[EM] Revised: Instant Pairwise Elimination (IPE)

[VoteFair]

On 1/22/2020 4:13 PM, Kristofer Munsterhjelm wrote:
 > So you should probably disregard what I've said about criterion
 > compliances.

OK. I removed
(at https://electowiki.org/wiki/Instant_Pairwise_Elimination)
these two criteria (which I have not found definitions for) because I 
don't know if IPE passes or fails them:

* Ranks equal

* Ranks greater than 2

Now the only criteria that Instant Pairwise Elimination (IPE) is 
indicated as passing are:

* Condorcet loser

* Resolvable

* Polytime

All other criteria are listed as "fail".

And I specified order N squared for summability (if that's a word).

 > So you can modify a pairwise matrix to the effect of eliminating a
 > candidate without needing more information than the pairwise matrix
 > itself -- as long as your criterion of who to eliminate only needs to
 > use the pairwise matrix.

The recent modification I made enables it to be calculated with just the 
pairwise matrix.

Interestingly IPE can be calculated "by hand" on paper just using the 
printed pairwise matrix and a calculator.  Each time a candidate is 
eliminated, the sums for each row and each column are updated by 
subtracting just one pairwise count in each row and each column.

Kristofer, thank you for your feedback!

As a reminder for anyone who is following this discussion, IPE is 
intended to have very low failure rates for the "fairness" criteria, 
even though there are some cases that cause it to fail almost all the 
criteria.  Its advantages will become clearer when failure rates are 
used instead of checklists.

Also keep in mind that the method is designed to be easy to understand 
for non-math folks.

And admittedly I named it in a way to help clarify that IRV is not the 
only "instant" runoff/elimination method.  :)

Richard Fobes



On 1/22/2020 4:13 PM, Kristofer Munsterhjelm wrote:
> On 16/01/2020 01.36, VoteFair wrote:
>> On 1/14/2020 2:34 PM, Kristofer Munsterhjelm wrote:
>>> The fallback sounds a lot like Raynaud(PO). If I'm right, then it should
>>> always pick the same winner as Raynaud(PO) does.
>>
>> Highlight from below: The fact that the Raynaud method passes the
>> Condorcet criterion suggested to me that the two methods could not be
>> the same.
>>
>> Apparently my description of the "fallback" method was not clear. So, I
>> added an example to the Electowiki article:
>>
>>   https://electowiki.org/wiki/Instant_Pairwise_Elimination#Example
>>
>> Notice that it looks at columns in the pairwise matrix and sums the
>> pairwise counts in each column -- and looks for the highest sum.
>>
>> If there is a tie, then it looks at rows in the pairwise matrix (and
>> sums each row and looks for the smallest sum).
>>
>> I believe -- but I could be wrong -- that Instant Pairwise Elimination
>> (IPE) fails the Condorcet criterion. (This would only happen when there
>> is a Condorcet cycle.)
>
> I see; I thought your fallback was to eliminate the candidate who's on
> the losing end of the strongest pairwise victory (by anyone over anyone
> else). The fallback as I thought it would have reduced IPE to
> Raynaud(PO), but the fallback you've given here does not.
>
> So you should probably disregard what I've said about criterion
> compliances. Or you could change the fallback to eliminate the candidate
> on the losing end of the strongest pairwise victory :-) It wouldn't take
> much to get you all the way to Condorcet.
>
> Though I would make one qualification to my previous sentence. IPE as
> you've defined it here is still O(n^2) summable. Consider what happens
> when you eliminate a candidate X:
>
> - Every pairwise contest between candidates not involving X stay the same.
> - Every pairwise contest including X is removed from the pairwise matrix.
>
> So you can modify a pairwise matrix to the effect of eliminating a
> candidate without needing more information than the pairwise matrix
> itself -- as long as your criterion of who to eliminate only needs to
> use the pairwise matrix.
>
> For instance, Schulze can be stated like this:
>
> 1. Find the smallest group of candidates who collectively beat or tie
> everybody else.
> 2. Eliminate every candidate not in that group. If only one candidate
> remains, he's the winner.
> 3. Replace the weakest pairwise defeat (between uneliminated candidates)
> with a tie.
> 4. Go to 1.
>
> And this can be done with only the pairwise matrix.
>