For example, let's see if I can formulate an automatic check for "Independence of Irrelevant Alternatives".
For any set of voters who vote honest preferences about some number of candidates, removing any combination of the non-winning choices shall not change the winner.
Combinatorics:
For C choices and V voters. There are factorial(C) possible ranking preferences for a voter. Multiply that by V voters and you get "any set of voters". For 20 voters and 5 choices that's 5*4*3*2*20 = 2400 possible sets of voters. Note that making the number of voters large is relatively easier than making the number of choices large. Choosing 1,2 or 3 of the 4 non-winning choices to discard results in 4 + 4*3/2 + 4*3*2/6 = 14 combinations of the non-winning to be run in a trial election. Add one more for the base case of having all 5 choices in and that's 15 * 2400 = 36000 trial elections to run. Easily computable.
Of course, what that exhaustively proves is only the 20 voters, 5 choices case. Is that good enough? How about 100 voters with 2,3,4,5 or 6 choices? That could probably still be run in a day.
Brian Olson http://bolson.org/ ---- Election-methods mailing list - see http://electorama.com/em for list info
