[EM] More BTR-IRV

[Dave Ketchum]

Seems "Condorcet compliant" is not a warranty of value all by itself.

Looking at an easily findable example I see:
      19 voters
      6 rank A top - more than any other.
      A wins, per BTR-IRV - seems reasonable
      Add in 6 more voters ranking A top - total of 12 now
      C wins, per BTR-IRV.  HUH!!!  Look close and C is truly CW

Game is to discard candidates with the least votes on the ballots,  
comparing a pair at a time.  This will end at the CW if there is one.   
Order is based on how many top ranked each, which can be a different  
order - and the difference affecting who is last when there is no CW.

If we went for such as this we could expect the same kind of  
complaints as we write about IRV.

As to what I suggest, based on the N*N matrix, and, for shortening  
discussion, assuming no equal comparisons:
      For N-1 races, discard loser in each pair.  We thus find at  
least a cycle member.
      For N-1 races, see if this cycle member is loser to another.  If  
so we have two members of a cycle and can proceed to complete it; if  
not we have the CW.

Dave Ketchum

On Aug 14, 2010, at 9:41 PM, robert bristow-johnson wrote:
Re: [EM] it's been pretty quiet around here...
> On Aug 14, 2010, at 6:45 PM, Dave Ketchum wrote:
>> On Aug 14, 2010, at 2:18 PM, robert bristow-johnson wrote:
>>>
>>> the other method, BTR-IRV (which i had never thought of before  
>>> before Jameson mentioned it and Kristofer first explained to me  
>>> last May), is a Condorcet-compliant IRV method.  i wonder how well  
>>> or poorly it would work if no CW exists.  i am intrigued by this  
>>> method since it could still be sold to the IRV crowd (as an IRV  
>>> method) and not suffer the manifold consequences that occur when  
>>> IRV elects someone else than the CW.  does "BTR" stand for "bottom  
>>> two runoff"?  and who first suggested this method?  is it  
>>> published anywhere?  Jameson first mentioned it here, AFAIK.  the  
>>> advantage of this method is that is really is no more complicated  
>>> to explain than IRV, and it *does* resolve directly to a winner  
>>> whether a CW exists or not.  i am curious in how, say with a Smith  
>>> Set of 3, this method would differ from RP or Schulze.
>>
>> For Condorcet you have the N*N matrix and precinct summability but,  
>> unlike IRV, you better do nothing that involves going back to look  
>> at any ballots.
>
> i guess you're right.  i was just intrigued about this variant of  
> IRV that is Condorcet compliant.  but the actual method should be  
> precinct summable so that leaves BTR-IRV out.
> --
> r b-j                  rbj at audioimagination.com