[EM] IFNOP Method (was Re: Question about Condorcet methods)

[mrouse1 at mrouse.com]

Quoting Raphfrk:

>> I wonder if this is sorta similar to a Borda count in terms of Clone
independence.
***

A clone doesn't appear to help in the case of a circular tie, so a party
wouldn't have an incentive to run similar candidates. A "spoiler" clone is
possible, as your example below points out, though it requires a circular
tie.
***

>> For example, if voters, A,B,C  vote
>> A: A>B>C1>C2
>> B:  B>C1>C2>A
>> C: C1>C2>A>B

>> A>B>C1>C2>A
...
>> The lowest ranked box is 4,4 and it is empty.  The next lowest is (3,4)
and (4,4) and only 3,4 has votes in it.

>> C2>A, C1>C2 and A>B have pairs in that box.  The only pair left is
B>C1. Thus, B wins?
***

Actually, C1>A and B>C2 are also left, but you are correct that B wins
with this method. By ignoring A>B, you can keep the other five pair
comparisons (B>C1, B>C2 C1>A, C2>A, and C1>C2), and come up with one
unique order, B>C1>C2>A.
***

>> I am not 100% sure if I have understood you method correctly.  However,
by  weighting lower preferences lower, the C "Party" has reduced the
weighting of  A>B in ballot C and C2>A in ballot B.  Without the first
one, the votes become:

>> A: A>B>C1>C2
>> B: B>C1>C2>A
>> C: C1>C2>A
***

In ballot C, does A=B or is A>B?
***

>> The effect seems to be opposite to Borda, cloning can result in your
least favorite  winning.  That would likely lead to 2 party domination
if cloning hurts a cause.
***
Adding a clone of C allows the voter for B to show how much he or she
dislikes A, which is where this method fails IIA. It does obey (I think)
Local IIA though, since any spoiler would have to be a member of the Smith
set. If two candidates are close enough to be clones, and if there were
two other very strong candidates, there would be an incentive for the two
clones to join forces in certain circumstances.
***

>> Is your method equavalent (or very similar) to the following method.

>> 1) Voters rank N candidates
>> 2) M=N
>> 3) Ballots truncated to M rankings
***
Pairwise comparison are truncated, but only the ones necessary to resolve
a tie. In your example above, by ignoring the bottom-ranked A>B, you
resolve the election to B>C1>C2>A. It's unnecessary to ignore C1>A and
C2>A, which are also in the same box.
***

>> 4) Condorcet winner is elected, if existing
>> 5) Otherwise, reduce M by 1 and goto 3)
***

4 is correct. If in 5 you mean move up one box (in a 4-candidate race,
this means going in the order (3,4)(2,3)(2,4)(1,2)(1,3)(1,4) ) you are
correct.
***

>> What about this instead

>> 1) Voters rank N candidates and include a range score
>> 2) M=N-1
>> 3) Ballots reduced to at most M clear preferences, least strong
preferences equalised first
>> 4) Condorcet winner is elected, if existing
>> 5) Otherwise, reduce M by 1 and goto 3)
***

I think Range-voting variants are interesting, though as votes become more
strategic they become more Approval-like (I like Approval a lot for its
simplicity and generally good behavior).
***

>> Ok, so I could fill in the following ballot:

>> A1:  1 (99)
>> A2:  2 (97)
>> B:  3 (90)
>> C1: 4 (0)
>> C2: 5 (1)

>> The voter  messed up the ballot for C1 and C2 and also didn't rank E1
or E2.

>> In effect, that preference will be equalised first due to the
contradiction (shown in bold below).

>> So, my ballot would effectively be
>> Round 1:  A1>A2>B>C1>C2>E1=E2 (from ballot)
>> Round 2:  A1>A2>B>C1>C2>E1=E2  (already has 1 equality so no change
required)
>> Round 3:  A1>A2>B>C1=C2>E1=E2  (-1 point difference between C1 and C2)
>> Round 4:  A1>A2>B>C1=C2=E1=E2  (1  point difference between  C2 and '0')
>> Round 5:  A1=A2>B>C1=C2=E1=E2 (2 point difference between  A1 and A2)
>> Round 6:  A1=A2=B>C1=C2=E1=E2 (10 point difference between A2 and B2)
>> Round 7:  A1=A2=B=C1=C2=E1=E2 (90 point difference between B and C1)
***

If I'm reading it right, Round 6 is equivalent to Approval voting. If you
can, could you show how this method would resolve a Condorcet cycle (you
can use a different example if there's an easier one).

I'm trying IFNOP with four examples from the Schulze method's page on
Wikipedia. I'll post them in the next couple of days.

Michael Rouse
mrouse1 at mrouse.com