# 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
N10 CDAB
N11 DACB
N12 DCAB

N13 ABDC
N16 BDAC
N17 DABC
N18 DBAC

N19 ABCD
N20 ACBD
N21 BACD