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# Typos (Brief. Concludes letter series)

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I'd like to correct a few typos in my previous letters. This
will really be the conclusion of this series of letters.
If anyone's interested in this matter of how to solve circular
ties, then it wouldn't make sense not to correct the typos.

A. When discussing my example, using debian's current circular
tie solution, the example where the middle candidate, B, gets
eliminated: When I said that A beats B pairwise, I meant to
say instead that A beats C pairwise.

B. Typos in "Criterion compliance demonstrations":

1. Neart he bottom of the 3rd page of the letter, under the
heading "Why SSD meets BC:", there were typos. I meant to say:

"If Z beats B, and B has no beatpath to Z as strong as B's defeat
by Z, then every cycle containing B & Z has a defeat that's weaker
than B's defeat by Z."

That's the only typo in that demonstration. But I should add
that an alternative to this demonstration that SSD meets BC is
the demonstration in my last letter "Why SSD is a Cycle Condorcet
method". When SSD has been shown to be a Cycle Condorcet method,
then, since the Cycle Condorcet methods have been shown to meet
BC, that means that SSD meets BC. I personally prefer that
line of demonstration, since it includes SSD in a larger general
category, so that a separate proof for SSD isn't needed (not needed,
but included under "Why SSDS meets BC").

C. Typos in "Why SSD is a Cycle Condorcet method":

1. In the 2nd paragraph of that letter, I said "That's the same
as saying that no member of the Schwartz set has an unreturned
beatpath to him, meaning that if A has a beatpath to B, then
B must have one to A."

Though that's a correct statement of what it means for B to not
have an unreturned beatpath to him, that isn't what I meant to
say. I meant an unreturned _defeat_, an unreturned _one-step_
beatpath. Though it can be shown, using only a little more space,
that no member of the Schwartz set has an unreturned beatpath
to him, I don't need that result here, and I prove the weaker
result that no member of the Schwartz set has an unreturned
defeat. In other words: If A beats B, then B has a beatpath to
A.

2. In the last paragraph on the 1st page of that letter, there's
a typo: The clause "...since neither B nor any S2 member beats
A or any S2 member." should be replaced by:

"...since no S2 member beats any S1 member."

3. In the middle paragraph on the last (2nd) page of that
letter the phrase "unreturned beatpath" should be replaced with
"unreturned defeat". As I said, saying that any candidate in
the Schwartz set can't have an unreturned defeat means that if
any B in the Schwartz is beaten by some A, then B must have a
beatpath to A.

That means that any defeat in the Schwartz set is in a cycle, and
the rest of that page is correct.

***

Sorry about the typos, and the necessity of writing again to
correct them. Now this series of letters is truly concluded.
But let me know if there are any questions, objections,
disagreements, etc.

Mike Ossipoff

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