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

[Dave Ketchum]

Remember that it better work with multiple polling places, such as for 
governor.  Condorcet arrays have no problem with this, for the arrays are 
simply summed.

Remember that Condorcet and IRV promise, usually, to agree as to winner. 
There are two exceptions:
      When IRV does not completely scan ballots, what was missed could 
matter, and thus change the winner.
      Condorcet cycles have no equivalent in IRV.

Thus there should not be a third variety of counting, unless you can sell 
such as worth the headaches.

DWK

On Sun, 15 Oct 2006 19:31:42 -0700 (PDT) mrouse1 at mrouse.com wrote:
> 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
-- 
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  Dave Ketchum   108 Halstead Ave, Owego, NY  13827-1708   607-687-5026
            Do to no one what you would not want done to you.
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