[EM] Idiots and information

[Paul Kislanko]

Let's go back to the original post. Mr Smith called me an idiot for pointing
out that his claim that approval ballots contain as much information as
ranked ballots or range ballots do. 

I point out that given a range ballot I can create a ranked ballot, and
given a ranked ballot (truncation allowed, equivalent to assigning a zero
for a range) I can create the approval equivalent.

Now, in a 3 alternative ballot with alternatives A, B, and C, I approove B
and C. Knowing only that, Mr Smith asserts their is as much information as
there would be if I'd ranked the candidates. I ask him publicly to derive
from my approval of B and C which one of them I'd prefer, using only the
knowledge that I approve both of them.

He can't do that, but he calls me an idiot.

That ranked ballots provide more informations  than approval ballots is not
a myth, it is a fact. Mr. Smith can evidently  tell from my {B C} > {A} what
my preference between B and C is. If he can't provide an algorithm for that,
his assertion that my explicitly telling him provides no new information is
certainly not correct.

If "ranked ballots provide more information than approval ballots" is a
MYTH, then Mr. Smith should be able to decide from {B C} > {A} which of {C
B} is preferred by the approval voter over the other.

-----Original Message-----
From: election-methods-bounces at lists.electorama.com
[mailto:election-methods-bounces at lists.electorama.com] On Behalf Of Warren
Smith
Sent: Saturday, June 06, 2009 6:36 PM
To: election-methods
Subject: [EM] Idiots and information

Since Paul Kislanko is now CCing my emails to him to this public
bulletin board without
my permission, I will do the same with his at the end of this.

I really do not want to get into a long diatribe of explaining well
known material,
but just this once I will.

Kislanko seems to think it is impossible to, in 3 bits, specify a
permutation (rank ordering) of 3 candidates. E.g. if it is
Kennedy>Bush>Gore, this can be specified
by the 3 bits 010, meaning the 5th out of the 6 possible permutations
ordered in lexicographic order.  [101 is 4 in binary, and the
possibilities are 0,1,2,3,4,5
of which this is the 5th.]

Similarly three such independent rank orderings can be encoded in 8
bits, for a per ranking average of 8/3.  Note to Kislanko:
8/3=2.666... is not the same as 8*3=24.

Now I do not want to hear any more on this.  Kislanko has repeatedly
made false statements and is unaware of the most basic, trivial, and
well known ideas in information theory dating back to its founder
Shannon who already said all this stuff in the 1940s.  K seems to
think 8/3=8*3, there is no way to do anything without writing down a
table, etc etc. All nuts. That's fine.  There are plenty of other
ignorant people besides Kislanko, and I hold no grudge against them.
But when he accuses me of being an idiot and insists on his rightness
in making these ludicrous assertions, he goes over the line.
I would like his apology both to me and to this bulletin board for
wasting all our time.

Now here are the Kislanko emails to me reprinted (as is his style)
without his permission.
Emails #2, 4, 5, 6 are wrong.  Email #1 is mostly right.  Email #3 is
irrelevant.
He sent me further emails but I deleted them without reading.

----------

EMAIL #1
Warren, I am embarrassed for you.

Approval does not provide more information than ranked ballots and to play
with combinatorics the way you do is to show a lack of understanding.
Approval can be encoded in N bits, 1 bit per alternative (and yes, there are
2^N possible ballots, but that's irrelevant.) Ranked ballots require log N *
N (where the log is to base 2.)

EMAIL #2
The number of possibilities with ranked ballots is ((log N)*N)! which is >
2^N as you note. So more info in ranked ballot represeetntations.

Tactical or not, you can't express more info in approval representation than
you can in ranked ballots.

EMAIL #3
Just consider this.

How do you break a tie when the only info available is what you can get from
an approval ballot?

How do you do so if there are ranked ballots?

EMAIL #4
'll try again.

An approval ballot for 3 candidates can be encoded in 3 bits. 000 = no
approvals, 111 = approva all, 100 = approva A, 110 = approve A and B, etc.

A ranked ballot requires two bits per candidate. 01, 10, 11 for 1st, 2nd,
3rd, for 6 bits, you can't do it in fewer.

6 > 3 means more information in the ranked ballot representations.

EMAIL #5
Evidently logic and math IS hard for you.

Your example of how to encode the 6-choice ballot in the same number of bits
ignores the fact that in order to do so you need a table. Youo can NOT
reproduce a ranked ballot from a 3-bit number WITHOUT adding to the
information the 3x2 bit table to EVERY ballot.

Don't call someone an idiot if you can't figure out that 24>3.

You claimed a ranked ballot for three candidates can be reconstructed from a
3-bit number. Then you added 6x3 = 18 bits "on the side" to show it was
possible. I say all you did was show that you could reconstuct it using 21
bits, which is even worse than the 6 bits I suggested.

Don't call someone an IDIOT if *YOU* do not know what you're talking about.

EMAIL #6
I know enough about data compression to know what logic and math is behind
it.

It is evident YOU don't know anything about logic.

Exract a ranked ballot from '011', please. Tell me how you did it.

-- 
Warren D. Smith
http://RangeVoting.org  <-- add your endorsement (by clicking
"endorse" as 1st step)
and
math.temple.edu/~wds/homepage/works.html
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