the comparison "A > B" degrade B's value and the comparison "C > D" advanced C's value. This results in C overcoming B. Thus, the minimax method fails the participation criterion. Ranked pairs This example shows that the ranked pairs method violates the participation criterion. Assume four candidates A, B, C and D with 26 potential voters and the following preferences: The four voters with preferences A > B > C > D are unconfident whether to participate in the election. Voters not participating Assume the 4 voters do not show up at the polling place. The preferences of the remaining 22 voters would be: The results would be tabulated as follows: The sorted list of victories would be: Result: A > D, B > C and D > B are locked in (and the other three can't be locked in after that), so the full ranking is A > D > B > C. Thus, A is elected ranked pairs winner. Voters participating Now, consider the 4 unconfident voters decide to participate: The results would be tabulated as follows: The sorted list of victories would be: Result: A > D, B > C and C > D are locked in first. Now, D > B can't be locked in since it would create a cycle B > C > D > B. Finally, B > A and C > A are locked in. Hence, the full ranking is B > C > A > D. Thus, B is elected ranked pairs winner. Conclusion By participating in the election the four voters supporting A would change A from winner to loser. The clear victory of D > B was essential for A's win in the first place. The additional votes diminished that victory and at the same time gave a boost to the victory of C > D, turning D > B into the weakest link of the cycle B > C > D > B. Since A had no other victories but the one over D and B had no other losses but the one over D, the elimination of D > B made it impossible for A to win. Thus, the ranked pairs method fails the participation criterion. Schulze method This example shows that the Schulze method violates the participation criterion. Assume four candidates A, B, C and D with 25 potential voters and the following preferences: The