But is that the only monotonic clone independent method? The method I describe elects D instead of A in accordance with D>A. But I don't see why it would violate clone independence.
Consider the matrix in which the elements are the number of times the column candidate has defeated the row candidate. we begin with ABCD WA1000 WB0100 WC0010 WD0001 D>B WA1000 LB0101 WC0010 WD0001 C>B WA1000 LB0111 WC0010 WD0001 A>C WA1000 LB1111 LC1010 WD0001 B>A and C>D WA2110 LB3232 LC2121 WD0011 D>A LA2132 LB3265 LC2143 WD0011 D wins -- Hallo, in section 5 stage 3 of my paper, I explain how Tideman's ranked pairs method can be used (without having to sacrifice monotonicity, independence of clones, reversal symmetry, or any other important criterion) to resolve situations where the Schulze winner is not unique: http://m-schulze.webhop.net/schulze1.pdf Example: There are four candidates. The defeat D > B has strength 5. The defeat C > B has strength 4. The defeat A > C has strength 3. The defeat B > A has strength 2. The defeat C > D has strength 2. The defeat D > A has strength 1. The strongest path from A to B is ACB of strength 3. The strongest path from A to C is AC of strength 3. The strongest path from A to D is ACD of strength 2. The strongest path from B to A is BA of strength 2. The strongest path from B to C is BAC of strength 2. The strongest path from B to D is BACD of strength 2. The strongest path from C to A is CBA of strength 2. The strongest path from C to B is CB of strength 4. The strongest path from C to D is CD of strength 2. The strongest path from D to A is DBA of strength 2. The strongest path from D to B is DB of strength 5. The strongest path from D to C is DBAC of strength 2. A beats B via beatpaths 3:2. A beats C via beatpaths 3:2. C beats B via beatpaths 4:2. D beats B via beatpaths 5:2. So the Schulze method does not give a complete order, but only a partial order: A > B, A > C, C > B, D > B. This partial order can be completed to three complete orders: O1 = ACDB, O2 = ADCB, O3 = DACB. In section 5 stage 3 of my paper, I suggest that the complete order Ox is stronger than the complete order Oy if and only if the strongest pairwise defeat, that is in Ox and not in Oy, is stronger than the strongest pairwise defeat, that is in Oy and not in Ox. The strongest pairwise defeat, that is in O1 = ACDB and not in O2 = ADCB, is C > D of strength 2. The strongest pairwise defeat, that is in O2 = ADCB and not in O1 = ACDB, is D > C of strength -2. Therefore, the complete order O1 = ACDB is stronger than the complete order O2 = ADCB. The strongest pairwise defeat, that is in O1 = ACDB and not in O3 = DACB, is C > D of strength 2. The strongest pairwise defeat, that is in O3 = DACB and not in O1 = ACDB, is D > A of strength 1. Therefore, the complete order O1 = ACDB is stronger than the complete order O3 = DACB. Therefore, the final order is O1 = ACDB. Markus Schulze
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