# [EM] Non-monotonicity of Zavist-Tideman

Markus Schulze markus.schulze at alumni.tu-berlin.de
Fri Nov 21 15:58:03 PST 2003

```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

```