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

[mrouse1 at mrouse.com]

Dave Ketchum wrote:
> DO NOT DO any switching such as you describe below.
> Even if it is far down in a voter's ranking, it is what this
> voter said about this pair. If this pair is far down in the
> list, there are many candidates this voter has ranked as
> better.

You're right, flipping the votes shouldn't be an option, because you could
make your preferred candidate do worse simply by voting (since you'd have
one more vote to flip). I therefore give you IFNOP (Ignore Fewest Number
of Preferences) Method.

(Note: someone may have already come up with this, I just gave it a name
for my own convenience.)

This is a work in progress, and it's definitely more complicated than it
needs to be. In addition, I know I've made some (perhaps many) mistakes,
but I'm coming off a caffeine/sugar high and I'm a bit tired to re-read
everything, so I will put this out there for comment. Feel free to rip it
apart -- I'm really just trying to get a feel for it, to see if something
interesting might come out of it.

BTW, if the formatting turns all weird, it's because of the annoying
webmail program I'm using.

IFNOP method

1. Each voter ranks as many candidates as desired.

2. If there is a Condorcet order (or a Condorcet winner in a single-winner
race), congratulations! No more work is necessary.

3. If there isn't a strict preference order (one or more circular ties),
create an NxN matrix, where N is the number of candidates. (Actually, only
half of the boxes would be needed -- the final result will be shaped
somewhat like a triangle.)

4. For every pair that is part of a circular tie, place the number and the
preference in the corresponding box (ties=0). For example, if 50 people
put A in the 7th position and B in the 12th position on their ballot, then
put (50 A>B) in box (7,12). Preferences from other voters can be in the
same box (B>C could be in (7,12)), and the same preference (A>B in this
example) can be in multiple boxes.

5. If a person ranks two or more candidates the same, they are assumed to
be in the closest available ranking to any candidate they are compared to.
For example, in the order A>B>C=D>E, C and D are both ranked 4th when
compared with E (you would have C>E and D>E in the box (4,5), and both are
ranked 3rd when compared with A and B (A>C, A>D in box (1,3), and B>C, B>D
in box (2,3) respectively). For this ballot, there is no score added for
C>D or D>C.

6. Ignore preferences one group at a time, starting with the lowest-ranked
ballots and working your way up. With this restriction, start with the
ballots closest together and work your way apart. The order is determined
when the circular tie is broken. (Basically, try ignoring one group, and
if it doesn't break the tie stop ignoring it and try the next group. If
none of the groups in the box are sufficient to break the tie, add the
next box and try again.)

Here is how it would work out. Let's say you have six candidates, with 1
as the top candidate and 6 as the bottom candidate. You would process the
boxes in the following order:

(5,6)
(4,5)
(4,6)
(3,4)
(3,5)
(3,6)
(2,3)...

More generally, let's say you have a total of X candidates. Setting 1 as
the top candidate and X as the bottom candidate, you would check the boxes
in the following order:

(X-1,X)
(X-2,X-1)
(X-2,X)
(X-3,X-2)
(X-3,X-1)
(X-3,X)
(X-4,X-3)
...
(1,X-1)
(1,X)

Beginning with the first box on the list ((X-1,X) in the example above),
try ignoring each group of preferences inside and see if the circular tie
is broken. If there is a unique way to resolve a particular tie, then this
determines the final order.

If there are several ways to resolve a tie (say if both A>B and B>C were
in the current box, and ignoring either one broke the circular tie
A>B>C>A) pick the order that requires the smallest number and lowest level
"ignores" necessary. (There are multiple ways to do this, but it's not
really necessary for this brief description.) If it's a true tie, pick one
order randomly.

Continue up the list until all the ties are resolved. If it's a
single-winner race, you can stop after the top circular tie is broken.

The thought behind this method is, I prefer (by definition) my
highest-ranked candidates more than my lowest-ranked candidates. If there
is a preference that needs to be removed to break a tie, I prefer it to be
at the bottom of my list -- better to ignore the difference between
candidate #99 and #100 than between candidates #1 and #2, even though both
are a single position apart. In addition, it is more important to keep
preferences that are far apart over those that are close together -- I can
tolerate a change between #1 and #2 more than I can a change between #1
and #100.

The method above *does* put more weight between #1 and #2 than between #2
and #100, but (especially in a single-winner race), maximizing my
highest-rated candidate is the most important to me -- plus, it's
impossible to pick a cutoff or scaling function that would work for every
voter in every election.

Of course, this method fails IIA, and is open to compromising. On the
other hand, bullet voting would be ineffective -- the top candidate vs.
every other one in the (1,2) box, rather than with a wider order -- and
burying would be most effective where it is least useful, in a race with
two strong candidates and a majority of weaker candidates.

To see why burying wouldn't be as effective as with some systems, assume
you have three strong candidates A,B,C.  If you prefer A, and you *think*
C is stronger than B, you can rank A...B,C. However, the last two ranks
are the very first ones we check for "ignorable" orders. If there is a
three-way tie between A>B>C>A, and if ignoring B>C breaks the tie, then C
would have to be the winner. If you gave both B and C the same ranking at
the bottom of the list, they are more likely to be ignored because of
section 5 above. The best strategy would be to bury the biggest threat and
raise the second-biggest threat as high as possible, to push their rank
comparison later in the list. This is more difficult than simply burying
all threats.

As I mentioned above, this is more brainstorming than anything -- I'm just
interested if any useful method might come out of it. Comments are
welcome, of course. I apologize for any confusing sections or errors, and
I *especially* apologize if someone else has already came up with it
(which happens with depressing frequency).

Michael Rouse
Mrouse1 at mrouse.com