Definition of Pairwise (to Bruce)

[Mike Ossipoff]

Bruce Anderson writes:
> 
> Define the Unanimity voting method as follows.  Let v be the number of voters 
> and let c be the number of candidates on the ballot.  If there is a candidate, 
> say i, such that the sum/j of p(i,j) = v(c-1), then Unanimity chooses that 
> candidate as its unique winner; otherwise (i.e., if there is no such candidate), 
> then Unanimity chooses every candidate on the ballot as being tied as its 
> winners.

When Bruce says that sum/j of p(i,j) = v(c-1), this is his way of saying
that i is everyone's 1st choice, if I interpret the formulese correctly.

But, in letters to Bruce, I've defined a "Unanimity" count method
that is different from the one he defines in the message quoted 
here.

> 
> If pairwise//nonpairwise methods are called pairwise, then wouldn't 
> "Unanimity//Plurality" and "Unanimity//Hare" also be pairwise?  Is this what you 
> want?  I sure would not want to have to explain to anyone why 
> Unanimity//Plurality and Unanimity//Hare are considered as being "pairwise 
> methods," since they are virtually identical to Plurality and Hare, 
> respectively.

Yes, so maybe that means that we have to either say that Hare is
a Pairwise method, or that Smith//Condorcete//Pluralitty &
Copeland//Plurality aren't Pairwise methods.

But there's something quite bogus about Bruce's Unanimity/Hare
& Unanimity//Plurality, since their results would be identical
to Hare & Plurality. Maybe we'd have to specify, in our definition
of a Pairwise method, then, that its results must not be identical
to those of non-Pairwise tie-breakers that it uses. 

Can we add that clause to our definitions of Pairwise methdods
& be done with it?



> 
> Again, let me emphasize that it's not at all important to me either way.  I 
> really don't care that much how "pairwise method" is defined.  What I do care 
> about is that each and every definition is precisely stated.

Ok, so can we please add the above clause to the definitions & dispense
with the subject?

> 
> Bruce
> .-
> 


--