Editing Schulze method (section)

== Ties and alternative implementations ==
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 ''d''s are defined, the Schulze ranking has no cycles, and assuming the ''d''s are unique it has no ties.<ref name="schulze20113">Markus Schulze, "[[doi:10.1007/s00355-010-0475-4|A new monotonic, clone-independent, reversal symmetric, and condorcet-consistent single-winner election method]]", Social Choice and Welfare, volume 36, number 2, page 267–303, 2011. Preliminary version in ''Voting Matters'', 17:9-19, 2003.</ref>

Although ties in the Schulze ranking are unlikely, they are possible. Schulze's original paper recommended breaking ties by [[random ballot]].<ref name="schulze20113" />

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.
{| class="wikitable" style="text-align:center"
|+ Initial results table
|-
! !! E !! A !! C !! B !! D
|-
! E
| || style="background:#dfd;"|1 || style="background:#fdd;"|−3 || style="background:#dfd;"|9 || style="background:#dfd;"|17
|-
! A
| style="background:#fdd;"|−1 || || style="background:#dfd;"|7 || style="background:#fdd;"|−5 || style="background:#dfd;"|15
|-
! C
| style="background:#dfd;"|3 || style="background:#fdd;"|−7 || || style="background:#dfd;"|13 || style="background:#fdd;"|−11
|-
! B
| style="background:#fdd;"|−9 || style="background:#dfd;"|5 || style="background:#fdd;"|−13 || || style="background:#dfd;"|21
|-
! D
| style="background:#fdd;"|−17 || style="background:#fdd;"|−15 || style="background:#dfd;"|11 || style="background:#fdd;"|−21 || 
|}
The first drop (A's loss to E by 1 vote) does not help shrink the Schwartz set.
{| class="wikitable" style="text-align:center"
|+ First drop
|-
! !! E !! A !! C !! B !! D
|-
! E
| || style="background:#dfd;"|1 || style="background:#fdd;"|−3 || style="background:#dfd;"|9 || style="background:#dfd;"|17
|-
! A
| style="background:#ddd;"|−1 || || style="background:#dfd;"|7 || style="background:#fdd;"|−5 || style="background:#dfd;"|15
|-
! C
| style="background:#dfd;"|3 || style="background:#fdd;"|−7 || || style="background:#dfd;"|13 || style="background:#fdd;"|−11
|-
! B
| style="background:#fdd;"|−9 || style="background:#dfd;"|5 || style="background:#fdd;"|−13 || || style="background:#dfd;"|21
|-
! D
| style="background:#fdd;"|−17 || style="background:#fdd;"|−15 || style="background:#dfd;"|11 || style="background:#fdd;"|−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.
{| class="wikitable" style="text-align:center"
|+ Second drop, final
|-
! !! E !! A !! C !! B !! D
|-
! E
| || style="background:#efe;"|1 || style="background:#eee;"|−3 || style="background:#efe;"|9 || style="background:#efe;"|17
|-
! A
| style="background:#ddd;"|−1 || || style="background:#dfd;"|7 || style="background:#fdd;"|−5 || style="background:#dfd;"|15
|-
! C
| style="background:#dfd;"|3 || style="background:#fdd;"|−7 || || style="background:#dfd;"|13 || style="background:#fdd;"|−11
|-
! B
| style="background:#fdd;"|−9 || style="background:#dfd;"|5 || style="background:#fdd;"|−13 || || style="background:#dfd;"|21
|-
! D
| style="background:#fdd;"|−17 || style="background:#fdd;"|−15 || style="background:#dfd;"|11 || style="background:#fdd;"|−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.