Schulze Method
This example shows that the Schulze method violates the Independence of irrelevant alternatives criterion. Assume four candidates A, B, C and D and 12 voters with the following preferences:
| # of voters | Preferences |
|---|---|
| 4 | A > B > C > D |
| 2 | C > B > D > A |
| 3 | C > D > A > B |
| 2 | D > A > B > C |
| 1 | D > B > C > A |
The pairwise preferences would be tabulated as follows:
| d | d | d | d | |
|---|---|---|---|---|
| d | 9 | 6 | 4 | |
| d | 3 | 7 | 6 | |
| d | 6 | 5 | 9 | |
| d | 8 | 6 | 3 |
Now, the strongest paths have to be identified, e.g. the path D > A > B is stronger than the direct path D > B (which is nullified, since it is a tie).
| d | d | d | d | |
|---|---|---|---|---|
| d | 9 | 7 | 7 | |
| d | 7 | 7 | 7 | |
| d | 8 | 8 | 9 | |
| d | 8 | 8 | 7 |
Result: The full ranking is C > D > A > B. Thus, C is elected Schulze winner and D is preferred over A.
Read more about this topic: Local Independence Of Irrelevant Alternatives, Voting Theory, Examples
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