# [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 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.

Anders

```