Hallo, in March 2003, Steve Eppley proposed 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) n = j 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 ---- Election-methods mailing list - see http://electorama.com/em for list info
