utility concepts.

[Juho Laatu]

Condorcet methods can work also without measuring preference strengths. All fine as long as we have a Condorcet Winner (this is the most common outcome). After that, finding the correct winner becomes difficult. Maybe ties and lotteries are ok then.

BR, Juho

Post by Curt

Post by Juho Laatu
If you observe the strength of comparisons from utility point of view, that could make those 66:33 preferences either stronger or weaker. But my argumentation (in favour of considering the non Smith Set members as potential winners) is based on plurality only, not on utilities.

I view a 66:33 Smith Cycle as having *no* additional useful information over a 60:40 Smith Cycle or a 50:49 Smith Cycle. The only way to see a 66:33 Smith Cycle as being “stronger” than the other cycles is to imagine that it corresponds to a strength of preference or depth of division in the electorate, when there is no data to support that.

Post by Juho Laatu
(In some sense pure ordinal methods make the one-man-one-vote assumption and behave as if utilities correspond directly to the plurality difference (of some other function of the pluralities).)

I would argue that they actually don’t behave that way, and should not be looked at this way. It assumes probability, from an imagined distribution that is not that predictable in reality. Electorate preference is not evenly distributed. It clusters all over the place in ways that are not predictable. There is nothing about a 66:33 cycle that makes “d” a more appropriate winner than a 60:40 or 50:49 cycle. To argue otherwise is to imagine a correspondence with utility.
Number of voters in a margin is meaningless to Condorcet methods. All that matters is whether a majority is reached. Beyond that, it is impossible to characterize that majority.
Incidentally, this does point to what I believe *is* a major downside of a Condorcet method - it’s by definition entirely unsuitable for figuring proportional representation, because that again would imbuing vote margins with utility concepts.
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[robert bristow-johnson]

---------------------------- Original Message ----------------------------

Subject: Re: [EM] smith/schwartz/landau

From: "Curt" <***@museworld.com>

Date: Wed, March 28, 2018 1:08 pm

To: "EM" <election-***@lists.electorama.com>

--------------------------------------------------------------------------

Post by Curt
Number of voters in a margin is meaningless to Condorcet methods. All that matters is whether a majority is reached. Beyond that, it is impossible to characterize that majority.

well, we have an ancillary concept in elections that we sometimes call "mandate".  seems to me
that margin has something to do with an election winner's mandate.

Post by Curt
Incidentally, this does point to what I believe *is* a major downside of a Condorcet method - it&rsquo;s by definition entirely unsuitable for figuring proportional representation, ...

are we discussing single winner
elections here or multi-winner elections?

Post by Curt
... because that again would imbuing vote margins with utility concepts.

if we accept the premise of One-Person-One -Vote, which assigns equal utility to each voter's preference, seems to me that margins are directly related to overall
utility for the voting population.

--



r b-j                         ***@audioimagination.com



"Imagination is more important than knowledge."

 
 
 
 

[Kristofer Munsterhjelm]

Post by Curt
Incidentally, this does point to what I believe *is* a major downside of
a Condorcet method - it’s by definition entirely unsuitable for figuring
proportional representation, because that again would imbuing vote
margins with utility concepts.

The Condorcet property (or even the majority property) is incompatible
with proportional representation. The usual example is something like:

51: X1>X2>X3>X4>X5>X6>X7
49: Y1>Y2>Y3>Y4>Y5>Y6>Y7

Suppose we want to elect four winners. The proportional outcome is to
have two Xes and two Ys, probably X1 X2 Y1 Y2. But applying the majority
criterion forces the election of X1, and after X1 is out of the picture,
it forces the election of X2, and so on until four X-es have been elected.

Majority and Condorcet simply are the wrong tools for the job of PR.
That said, there are methods that pass Condorcet when there's only one
candidate to elect, and pass the Droop proportionality criterion when
there's more than one to elect. One example is CPO-STV, and another is
Schulze STV.
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[VoteFair]

Another proportional method that uses a Condorcet method (Kemeny) as its
foundation is VoteFair ranking.

Specifically VoteFair representation ranking handles two winners per
district, and VoteFair partial proportional ranking handles the filling
of (extra) nationwide (or statewide) seats.

The "partial" aspect can be adjusted to become fully proportional, just
by adding more nationwide seats.

Richard Fobes

Post by Kristofer Munsterhjelm

Post by Curt
Incidentally, this does point to what I believe *is* a major downside
of a Condorcet method - it’s by definition entirely unsuitable for
figuring proportional representation, because that again would imbuing
vote margins with utility concepts.

The Condorcet property (or even the majority property) is incompatible
51: X1>X2>X3>X4>X5>X6>X7
49: Y1>Y2>Y3>Y4>Y5>Y6>Y7
Suppose we want to elect four winners. The proportional outcome is to
have two Xes and two Ys, probably X1 X2 Y1 Y2. But applying the majority
criterion forces the election of X1, and after X1 is out of the picture,
it forces the election of X2, and so on until four X-es have been elected.
Majority and Condorcet simply are the wrong tools for the job of PR.
That said, there are methods that pass Condorcet when there's only one
candidate to elect, and pass the Droop proportionality criterion when
there's more than one to elect. One example is CPO-STV, and another is
Schulze STV.
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[Kristofer Munsterhjelm]

Thanks to Kristofer for explaining my “beats” vs “beats or ties” confusion.
For anyone interested, here is the software package of me using scala to
compute Smith and Schwartz sets. It’s not super-advanced, but it at
least avoids mutable variables. In the future I may try to use more
expressive FP concepts, and pull in one of the faster Schwartz
algorithms. I don’t entirely understand the graph algorithms yet.
https://github.com/tunesmith/condorcet-counter
I opined a bit in the README but that’s not really the point of the
project. I just wanted an easy way to identify Smith and Schwartz sets
for myself.

As Kevin pointed out, the Schwartz set is actually the union of all
minimal such sets, because (unlike for Smith) that's not the same thing
as just one such minimal set.

At https://wiki.electorama.com/wiki/Beatpath_example_12 there is an
example where the Schwartz set has two components ({A,B,C,D} and {E}).
Could you test your software on it?

-km
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[Curt]

Hi Kristofer,

Thank you for the test case, I’ve added it locally.

I had deja vu on this example because I came across it once before and it looked wrong to me at first. Here, I input the ballots and I got {E} as the single-member Smith and Schwartz set.

But then I realized the example has an alternate ballot style where

1:A>B

is supposed to be interpreted as

1:A>B>C=D=E=F=G=H=I=J=K=L

My default ballot counter does not do that, it assumes no opinion over the rest. I now have a todo to supply a flag to support both ballot styles. :)

I adjusted the ballot input to manually specify the ties, and I got a Schwartz Set of {A, B, C, D, E}. (And then {F, G}, and then {H}, and then {I, J, K, L}).

It identifies the Smith Set as all candidates.

My Schwartz implementation is based off the Floyd-Warshall algorithm here: https://wiki.electorama.com/wiki/Maximal_elements_algorithms
My implementation:
https://github.com/tunesmith/condorcet-counter/blob/master/src/main/scala/com/keenworks/vote/condorcet/set/SchwartzService.scala

It looks like that pseudocode doesn’t try to identify components within the Schwartz Set. But looking at the tally I see how E is the Weak Condorcet Winner. Is there pseudocode that identifies these minimal sets separately?

I will have to update my mental definition of Schwartz Set and update y docs; I would have thought E would be the Schwartz Set, although I see looking at the tally how that isn’t true. But I wish my code to identify the WCW(s) from the Schwartz Set, just like it identifies any CW from the Smith Set.

Curt

Thanks to Kristofer for explaining my “beats” vs “beats or ties” confusion.
For anyone interested, here is the software package of me using scala to compute Smith and Schwartz sets. It’s not super-advanced, but it at least avoids mutable variables. In the future I may try to use more expressive FP concepts, and pull in one of the faster Schwartz algorithms. I don’t entirely understand the graph algorithms yet.
https://github.com/tunesmith/condorcet-counter
I opined a bit in the README but that’s not really the point of the project. I just wanted an easy way to identify Smith and Schwartz sets for myself.

As Kevin pointed out, the Schwartz set is actually the union of all minimal such sets, because (unlike for Smith) that's not the same thing as just one such minimal set.
At https://wiki.electorama.com/wiki/Beatpath_example_12 there is an example where the Schwartz set has two components ({A,B,C,D} and {E}). Could you test your software on it?
-km

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