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

[Juho]

Maybe there is some potential in doing the IRV style "never  
considering all the given opinions" in some better way. I don't have  
any opinion yet on if this is that case but maybe something can be  
found.

Juho Laatu



On Oct 17, 2006, at 10:47 , Dave Ketchum wrote:

> 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
> -- 
>   davek at clarityconnect.com    people.clarityconnect.com/webpages3/ 
> davek
>   Dave Ketchum   108 Halstead Ave, Owego, NY  13827-1708    
> 607-687-5026
>             Do to no one what you would not want done to you.
>                   If you want peace, work for justice.
>
>
> ----
> election-methods mailing list - see http://electorama.com/em for  
> list info

Send instant messages to your online friends http://uk.messenger.yahoo.com