Hello,

I'm developing a calculator webpage for LNHarm methods so I've been trying to
think of things I may have missed.

#1: MDDTR (the method where you do majority defeat disqualification if possible,
and then elect the remaining candidate who received the most first preferences)
satisfies LNHarm. This isn't that hard to see, but I never noticed it for some
reason.

It does fail Plurality, though, like MDDA or MMPO do. It isn't quite as bad as
MMPO, but is not strictly better:

8: A
6: D>B>C
5: C
4: B>C

MMPO elects B, which is allowed, while MDDTR elects D, which is not.

#2: There is a way of resolving truncation under Borda which satisfies LNHarm. Just
award a candidate a point (on a given ballot) for every candidate that they are
ranked equal to or above. Crucially, this means points are awarded for truncated
candidates.

This satisfies LNHarm because adding a new preference does nothing but reduce the
scores of the candidates left bottom-ranked. It also fails Plurality, as one might
expect from such a method:

9: A
8: B
7: C>D

C wins.

#3: I was trying to remember the details of my 2007 method SPST, "Strongest Pair
with Single Transfer." This was an attempt to mix solid coalitions and a pairwise
contest in one method.

After messing around with SPST, I discovered that I can simplify and somewhat
improve it by, actually, getting rid of the solid coalition idea. I call the new
method "RUE FPP" for "runner-up exception" FPP:

If both the second- and third-place candidates on first preferences possess a full
pairwise majority over the first-preference winner, then elect the pairwise winner
between the second- and third-place candidates. Else elect the first pref winner.

This (seemingly) can't be generalized further, for example to bring in the
fourth-place candidate. We also can't weaken the "full majority" requirement, or
the requirement that both runners-up must beat the FPP winner this way.

SPST is strictly a middle point between FPP and RUE: The SPST winner is never
unique among the three. So I'll discard it.

One might look at RUE and say that all this does is enforce mutual majority for the
3-candidate case. It's at least true that it does this better than IFPP.

It is a little better than this though. A mutual majority need not exist at all,
and if there is one, it doesn't need to be limited to two candidates:

49: A>C
18: B>E>D>C
17: C>B>D
16: F>D>B>C

Here you have B, C, and D with majorities over A. The (potential) mutual majority
fails to be a solid coalition due to rankings for candidates E and F.

RUE elects C, the CW. DSC in contrast can't see the majorities and elects A.

Notice also that the 49 A voters are empowered to choose between B and C. That's
something they can't do in TTR or IRV.

In terms of properties, it is difficult to argue for RUE over IRV or top-two
runoff, though. The main thing is just if you want the third-place candidate to be
able to win (in a 3-candidate race). (And fewer monotonicity issues, I suppose.)

Kevin

Binomial STV is a later-no-harm method. It is also monotonic. It is a both elective and exclusive rational count.

On 11 May 2022, at 2:26 am, Kevin Venzke stepjak@yahoo.fr wrote:

Hello,

I'm developing a calculator webpage for LNHarm methods so I've been trying to
think of things I may have missed.

#1: MDDTR (the method where you do majority defeat disqualification if possible,
and then elect the remaining candidate who received the most first preferences)
satisfies LNHarm. This isn't that hard to see, but I never noticed it for some
reason.

It does fail Plurality, though, like MDDA or MMPO do. It isn't quite as bad as
MMPO, but is not strictly better:

8: A
6: D>B>C
5: C
4: B>C

MMPO elects B, which is allowed, while MDDTR elects D, which is not.

#2: There is a way of resolving truncation under Borda which satisfies LNHarm. Just
award a candidate a point (on a given ballot) for every candidate that they are
ranked equal to or above. Crucially, this means points are awarded for truncated
candidates.

This satisfies LNHarm because adding a new preference does nothing but reduce the
scores of the candidates left bottom-ranked. It also fails Plurality, as one might
expect from such a method:

9: A
8: B
7: C>D

C wins.

#3: I was trying to remember the details of my 2007 method SPST, "Strongest Pair
with Single Transfer." This was an attempt to mix solid coalitions and a pairwise
contest in one method.

After messing around with SPST, I discovered that I can simplify and somewhat
improve it by, actually, getting rid of the solid coalition idea. I call the new
method "RUE FPP" for "runner-up exception" FPP:

If both the second- and third-place candidates on first preferences possess a full
pairwise majority over the first-preference winner, then elect the pairwise winner
between the second- and third-place candidates. Else elect the first pref winner.

This (seemingly) can't be generalized further, for example to bring in the
fourth-place candidate. We also can't weaken the "full majority" requirement, or
the requirement that both runners-up must beat the FPP winner this way.

SPST is strictly a middle point between FPP and RUE: The SPST winner is never
unique among the three. So I'll discard it.

One might look at RUE and say that all this does is enforce mutual majority for the
3-candidate case. It's at least true that it does this better than IFPP.

It is a little better than this though. A mutual majority need not exist at all,
and if there is one, it doesn't need to be limited to two candidates:

49: A>C
18: B>E>D>C
17: C>B>D
16: F>D>B>C

Here you have B, C, and D with majorities over A. The (potential) mutual majority
fails to be a solid coalition due to rankings for candidates E and F.

RUE elects C, the CW. DSC in contrast can't see the majorities and elects A.

Notice also that the 49 A voters are empowered to choose between B and C. That's
something they can't do in TTR or IRV.

In terms of properties, it is difficult to argue for RUE over IRV or top-two
runoff, though. The main thing is just if you want the third-place candidate to be
able to win (in a 3-candidate race). (And fewer monotonicity issues, I suppose.)

Kevin

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