[EM] Does MAM use the Copeland method?

[Steve Eppley]

Hi,
Gervase L asked:
> Just a quick question that should clear up my understanding 
> of MAM.  Is it the same as Copeland (i.e. count each
> candidate's number of wins) except that any pairwise wins
> that are inconsistent with the Rank Pairs ranking are
> dropped before the Copeland score is tallied up? 

No, that's not MAM. 

The question itself isn't very clear, since some people
use the name Ranked Pairs as a synonym for MAM, while 
others, including me, use the name Ranked Pairs to refer
only to Tideman's 1987 method and Zavist-Tideman's 1989
method.  Those two methods differ from MAM in several 
ways, the most important difference being that MAM 
measures the size of each majority primarily by the 
number of votes that rank the majority's more-preferred 
candidate over their less-preferred candidate, whereas
Tideman and Zavist-Tideman subtract from that the size
of the opposing minority (that is, the number of votes 
that rank the majority's less-preferred candidate over 
their more-preferred candidate).  Because of this, MAM 
satisfies some criteria that Ranked Pairs does not.

MAM constructs the order of finish of the candidates
a piece at a time, by considering the majorities one at 
a time from largest to smallest, adopting into the finish 
order each majority's preference that does not conflict 
(cycle) with the partial order of finish already 
constructed.  The winner is the candidate that tops 
the order of finish so constructed. (In a multiwinner 
context, the N winners are the N candidates that top 
the order of finish.)

Here's a simple example to illustrate:

   9 voters, 3 candidates

   The votes:
      4: A>B>C
      3: B>C>A
      2: C>A>B

   There are three majorities:
      7: B>C
      6: A>B
      5: C>A

   MAM first adopts the preference of the largest majority,
   for B over C.  Thus B finishes over C.  Then MAM adopts
   the preference of the second largest majority, for 
   A over B.  Thus A finishes over B.  Since MAM is
   constructing an ordering and A finishes over B and 
   B finishes over C, this implies A finishes over C.  
   So when MAM considers the preference of the third
   largest majority, for C over A, it is not adopted 
   since it has already been established that A finishes
   over C.  Thus the complete order of finish is A>B>C. 
   Since A tops the finish order, A is elected.

I've oversimplified a bit.  When the number of voters is
small, it's not implausible that two or more majorities 
will be the same size and it's not implausible that one 
or more pairings won't have a majority (in other words, 
a pairing may be a "tie.")  For more details about MAM, 
see my web pages at <www.alumni.caltech.edu/~seppley>.

--Steve