[EM] Fixing IRV

[Richard Moore]

So in effect, if for a non-elimination method we can always 
find one candidate who can be eliminated without affecting 
the rankings of the other candidates, then we can construct 
an elimination method that is equivalent to the 
non-elimination method. But that is not really an 
elimination method in the sense of the theorem I was 
referring to. I did find the reference I was looking for,
in Lorrie Cranor's dissertation (

Monotonicity

             A voting system is monotonic if when a voter 
raises the valuation for a winning al-
ternative it remains a winning alternative, and when a voter 
lowers the valuation for a los-
ing alternative it remains a losing alternative. All voting 
systems that eliminate alternatives
prior to selecting a winner violate monotonicity [86].



Blake Cretney wrote:
> On Wed, 08 Aug 2001 17:36:27 -0700
> Richard Moore <rmoore4 at home.com> wrote:
> 
> 
>>Markus Schulze wrote:
>>
>>>Dear Richard,
>>>
>>>you wrote (7 Aug 2001):
>>>
>>>
>>>>Actually, IIRC, there is a proof somewhere in the literature
>>>>that elimination methods are not monotonic. Does anyone recall
>>>>the theorem I mentioned above about elimination and monotonicity?
>>>>
>>>>
>>>Some elimination methods are monotonic (e.g. Ranked Pairs).
>>>
>>Since when is RP considered an elimination method?
>>
> 
> RP gives you a complete ordering of the candidates.  This ordering
> gives you a lowest candidate.  So, you might suggest a method,
> RP-elimination, that finds the candidate when you successively
> eliminate the lowest RP-ranked candidate.  Who is that candidate? 
> Turns out, it's the same as the normal RP winner.  So, RP-elimination
> = RP.  Since RP is monotonic, RP-elimination must be as well.
> 
> 
>>If we have
>>
>>6 
>>ABC
>>5 
>>CAB
>>4 
>>BCA
>>
>>then RP gives the following rankings for the pairwise contests:
>>
>>11 
>>A>B
>>10 
>>B>C
>>9 
>>C>A
>>
>>with A as the winner. If RP were done as an elimination 
>>method, then B would be eliminated following the highest 
>>pairwise defeat. We then wouldn't bother comparing B and C, 
>>and we would get
>>
> 
> The complete RP ranking is A>B>C.  So, you first eliminate C.  This
> gives A vs. B.  A is the RP winner between them. 
> 
> Now, you might rightly state that although I could define RP as an
> elimination method, it would be ridiculous to do so.  Nevertheless,
> since I could, it follows that there can't be a proof that no
> elimination method is monotonic, since this isn't technically true.
> 
> ---
> Blake Cretney
> 
> 
>