I have put online some REDLOG & REDUCE code that will simplify fully
general non-convex polytopes.
Simplification involves the removing of redundant flats specifying faces
(i.e. half-flats, terms that are inequalities).
It seems to handle complex numbers and also inequalities about quadratics
and so one.
Mr Catchpole has worded his theorem below in a 'we shadow a known solution
to give a new known solution' form, and that was followed up and the rule
was discarded for implying FPTP.
I wrote a computer subroutine to cast (P1) B-wins shadows. It might have a
bug, since I found B-must-win region/shadow that intersected with a
B-must-lose shadow/region. Here is the code to cast shadows of B-wins
regions using a simplex (P1) light source:
------------------------------
kknoa := (0<ab)and(0<ac)and(0<ad)and(0<b)and(0<c)and(0<d);
procedure bwsh(bw2); begin % cast shadow of B-wins
scalar f,kon;
clear r,u,v,w,x,a,b,c,d,ab,ac,ad,tab,tac,tad,tb,tc,td;
bw2 := sub({a=ab+ac+ad},bw2); bwshf := {};
kon:=(0<=ab+r-w-x)and(0<=b-r-u-v)and(0<=r)and(0<=u)and(0<=v)and(0<=w)and(0<=x);
f := rldnf(kon and sub({tab=ab,tac=ac,tad=ad,tb=b,tc=c,td=d},
sub({ab=tab+r-w-x,ac=tac+w,ad=tad+x,b=tb-r-u-v,c=tc+u,d=td+v}, bw2)));
bwshres := rlsimpl rlqe ex({r,u,v,w,x}, f);
write "Finished, the answer is in bwshres. Checking...";
if 'true = rlqe rlex (kknoa and bw2 and not bwshres) then begin
write "BWSH FAILED CHECK", rlqe rlex (kknoa and bw2 and not bwshres);
bwshf := gs3 (kknoa and bw2 and not bwshres);
end else
return rlsimpl sub({ab=a-ac-ad}, kknoa and bwshres);
end;
------------------------------
The papers are only these 6: AB, AC, AD, B, C, D. (1 winner) 4 candidates.
Notes: sub({x=y},z) replaces all x variables with y.
rlex creates ("There Exists") for all free variables.
rlqe rlex returns true if the region is not empty.
rlsimpl is REDLOG's fast simple simplifier.
rldnf expands the equation.
Casting shadows using simplex light sources is obviously very easy in
REDLOG.
(REDUCE is used in mathematics departments apparently.
REDLOG is LISP code package for REDUCE. REDUCE runs in LISP.)
-------------------------------------------------------------------------
A rule of Mr Catchpole is examined and rejected for requiring that a
method (where the papers are constrained to be just 3) is FPTP...
At 15:39 08.12.99 , David Catchpole wrote:
>Can I call 'em the Metameucil systems, can I? Can I?
>
...
>
>The addition (removal) of a candidate does not, for any other candidate,
>increase (decrease) the probability of that other candidate winning.
>
That is clear enough. It is not something people will agree is desirable
since it does not match what happens in elections. In reality with
actual preferential voting where preferences lists are not full, and
people don't mind to what happens to all the candidates for which no
preference was expressed. They are only interested in the candidates that
they vote for.
Perhaps election methods could allow them to cast votes having a power of
10 votes, and allow them to cast upto 10 positive votes or upto 10 papers
with negative votes to help some candidates lose.
Add a constraint. Let the papers be just these, and find the 1 winner solution:
AB a
B b
C c
A wins = (b<a) and (c<a)
B wins = (A loses) and y(a,b,c)
C wins = (A loses) and not y(a,b,c)
Any method with those three papers can be represented as coloured regions
inside a triangle, say with AB at the top, B at the left, and C on the
right.
The region for B on the lower left is coloured red, and C's region is coloured
green.
Where A wins is the same as in FPTP (say) (otherwise the method makes
outcomes for a candidate to be affected by subsequent preferences.
It can be asserted that the method must be not worse than STV).
All that is needed to define a method then, is the curve or piecewise
line segment between the centre point, and the midpoint in the middle
of the bottom edge.
Consider a point on the bottom edge of the triangle, say this
a=0, b=0.501, c=0.499
What happens when papers for AB are added?. If proportionality is the aim
then the winner would flip from B to C. That certainly happens in
1 winner STV and IFPP, so proportionality is some achievable. STV goes
bad near the centre (B's region is too large) and Mr Catchpole's rule
is found wrong very close to the bottom edge, so ignoring IFPP and using
STV is OK for the argment that follows.
This was written (repeated) [might have seen this rule before]:
Rejected 'Metameucil' or 'Regularity'(?) rule:
-------------------------------------------------------------------------
>The addition (removal) of a candidate does not, for any other candidate,
>increase (decrease) the probability of that other candidate winning.
-------------------------------------------------------------------------
Probability probably means a blurring or fuzzifying of the B-C boundary.
If so, the boundary between B (coloured red) and C (green, on the lower
right), would be a blurry strip coloured red-pink-yellow-yellow_green.
If papers for AB are added then the point that is the midpoint of the
bottom B--C edge of the triangle, is dragged upwards
Hence, the divide between the regions of B and C is a line. There does
not seem to be any involvement of probability at all.
A possibility is that the author of the rule had never tested a method
against it ever.
FPTP makes it past the rule. Perhaps David could tell us if FPTP is the
only method that is passed by the rule, for all numbers of candidates
(or maybe the method no longer needs to be considered(?)).
>Now, say you know how the system works for a two-candidate contest- for
>simplicity, assuming no dependence on the number of voters- that is, you
>know the function relating the ratio of probabilities of those two
>candidates winning to the ratio of voters ranking one over the other to...
>oh, blah, you get the idea.
>Now, let's make it a three candidate election. It's easy enough to see...
...
The rule seems to be just the sort of rule that would propogate a
requirement that the method be 'similar to FPTP', throughout all election
problems involving a larger numbers of papers or candidates.
FPTP =? "The family of "regular" probabilistic (stochastic) electoral systems"
-------------------------------------
Note to Mr Davison: in replying to my message, you quoted the message I had
replied to:
At 00:12 02.12.99 , Donald E Davison wrote:
>Greetings EM list,
>
> Craig Carey asked about someone from the Freedom Party to comment on MMP.
> The following letter is about MMP and by Bill Frampton, Vice President
>of the Freedom Party.
Also: New Zealand got a better electoral system (MMP) and wanted one. While
the government did change in some elections in the past, and at other times
it did not, FPTP was insulating politicians from public political opinion.
The public had an opinion that it didn't matter which party it was.
Proportionality has improved that. The choice of MMP allowed a near minimal
change (with the electorates did drop in number from 90-something to 60).
That's one viewpoint.
(Errata: I said that the NZ National party was losing for no good reason.
There must have been a good reason given how much they lost.)
What was a little interesting was how the "Green" party so closely lost in
Coromandel (near Auckland city on the peninsular on the east): the National
party tried to win there (and did) and the leader of the Green party might
have won that seat (and hence seats in parliament too [which they never
got since under the 5% threshhold]), but the Labour party never withdraw
their candidate from the Coromandel seat. They withdrew one in another seat
so they must have intended to risk havign the Greens lose. The labout party
got a really quite small number of votes in Coromandel and prompted the
Green party to lose. The Green party didn't exist as a separate party last
election (1996). Labour can deal with the right using its coalition majority.
Mr G. A. Craig Carey, [EMAIL PROTECTED]
Auckland, New Zealand.
Snooz Metasearch: <http://www.ijs.co.nz/info/snooz.htm>