[EM] MAM non-deterministic when tied losers make cycle?

[matt matt]

Regarding step 2. below, what if there is a three way tie like this:

winner loser     strength
D      C               20
B      D               20
C      B               17
A      C               17
A      D               17
B      A               14

There is a cycle among the three losers BCD: C>B, B>D, and D>C.    
Under this scenario your procedure appears to be non-deterministic.

Steve Eppley wrote: 

In the meantime, you can compare my tiebreak method with Zavist's. (My tiebreaker was inspired by Zavist's; I don't claim credit for an independent development there.  MAM was mostly an independent development, though.)  Here's how MAM sorts the majorities (after constructing a strict "tiebreaking" ordering of the candidates, call it T, using the Random Voter Hierarchy procedure):    

   For all pairs of candidates, for instance x and y, 
   let #xy denote the number of votes that rank x over y.

   For all pairs of majorities, for instance x>y and z>w, 
   x>y precedes z>w if and only if at least one of the 
   following conditions holds:

      1.  #xy > #zw.
      2.  #xy = #zw and #wz > #yx.
      3.  #xy = #zw and #wz = #yx and T ranks w over y.
      4.  #xy = #zw and #wz = #yx and w = y and T ranks x over z. 

Condition 1 makes MAM a "winning votes" method.  
Conditions 3 and 4 define the "tiebreaker" when two majorities
are the same size.  Since T is a strict ordering of the candidates, the definition above generates a strict ordering of the majorities.  (There's a subtle difference in how Zavist's tiebreaker works.)  




________________________________________________________________
Sent via the WebMail system at tidalwave.net