[EM] Did Mr May bungle 2 candidate SNTV election maths to hide the fact of 5 papers?

[Craig Carey]

At 2005-01-02 08:44 -0500 Sunday, Eric Gorr wrote to Election-Methods-List

>Craig Carey wrote:
>>At Sat Jan 1 16:08:58 PST 2005 , Marcus Shulze wrote:
>> | Hallo,
>> 
>> That is an incorrect spelling of "hello".
>
>Yes, it is.
>
>It's German.
>

I could not find text of 1952 that is by Mr May in 1952.

--- -- -

Here is a derivation of the solution of the 2 candidate 1 winner election.
This is not the only way to derive the solution.

Producing political elections in the IFPP style involves:
 [1] Embedding a method (First Past The Post or SNTV)
 [2] Casting shadows using the strict fairness axioms.
 [3] Using a proportionality aim.


Problem: solve this 1 winner election:

   a0  (A)
   ab  (AB)
   b0  (B)
   ba  (BA)

My P2 axiom P2 is in the set of axioms. It says that there is no change
when adding or subtracting these papers:

  (),
  -2*() + (A) + (B),
  -1*(AB) + (A),
  -1*(BA) + (B).

To the original election, add these papers:

   ab*((AB)-(A)) + ba*((B)-(BA))

Then this election has exactly the same winners:

a0+ab  (A)
b0+ba  (B)

Next, make it an axiom that First Past the Post is embedded.

The winners are immediately found to be:

 (b0+ba < a0+ab) implies A wins
 (a0+ab < b0+ba) implies B wins

Undoing the step of adding papers, leaves the winners unchanged.

Conclusion: these axioms were not used:
 * Right number of winners
 * Truncation Resistance, Later No Harm.
 * Monotonicity (Mono-Raise-Random) & Generalized Monotonicity
 * A 'Power<=1' rule
 * Proportionality. However, proportionality could be added
   with embedding removed.

Mr Woodall defines a rule Symmetric Completion rule in Issue 3
of Voting Matters:

   http://www.mcdougall.org.uk/VM/ISSUE3/P5.HTM

_____________________________________________________________________
...
| If one were using this For example, suppose that there are five
| candidates, A, B, C, D, E. Then
|
|     * the symmetric completion of a ballot marked (AB) consists
|       of six ballots, each with weight 1/6, completed in the six
|       different possible ways: that is, (ABCDE), (ABCED), (ABDCE),
|       (ABDEC), (ABECD) and (ABEDC);
|
|     * the symmetric completion of a ballot marked (A) consists of
|       24 ballots, each with weight 1/24, completed in the 24
|       different possible ways; and so on.
|
| Symmetric-completion. A truncated preference listing should be
| treated in the same way as its symmetric completion.
...
_____________________________________________________________________


Rule P2 is defined (by me) to fill in only the next empty preference.

Mr Woodall's Symmetric Completion is defined to fill in all of the
 subsequent empty preference-places in the ballot paper.

It seems that the two rules are the same.

The Symmetric Completion rule says that there is no change to all winners
 when these expressions are added:

[eq1]:  -6*(AB) + (ABCDE) + (ABCED) + (ABDCE) + (ABDEC) + (ABECD) + (ABEDC)

[eq2.1]:  -2*(ABC) + (ABCDE) + (ABCED)
[eq2.2]:  -2*(ABD) + (ABDCE) + (ABDEC)
[eq2.3]:  -2*(ABE) + (ABECD) + (ABEDC)

[eq3.1]:  -1*(ABCD) + (ABCDE)
[eq3.2]:  -1*(ABCE) + (ABCED)

Modify [eq2] using [eq3].
So no winners change when this is added/subtracted:

       -2*(ABC)  + (ABCDE) + (ABCED);
<-->   -2*(ABC)  + (ABCD) + (ABCE);

That is a P2 rule. Now modify [eq1] using [eq2]. No winner changes on
 adding these:

     -6*(AB) + [(ABCDE)+(ABCED)] + [(ABDCE)+(ABDEC)] + [(ABECD)+(ABEDC)]
<--> -6*(AB) + 2*(ABC) + 2*(ABD) + 2*(ABE)
<--> -3*(AB) + (ABC) + (ABD) + (ABE)             [a P2 change]

Similarly for the 24*(A) term.

Mr Woodall's Symmetric Completion ought have these 2 subrules added to it:

   All win-lose statuses do not change when these are added or removed:

     (A)+(B)+(C)+(D),  ().

The P2 and Symmetric Completion rules are indifferent to how many winners
there are.

The zero and two winner cases are solved in the same way.

This mailing list possibly has people that would pride themselves on how
they are unable to solve the zero winner "roll call" election.

Mr Heitzig is a name that pops up: <<my principles suck and are irrelevant
in all 2 or 0 winner elections>>. So they fix it, right ?. No.

I guess Mr Schulze can't solve 0 winner 1 candidate elections using the
exact same axioms that are used for 3 candidate 1 winner theories.


---

I sent this text to Mr Venkze today. I have no idea why Mr Venkze was
sending crap to me; i.e this text ("if more than half the papers ...")


| ...
| >There are some axioms that IFPP doesn't meet. Mike Ossipoff has one
| >that says if more than half the voters rank X relatively above Y, and
| >do not list Y at all, then Y can't win.
|
| Here are some problems with that idea that you attributed to Mr OSSIPOFF
|
|  * Suppose there are no voters. He booby trapped the definition so that
|    it says nothing if there are no voters. There is zero tolerance for
|    that in general (except at the Election Methods List).
|
|  * It is incredible that the author holds the false belief that there
|    is one vote for each person.
|
|  * Also it can't handle these 2 cases: fraudulent remarking of papers,
|    and the case of some people casting a number of votes that is not
|    equal to one.
|
|  * Suppose that some of the counts are negative. Surely OSSIPOF does
|    not expect us to observe the word "voters" and then conclude that the
|    counts are non-negative. Prohibiting negative numbers is done by the
|    voting system.
|
|  * It lacks the differential/comparative form that acceptable fairness
|    axioms would have.
|
|  * The words "rank X relatively above Y" fail to the summation and
|    also fail to define the weighting numbers.
|
|  * A statement saying "do not list y at all" is probably inconsistent
|    with the corollary saying that the presence or absence of the last
|    preference makes no difference.
|
...
|  * The word "relatively" seems to have no meaning. I.e. what it means
|    seems to be unknown.
|
|  * If the number of winners equals the number of candidates then MIKE
|    OSSIPOFF has got a rule that is inconsistent with the axiom
|    requiring the the number of winners be correct.
|
|  * As usual, MIKE won't fix his definitions.
|
| ---
|
| It seems suspect that you quoted Mr OSSIPOFF and yet didn't find a
| fault in his definition.
|


We already know that Mr SCHULZE is no good as a public relations matters.
What is the first public interest argument that MR SHCULZE would use when
reversing some PR (public relations) mishap ?.


G. A. Craig Carey