[EM] STV: who could ask for anything more?

Craig Carey research at ijs.co.nz
Mon Jul 14 09:15:02 PDT 2003

Below, a reply to Mr Venkze:
   Ballot papers inside of ballot papers; Approval vs monotonicity


At 2003-07-13 19:14 +0100 Sunday, James Gilmour wrote:
>> 	What are the flaws of STV?
>Like several other systems, it is non-monotonic.
>Also, it can exclude a "Condorcet winner" when exclusions have to be

STV picks the wrong winners. E.g. a running average of 50% of the
councillors being the wrong people.

The Condorcet winner is a wrong test. I do not suspect that it is
friendly with a whole class of STV-like methods around that ideas of
P2, monotonicity, truncation resistance, 0<=power<=1.

It is not known that it can be worded up as more desirable than
rights. I.e. on the back of a ballot papers. Text on the back of
ballot papers looks bad if talking about the power of every other
voter, and pairwise involves just such a thing.


At 2003-07-13 04:46 -0400 Sunday, James Green-Armytage wrote:
>A small point I forgot to mention, in the category of STV vs. party list:
>	Open and closed lists aside, in list PR there is also the possibility
>that you will cast a vote for a party who doesn't get any seats at all.
>This is a wasted vote, basically, and so voters are given some incentive
>to avoid voting for parties they think might be a "lost cause".

The idea is perhaps, to change the party vote from being:
 * circles with felt pen crosses or ticks, to:
 * circles with integers (or one cross or tick).

That sound like a good idea. With it, small parties might have better
data on their own outcome weeks before the election since the public is
doing less tactical voting relying on recent data.

The party vote would be preferential and lack the bad behaviour of the
Alternative Vote (non-monotonicity).

It could be made to reduce vote wasting for all parties that is caused by
rounding the party vote percentage to a number of seats.

(PS. In New Zealand, the threshhold vanishes on a political party if they
can win a single electorate. For some, there is proportionally under the


I regard the following as unimportant.

At 2003-07-14 01:40 +0200 Monday, Kevin Venzke wrote:
> --- Craig Carey <research at ijs.co.nz> a écrit : > 
>> >Pass: Approval, Average Rating, Black, Borda, Bucklin, Kemeny-Young ,
>> >Median Rating, Minmax, Ranked Pairs, Schulze, Smith//Minmax,
>> >Sum of Defeats 
>> Since quite a lot of those methods are very unimportant, you could
>> proved you proved monotonicity of 2 or 3 for me.
>> I request the reasoning for the 10 statements that those methods are
>> passed. Approval is actually failed and I guess have conceded that
>> to the fullest possible extent and never got around to updating
>> your website. The wikipedia.org website got fixed somewhat over that.
>Are you saying Approval is not monotonic?  That would be interesting news to
>many of us.

How many ?.


I now request the monotonic rule, testing Approval, that you appeared to
imply that you hold in your question. (Please don't lose it before
delivering it. If I caught you without an answer, it seems that Dan Keshet
of www.wikipedia.com, has a *modified* monotonicity rule that tests
the "Approval" method. 


I am (now) saying that the checkbox-y Approval method is not monotonic
in the sense that it is a type of preferential voting method that can't
be tested by that rule. The rule has to consider 2 preferences and
know which is before and which is after.

There is some chance I could lose this.

The way to solve the question is to use the public definition, that
is in journals or books that have not got their definition too narror
or got it wrong. Then the understanding is replaced with symbols
and the understanding is discarded. Then changes to symbols that
don't disagree with the old meaning can be made.

That follows from some principle about the freedom of variables.


To check Approval (checkboxes only now), there is a enhanced
monotonicity (if it exists).

This modified monotonicity presumably would be putting checkboxes
in 2 groups.

The monotonicity rule has checks the ordering of 2 preferences.
The preferences are coded as symbols. E.g. "x", "y", and they
name a candidate, or they can be compared with the symbol of a
candidate, and then they are of a Boolean type.

So their type is changed into a set type.

Now Approval papers are preferential papers with the preferences
being sets. Also for that checkboxy method, their would be only
one preference since there are 2 states for each checkbox.

Now allow more preferences on the paper (so the method isn't
Approval now). All preferences would have to be sets.


Suppose a paper looks like this:

  = ({A},{B},{C,D,E})   % means:  1st={A}, 2nd={B}, 3rd={C,D,E}

Suppose also that the candidates are: A,B,C,D,E.

That would remind EMers of the "A > B > C=D=E" notation.

I assumed that readers wanted   X=Y=Z,

where X,Y,Z satisfy these equalities:

   X = ({A},{B},{C,D,E})
   Y = [ ({A},{B},{C,D}) + ({A},{B},{C,E}) + ({A},{B},{D,E}) ] / 3
   Z = [ ({A},{B},{C}) + ({A},{B},{D}) + ({A},{B},{E}) ] / 3

Any problem there can be removed by erasing a memory of the
"A > B > C=D=E" notation. It is only a notation or definition.

Suppose weights are attached to checkboxes on an Approval paper.
Now the preferences of ballot papers would seem to need to be
ballot papers themselves since there are numbers inside of sets.

Actually, for the mixed approval and dissapproval paper, there was
a desire to have the number +/- 1 with the preferences:

 ((+1,A), (+1,B), (-1,C))

But for the Approval need, this would be the form.

 ({+1(A)}, {+1(B)}, {-1(C)})

That has its list of sets of candidates changed into sets of
 papers. There might be good arguments to have +1 and -1 instead
of papers. since if the paper is this

 ({+2(A,D)}, {+2(B,D)}, {-3(D)})

then it is not yet defined how the base 2 satisfaction number of
(ADE) would rise when the 1st 2 preferences of this are swapped:

 ({+2(A,D)}, {+1(B,D)}, {-3(D)})

A meta-principle prohibits principles that are unusable for all
possible numbers of winners.

That notation can be ignored while optimizers are missing and all
ideal IFPP 4 candidate methods are missing.


Monotonicity is out of date in my opinion. To contradict, say why
it should not be merged with the rule of Truncation Resistance (TR) ?

  (TR is approx: shuffles of later preferences (on the
  last-preferences side) do not have an affect on the win-lose state
  of earlier preferences).

I expect that you (Mr Venkze) should be losing the argument that Approval
is tested by monotonicity if he does not send in old quoted EM
message text that shows that sets for all preferences were considered
to be worth having. The idea of freedom for the meanings of symbols
was being rejected here anyway.

If monotonicity checks Approval then it would check methods using
negative votes at non-integral spots. If monotonicity does not check
at -(1+sqrt(5))/2 then the principle of freedom permitting Approval
to be tested being maybe/probably rejected.

Montonicity a rule for the public. P1 is easier to use. We could
find out if monotonicity is a thing for the publishers to the masses,
in the opinion of the list, by finding out that they would deleted
more weight from the 1-preference papers (X) than that paper
started out with.

The candidates are A,B:
The method is 1 winner FPTP.
The papers are:
   (B)  -1 votes
So the winner is A.
Now take off (1+sqrt(5))/2 votes:
   (B)  -1 votes
   (A) -(1+sqrt(5))/2 votes

That passes monotonicity OK. I have spotted obstinacy at this
mailing list over Real valued weights for preferences.


Approval was always believed to fail the One Man One Vote ideal.
I was not showing that since my test only handled preferential

Suppose the candidates are: A,B,C,D,E; and the Approval paper is:
        (A:[1], B:[1], C:[1], D:[0], E:[0])

or this ({A,B,C}) [An Approval paper with 3 marks].

Then the multiwinner power rule says appproximately that the
fragments can reproduce the same base 2 satisfaction.

Have the given list be (A).
After fragmenting up the paper ({A,B,C}), then this has to hold
if A wins, i.e. (A in Winnners(S + 1({A,B,C})))

   (Exists x,y,z)[(A in Winners(S + x({A}) + y({B}) + z({C})))]

If A was winning then A can be made to still win:

   (Exists x=1,y=z=0)[(A in Winners(S + ({A}))]

If A was losing then presumably the empty paper can be a fragment
so that can be reproduced too.

So it looks like 1 winner Approval would pass that
One Man One Vote style power rule. 

(Multiwinner Approval would fail a no-corruption rule as hoped.)

There is a possibility that the rule is not right or that another
rule is needed that has a feature of rejecting the Approval Vote.

Craig Carey

Ada 95 Computer Language Shootout: http://dada.perl.it/shootout/

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