Condorect sub-cycle rule
Markus Schulze
schulze at sol.physik.tu-berlin.de
Thu Jan 1 11:13:37 PST 1998
Dear participants,
in one of my last e-mails (04 Oct 1997), I introduced
Tideman's method. I mentioned two possible explanations
of Tideman's method: Tideman's original formulation
(successively locking resp. skipping the worst defeats)
and my own formulation (calculating the strength of defeats
via beat-paths).
Then I got an e-mail (14 Nov 97) from Mike Ossipoff, who
believed, that my beat-path-method and Tideman's original
method are different things. He was right. My
Condorcet(beat-path) method and Tideman's method are
different things.
At first, it seemed to me obvious, that both methods are
identical. But now, I found examples where both methods
lead to different results.
************************************************************
Example:
26 voters vote C > A > B > D.
20 voters vote B > D > A > C.
18 voters vote A > D > C > B.
14 voters vote C > B > A > D.
08 voters vote B > D > C > A.
07 voters vote D > A > C > B.
07 voters vote B > D.
A:B=51:49
A:C=45:48
A:D=58:42
B:C=35:65
B:D=75:25
C:D=40:60
As A > B > D > C > A, every candidate is in the Smith set.
***
Tideman's method:
Step1: The worst defeat is B:D=75:25. Thus, B > D is locked.
Step2: The worst defeat, that hasn't been locked or skipped
yet, is B:C=35:65. Thus C > B is locked.
Step3: The worst defeat, that hasn't been locked or skipped
yet, is C:D=40:60. But D > C would create a cycle C > B > D > C.
Thus, D > C is skipped.
Step4: The worst defeat, that hasn't been locked or skipped
yet, is A:D=58:42. Thus A > D is locked.
Step5: The worst defeat, that hasn't been locked or skipped
yet, is A:B=51:49. Thus A > B is locked.
Step6: The worst defeat, that hasn't been locked or skipped
yet, is A:C=45:48. Thus C > A is locked.
Thus, I got:
B > D, C > B, A > D, A > B, C > A.
C is only candidate, who is not dominated. Thus, C wins.
***
My method:
A has 58 votes against B via the beat-path A > D > C > B.
A has 58 votes against C via the beat-path A > D > C.
A has 58 votes against D via the beat-path A > D.
B has 48 votes against A via the beat-path B > D > C > A.
B has 60 votes against C via the beat-path B > D > C.
B has 75 votes against D via the beat-path B > D.
C has 48 votes against A via the beat-path C > A.
C has 65 votes against B via the beat-path C > B.
C has 65 votes against D via the beat-path C > B > D.
D has 48 votes against A via the beat-path D > C > A.
D has 60 votes against B via the beat-path D > C > B.
D has 60 votes against C via the beat-path D > C.
Thus, I got via beat-paths:
A:B=58:48
A:C=58:48
A:D=58:48
B:C=60:65
B:D=75:60
C:D=65:60
Candidate A wins every pairwise comparison via beat-paths.
Thus, A wins.
************************************************************
To my opinion, my Condorcet(beat-path) method is better than
Tideman's method, because my method meets GMC:
"X >> Y" means, that a majority of the voters prefers
X to Y.
"There is a majority beat-path from X to Y," means,
that X >> Y or there is a set of candidates
C[1], ..., C[n] with X >> C[1] >> ... >> C[n] >> Y.
A method meets the "Generalized Majority
Criterion" (GMC) if and only if:
If there is a majority beat-path from A to B, but
no majority beat-path from B to A, then B must not
be elected.
***
It is necessary to use the above mentioned definition
of GMC, because the usual definition of GMC is not compatible
to the Generalized Independence from Twins Criterion (GITC).
If the usual definition of GMC was used (i.e.: An alternative
that has a majority against must not be elected unless every
alternative has a majority against it.), it could happen, that
twins beat eachother so badly that they would disqualify
eachother by GMC.
***
In the example above, A should be elected, because A is the
only candidate, who is not rejected by a majority of the
voters!
It remains to be demonstrated, that my method meets Twin
Criteria.
Suppose, B is not a twin of A. Then the introduction of
twins cannot change the maximum beat-path from A to B,
because all the twins from a set of twins have the same
defeat against non-twins. Thus, the maximum beat-path
from A to B cannot be increased by adding twins.
Markus Schulze
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