Criterion compliance demonstrations
This is the letter that I said I'd send about demonstrations of
compliances with the defensive strategy criteria.
First let me point out that SD, DCD, & Tideman all have one
thing in common, any defeat that they drop is the weakest defeat
in some cycle.
When DCD drops the weakest defeat in each cycle, that's obvious.
When SD drops the weakest defeat that's in a cycle, that
defeat must also be the weakest defeat in any cycle that it's
SD, DCD & Tideman have a symmetrical relation.
DCD just simultaneously drops each cycle's weakest defeat, which
means that it drops every defeat that's the weakest defeat in a
SD could be worded as saying: Drop the weakest defeat that is
the weakest defeat in some cycle that it's in. Repeat till
there's an unbeaten candidate.
Tideman iterates down from the top, saying:
Drop the strongest defeat that is the weakest defeat in some
cycle. Repeat till all cycles are solved.
[It goes without saying that, when the iteration is from the
top, "the weakest defeat in some cycle" means "the weakest defeat
in some cycle with undropped defeats].
I haven't recommended Tideman because it's motivation &
aren't, to me, as obvious & natural as those of DCD & SD & SSD.
So all three of those methods, DCD, SD, & Tideman, drop
defeats that are the weakest in some cycle to which they
belong. SD starts with the weakest such defeat. Tideman starts
with the strongest such defeat. And DSC simply drops all such
These symmetrically-related methods I call the cycle Condorcet
versions. I'll later show that SSD can be counted among them,
but that isn't important now.
First, I'd like to show that any method that drops a defeat only'
if it's the weakest defeat in some cycle automatically meets all
of the 5 defensive strategy criteria.
If no unfelt preferences are voted, then there's no way that a CW
can have a majority defeat. By assumption the CW beats B by
a majority. If B isn't in a cycle with the CW, then its majority
defeat by the CW can never be dropped, since the method in use
won't drop a defeat that isn't the weakest defeat in some cycle.
If B is in a cycle with the CW, then, because the CW can't have
a majority defeat, then B's defeat by the CW can't be the weakest
defeat in that cycle; the CW's defeat in that cycle is weaker.
So B's defeat won't be the one that is dropped, and B can't win.
The CW will become unbeaten before B could.
2. GSFC (This demonstration is similar to the above).
If no unfelt preferences are voted, then there's no way that
a member of the sincere Smith set can have a majority defeat against
him (because, compared to any other candidate, more people prefer
a sincere Smith set candidate to him than vice versa).
A member of the sincere Smith set has a majority against B, who
isn't in the sincere Smith set.
Give the name Z to the sincere Smith set member who beats B
by majority. As before, if B isn't in a cycle with Z, then
Z's majority defeat of B isn't in a cycle and can't be dropped.
If B is in a cycle with Z, then all of the defeats of sincere
Smith set members by nonmembers that are part of that cycle are
weaker parts of that cycle than B's defeat by Z. If we only
drop defeats that are the weakest in some cycle, then Z some
member of the sincere Smith set will become unbeaten before
3. WDSC, SDSC, & SrDSC:
A majority of all the voters prefer A to B. Say they all rank
A over B. B has a majority defeat. Now, say that the members of
that majority refuse to include B in their ranking. That means
that B can't have a majority defeat against anyone. That means
that if A & B are in a cycle, then B's defeat by A can't be the
weakest defeat in that cycle, because B doesn't beat anything
by majority. Some other member of the cycle will become unbeaten
before B can.
This argument also applies to SDSC & SrDSC, since the members
of that majority achieved their goal merely by not ranking B.
They didn't have to vote anyone equal to or over a more-liked
candidate, or violate the terms of SrDSC.
[The fact that they have refused to vote for B and also for those
less-liked than B doesn't count as voting a less-liked candidate
equal to a more-liked one; instead, it could be said that these
voters have equally _not_ voted for B and those less-liked
So the Cycle Condorcet methods meet all 5 defensive strategy
Before I show that SSD meets them too, let me introduce another
criterion, a generalization of the 5 defensive strategy criteria:
Beatpath Criterion (BC):
If X beats Y, and if Y's strongest beatpath to X is weaker than
X's defeat of Y, then Y shouldn't win.
Definition of a beatpath:
X has a beatpath to Y if either X beats Y, or if X beats someone
who has a beatpath to Y.
The strength of a beatpath is the strength of its weakest defeat.
Since Y doesn't have any beatpath to X that is as strong as X's
defeat of Y, that means that if Y is in a cycle with X, then
Y's defeat by X can't be the weakest defeat in that cycle.
That means that any method that only drops a defeat that is
the weakest defeat in some cycle will never drop Y's defeat by
X before it breaks the cycle elsewhere, making someone else
unbeaten. Therefore, the Cycle Condorcet versions meet BC.
To show that methods that meet BC meet those 5 criteria that the
Cycle Condorcet methods meet: The Cycle Condorcet methods meet
the defensive strategy criteria because, in each case, B's
majority defeat isn't the weakest defeat in a cycle with the
candidate who beats B (call him Z). If any cycle including B
& Z has a weaker defeat than B's defeat by Z, then B doesn't
have a beatpath to Z as strong as Z's defeat of B, and so
B can't win using any method that meets BC.
Why SSD meets BC:
If Z beats B, and Z has no beatpath to B as strong as B's defeat
by B, then every cycle including B & Z has a defeat that's weaker
than B's defeat by Z. If B isn't in the Schwartz set then
his defeat can't be dropped & he can't win.
If B is in the Schwartz set, then anything that beats him must
be in the Schwartz set too, since the Schwartz set is an unbeaten
set, not beaten from without. If some candidate is in the Schwartz
set, then anyone who beats him must also be in the Schwartz set.
So, anyone in a cycle containing B & Z must be in the Schwartz set.
According to BC's premise, any cycle containing B & Z has a
defeat weaker than B's defeat by Z (from the paragraph before
last). That means that B's defeat by Z isn't the weakest defeat
in the Schwartz set. Any cycle containing those 2 candidates has
a weaker defeat elsewhere. When those defeats have all been dropped,
as SSD repeatedly drops the Schwartz set's weakest defeat,
some candidate other than B will have been made unbeaten. That
could happen the first time a defeat is dropped, but it has to
happen by the time that the weakest defeat is dropped from each
cycle containing B & Z. So someone other than B will be the
undefeated before B is, and B can't win.
So SSD meets BC, and therefore meets all 5 of the defensive
Schulze meets BC for obvious reasons. If there's no beatpath
from B to Z as strong as Z's defeat of B, that means that Z
has a (1-defeat) beatpath to B stronger than any beatpath from
B to Z, meaning that B can't be the candidate with a Schulze win
over each one of the other candidates, and that B can't win.
I've shown that the Cycle Condorcet methods and SSD &
Schulze meet BC, and therefore meet all 5 of the defensive
Now, in my last of these messages, I'd like to show, as I said
I would, that SSD can be counted as one of the Cycle Condorcet
versions, because if a defeat is among the Schwartz set members,
it is in a cycle. That will be the next & last of this series
Get Your Private, Free Email at http://www.hotmail.com