[EM] Circular tie solutions should be obvious
DEMOREP1 at aol.com
DEMOREP1 at aol.com
Tue Feb 15 18:33:39 PST 2000
Forwarded from Mike Ossipoff <nkklrp at hotmail.com>--
What's the obvious solution to circular ties?
Well, if the people say that X is better than Y, then we want
to honor that vote result. If more vote X over Y than vice-versa.
Of course, when there's a circular tie, we can't honor all of
the pair-defeats. Any candidate we elect, we'll be ignoring,
disregarding, overruling some voters who said that Y isn't
as good as X, when they made X beat Y pairwise. We don't do
that lightly. What should we do? Minimize how many voters we
ignore or overrule. We have to ignore or overrule one or
more pair-defeats; for the purposes of the count procedure,
I'll call that "dropping" that pair-defeat. We delete it from
our pair-defeat table & our arrow-diagram.
To minimize the number of voters we overrule when getting rid
of cycles, we want to drop weak defeats instead of stong ones.
(Weak defeats are defeats with few people voting the defeater
over the defeated).
That meets your obviousness standard. What's an obvious rule, then,
when there's a circular tie? How's this for obvious?:
"Get rid of cycles by dropping the weakest defeat in each cycle".
This extremely obvious & simple rule could be called:
"Drop Contradicted Defeats" ("DCD").
It can't get more obvious than that. If, as can happen with
that rule, 2 candidates become unbeaten, then the rule can be
re-applied to those 2 candidates, starting all over with the
original pair-defeats. So if 2 tie in that way, then the
winner is the one who beats the other.
If there are more than 2 candidates undefeated after the rule
is applied to the whole candidate set, then re-apply the rule
to those candidates only, starting with the original set of
Yes, it's a nuisance to have to talk about a tiebreaker for
the rule. But at least re-application of the rule to the
tied candidates only, starting all over with the original
set of pair-defeats, is the obvious way to solve the tie that
we have if the above rule (the one in quotes) gives us 2 or
more undefeated candidates.
But with a slightly wordier rule we could avoid the need for
any tiebreaker under big-election conditions (no pair-ties or
equal defeats). This rule is also completely obvious:
"Drop the weakest defeat that is in a cycle with defeats that are
all stronger than it is". Repeat that until there's an unbeaten
This has been called "Sequential Dropping" ("SD").
SD differs from DCD in that SD starts with the weakest instead
of dropping simultaneously, and stops when it makes one
undefeated candidate. Those are obvious rules. SD, like DCD,
is an obvious solution to a circular tie.
When there are no pair-ties or exactly equal defeats (there
won't be any in public elections, due to the great number of
voters), SD is equivalent to Schulze's method. SD gives the
same outcome as Schulze's method under those conditions.
Schulze, DCD, & SD all meet all of the defensive strategy criteria
that I consider important. And, at least when there are no pair-ties
or equal defeats, these methods meet the Clone Criterion too.
The simplest & most obvious rank-count rules can also be the
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