[EM] "None of ..." -- Negative votes

Craig Carey research at ijs.co.nz
Wed Feb 23 09:34:34 PST 2000

Voting against candidates using negative votes.

At 23:05 23.02.00 , Donald E Davison wrote:
>Monday 2/21/00, Stephen Todd wrote:
>     If a voter gives his first preference to an actual candidate, he can
>then go on and give a second or subsequent preference to the [None of the
>Above]NA option, but cannot then indicate further preferences beyond NA.
>In this latter case, the voter is saying, in effect, 'None of the Rest'.
>Tuesday 2/22/00, Stephen writes:
>     We have never had a NOTA option for any public election in NZ, so I
>was not exactly sure of my ground when I suggested a 'None of the Rest'
>(NOTR) option for STV elections yesterday.  Do you think it is consistent
>with the rationale behind the NOTA option (as it is understood in the US)
>to also allow voters to indicate a lower preference for that option (in
>effect, saying None of the Rest)?

[A-start: shifted from below:]

>    I am supporting this None concept for the purpose of allowing the
>voters to reduce the number of seats when there is a shortage of candidates
>for any one election. I am not trying to reject any certain candidates. I
>want the voter to be able to vote for his most preferred preference and
>also vote to reduce the number of seats. For example, in Plurality-at-Large

If the voters are going to be allowed to reduce the number of winners,
 then why can they only reduce the number of winners by one?

>the voter can select candidates with most of his votes, but he can also
>indicate with some votes that he feels the number of seats should be less
>for this election.

A separate question could ask that.

>    This cannot be done in the methods of SNTV, Bottoms Up, nor Choice
>Voting. That is why I inserted the vote on reducing the seats to one half
>of available candidates for those three methods. Of course, another
>percentage could be used, like say, sixty percent of available candidates.


>Dear Steve, I like the term `None of the Rest' better than None of the
>Above. None of the Rest gives the voter some degree in selecting some
>candidates, while None of the Above indicates that the voter is rejecting
>all candidates. It is too much like `All or Nothing'.
>Steve: If [the] 5th preference were given for the NOTA option (and that
>preference was eventually activated), it would allow that part of the vote
>to remain in the election (to help the NOTA option attain the quota),
>rather than go straight to Non-transferable as it would if the voter were
>to stop numbering the candidates at his 4th preference.
>    It seems to me that voters would want to have more options available to
>Donald: Yes, I agree.
>Steve: Do you think the NOTR option (later preferences) is a valid
>extension of the NOTA option (unique first preference), having regard for
>the US understanding of the rationale for the NOTA option?
>Best regards
>Donald: I think the concept of the term NOTR would fly much better than
>NOTA. Most voters do not like the idea of giving up their vote for one
>candidate just so they can reject other candidates.
>    My current position is to use the term None of the Rest in place of
>NOTA. Most voters are not using NOTA at the present time so it should be
>possible to introduce NOTR.

Isn't this the NOTR method undefined?

If a NOTR preferences helps candidates of subsequent preferences lose,
 then shouldn't it also lessen the voter's ability to make candidates of
 earlier preferences win?. Otherwise a voter could get more power than
 that voter ought have.

[text here was shifted above]


If an election should have 10 winners and voters can only get the number
 of winners down to 9, but not down to 8, then why even give voters any
 option at all, to alter the number of winners. Two types of NOTA/NOTR
 would allow voters to reduce the winners by 2, but very few will know
 which "no further preferences specified" pseudo-candidate is the one to
 vote for.

A better idea is to use negative votes. If any candidate gets less than
 some threshhold then the candidate could lose, but if everyone votes
 against candidates not wanted, then why should none win.

If a voter had both positive and negative votes, it would be required
 they do not get more power than they are due.

Suppose a voter was allowed to cast 1.00 votes, and that voter could
 spread the votes across any number of voting papers.

This following example shows that it would be wrong to say that the sum
 of the squares of the weights of all the votes of that voter had to not
 exceed 1.00.
     Example: the voter seeks to have candidate B beat candidate C. All
        the other candidates are sure to lose. With a sum of squares
        scheme, the voter could cast +0.70710678 of a vote with a paper
        marked (B), and -0.70710678 of a vote on a paper marked (C).
      If that is done, then the voter would have a voting power of about
       41.4% more than that voter ought have.
It seems most plausible that the constraint on the votes that a voter
 can cast, should be a constraint on the sum of the absolute values of
 the voter's votes. What to put on the voting paper is a problem.

In large elections, voters could be only allowed to cast a single vote
 and plus they have to say whether it is for or against....

This is perhaps ideal, but STV might not do this:

Four plausible statements:

If candidate C wins (-20 ACDE, S) then C must also win (-20 ADCE, S).
If candidate C wins (-20 ADCE, S) then C must also win (-20 ADEC, S).
If candidate C wins (-20 ABC,  S) then C must also win (-20 AB,   S).
If candidate C wins (-20 C, S) then C must also win (-10 C, -2 B, S).

For P1 (all 4 of hte above), and for monotonicity (the 1st 3), the words
 "win" and "lose" need to be swapped apparently.

A check of STV against P2 with votes allowed to be negative.

P2: Suppose the papers are {A,B,C}. Suppose some C supporters voted
  against A then B, instead of voting for C:

        (a0:A, -x:AB,       b:B, c:C)
        (-2t:A, t:AB, t:AC,         )
        ( u:A,              u:B, u:C)

  A.  a0-2t+u,              a = a0-x
  AB  -x+t
  AC  t
  B.  b+u
  C.  c+u
(a<b)(a<c) => Winners = (if (c+u+t)<(b+u+t-x) then {B} else {C})
(b<a)(b<c) => Winners = (if c<a   then {A} else {C})
(c<a)(c<b) => Winners = (if b<a   then {A} else {B})

For the Alternative Vote, the set of winners is independent of the
 variables t and u, irrespective of their signs. So the Alternative Vote
 passes this particular P2 test.
In IFPP, a+u and b+u and c+u are compared against he quota (a+b+c+3u)/3,
 so IFPP passes this too.

IFPP with 3 candidates and 2 winners passes P2, since the losers of that
 election are the winners of the 1 candidate case which received negated

-------------------------------- STV --------------------------------

Below, the 3 candidate 2 winners STV formulae is tested against this
 P2. STV will pass if the STV formula for the set of winners does not
 depend on either t or u. An aim is to decide which seems to be the
 better for winners: Hare-STV, Droop-STV.

A 3 candidate 2 winner STV case.

  A.  a0-2t+u,              a = a0-x
  AB  -x+t
  AC  t
  B.  b+u
  C.  c+u

"Quota" = Q = (a+b+c+3u)/K.     "Sum" = S = (a+b+c+3u).
K=3 in "Droop-STV", and K=2 in "Hare-STV".
Since K<=3, this is false: (Q<a+u)&(Q<b+u)&(Q<c+u).

Case (c+u<Q)(Q<b+u):  Winners = (if a<c then {B,C} else {A,B})
Case (b+u<Q)(Q<c+u):  Winners = (if a<b then {B,C} else {A,C})
Case (Q<b+u)(Q<c+u):  Winners = {B,C}
Case (b+u<Q)(c+u<Q)(a+u<Q): All counts are under the quota.
   This case does not occur with the Droop-STV quota.
   The winner is given by the Alternative vote, which was already found
   to find a set of Winners that was independent of t.

Case (b+u<Q)(c+u<Q)(Q<a+u):
   Surplus for A: S = (a+u)-Q = a+u-(a+b+c+3u)/K
   After transferring papers away from candidate A, and then ignoring
   A's bundles of papers, the papers & counts become:
     B   (b+u) + T.(-x+t)
     C   (c+u) + T.t
   Let the difference be YBC, YBC = (0 < c - b + x.T)
                      Winners = (if YBC then {A,C} else {A,B})

   Case "T1": Transfer Value = T = S/a = (a+b+c+3u)/a   ("Standard STV")
      YBC = 0 < c - b + x(a+b+c+3u)/a
          = 0 < a.c - a.b + x(a+b+c+3u)      (provided 0<a)
      The set of winners is independent of t.
      I much suspect that YBC (which decides whether the winners are
      {A,B} or {A,C}) is not dependent on u, despite it containing u.

   Case "T2": Transfer Value = T = min(1, S/((t-x)+t)
                                    ("Irish E.U. more-proportional STV")
      The set of winners is not independent of both u and t.
      Case 0<a+b+c+3u: independent of t iff 0<=(2t-x)<=(a+b+c+3u)
      Case a+b+c+3u<0: independent of t iff (a+b+c+3u)<=(2t-x)<=0
      The set of winners could easily be dependent on t.

[All cases have been covered, but not in detail]

Conclusion: It seems to me that standard "T1" STV, with 3 candidates
 passes "P2", in these 4 cases: Droop and Hare quotas (i.e. 1/3 and 1/2),
 and 1 and 2 winners.

The Irish European Community version of STV (a system in which Mr Paisley
 won once), seems a less promising method for elections where voters can
 cast negative votes.

If too many try to vote out the same candidate with the 1st preference,
 then method will attempt to remove candidates chosen by subsequent

It would be simple enough to use negative votes in a First Past the Post
 election. What if 35% vote with negative votes and only half of the
 candidates get more than 0.


At 04:08 10.02.00 , Craig Carey wrote:
>TR := rlqe rlall ((simpu.simpv) => (ta => (wua = wva)))
> 10 free variables. I.e. it is a  "(For All)" operator (existential
> logic), that is true iff the polytope is not empty.

This is more true:
 TR is true iff the region "(simpu & simpv)" is inside the region
   "(ta => (wua = wva))".

G. A. Craig Carey

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