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

[Ross Hyman]

Markus is right.

One way of retaining monotonicity, I think, is to replace the Sets with objects that record the number of times that a A has beaten B.

Then for the pair ordering
A>C, B>C
C>D
D>A, D>B, A>B

Affirming A>C and B> C
A(W):A(W)
B(W):B(W)
C(L):A(W),B(W),C(L)
D(W):D(W)

Affirming C>D
A(W):A(W)

B(W):B(W)

C(L):A(W),B(W),C(L)

D(L):A(W), B(W), C(L), D(L)

Affirming D>A, D>B, A>B
A(W):A(W) [A(W), B(W), C(L), D(L)]


B(W):B(W) [A(W), B(W), C(L), D(L)]A(W)


C(L):A(W),B(W),C(L)[A(W), B(W), C(L), D(L)][A(W), B(W), C(L), D(L)]


D(L):A(W), B(W), C(L), D(L)[A(W), B(W), C(L), D(L)][A(W), B(W), C(L), D(L)]
Now remove equal numbers of A from B and B from A.
A(W):A(W) [A(W), C(L), D(L)]



B(W):B(W) [A(W), B(W), C(L), D(L)]



C(L):A(W),B(W),C(L)[A(W), B(W), C(L), D(L)][A(W), B(W), C(L), D(L)]



D(L):A(W), B(W), C(L), D(L)[A(W), B(W), C(L), D(L)][A(W), B(W), C(L), D(L)]
B is reclassified a Loser
A(W):A(W) [A(W), C(L), D(L)]




B(L):B(L) [A(W), B(L), C(L), D(L)]




C(L):A(W),B(L),C(L)[A(W), B(L), C(L), D(L)][A(W), B(L), C(L), D(L)]




D(L):A(W), B(L), C(L), D(L)[A(W), B(L), C(L), D(L)][A(W), B(L), C(L), D(L)]
A wins







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