[EM] MMV and resolvability

Anders Kaseorg andersk at MIT.EDU
Sat Dec 7 03:34:03 PST 2013

Maximum Majority Voting exhibits a silly failure of Tideman’s 
resolvability criterion in the following four-candidate election:
  3: A > B > C > D
  1: B > D > A > C
  2: C > D > B > A
  1: D > A > B > C

According to my reading of 
http://radicalcentrism.org/resources/maximum-majority-voting/, and also 
according to Eric Gorr’s calculator at http://condorcet.ericgorr.net/, MMV 
proceeds as follows.  The defeats are sorted:
  5/2: A > C, B > C, C > D
  4/3: A > B, B > D, D > A
In the first step, MMV affirms the three 5/2 defeats.  In the second step, 
it ignores all three 4/3 defeats because they form a cycle.  The final 
result is A = B > C > D.

However, there’s no way to add another ballot to make B the unique winner, 
in violation of resolvability.

This failure is silly because, if the three 4/3 defeats had been sorted in 
_any_ strict order, D > A would have been ignored for already forming a 
cycle with the 5/2 defeats, and the final result would unambiguously be A 
> B > C > D.

This feels like an oversight in the MMV definition, and although it’s 
highly unlikely to matter in practice, fixing it is simple enough.  At 
each step, when considering a set of equally strong defeats, we should 
immediately discard any defeats that would complete a cycle with strictly 
stronger affirmed defeats, so that they are not considered for use in 
cycles with equally strong defeats.


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