Condorect sub-cycle rule
djmarsay at dra.hmg.gb
Mon Sep 29 08:06:29 PDT 1997
Thank you for your responses. You inspired me to look through
the archives more thoroughly. I still don't understand the sub-cycle
rule, though. First, do you have a reference to Tideman, MIIAC and
I wont attempt the original example. Here's a simpler one.
Consider options A,B,C,D, where A>B>>C>>>A, D>B>>C>>>D, and A~D,
where more >s indicates a larger majority. Then ABCD is a cycle with
symmetric sub-cycles ABC, DBC. Mike's sub-cycle rule that you quote
allows to to pick either sub-cycle arbitrarily. You claim that this
makes no difference.
The way I see it, if I start with the sub-cycle ABC I have
A>B>>C>>>A, for which A>B is the weakest link, and the
Condorcet-winner is B. Thus I delete A and C, leaving:
D>B, so D is the overall winner.
Now I repeat, but with the other sub-cycle. By symmetry, A is now the
The only way around this that I can see is to look at the whole
I understand from a DEMOREP1 posting of 26 July 1996 that Young
cites Condorcet as considering all cycles together. This gives B.
However, if one considers all cycles together, then a strong
sub-cycle A>>>B>>>C>>A will effectively destroy all information of a
lesser strength. This seems undesirable.
It seems to me that one needs to consider all sub-cycles of the same
cycle together. This notion is given a meaning as follows:
Note that if a relation has cycles, then there exists a unique set of
basis cycles. These are disjoint connected cycles, such that every
cycle lies within a basis cycle.
(To construct, choose any cycle and put it into the set. For each
remaining cycle, r, in turn, for each cycle s in the set that it
intersects, add s to r, and put r in the set).
Now, where the basic Condorcet procedure gives cycles, form the basis
cycles. These correspond to Condorcet's "impossible opinions". For
each basis cycle, find the least threshold T such that discounting
all members of the cycle with Condorcet-plurality of less than T
removes the cycles (creates possible opinions).
This sub-cycle rule is now rather like the one I
Sorry folks, but apparently I have to do this. :-(
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