The Schulze method (/ˈʃʊltsə/) is a single winnerranked-choice voting rule developed by Markus Schulze. It is also known as the beatpath method. The Schulze method is a Condorcet method, which means it will elect a majority-choice candidate if one exists; in other words, if most people rank A above B, A will defeat B (whenever this is possible).
Schulze's method is based on the idea of breaking cyclic ties by using indirect victories. The idea is that if Alice beats Bob, and Bob beats Charlie, then Alice (indirectly) beats Charlie; this kind of indirect win is called a beatpath.
Schulze's method uses ranked ballots with equal ratings allowed. There are two common (equivalent) descriptions of Schulze's method.
Beatpath explanation
The idea behind Schulze's method is that if Alice defeats Bob, and Bob beats Charlie, then Alice "indirectly" defeats Charlie; this kind of indirect win is called a 'beatpath'.
Every beatpath is assigned a particular strength. The strength of a single-step beatpath from Alice to Bob is just the number of voters who rank Alice over Bob. For a longer beatpath, consisting of multiple "beats", the strength of a beatpath is as strong as its weakest link (i.e. the beat with the smallest number of winning votes).
We say Alice has a "beatpath-win" over Bob if her strongest beatpath to Bob is stronger than all of Bob's beatpaths to Alice. The winner is the candidate who has a beatpath-win over every other candidate.
Markus Schulze proved that this definition of a beatpath-win is transitive; in other words, if Alice has a beatpath-win over Bob, and Bob has a beatpath-win over Charlie, Alice has a beatpath-win over Charlie.[1]: §4.1 As a result, the Schulze method is a Condorcet method, providing a full extension of the majority rule to any set of ballots.
Iterative description
The Schulze winner can also be constructed iteratively, using a defeat-dropping method:
Draw a directed graph with all the candidates as nodes; label the edges with the number of votes supporting the winner.
If there is more than one candidate left:
Check if any candidates are tied (and if so, break the ties by random ballot).
The winner is the only candidate left at the end of the procedure.
Example
In the following example 45 voters rank 5 candidates.
Number of voters
Order of preference
5
ACBED
5
ADECB
8
BEDAC
3
CABED
7
CAEBD
2
CBADE
7
DCEBA
8
EBADC
The pairwise preferences have to be computed first. For example, when comparing A and B pairwise, there are 5+5+3+7=20 voters who prefer A to B, and 8+2+7+8=25 voters who prefer B to A. So and . The full set of pairwise preferences is:
Matrix of pairwise preferences
20
26
30
22
25
16
33
18
19
29
17
24
15
12
28
14
23
27
21
31
The cells for d[X, Y] have a light green background if d[X, Y] > d[Y, X], otherwise the background is light red. There is no undisputed winner by only looking at the pairwise differences here.
Now the strongest paths have to be identified. To help visualize the strongest paths, the set of pairwise preferences is depicted in the diagram on the right in the form of a directed graph. An arrow from the node representing a candidate X to the one representing a candidate Y is labelled with d[X, Y]. To avoid cluttering the diagram, an arrow has only been drawn from X to Y when d[X, Y] > d[Y, X] (i.e. the table cells with light green background), omitting the one in the opposite direction (the table cells with light red background).
One example of computing the strongest path strength is p[B, D] = 33: the strongest path from B to D is the direct path (B, D) which has strength 33. But when computing p[A, C], the strongest path from A to C is not the direct path (A, C) of strength 26, rather the strongest path is the indirect path (A, D, C) which has strength min(30, 28) = 28. The strength of a path is the strength of its weakest link.
For each pair of candidates X and Y, the following table shows the strongest path from candidate X to candidate Y in red, with the weakest link underlined.
Strongest paths
To
From
A
B
C
D
E
A
—
A-(30)-D-(28)-C-(29)-B
A-(30)-D-(28)-C
A-(30)-D
A-(30)-D-(28)-C-(24)-E
A
B
B-(25)-A
—
B-(33)-D-(28)-C
B-(33)-D
B-(33)-D-(28)-C-(24)-E
B
C
C-(29)-B-(25)-A
C-(29)-B
—
C-(29)-B-(33)-D
C-(24)-E
C
D
D-(28)-C-(29)-B-(25)-A
D-(28)-C-(29)-B
D-(28)-C
—
D-(28)-C-(24)-E
D
E
E-(31)-D-(28)-C-(29)-B-(25)-A
E-(31)-D-(28)-C-(29)-B
E-(31)-D-(28)-C
E-(31)-D
—
E
A
B
C
D
E
From
To
Strengths of the strongest paths
28
28
30
24
25
28
33
24
25
29
29
24
25
28
28
24
25
28
28
31
Now the output of the Schulze method can be determined. For example, when comparing A and B,
since , for the Schulze method candidate A is better than candidate B. Another example is that , so candidate E is better than candidate D. Continuing in this way, the result is that the Schulze ranking is , and E wins. In other words, E wins since for every other candidate X.
Implementation
The only difficult step in implementing the Schulze method is computing the strongest path strengths. However, this is a well-known problem in graph theory sometimes called the widest path problem. One simple way to compute the strengths, therefore, is a variant of the Floyd–Warshall algorithm. The following pseudocode illustrates the algorithm.
# Input: d[i,j], the number of voters who prefer candidate i to candidate j.
# Output: p[i,j], the strength of the strongest path from candidate i to candidate j.
for i from 1 to C
for j from 1 to C
if i ≠ j then
if d[i,j] > d[j,i] then
p[i,j] := d[i,j]
else
p[i,j] := 0
for i from 1 to C
for j from 1 to C
if i ≠ j then
for k from 1 to C
if i ≠ k and j ≠ k then
p[j,k] := max (p[j,k], min (p[j,i], p[i,k]))
When allowing users to have ties in their preferences, the outcome of the Schulze method naturally depends on how these ties are interpreted in defining d[*,*]. Two natural choices are that d[A, B] represents either the number of voters who strictly prefer A to B (A>B), or the margin of (voters with A>B) minus (voters with B>A). But no matter how the ds are defined, the Schulze ranking has no cycles, and assuming the ds are unique it has no ties.[2]
Although ties in the Schulze ranking are unlikely, they are possible. Schulze's original paper recommended breaking ties by random ballot.[2]
There is another alternative way to demonstrate the winner of the Schulze method. This method is equivalent to the others described here, but the presentation is optimized for the significance of steps being visually apparent as a human goes through it, not for computation.
Make the results table, called the "matrix of pairwise preferences", such as used above in the example. Then, every positive number is a pairwise win for the candidate on that row (and marked green), ties are zeroes, and losses are negative (marked red). Order the candidates by how long they last in elimination.
If there is a candidate with no red on their line, they win.
Otherwise, draw a square box around the Schwartz set in the upper left corner. It can be described as the minimal "winner's circle" of candidates who do not lose to anyone outside the circle. Note that to the right of the box there is no red, which means it is a winner's circle, and note that within the box there is no reordering possible that would produce a smaller winner's circle.
Cut away every part of the table outside the box.
If there is still no candidate with no red on their line, something needs to be compromised on; every candidate lost some race, and the loss we tolerate the best is the one where the loser obtained the most votes. So, take the red cell with the highest number (if going by margins, the least negative), make it green—or any color other than red—and go back step 2.
Here is a margins table made from the above example. Note the change of order used for demonstration purposes.
Initial results table
E
A
C
B
D
E
1
−3
9
17
A
−1
7
−5
15
C
3
−7
13
−11
B
−9
5
−13
21
D
−17
−15
11
−21
The first drop (A's loss to E by 1 vote) does not help shrink the Schwartz set.
First drop
E
A
C
B
D
E
1
−3
9
17
A
−1
7
−5
15
C
3
−7
13
−11
B
−9
5
−13
21
D
−17
−15
11
−21
So we get straight to the second drop (E's loss to C by 3 votes), and that shows us the winner, E, with its clear row.
Second drop, final
E
A
C
B
D
E
1
−3
9
17
A
−1
7
−5
15
C
3
−7
13
−11
B
−9
5
−13
21
D
−17
−15
11
−21
This method can also be used to calculate a result, if the table is remade in such a way that one can conveniently and reliably rearrange the order of the candidates on both the row and the column, with the same order used on both at all times.
Satisfied and failed criteria
Satisfied criteria
The Schulze method satisfies the following criteria:
^ abc Approval voting, score voting, and majority judgment satisfy IIA if it is assumed that voters rate candidates independently using their own absolute scale. For this to hold, in some elections, some voters must use less than their full voting power despite having meaningful preferences among viable candidates.
^Majority Judgment may elect a candidate uniquely least-preferred by over half of voters, but it never elects the candidate uniquely bottom-rated by over half of voters.
^Majority Judgment fails the mutual majority criterion, but satisfies the criterion if the majority ranks the mutually favored set above a given absolute grade and all others below that grade.
^ abcIn Highest median, Ranked Pairs, and Schulze voting, there is always a regret-free, semi-honest ballot for any voter, holding all other ballots constant and assuming they know enough about how others will vote. Under such circumstances, there is always at least one way for a voter to participate without grading any less-preferred candidate above any more-preferred one.
^ abA variant of Minimax that counts only pairwise opposition, not opposition minus support, fails the Condorcet criterion and meets later-no-harm.
^A randomly chosen ballot determines winner. This and closely related methods are of mathematical interest and included here to demonstrate that even unreasonable methods can pass voting method criteria.
^Where a winner is randomly chosen from the candidates, sortition is included to demonstrate that even non-voting methods can pass some criteria.
The main difference between the Schulze method and the ranked pairs method can be seen in this example:
Suppose the MinMax score of a set X of candidates is the strength of the strongest pairwise win of a candidate A ∉ X against a candidate B ∈ X. Then the Schulze method, but not Ranked Pairs, guarantees that the winner is always a candidate of the set with minimum MinMax score.[2]: §4.8 So, in some sense, the Schulze method minimizes the largest majority that has to be reversed when determining the winner.
On the other hand, Ranked Pairs minimizes the largest majority that has to be reversed to determine the order of finish, in the MinLexMax sense.[citation needed][4]
In other words, when Ranked Pairs and the Schulze method produce different orders of finish, for the majorities on which the two orders of finish disagree, the Schulze order reverses a larger majority than the Ranked Pairs order.
History
The Schulze method was developed by Markus Schulze in 1997. It was first discussed in public mailing lists in 1997–1998[5] and in 2000.[6]
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