No, no no no NO. Now you're introducing counting algorithms. Which have to have pre-processed the ballots to produce the summary in a compacted format.
You MUST consider how to use ONE ballot to represent A>B>C in a three-candidate race. You cannot do it with less than 6 bits. ONE ballot. Three candidates. Express ONE VOTER's choices with the minimum number of bits. -----Original Message----- From: [email protected] [mailto:[email protected]] On Behalf Of Kristofer Munsterhjelm Sent: Saturday, June 06, 2009 2:18 PM To: Paul Kislanko Cc: [email protected] Subject: Re: [EM] Some myths about voting methods Paul Kislanko wrote: > The number of possible votes is not the same as the amount of information in > a single ballot. With 3 candidates, there are indeed 8 possible ballots, but > any one ballot can be encoded in 3 bits, since any particular choice > requires only that many to represent it. > > Ranked ballots require 2 bits per alternative (01 = 1st, 10 = 2nd, 11 = 3rd) > so the minimum ballot representation is six bits, twice as much information > as is contained in an approval ballot. That argument is not true, as such. All you need to store a ballot in a way that can be recognized is an enumeration. In the worst case, when you only have the enumeration, recovering the actual ballots will be very slow, but it's feasible. In the three candidate case, you can map Approval ballots to integers (and back) by simply using the Approval vote as a bitfield representation, i.e. 0 <-> 000 1 <-> 001 2 <-> 010 3 <-> 011 and so on. For ranked votes, it's no different: 0 <-> A > B > C 1 <-> A > C > B 2 <-> B > A > C 3 <-> B > C > A 4 <-> C > A > B 5 <-> C > B > A So if a hundred people voted A > B > C and ten voted B > A > C, you could compress that down to: (100, 0) (10, 2) Now, this does show that there's more raw information in an Approval ballot than in a ranked ballot (if no ties nor equal rank is permitted); however, it doesn't by itself mean that that information is useful. Of course, that objection can be turned right back at us for num candidates > 3, in which case it's ranked ballots that have the upper hand, raw information wise. For an extreme of how raw information may not actually make a difference, consider a ballot that's "rank the candidates, then specify a random number". Say the random number is used for a tie break, in some manner. Then there's a lot of raw information (the number of unique votes depend on the range of the random numbers), but the random number part only makes a difference in the case of a tie (which would happen infrequently). Also, if one permits ranked ballots to rank multiple candidates equally or to truncate (which we can treat as "equal rank last", unless there's an approval cutoff), Approval votes are a subset of the ranked votes permitted, and therefore, there will always be more ranked votes than Approval votes. Since the methods we are discussing here usually permit equal-rank, I don't see the strength of the argument regarding raw numbers of ballots. For Range ballots, say WLOG that it's integer Range, the minimum range is 1 and the maximum is x, and there are y candidates. Then there are x^y different Range ballots. For any predefined number of candidates, you can set x so that the number of unique Range ballots exceeds the number of unique preferential ballots. Keeping x static, there will, however, be another y at which the number of unique preferential ballots exceed the number of Range ballots (for that x) again. If one doesn't permit equal-rank or truncation, the numbers are y=2 x^2 = 2! x = 1.42 y=3 x^3 = 3! x = 1.81 y=4 x^4 = 4! x = 2.21 y=10 x^10=10! x = 5 y=18 y^18=18! x = 7.55 y=25 x^25=25! x = 10 y=100 x^100=100! x = 38 etc. For equal rank, the relevant count is the number of total preorders, and so: y=2 x^2 = 3 x = 1.3 y=3 x^3 = 13 x = 2.35 y=4 x^4 = 75 x = 2.94 y=10 x^10 = 102247563 x = 6.32 y=18 x^18 = 3385534663256845323 x = 10.70 and so on. ---- Election-Methods mailing list - see http://electorama.com/em for list info ---- Election-Methods mailing list - see http://electorama.com/em for list info
