Bug#727708: TC resolution revised draft

[Anthony Towns <aj@erisian.com.au>]

On 2 February 2014 04:05, Uoti Urpala <uoti.urpala@pp1.inet.fi> wrote:
> On Sat, 2014-02-01 at 17:10 +0000, Ian Jackson wrote:
>> Sébastien Villemot writes ("Bug#727708: TC resolution revised draft"):
>> > P1: DT > UT > DL > UL

> So his example was one where the D/U and L/T choices were independent
> for every voter,

Formally, there isn't a way to vote for an arbitrary partial order by
ranking options. ie, you can't vote for:

  DT > UT
  DL > UL
  UT > UL
  DT > DL

without inadvertently and insincerely expressing further preferences.

Err, yes you can:

  DT > UT = DL > UL

works fine. In which case the votes would be:

  P1: DT > UT = DL > UL
  P2: DL > UL = DT > UT
  P3: UT > DT = UL > DL
  P4: UT > DT = UL > DL

and the pairwise defeats are:

  DT > DL : 3 vs 1
  UT > UL : 3 vs 1
  DT > UL : 1 vs 0
  UT > DL : 2 vs 1

  UT = DT : 2 vs 2
  UL = DL : 2 vs 2

Transitive defeats are then just:

  DT t. defeats DL, UL
  UT t. defeats DL, UL

Schwartz set: { DT, UT }

There aren't any defeats in the schwartz set, so P1 uses a casting
vote to choose which of DT, UT is the winner, presumably DT.

Compare that to the corrected example's potential results:

  combined: UT wins
  D/U first: draw, tie break = D, T wins, so DT
  L/T first: T wins, draw, tie break = D, so DT

So I think you can put the difference down to either people expressing
insincere preferences, or that the additional, sincerely held,
preferences expressable via the combined vote having an effect on the
outcome, which shouldn't be surprising.

Cheers,
aj