4 choice circular tie
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Tue Jun 23 14:04:42 PDT 1998
Below is the 4 choice general circular tie case with each choice grouped last
(6 each) (assuming no truncated votes). N number= a number of votes
Perhaps it will enlighten as to how to get a majority consensus winner in all
cases (3 to M(any) choices in a circular tie).
The 3 choice general case with only 6 possible types of votes (assuming no
truncated votes) is not complex enough.
Assume A>B>C>D>A or some other type of circular tie. Each type of circular
tie means that various combinations of the N values are related (such as-- if
A > B, then the sum of N7 to N12, N13, N14, N17, N19, N20, N23 is greater than
the sum of N1 to N6, N15, N16, N18, N21, N22, N24) (i.e. the sum of 12 values
each)).
Thus, there is some major complexity if all 24 types of votes are used in
examples (which partially helps to explain why the single winner problem has
been around since the days of Condorcet and Borda 1770's-1780's). Throw in
the various criterion of various folks and the complexity becomes
overwhelming. Perhaps someone might post the various criterion that have
been thought about. I will only observe that the tiebreaker must deal with
how the voters actually vote whether or not they are sincere (noting that
polls will be taken before the votes are cast).
N1 BCDA
N2 BDCA
N3 CBDA
N4 CDBA
N5 DBCA
N6 DCBA
N7 ACDB
N8 ADCB
N9 CADB
N10 CDAB
N11 DACB
N12 DCAB
N13 ABDC
N14 ADBC
N15 BADC
N16 BDAC
N17 DABC
N18 DBAC
N19 ABCD
N20 ACBD
N21 BACD
N22 BCAD
N23 CABD
N24 CBAD
Anybody want to do the 5 choice general case (120 types of votes with 60 types
of votes where each choice beats each other choice) ?
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