IRV failing monotonicity criterion I am looking for the simplest possible example of instant runoff voting failing the monotonicity criterion. By “simplest possible” I mean the scenario with the fewest number of candidates $(3)$ and the fewest number of votes switched $(1?,  2?)$ that still demonstrates the effect. I'm guessing that the result is the same, but for the sake of specificity, let's say we want the case where increasing support for a candidate causes them to lose the election in round two (as opposed to decrease support causing them to win).
The wikipedia article (https://en.wikipedia.org/wiki/Monotonicity_criterion#Instant-runoff_voting_and_the_two-round_system_are_not_monotonic) uses an example with 3 candidates and 2 votes being switched, but with $100$ total votes. I’m wondering what the minimum number of total votes needed is. Also, is there a general proof to show that the solution really is the minimum?
Thanks in advance for your insights!
-James
 A: There's a simpler one on RangeVoting.org's (non)Monotonicity and Instant Runoff Voting :

The original ballots are:


*

*C > B > A (5 voters)

*A > C > B (4 voters)

*B > A > C (8 voters)


As shown, A gets eliminated first, all of A's 2nd choices go to C, and C wins with 9.
If two of the B > A > C  voters cross the line and switch to A > B > C, decreasing their support for B:

the ballots become:


*

*C > B > A (5 voters)

*A > C > B (4 voters)

*B > A > C (6 voters)

*A > B > C (2 voters)


then C is eliminated, all of C's 2nd choices go to B, and B wins with 11.
So 2 voters decreasing their support for a candidate caused that candidate to win.
For this cyclic example to work, these criteria must be met (where the numbers are the first or second scenario):


*

*CBA > ACB

*BAC1 > ACB

*ACB + CBA > BAC1

*BAC1 = BAC2 + ABC2

*BAC2 > CBA

*ACB + ABC2 > CBA

*CBA + BAC2 > ACB + ABC2


I wrote a quick computer program to try every combination and it didn't find any simpler solution (for these conditions).
