[EM] Automatic Criterion Checking
kislanko at airmail.net
Tue Nov 16 15:25:24 PST 2004
This is in fact, one of the great questions of mathematical logic.
Unfortunately, 20 years before Arrow's theorem, there was Godel's Theorem,
which proves that such an "axiom-checker" is as impossible as squaring the
circle or simultaneously finding the position and momentum of an electron.
You can "for example" all you want, but for any logical system that's at
least as complex as arithmetic, there is no way to prove consistency without
applying something "outside" the system. There's a symmetry in Godel's proof
and Arrow's - the only way to satisfy the other three of Arrow's criteria is
to have a Dictator who specifies how everybody else should vote to satisfy
the other criteria.
> -----Original Message-----
> From: election-methods-electorama.com-bounces at electorama.com
> [mailto:election-methods-electorama.com-bounces at electorama.com
> ] On Behalf Of bql at bolson.org
> Sent: Tuesday, November 16, 2004 5:00 PM
> To: election-methods-electorama.com at electorama.com
> Subject: [EM] Automatic Criterion Checking
> I wish we had nice clean definitions of our favorite criteria
> that were
> amenable to automatic checking. Then we just implement any
> new method in a
> few lines of code, and run the checker. In most cases I believe the
> computations could be completed in a few hours or a few days
> on any of our
> common home computers.
> 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.
> 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
> 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
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