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Re: Tideman's Tiebreaker  Markus Schulze
 Nov 22, 2003 06:00 PST 

Hallo,

in March 2003, Steve Eppley proposed to the Election Methods mailing
list a new tie-breaking strategy for Ranked Pairs. (Actually, as far
as I remember correctly, this tie-breaking strategy has already been
proposed in Sep. 2001 by Rob LeGrand to the Ranked Pairs mailing list.)

Suppose that "pos[i]" is the position of candidate i in the
"Tie-Breaking Ranking of the Candidates" (TBRC). Then "pos[i] < pos[j]"
means that "candidate i is ranked higher than candidate j in the TBRC".

Eppley's tie-breaking strategy says that when ij and mn each have the
same strength then ij should be considered first if and only if one of
the following conditions is met:

   1) pos[n] < pos[j].
   2) j = n and pos[i] < pos[m].

So when the TBRC is ABCDEFG then this tie-breaking strategy orders
defeats of equal strength as follows:

AG, BG, CG, DG, EG, FG, AF, BF, CF, DF, EF, GF, AE, BE,
CE, DE, FE, GE, AD, BD, CD, ED, FD, GD, AC, BC, DC, EC,
FC, GC, AB, CB, DB, EB, FB, GB, BA, CA, DA, EA, FA, GA.

A problem of Eppley's tie-breaking strategy is that it isn't reversal
symmetric. Therefore, I propose the following strategy. When ij and
mn each have the same strength then ij should be considered first if
and only if at least one of the following conditions is met:

   1) pos[i] < pos[j] and pos[n] < pos[m].
   2) pos[i] < min ( pos[j], pos[m], pos[n] ).
   3) pos[n] < min ( pos[i], pos[j], pos[m] ).
   4) i = m and pos[n] < pos[j].
   5) j = n and pos[i] < pos[m].

So when the TBRC is ABCDEFG then this tie-breaking strategy orders
defeats of equal strength as follows:

AG, AF, AE, AD, AC, AB, BG, BF, BE, BD, BC, CG, CF, CE,
CD, DG, DF, DE, EG, EF, FG, GF, FE, GE, ED, FD, GD, DC,
EC, FC, GC, CB, DB, EB, FB, GB, BA, CA, DA, EA, FA, GA.

Markus Schulze
	
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