[EM] ranked pair method that resolves beath path ties.

[Ross Hyman]

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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