# [EM] Correction. Big CS fault?

Craig Carey research at ijs.co.nz
Fri Dec 20 05:15:08 PST 2002

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Hi Forest.

I find your message incomprehensible and most of the problem is with the
English words. This is not a message that is sent merely to create an
opportunity for you to subsequently write nothing. The geometry of
voting is somewhat really simple. There is plain space and a dimension
for each kind of paper. It could be complex if there was a tricky way
of reducing the dimension. Mainly what happens is that less kinds of
papers are used. The dimension equals the numbers of kinds of papers.

At 2002\12\18 18:02 -0800 Wednesday, Forest Simmons wrote:
>Here's the version of Candidate Space (CS) that I like the best now:
>
>The ballots must have some way of determining favorite, so must have at

There still is no formula defining the "favorite" (favourite). That
whole paragraph has no mathematical meaning.

>least the expressivity of Majority Choice ballots.
>

I don't know what a "Majority Choice ballot" is. Do you mean a
papers with only one preference showing on it ? (i.e. a weighted
preference list having a list that is of a length equal to 1 ?)

>[The favorite on the expressive side of the ballot must have maximal
>positive instrumentality in the instrumental aspect of the ballot, as
>well.]
>

What are the sides of a ballot ?. The preference list (A,B,C) has not
got 2 sides but it has 3 preferences. I want to ask why you don't

>Step 1.  Each candidate has a voter profile vector; the ith component is
>the rating given this candidate by ballot number i.
>

This voter profile of candidates is maybe a vector that has a length
equal to number of different kinds of papers.

It might seem that in a 6 candidate election, the paper (ABC) is more
about A,B,C, than about D,E,F. But it can be expanded out like this:

1(ABC) = ((ABCDEF) + (ABCDFE) + (ABCEDF) + (ABCEFD) + (ABCFDE) + (ABCFED))/6

So every single paper is a paper that candidates can hold an interest
for. Another way to get to the same result of candidates wanting to know
about papers that do not name them, is to note that when there are 2
candidates, A and B, then B can win even though not named by ever paper,
e.g. when the papers are just the single paper, -10(A).

So what is the "voter profile vector". Is there only one
"voter profile vector" and all candidates have the same vector, and it
merely the vector that can say what the weights/counts are.

What is the "i" that was named "ballot number" ?. Are the number of
ballots greater than the number of kinds/types of ballot papers?.

E.g. 1(AB) + 1(AB) is 2 ballot papers, but 1 kind of ballot paper that
has a weight of 2.

>[Convert rankings to ratings if necessary.]
>
>Step 2. The distance from each candidate to every other candidate is
>measured by comparing their voter profile vectors.
>
...

The word comparing normally means to use a relational operator like
"less than". That returns a Boolean value. Either the "Step 2" sentence
is a mistake or the geometry is not Euclidean. Anyway, again there is
the suggestion that somehow candidates are mixed in with ballot papers
to form the space, and so far there was not a mention of real
numbers. In text I deleted below, a since or cosine was calculated.
Thus the axes are labeled with real numbers. The just one source of
real numbers: the counts/weights of the kinds of ballot papers.

If there is to be a space (and Euclidean would be preferable) then we
would all start a personal search for some real numbers to use. That
whole process of inquiry stops when a decision is made to make the
real numbers of the space, be the weights each kind of ballot paper.
E.g. if the election is this

x (A)
y (BA)

then there can be a 2-D space.

We don't use ideas from Condorcet since that is intolerably presumptuous,
unjustified, wrong, and unacceptable to persons that want a method to be
politically tolerable and like some ideal debugged STV method.

>Addendum: If a voter lists two favorites, then he contributes 1/2 to the
>count of each in the Condorcet Step (4).

It is really hard to be sure if that is true since one voter might
have had the power of two voter that can only use a single FPTP paper,
when actually casting those two ballot papers. Until you say otherwise,
Mr Forest Simmons, the rather undefined and apparently irrelevant
mathematical thing you name a "voter" might be able to have a
relationship with any number of preference lists of weight 1.

>Step 4. If a candidate is the favorite of 7526 voters, then that
>candidate's preference ballot is counted 7526 times in a Condorcet
>election.

What is the point of restricting the statement to Condorcet elections.
If the intent was to define a geomotric interpretation, then the
voting method need not be specified.

I can't make a conclusion here.

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