Tiebreakers, Subcycle Rules

[Markus Schulze]

Dear participants,

usually, if a tiebreaker doesn't lead to a unique
winner but to a set of potential winners, the
tiebreaker is restarted among the potential
winners.

I have made the observation, that mostly the Schulze
method is decisive. But in those cases, in which the
Schulze method is not decisive, it is very often
not possible to get a further reduction of the set
of potential winners by restarting the tiebreaker
among the potential winners. In those cases, which I
had observed, a further reduction of the set of
potential winners was possible only if the
beat-path Schwartz set of the potential winners was
used to restart the tiebreaker. The aim of this e-mail
is to present an example where the Schulze method is
indecisive.

Example (C.G. Hoag and G.H. Hallett, "Proportional
Representation," page 502, 1926):

   3 voters vote A > B > C > D.
   2 voters vote D > A > B > C.
   2 voters vote D > B > C > A.
   2 voters vote C > B > D > A.

   The matrix of defeats looks as follows:

   A:B=5:4
   A:C=5:4
   A:D=3:6
   B:C=7:2
   B:D=5:4
   C:D=5:4

   As A > B > C > D > A, every candidate is in the
   Smith set.

   Via beat-paths, the matrix of defeats looks as
   follows:

   A:B=5:5 via beat-paths
   A:C=5:5 via beat-paths
   A:D=5:6 via beat-paths
   B:C=7:5 via beat-paths
   B:D=5:5 via beat-paths
   C:D=5:5 via beat-paths

   There is no candidate, who wins against every other
   candidate via beat-paths. Thus, there is no Schulze
   winner. The beat-path Smith set (i.e. the smallest
   set of candidates, such that every candidate in this
   set wins against every candidate outside this set
   via beat-paths) consists of all four candidates,
   because A=B=D=C=A.
   Thus: It is not possible to get a further reduction
   of the set of potential winners simply by restarting
   the tie-breaker among the potential winners.
   The only way to get a further reduction of the
   set of potential winners (without violating clone
   criteria, monotonicity or other desired criteria),
   that I have found, is to calculate the beat-path
   Schwartz set of the potential winners. In the example
   above, the beat-path Schwartz set consists of the
   candidates B and D. If the tiebreaker is restarted
   among the candidates B and D, candidate B wins the
   election, because candidate B wins against
   candidate D 5:4.

Markus Schulze