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a simple, great ISDA-compliant method I've never seen mentioned

FE
Filip Ejlak
Mon, Apr 17, 2023 4:19 PM

Hello, everyone, new to the list here,

what I really like about First preference Copeland is that it implies a
nice method for calculating a sort of "Condorcet score": just count 1st
preferences of all candidates (including candidate X) that do not beat X
pairwise. Although it is highly resistant to burial, FPC is not cloneproof
and fails ISDA.

However, there is an easy way to fix these two problems: just do sequential
loser elimination using FPC score!
(And when there is a tie at the bottom, break it by applying the same
method to the candidates involved in the tie and finding the loser.)

This method could be named FPCE (First Preference Copeland Elimination) or
ISCR (Instant Simmons-Copeland Runoff). It is cloneproof (because even if
there are clones which hurt one another, eventually there is only one of
them left) and ISDA-compliant (because all candidates outside the Smith set
get eliminated before any Smith set member is eliminated). I think its
burial resistance might be weakened, though, as it fails DMTBR.

Could anyone help with proving/disproving monotonicity, or even LIIA? I
applied the method to the 5-candidate Smith set from the Wikipedia article
about Schulze and there turned out to be an LIIA-compliant order: ADCEB.
I'm curious if it was by accident or not.

Filip Ejlak

Hello, everyone, new to the list here, what I really like about First preference Copeland is that it implies a nice method for calculating a sort of "Condorcet score": just count 1st preferences of all candidates (including candidate X) that do not beat X pairwise. Although it is highly resistant to burial, FPC is not cloneproof and fails ISDA. However, there is an easy way to fix these two problems: just do sequential loser elimination using FPC score! (And when there is a tie at the bottom, break it by applying the same method to the candidates involved in the tie and finding the loser.) This method could be named FPCE (First Preference Copeland Elimination) or ISCR (Instant Simmons-Copeland Runoff). It is cloneproof (because even if there are clones which hurt one another, eventually there is only one of them left) and ISDA-compliant (because all candidates outside the Smith set get eliminated before any Smith set member is eliminated). I think its burial resistance might be weakened, though, as it fails DMTBR. Could anyone help with proving/disproving monotonicity, or even LIIA? I applied the method to the 5-candidate Smith set from the Wikipedia article about Schulze and there turned out to be an LIIA-compliant order: ADCEB. I'm curious if it was by accident or not. Filip Ejlak
JF
James Faran
Mon, Apr 17, 2023 5:48 PM

It is my impression that all elimination methods are not monotonic, though I certainly can't prove that.

In this case, if I've understood it correctly, try

7 A>B>C
2 B>A>C
7 B>C>A
8 C>A>B

Pairwise we have a cycle: A>B>C>A, so FPC scores are A:9; B:8; C:7. Thus, the method eliminates C and A beats B and A wins.

Now suppose the two B>A>C voters increase their appreciation of A and vote A>B>C instead.

9 A>B>C
7 B>C>A
8 C>A>B

We've got the same pairwise cycle, but now A gets eliminated first and B wins.

Did I calculate correctly?  Is this the type of monotonicity you were asking about?

Jim Faran


From: Election-Methods election-methods-bounces@lists.electorama.com on behalf of Filip Ejlak tersander@gmail.com
Sent: Monday, April 17, 2023 12:19 PM
To: election-methods@lists.electorama.com election-methods@lists.electorama.com
Subject: [EM] a simple, great ISDA-compliant method I've never seen mentioned

Hello, everyone, new to the list here,

what I really like about First preference Copeland is that it implies a nice method for calculating a sort of "Condorcet score": just count 1st preferences of all candidates (including candidate X) that do not beat X pairwise. Although it is highly resistant to burial, FPC is not cloneproof and fails ISDA.

However, there is an easy way to fix these two problems: just do sequential loser elimination using FPC score!
(And when there is a tie at the bottom, break it by applying the same method to the candidates involved in the tie and finding the loser.)

This method could be named FPCE (First Preference Copeland Elimination) or ISCR (Instant Simmons-Copeland Runoff). It is cloneproof (because even if there are clones which hurt one another, eventually there is only one of them left) and ISDA-compliant (because all candidates outside the Smith set get eliminated before any Smith set member is eliminated). I think its burial resistance might be weakened, though, as it fails DMTBR.

Could anyone help with proving/disproving monotonicity, or even LIIA? I applied the method to the 5-candidate Smith set from the Wikipedia article about Schulze and there turned out to be an LIIA-compliant order: ADCEB. I'm curious if it was by accident or not.

Filip Ejlak

It is my impression that all elimination methods are not monotonic, though I certainly can't prove that. In this case, if I've understood it correctly, try 7 A>B>C 2 B>A>C 7 B>C>A 8 C>A>B Pairwise we have a cycle: A>B>C>A, so FPC scores are A:9; B:8; C:7. Thus, the method eliminates C and A beats B and A wins. Now suppose the two B>A>C voters increase their appreciation of A and vote A>B>C instead. 9 A>B>C 7 B>C>A 8 C>A>B We've got the same pairwise cycle, but now A gets eliminated first and B wins. Did I calculate correctly? Is this the type of monotonicity you were asking about? Jim Faran ________________________________ From: Election-Methods <election-methods-bounces@lists.electorama.com> on behalf of Filip Ejlak <tersander@gmail.com> Sent: Monday, April 17, 2023 12:19 PM To: election-methods@lists.electorama.com <election-methods@lists.electorama.com> Subject: [EM] a simple, great ISDA-compliant method I've never seen mentioned Hello, everyone, new to the list here, what I really like about First preference Copeland is that it implies a nice method for calculating a sort of "Condorcet score": just count 1st preferences of all candidates (including candidate X) that do not beat X pairwise. Although it is highly resistant to burial, FPC is not cloneproof and fails ISDA. However, there is an easy way to fix these two problems: just do sequential loser elimination using FPC score! (And when there is a tie at the bottom, break it by applying the same method to the candidates involved in the tie and finding the loser.) This method could be named FPCE (First Preference Copeland Elimination) or ISCR (Instant Simmons-Copeland Runoff). It is cloneproof (because even if there are clones which hurt one another, eventually there is only one of them left) and ISDA-compliant (because all candidates outside the Smith set get eliminated before any Smith set member is eliminated). I think its burial resistance might be weakened, though, as it fails DMTBR. Could anyone help with proving/disproving monotonicity, or even LIIA? I applied the method to the 5-candidate Smith set from the Wikipedia article about Schulze and there turned out to be an LIIA-compliant order: ADCEB. I'm curious if it was by accident or not. Filip Ejlak
KM
Kristofer Munsterhjelm
Mon, Apr 17, 2023 9:00 PM

On 4/17/23 19:48, James Faran wrote:

It is my impression that all elimination methods are not monotonic,
though I certainly can't prove that.

Just a nitpick: if the base method passes LIIA, then the elimination
method is equal to the base method and is monotone if the base method is.

But I suspect that every elimination method based on a positional
weighted method is nonmonotone, and it shouldn't be too hard to prove.
I'm not sure if every non-LIIA elimination method is nonmonotone.
(Possible proof strategy: suppose the base method fails LIIA, passes
majority, and also has the property that raising A may change the social
order from A>X>Y to A>Y>X; then if A, X, and Y are in a Condorcet cycle
(A>X>Y>A) and the base method passes majority, raising A would lead to X
being eliminated first, after which Y beats A pairwise and wins. Now
"simply" prove that strong monotonicity - that raising A should never
disturb X vs Y in the social outcome - is hard if not impossible to
pass, so that every universal domain method usually considered leads to
a nonmonotone elimination method.)

-km

On 4/17/23 19:48, James Faran wrote: > It is my impression that all elimination methods are not monotonic, > though I certainly can't prove that. Just a nitpick: if the base method passes LIIA, then the elimination method is equal to the base method and is monotone if the base method is. But I suspect that every elimination method based on a positional weighted method is nonmonotone, and it shouldn't be too hard to prove. I'm not sure if every non-LIIA elimination method is nonmonotone. (Possible proof strategy: suppose the base method fails LIIA, passes majority, and also has the property that raising A may change the social order from A>X>Y to A>Y>X; then if A, X, and Y are in a Condorcet cycle (A>X>Y>A) and the base method passes majority, raising A would lead to X being eliminated first, after which Y beats A pairwise and wins. Now "simply" prove that strong monotonicity - that raising A should never disturb X vs Y in the social outcome - is hard if not impossible to pass, so that every universal domain method usually considered leads to a nonmonotone elimination method.) -km