[EM] One Man One Vote in equation form; Power and rejecting Approval
Craig Carey
research at ijs.co.nz
Wed Jul 31 00:11:05 PDT 2002
Forwarded
>In this message I provide a definition of One Man One Vote that I
>wrote down earlier this year.
>
>
> >From:· Richard Moore <rmoore4 at c...>
> >Date:· Tue Jul 30, 2002· 2:20 pm
> >Subject: ·Re: What are we all about?, etc.
>http://groups.yahoo.com/group/election-methods-list/message/9924
>
> >It's simple to do this for IRV: If there are N candidates, and M
> >candidates are eliminated before there is a pairwise winner against
> >the remaining candidates, then it is possible that that winner
> >is a pairwise loser in M contests. This looks very bad for IRV if M =
> >N - 2 and N is large. IRV needs to test the winner against M
> >additional candidates.
> >
>
>The argument is based on an error which is that pairwise losing is somehow
>able to provide information on the right winner. Whatever the argument is,
>is rejected for using the idea of considering pairwise winners.
>
>My Ppos and Pneg equations take account of votes. Pairwise comparing loses
>information by producing Boolean values that then lead to tiny little
>"paradoxes". Some of the Condorcet fixes back out of the mistake. If your
>argument is valid then repaired Condorcet methods would be rejected too
>for paying too close an attention to "voters". [sic]. All we need then is
>to assume that the possibly false idea saying that the Condorcet method is
>like repaired Condorcet methods, and the idea that pairwise comparing is
>like Condorcet, and then I have an argument saying that the assumption
>that pairwise comparing is due respect, leads to a conclusion that it is
>not.
>
>The comment "looks very bad" is simply rejected. No method is found to be
>wrong by using a wrong test. It Mr Moore wants to cut the wrong parts of
>a pairwise comparing test out of it then that could be attempted and then
>the details posted up.
>
>------------------------------------------------------------
>
>In the following equations
>
>W is the method
>L is a truncated preference list
>"." is set intersection or Boolean "AND", and in this equation
>"#" returns the number of elements in a set or list.
>"\" means the complement of the set. The universe set is the set of all
>candidates. (Remember that candidates can win even when no paper names the
>candidate).
>a set "."-ed with a list, returns a set.
>
>H is a base election point.
>
>H+S is that H election that has added to it, the paper S that is having
>its power tested. (Actually S is a combination of papers).
>
>H+T is the H election that has added to it, a weighted combination of FPTP
>papers, where T is in that base of a simplex. What the "13Mar2002" formula
>does is check that T can get an outcome as good as what S got. It is not
>that obvious in this "P4" because the use of an inductive argument
>involving the deletion of digits on both sides of a "<=" comparing 2
>binary numbers smaller than 1, has driven some of the meaning into other
>formulae with different #L values.
>
>--------------------------------------------------------------------
>(P4 13Mar2002) =
>not (Exists L, t, H, S) [
>· [ ·(S perpendicular to #L FPTP axes for L)
>· · · (All k)(0 <= t.S[k])(abs((Sum k)S[k]) <= abs(t))
>· · · (W(H+S) is defined)
>· · · [ (0<t)({}=L.\W(H+S)) or (t<0)({}=L.W(H+S)) ]
>· ] and
>· · · (All T) [
>· · · · [ ·(T is a sum of #L weighted FPTP axes for L)
>· · · · · · (All k)(0 <= t.T[k])(t = (Sum k)T[k])
>· · · · ] implies [
>· · · · · · (W(H+T) is defined)
>· · · · · · [ (0<t)({} <> L.\W(H+T)) or (t<0)({} <> L.W(H+T)) ]]]]
>--------------------------------------------------------------------
>
>(1) The P4 rule is adjusted so that it excuses a method for imposing a
>need to vote for the wrong candidate. The aim is to find out if a paper
>can have more power than 1 FPTP paper, without checking to see if the
>paper has to name the candidate being advanced. [One Man One Vote
>overlooks insincerity.]
>To do that
>· ·"S perpendicular to #L FPTP axes for L"
>is rewritten as
>· ·"S does not contain FPTP papers"
>and
>· ·"T is a sum of #L weighted FPTP axes for L"
>is rewritten as
>· ·"T is a sum of FPTP papers"
>That changes the rule: let the new rule be called P5.
>
>(2) Also define a variable, p, called "power". "t" is the count of the
>FPTP papers added, and "(Sum k)S[k]" is the count of the sample tested
>paper. To find out what the power is, the weight of the T papers can be
>divided by p to give a value to compare against the tested paper.
>
>(3) Assume that the method W has one winner.
>Without effect, let the length of list L, equal 1. Let L = {g}:
>· · · ({}=L.\W(H+S))· = ·(g in W(H+S))
>· · · ({}=L.W(H+S)) · = -(g in W(H+S))
>· · · ({}<>L.\W(H+T)) = ·(g in W(H+T))
>· · · ({}<>L.W(H+T))· = -(g in W(H+T))
>
>--------------------------------------------------------------------
>P = not (Exists L, t, H, S) [
>· [ ·(S does not contain FPTP papers)
>· · · (All k)(0 <= t.S[k])(p.abs((Sum k)S[k]) <= abs(t))
>· · · (W(H+S) is defined)
>· · · [ (0<t)(g in W(H+S)) or (t<0).-(g in W(H+S)) ]
>· ] and
>· · · (All T) [
>· · · · [ ·(T is a sum of FPTP papers)
>· · · · · · (All k)(0 <= t.T[k])(t = (Sum k)T[k])
>· · · · ] implies [
>· · · · · · (W(H+T) is defined)
>· · · · · · [ (0<t)(g in W(H+T)) or (t<0).-(g in W(H+T)) ]]]]
>--------------------------------------------------------------------
>
>That can be split into 2 parts: P = Ppos & Pneg
>
>--------------------------------------------------------------------
>Ppos(p) = not (Exists g, t>0, H, S) [
>· [ ·(S does not contain FPTP papers)
>· · · (All k)(0 <= S[k])(p.(Sum k)S[k] <= t)
>· · · (W(H+S) is defined)
>· · · (g in W(H+S))
>· ] and
>· · · (All T) [
>· · · · [ ·(T is a sum of FPTP papers)
>· · · · · · (All k)(0 <= t.T[k])(t = (Sum k)T[k])
>· · · · ] implies [
>· · · · · · (W(H+T) is defined)
>· · · · · · (g in W(H+T)) ]]]
>--------------------------------------------------------------------
>Pneg(p) = not (Exists L, t<0, H, S) [
>· [ ·(S does not contain FPTP papers)
>· · · (All k)(S[k] <= 0)(t <= p.(Sum k)S[k])
>· · · (W(H+S) is defined)
>· · · not (g in W(H+S))
>· ] and
>· · · (All T) [
>· · · · [ ·(T is a sum of FPTP papers)
>· · · · · · (All k)(T[k] <= 0)(t = (Sum k)T[k])
>· · · · ] implies [
>· · · · · · (W(H+T) is defined)
>· · · · · · not (g in W(H+T)) ]]]
>--------------------------------------------------------------------
>
>Since Approval is expected to not fail any worse when checked with
>negative numbers, it can be assumed that W is fully defined:
>
>--------------------------------------------------------------------
>Ppos(p) = not (Exists g, t>0, H, S) [
>· [ ·(S does not contain FPTP papers)
>· · · (All k)(0 <= S[k])(p.(Sum k)S[k] <= t)
>· · · (g in W(H+S))
>· ] and
>· · · (All T) [
>· · · · [ ·(T is a sum of FPTP papers)
>· · · · · · (All k)(0 <= t.T[k])(t = (Sum k)T[k])
>· · · · ] implies [
>· · · · · · (g in W(H+T)) ]]]
>--------------------------------------------------------------------
>Pneg(p) = not (Exists L, t<0, H, S) [
>· [ ·(S does not contain FPTP papers)
>· · · (All k)(S[k] <= 0)(t <= p.(Sum k)S[k])
>· · · not (g in W(H+S))
>· ] and
>· · · (All T) [
>· · · · [ ·(T is a sum of FPTP papers)
>· · · · · · (All k)(T[k] <= 0)(t = (Sum k)T[k])
>· · · · ] implies [
>· · · · · · not (g in W(H+T)) ]]]
>--------------------------------------------------------------------
>
>(Largest Power) = (Largest Power of a Paper in method W) =
>· · ·(The least positive p such that (All r>p)[ Ppos(r) . Pneg(r) ]
> [(or "All r>=p")]
>
>('One Man One Vote' test pass mark) = ((Largest Power) <= 1).
>
>There is a definition of One Man One Vote that is a weaker test than
>it, but maybe exactly right when the number of preferences on each
>ballot paper equals two or less.
>
>If there is a problem with it, then do identify the exact parts of the
>equation that are permitting any such concern to arise.
>
>The equations and Approval can be very easily checked with a QE solver,
>(a Quantifier Eliminator system), e.g. REDLOG of Germany.
>
>Thus... One Man One Vote rejects Approval, and also One Man One Vote
>is clarified up. Persons that like Approval but who ignore the
>equations, likely could not be mathematicians inside of the topic of
>preferential voting (or maybe some similar topic).
>
>
>---------------------------------------------------------------------
>
>The same message of Mr Moore
>
> >From:· Richard Moore <rmoore4 at c...>
> >Date:· Tue Jul 30, 2002· 2:20 pm
> >Subject: ·Re: What are we all about?, etc.
> >...
> >
> >However, Alex was merely developing an idea (in the collaborative
> >environment of the list), not submitting it for publication in a
> >mathematical journal, so most of Craig's pedantic comments in that
> >post are uncalled for at this stage.
>
>I can't see how Alex's ideas threaten blacks in USA. But Appoval
>could since it presumably disadvantages uneducated persons.
>
>Alex's ideas were unfixable and vague.
>Something to be swept under a mat perhaps.
>
> >
> >4. Craig's anti-approval diatribe (July 27) fails to point out any
> >real problems. Examples presented on the EM list and elsewhere lack a
>
>The next step could be to say that he doesn't see a fault in my
>equations that define power.
>
> >sufficient number of candidates for his taste. Yet the existence of
> >small examples does not imply failure with large numbers of
> >candidates. Will Craig point out how increasing the number of
>
>What a dud comment: once a method is failed by a Boolean-valued test then
>the testing can stop since any further "AND"-ing with "False" won't alter
>that fail mark.
>
> >candidates would make approval deteriorate?
> >
>
>Is there anything else you want, Mr Moore, in addition to having equations
>testing the reject Approval method, to have a nature that varies with
>"time" ?.
>
>Of course increasing numbers of candidates make Approval deteriorate.
>
>I will check back every 10 years and see if the Approvalists have
>cast comprehensible doubts on my power equation that is continued in
>this message. You can test the method yourself, allowing 1 winner,
>any number of candidates, and letting no paper have 3 or more
>preferences. Some of the Approvalists would live a long time and may
>want to investigate the method. Probably not many.
>
>--
>
> >I would like Craig to present an argument, at least as coherent as the
> >one above, for why Approval fails if the number of candidates is >large,
>and to state what test it is failing.
> >
>
>Approval is therefore a method that no government would want to use.
>However the equation has not been used to test the method. There is
>no doubt it will fail. It does not reject Rob Richie's IRV method
>(maybe Alaska 's after a while) since it is not paying attention to
>the appalling "power<0" problems.
>
>That "One Man One Vote" is not designed by seeing what passes IFPP
>and what fails others methods.
>
>Readers can see how the Approvalist's undo the damage to their movement
>that this new mathematical formula is providing.
>
>
>
>=======================================================================
>
>References
>
>[1] Use of the method on Borda:
>http://groups.yahoo.com/group/politicians-and-polytopes/message/173
>
>[2] An implementation of an older version of Ppos:
>
>--------------------------------------------------------------------------
> >167: procedure corrupt(mn); begin
> >167: zp1 := rlqe all({ta,tb,tc},
> >167: · ·( (0<=ta) and (0<=tb) and (0<=tc) and (t = ta+tb+tc) )
> >167: · ·impl ( ( · · ·isok(· · · Ha0+ta, Hab, Hac,
> >167: · · · · · · · · · · · · · · Hb0+tb, Hbc, Hba,
> >167: · · · · · · · · · · · · · · Hc0+tc, Hca, Hcb) ) and
> >167: · · · (stri nott meth(mn,a, Ha0+ta, Hab, Hac,
> >167: · · · · · · · · · · · · · · Hb0+tb, Hbc, Hba,
> >167: · · · · · · · · · · · · · · Hc0+tc, Hca, Hcb) ) ) );
> >167: zp2 := zp1 and (· (0 < t) · and ·(0<=Sab) and (0<=Sac) and
> >167: · · · · · · · · · · · · · · · · ·(0<=Sbc) and (0<=Sba) and
> >167: · · · · · · · · · · · · · · · · ·(0<=Sca) and (0<=Scb) and
> >167: · · · ( · · · · ·isok(· · · Ha0, Hab+Sab, Hac+Sac,
> >167: · · · · · · · · · · · · · · Hb0, Hbc+Sbc, Hba+Sba,
> >167: · · · · · · · · · · · · · · Hc0, Hca+Sca, Hcb+Scb) ) );
> >167: zp3 := gs2 stri zp2;
> >167: zp4 := zp3 and (p * (Sab+Sac + Sbc+Sba + Sca+Scb) <= t) and
> >167: · · · (stri · · ·meth(mn,a, Ha0, Hab+Sab, Hac+Sac,
> >167: · · · · · · · · · · · · · · Hb0, Hbc+Sbc, Hba+Sba,
> >167: · · · · · · · · · · · · · · Hc0, Hca+Sca, Hcb+Scb) );
> >167: p5p := rlqe all({t,
> >167: · · · · · · · · Sab, Sac, · · ·Sbc, Sba,· · · Sca, Scb,
> >167: · · · · · ·Ha0, Hab, Hac, Hb0, Hbc, Hba, Hc0, Hca, Hcb }, nott stri zp4);
> >167:
> >167: if 'list = part(p5p,0) then begin · write "REDLOG: p5p = "; write p5p;
> >167: end else begin
> >167: · ·p5p := gs2 p5p;
> >167: · ·write "CORRUPT: p5p = ", p5p;
> >167: end;
>[the other test I call the dual follows immediately after]
>
>From: http://groups.yahoo.com/group/politicians-and-polytopes/message/172
>Sample output showing Borda has a max power value of 2.0, which implies
>that that method fails One Man One Vote since 2>1.
>
> >% CLASSICAL BORDA:
> >x4 := 4/3;
> >x2 := 2/3;
> >
> >> · · · · · · · · ·2
> >>CORRUPT: p5p = 3*p ·- 4*p >= 0 and p > 0
> >>
> >> · · · · · · · · ·2
> >>CORRUPT: p5m = 3*p ·- 4*p >= 0 and p > 0
>
>"nott" means 'not'
>"stri" makes inequalities strict
>"lax" makes inequalities lax
>"rlqe all(Vars, Expr)" is "(All vars...)(Expr(vars))
>"rlqe ex(Vars, Expr)" is "(Exists vars...)(Expr(vars))
>"gs2" simplifies.
>T*,H*,S*,t are symbols for Reals that do not contain a single Real
>If "p5p" is false then the method
>
>--------------------------------------------------------------------------
If the formula is modified then the "p" power number can be negative and
the power of the so called IRV (STV) method could be precisely checked.
Here is a question:
-----------------------------------------------------------------
Is the most negative power of any 3 candidate STV ballot paper
in any given election, ever less than -1/2 ?.
-----------------------------------------------------------------
>Craig Carey
>Wed 31 July 2002
The headers (resent after being posted while I was not subscribed):
>To: election-methods-list at eskimo.com,
> politicians-and-polytopes at yahoogroups.com
>From: Craig Carey <research at ijs.co.nz>
>Date: Wed, 31 Jul 2002 18:52:41 +1200
>Subject: One Man One Vote in equation form; Approval is rejected
>Reply-To: politicians-and-polytopes at yahoogroups.com
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