Schulze method: Difference between revisions

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:# For all i = 1,...,(n-1): d[C(i),C(i+1)] > d[C(i+1),C(i)].

:# For all i = 1,...,(n-1): d[C(i),C(i+1)] ≥ p.

:# For at least one i = 1,...,(n-1): d[C(i),C(i+1)] = p.

p[A,B], the ''strength of the strongest path'' from candidate A to candidate B, is the maximum value such that there is a path from candidate A to candidate B of that strength. If there is no path from candidate A to candidate B at all, then p[A,B] : = 0.

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Candidate D is a ''potential winner'' if and only if p[D,E] ≥ p[E,D] for every other candidate E.

=== Remark ===

It is possible to prove that p[X,Y] > p[Y,X] and p[Y,Z] > p[Z,Y] together imply p[X,Z] > p[Z,X]. Therefore, it is guaranteed (1) that the above definition of "''better''" really defines a transitive relation and (2) that there is always at least one candidate D with p[D,E] ≥ p[E,D] for every other candidate E.

=== Implementation ===

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Candidate E is a potential winner, because p[E,X] ≥ p[X,E] for every other candidate X.

As 25 = p[E,A] > p[A,E] = 24, candidate E is ''better'' than candidate A.

As 28 = p[E,B] > p[B,E] = 24, candidate E is ''better'' than candidate B.

As 28 = p[E,C] > p[C,E] = 24, candidate E is ''better'' than candidate C.

As 31 = p[E,D] > p[D,E] = 24, candidate E is ''better'' than candidate D.

As 28 = p[A,B] > p[B,A] = 25, candidate A is ''better'' than candidate B.

As 28 = p[A,C] > p[C,A] = 25, candidate A is ''better'' than candidate C.

As 30 = p[A,D] > p[D,A] = 25, candidate A is ''better'' than candidate D.

As 29 = p[C,B] > p[B,C] = 28, candidate C is ''better'' than candidate B.

As 29 = p[C,D] > p[D,C] = 28, candidate C is ''better'' than candidate D.

As 33 = p[B,D] > p[D,B] = 28, candidate B is ''better'' than candidate D.

Therefore, the Schulze ranking is E > A > C > B > D.

==== Example 2 ====

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Candidate D is a potential winner, because p[D,X] ≥ p[X,D] for every other candidate X.

As 18 = p[D,A] > p[A,D] = 17, candidate D is ''better'' than candidate A.

As 18 = p[D,B] > p[B,D] = 17, candidate D is ''better'' than candidate B.

As 18 = p[D,C] > p[C,D] = 17, candidate D is ''better'' than candidate C.

As 20 = p[A,B] > p[B,A] = 19, candidate A is ''better'' than candidate B.

As 20 = p[A,C] > p[C,A] = 19, candidate A is ''better'' than candidate C.

As 21 = p[C,B] > p[B,C] = 19, candidate C is ''better'' than candidate B.

Therefore, the Schulze ranking is D > A > C > B.

==== Example 3 ====

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Candidate B is a potential winner, because p[B,X] ≥ p[X,B] for every other candidate X.

As 19 = p[B,A] > p[A,B] = 18, candidate B is ''better'' than candidate A.

As 19 = p[B,C] > p[C,B] = 18, candidate B is ''better'' than candidate C.

As 19 = p[B,D] > p[D,B] = 18, candidate B is ''better'' than candidate D.

As 19 = p[B,E] > p[E,B] = 18, candidate B is ''better'' than candidate E.

As 20 = p[A,C] > p[C,A] = 19, candidate A is ''better'' than candidate C.

As 21 = p[A,D] > p[D,A] = 19, candidate A is ''better'' than candidate D.

As 21 = p[A,E] > p[E,A] = 19, candidate A is ''better'' than candidate E.

As 20 = p[D,C] > p[C,D] = 19, candidate D is ''better'' than candidate C.

As 30 = p[D,E] > p[E,D] = 19, candidate D is ''better'' than candidate E.

As 20 = p[E,C] > p[C,E] = 19, candidate E is ''better'' than candidate C.

Therefore, the Schulze ranking is B > A > D > E > C.

==== Example 4 ====

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Candidate B and candidate D are potential winners, because p[B,X] ≥ p[X,B] for every other candidate X and p[D,Y] ≥ p[Y,D] for every other candidate Y.

As 7 = p[B,C] > p[C,B] = 5, candidate B is ''better'' than candidate C.

As 6 = p[D,A] > p[A,D] = 5, candidate D is ''better'' than candidate A.

Possible Schulze rankings are B > C > D > A, B > D > A > C, B > D > C > A, D > A > B > C, D > B > A > C, and D > B > C > A.

== The Schwartz set heuristic ==