User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting

GIF for negative counting. Click on the image and then the thumbnail of the image to see the animation.

The negative counting approach is an alternative method of doing pairwise counting. It is faster, depending on implementation, when voters don't rank all of the candidates (or when they rank multiple candidates last), because it takes advantage of the fact that in most ranked voting elections, voters are assumed to prefer every candidate they ranked over every candidate they left unranked.

A simple way of describing it is that, if you know that 5 voters ranked a candidate (A), and 3 of them ranked A below some other candidate (B), then 2 voters must have ranked A above or equal to B.

Description

There are two steps to the negative vote-counting approach: the vote-counters must capture certain pieces of information, and then some math is done on this information to find the final result.

1. Vote-counting: The precinct vote-counters count the following values for a given candidate:
   1. The number of voters who ranked/rated/marked a candidate on their ballot.
   2. In each head-to-head matchup, the number of voters who explicitly ranked that candidate below the other candidate ("explicitly" meaning they actually marked/ranked that candidate).
      • (This can be considered as, for a given ballot that ranks two candidates A and B as B>A, either:
        • Counting the number of votes for A<B. This yields a positive value.
        • Adding a negative vote (-1 votes) for A>B. This yields a negative value.)

1. Math: The final number of votes for the first candidate against the second candidate in each head-to-head matchup is then found by, if treating the second value as a positive number, subtracting the second value for the first candidate from the first value (addition is done instead of subtraction if the second value is treated as a negative number).
   •  This can be succinctly summarized as, for finding the vote totals in a matchup between candidates A and B:

     A>=*B = A - A<B

     * meaning that the number of ballots ranking A over or (explicitly) equal to B, is equal to the number of ballots ranking A, minus the number of ballots ranking A below B (i.e. B over A).

The number of ballots marking each candidate can be placed in the blank cell comparing themselves to themselves in the pairwise matrix i.e. for candidate A, the cell A>A would contain the number of voters ranking A.

Note: When using this approach, there is an important caveat to consider when dealing with voters who explicitly rank two candidates equally; see the #Dealing with equal-ranking section below.

Example

If the votes are:

10: A>B 5: B

then each candidate is explicitly marked on:

• A: 10 ballots
• B: 15 ballots

with the number of ballots explicitly ranking one candidate below the other being:

• A<B: 0 ballots
• B<A: 10 ballots

The math that shows the final number of votes in favor of each candidate is then:

• A>B is (10-0)=10 votes.
• B>A is (15-10)=5 votes.

This can be demonstrated as:

	A	B
A	10 ballots	0 votes
B	-10 votes	15 ballots

becoming:

	A	B
A	---	10 (votes)
B	5	---

Dealing with equal-ranking

The negative counting approach, depending on implementation, can require even more markings when equal-ranking is allowed and it is desired to have traditional pairwise vote totals. Any implementation of negative counting will give accurate information about who won, tied, or lost each matchup (i.e. the pairwise margins), however. This is because if there are 2 candidates A and B, with the (explicit) votes being:

2 A>B

1 B>A

5 A=B

then either it can be marked that A wins against B by:

• 2 votes to 1
• 7 votes to 6

This is because the voters who equally ranked A and B can be considered to, in the A vs B matchup, either be voting for:

• Neither of them (similar to choose-one FPTP voting; this is the traditional pairwise counting approach).
• Both of them (similar to Approval voting).

This is related to how, in Approval voting, if A has 30 approvals and B 20, and no other information is supplied, then it is impossible to know whether any of the 20 voters who approved B also approved A or not.

This can change who wins in certain Category:Pairwise counting-based voting methods; for example, in the winning-votes version of a Category:Defeat-dropping Condorcet method, not only does it matter who wins the matchup, but also exactly how many voters genuinely preferred the winner to the loser in each matchup.

Note: Voters who don't rank either of the candidates in a matchup are generally considered to equally rank them, but no implementation of negative counting would consider them to vote for both candidates in the matchup. Only voters who have marked both candidates can be counted that way.

Example of the two approaches to equal-ranking

Suppose a voter had ranked 9 of 10 candidates as their 1st choices, and the 10th candidate was unranked (i.e. implicitly ranked last).

At least 9 marks must be made in any approach to negative counting, to indicate that 1 voter has ranked each of the 9 candidates who are each one of the voter's 1st choice.

In addition:

Explicitly equal-ranked candidates both get a vote

No extra work needs to be done.

Equal-ranked candidates don't get votes

For each matchup, the following number of markings can be made for two candidates A and B:

• 2 markings can be made (1 negative vote for A>B and 1 for B>A).
  • Note: In essence, this approach involves counting the number of voters who explicitly ranked a given candidate below or equal to the other candidate, rather than only below.
• 1 negative marking can be made for the A vs B matchup in general, which is later interpreted as a negative vote for both candidates.

In this example, there are 0.5*(9*8)=0.5*72=36 matchups to count between equally-ranked candidates. Accordingly, at least either 36*2=72 or 36 markings must be made.

Dealing with last-place candidates

It is not necessary to make any markings for a candidate a voter ranked as their last choice, because this means they wouldn't vote for that candidate in any matchups.

Write-in candidates

This advice is less relevant when write-ins are allowed, however, because even if a voter ranks a candidate last among the candidates named on their ballot, they are still implicitly ranking that candidate above all of the write-in candidates they didn't rank on their ballot. So if last-ranked candidates aren't counted, then it may be necessary to modify how the calculation is done to avoid giving the impression that the vote totals are accurate for the matchups involving write-in candidates.

It is possible to make a special marking for a last-choice candidate that indicates they are not preferred over any of the on-ballot (regular) candidates, but that they are preferred over all write-in candidates. It would then only be necessary to record negative votes for matchups involving write-in candidates who are ranked above the last-choice candidate on some ballots. This would mean making at most two additional marks for every last-ranked candidate on a ballot, because in practice voters are only allowed to write in one candidate. This can be compared to the regular approach to dealing with write-ins at Pairwise counting#Dealing with write-in candidates.

Semi-negative counting procedure

When a voter ranks more than half of the candidates, it is possible to further reduce the number of marks counted. This is because when a voter ranks one candidate below more than half of the candidates, instead of counting negative marks in that candidate's matchups against all of those candidates, it would be faster to count positive (regular) marks for that candidate against the fewer than the half of the candidates that they're ranked above. Example of this approach with a voter ranking all of 5 candidates:

Negative counting values italicized, regular pairwise counting values normal, and semi-negative counting values bolded and big

Voter's ranking		A	B	C	D	E
1st (choice)	A	1 ballot	1	1	1	1
2nd	B	-1	1 ballot	1	1	1
3rd	C	-1	-1	1 ballot	1	1
4th	D	-1	-1	-1	1 ballot	1
5th	E	-1	-1	-1	-1	1 ballot

Counting work for this example:

• With regular counting, this ballot would require at least 10 marks to count.
  • The same applies to negative counting when last-ranked candidates aren't counted.
• With semi-negative counting, it only required 6 marks to count.

With semi-negative counting, it is possible to reduce the number of marks required to count a ballot by roughly half or greater, depending on which of the other two approaches it is compared to, and how many of the candidates are ranked by the voter. The positive and negative marks can either be stored as separate values, or they can be combined during the counting.

Comparison to other vote-counting procedures

It can also be useful to compare these results to the amount of vote-counting work that would be done in other voting methods.

• For FPTP, one mark is made per ballot.

• With Score voting, the number of candidates that a ballot gives an above-minimum score to is the number of marks that are made for that ballot.
  • Note that this will generally be strictly greater than with Approval voting, because a voter is likely to score more candidates than they would approve; the general trend is that the more gradations/allowed scores there are, the more likely a voter is to score more candidates.
  • It isn't possible to know for certain how voters would score the candidates when provided only ranked data. However, it is possible to guess; for example, if there were certain candidates known to be frontrunners, then strategic voters would likely do min-max voting among those frontrunners, and adjust their scores for candidates they preferred more or less than their preferred frontrunner(s) accordingly. In addition, it can be assumed some voters would do normalization.
• With RCV, there is some additional work involved in transporting the ballots to a centralized location, and in doing multiple rounds of counting, rather than one; ignoring that, however, at least one mark must be made per ballot (to indicate 1st choices), and then in each sequential round where votes are transferred, the number of transferred votes is the number of additional marks that must be made in that round.

Comparison to the regular pairwise counting approach

Verbal comparison between the regular approach and negative counting:

• The regular approach: The precinct vote-counters manually count all of the voter's preferences in each head-to-head matchup; in other words, a candidate is assumed to be preferred only in the matchups where the vote-counters mark them as being so.
  • This can be slow, and also can make it difficult to accommodate write-in candidates (see Pairwise counting#Dealing with write-in candidates), since the vote-counters won't know ahead of time who those candidates are, and thus won't be able to indicate preferences in those matchups.
• Negative counting approach: The vote-counters mark a candidate as being ranked on a ballot, assume the voter who ranked them prefers that candidate in every matchup, and then show which matchups this is not true for.

It's actually possible to visualize, in a pairwise counting matrix, what values have to be recorded for each approach; this is because the cells of the pairwise matrix used in each approach are disjoint.

Example with a voter voting A>B=C>D with 6 candidates:

Negative counting values italicized and smaller (with optional negative counting values italicized but regular-size), Regular approach values bolded and enlargened

	A	B	C	D	E	F
A	1 ballot	1	1	1	1	1
B	-1	1 ballot	(1 equal-ranking)	1	1	1
C	-1	(1 equal-ranking)	1 ballot	1	1	1
D	-1	-1	-1	1 ballot	1	1
E
F

The negative counting values can clearly be used to reproduce the values created by the regular approach (see the Math part of the #Description), whereas the regular approach leaves each of the cells used by the negative approach blank (i.e. with a value of 0). In either approach, the values in each cell can be added up per ballot; see the #Comprehensive example section.

The "(1 equal-ranking)" part indicates that one voter equally ranked B and C; this value is optional in the negative counting approach (and is never used in the regular approach), but can be included if it is desired to have traditional pairwise totals, by using it to give both B and C negative votes in the B vs C matchup (see the #Dealing with equal-ranking section).

Formula for counting the required number of marks to be made

The formula is a series (though it is only accurate when equal-ranking isn't allowed; when it is allowed, then depending on implementation, this series may provide an upper bound on the number of marks that are to be made):

• In negative counting, the series starts at 0 when 0 candidates are ranked, and then the formula is that the number of marks that must be made for a given number of candidates that were ranked on a ballot is the value in the series for a ballot that had ranked one less candidate than that given number, plus the number of candidates ranked on the ballot.
  • This is because for each additional candidate added (ranked below all candidates already on a ballot), one mark is made to indicate that they were ranked, and [number of candidates already on ballot] marks are made to indicate the voter's preference for all of those already-ranked candidates over the newly ranked candidate.
• In the regular approach, take whatever number of marks would be produced in negative counting for a given number of ranked candidates on a ballot (see above bullet point), and then subtract it from the number of candidates that are ranked on a ballot multiplied by the number of non-write-in candidates in the election.
  • This is because, if the negative counting approach is thought of as "count the number of times a candidate is "preferred to themselves" (i.e. the number of ballots they're ranked on) and the number of ballots that prefer them less than each other candidate", then the regular approach doesn't require counting either of those pieces of information, but does require counting the number of times a candidate is preferred over each of the other candidates
    • For example, with 10 candidates, for a ballot with one 1st choice that doesn't rank any of the other candidates, the negative counting approach only requires counting the fact that the 1st choice is "preferred to themselves", whereas the regular approach requires counting that the 1st choice is preferred over each of the other 9 candidates; so there are either 1 or 9 (10-1) marks to be made.
      • If this ballot also ranked a 2nd choice, the negative approach requires indicating that the 2nd choice is also "preferred to themselves", and preferred less than the 1st choice, while the regular approach requires indicating that the 2nd choice is preferred over each of 8 candidates i.e. all of the 10 candidates that aren't the 1st choice or 2nd choice.

Here are some examples for the first numbers in each series (with the upper bound bolded):

Number of marks required in each vote-counting approach when equal-ranking isn't allowed ("N" refers to total number of non-write-in candidates in election)

Number of candidates ranked ("R")	Regular approach	Negative counting	Semi-negative counting
1	N-1 [0, 1+]	1	1
2	2N-3 [1, 3+]	[1, 3]	[1, 2, 3]
3	3N-6 [3, 6+]	[3, 6]	[2, 4, 5, 6]
4	4N-10 [6, 10+]	[6, 10]	[4, 6, 8, 9, 10]
5	5N-15 [10, 15+]	[10, 15]	[6, 9, 11, 13, 14, 15]

Notes:

• For regular counting, a lower bound has been provided (when N=R), and since there is no upper bound, instead the number of marks required when N is one higher than R is provided.
• For both negative and semi-negative counting, the numbers provided are a series that include both lower and upper bounds on the number of marks that have to be made, depending on N (which starts at R, and sequentially increases by one, up to twice that i.e. 2R), and how last-ranked candidates are counted.
  • For counting last-ranked candidates, see #Dealing with last-place candidates. If no marks are made for them, then a ballot that ranks all candidates requires the same number of marks as a ballot that ranks at least one less candidate i.e. all candidates except the last-ranked candidate(s). For example, a ballot that ranks 5 candidates when there are 5 candidates total can be thought of as ranking the top 4 candidates, and leaving the 5th candidate unranked.
  • Semi-negative counting's performance is always better than or equal to negative counting's performance when the same number of candidates run and the same number of candidates are ranked in both approaches. As an example for a voter who ranks 2 candidates:
    • When only those 2 candidates run, then in both approaches, at least 1 mark needs to be made (to count the 1st choice; the 2nd choice can be skipped because they are ranked last).
    • If 3 candidates run, then in negative counting, the 2nd choice is counted with 2 additional marks, so at least 3 marks are made; but in semi-negative counting, only 1 additional mark needs to be made (because 2nd choice is ranked above only 1 candidate), so at least 2 marks are made.
    • If 4 candidates run, then in both approaches, 3 marks are made (2nd choice is ranked above 2 candidates and below 1 candidate).
  • Note that regular pairwise counting's formula can be derived as follows: when only 1 candidate is ranked, the formula is N-1. When 2 candidates are ranked, it is the formula for when only 1 candidate is ranked plus another formula i.e. (N-1)+(N-2), which becomes 2N-3. For 3 candidates ranked, it is (2N-3)+(N-3), which is 3N-6, etc.

Note that negative counting is faster when voters rank only a few of all candidates, and potentially slower otherwise.

For example, a voter who votes A>B when there are 10 candidates can be assumed to vote for A and B in every matchup, except they don't prefer B>A:

• Usually, this would require manually marking each of those positive preferences, resulting in 9 marks to show A being preferred to all other candidates, and 8 marks to show B preferred to all candidates except A, for a total of 17 marks.
• But negative counting only requires 3 marks: 1 each for A and B to indicate they are preferred in every matchup, and 1 to indicate that this isn't the case for B>A.

Regular pairwise counting but done by counting first choices separately

Note that regular pairwise counting can have its required number of marks reduced, without using any negative numbers, by counting 1st choices separately from all other ranks; see the section above Pairwise counting#Uses for first choice information. The regular approach requires [(number of candidates)-2] less marks if using this modification i.e. a voter who ranks 2 candidates sequentially when there are 10 candidates only requires 1+8=9 marks rather than 9+8=17 marks to have their ballot counted, an [(10)-2=] 8-mark difference.

However, when voters equally rank multiple candidates 1st, then whether or not the modification is still applied can make a significant difference in overall number of marks.

Election example comparisons

It is possible to compare the number of marks that must be made in either approach for certain elections, because their full ballot set has been published. Any election using ranked or rated ballots can be used for this purpose. The calculations and numbers used in this section may be slightly off for some examples, though general conclusions (should) still hold; because of this, it is suggested that the reader apply a ~10% margin of error or more when considering how superior one pairwise counting approach is to another.

Note that when equal-ranking isn't allowed, only the number of voters who ranked a certain number of candidates needs to be known.

Burlington 2009 mayoral election

(Vote totals taken from . The numbers and calculations are slightly off for this analysis.) This was an RCV election.

Almost no voters used equal-ranking (i.e. because it wasn't allowed), so it will be ignored for this analysis. There were roughly 8,980 non-equal-ranking votes.

There were 5 on-ballot candidates in the election (6 if all write-ins are counted as one "supercandidate").

Number of Ballots Ranking This Many Candidates:

	1	2	3	4	5
	1481	1912	1799	873	2944
Cumulative:	1481 (16.4%)	3393 (37.6%)	5192 (57.6%)	6065 (67.3%)	9009 (100%)

Based on the above table:

• The negative counting approach would require at least 56,181 marks for any implementation.
  • This assumes last-ranked candidates weren't counted with any marks (which means write-in candidates' pairwise matchups wouldn't have been counted accurately).
  • The calculation is (1*1481 + 3*1912 + 6*1799 + 10*873 + 10*2944).
• The semi-negative approach would require at least 39,114 marks.
  • Calculation: (1*1481+3*1912+5*1799+6*873+6*2944). This is found by observing that, for example, when a voter ranked 2 candidates, it was most efficient to count their 2nd choice with the negative approach rather than the regular approach, but when they ranked their 3rd choice, it was faster to use the regular approach (i.e. mark that they're ranked above 2 candidates) rather than the negative approach (3 values, because the candidate is ranked below 2 candidates, and a 3rd mark has to be made to show that they're ranked by the voter).
• The regular approach would require at least 73,669 marks.
  • If counting 1st choices separately from all other ranks, then it would require 46,642 marks.
    • Calculation: (1*1481 + 4*1912 + 6*1799 + 7*873 + 7*2944)
  • This is if miscounting write-ins' matchups, which is the usual way to treat them.
  • Calculation: (4*1481 + 7*1912 + 9*1799 + 10*873 + 10*2944). (This is with number of candidates, N, being 5.)

Non-pairwise methods

Also of interest may be the comparison to non-pairwise counting methods. The following are rough estimates for the number of marks:

• FPTP: ~8,980 marks
• RCV: 12,309 marks (8,980 1st choices + 3,329 vote transfers)
• Approval voting: ~20,000 marks (one possible upper bound on number of approvals is ~20,000 (i.e. each voter approves an average of ~2 candidates))
• Score voting: ~20,000 to ~30,000 marks

Evanston, IL 2020 Democrat endorsement

(Vote totals from ) This was an RCV election with no equal-ranking.

There were 249 ballots.

Number of Ballots Ranking This Many Candidates:

	1	2	3	4	5	6	7	8	9	10	11
	48	35	22	39	25	16	10	6	4	4	40
Cumulative:	48 (19.2%)	83 (33.3%)	105 (42.1%)	144 (57.8%)	169 (67.8%)	185 (74.2%)	195 (78.3%)	201 (80.7%)	205 (82.3%)	209 (83.9%)	249 (100%)

• Negative counting approach requires at least 4,482 marks.
  • Calculation: (48 + 3*35 + 6*22 + 10*39 + 15*25 + 21*16 + 28*10 + 36*6 + 45*4 + 55*4 + 55*40)
• Semi-negative approach requires at least 3,208 marks.
  • Calculation: (48 + 3*35 + 6*22 + 10*39 + 15*25 + 20*16 + 24*10 + 27*6 + 29*4 + 30*4 + 30*40)
• Regular approach requires at least 8,223 marks.
  • If 1st choices are counted separately, then this requires 5,982 marks.
    • Calculation: (1*48 + 10*35 + 18*22 + 25*39 + 31*25 + 36*16 + 40*10 + 43*6 + 45*4 + 46*4 + 46*40)
  • Calculation: (10*48 + 19*35 + 27*22 + 34*39 + 40*25 + 45*16 + 49*10 + 52*6 + 54*4 + 55*4 + 55*40)

Non-pairwise methods

• RCV: 354 marks (249 voters' 1st choices + 105 votes transferred throughout)

Connection to cardinal methods

Also see rated pairwise preference ballot.

This approach can be considered an Approval voting-based or cardinal approach, because when using the Approval-style approach for equal-rankings (explicit equal-rankings are counted as a vote for both candidates in the matchup), then each voter that votes Approval-style (i.e. explicitly ranks some candidates 1st and leaves all other unranked, which is implicitly treated as ranking them last) will have their ballot counted like an Approval ballot (i.e. the preference for each approved candidate on the ballot will be counted with one mark per candidate, and no marks will be used to count disapproved candidates). Further, in some implementations, it allows a voter to give support to both candidates in a matchup, just like in cardinal methods.

Using with strength of preference

Negative vote-counting can be used to count weak pairwise preferences (i.e. if a voter only wants to give 0.4 votes in a matchup, rather than 1 vote; see Rated pairwise preference ballot#Implementations) by counting only a "partial ballot" marking a candidate, and partial (i.e. weighted or fractional) negative votes in certain matchups. In other words, it is treated as if only a partial voter or ballot supported a candidate (see KP transform).

Notes

Negative vote-counting approach for pairwise counting (Note: Regular approach may be better in some use cases; see cited discussions in text to the left).

An alternative way to do the negative approach, which is more similar to the regular approach, is to, when candidate B is explicitly ranked below A on a ballot, instead of counting -1 votes for B>A, count 1 vote for A>B, and later on, when the math is done, the number of votes for B>A is the number of ballots ranking B minus the number of votes for A>B. In other words, a part of the regular pairwise counting approach is used, but only in matchups where both candidates are explicitly ranked by the voter (i.e. a voter who voted A>B and left C unranked would have their vote for A>B counted, but not their vote for A>C, because later on it will be inferred that they must have preferred A to C by virtue of having ranked A but not C).

In practice, to make pairwise counting easier, voters could be provided with two or fewer ranks than the number of candidates, with equal-ranking being allowed so that voters could do preference compression. This way, a voter who would usually indicate a preference that would have to be counted between two candidates would have to indicate no preference between them instead.

Inspiration

Approval voting can be thought of as a Smith-efficient Condorcet method where, when a voter approves a candidate, they are assumed to vote for them in every head-to-head matchup (see Self-referential Smith-efficient Condorcet method). Further, approving a candidate can be thought of as ranking them 1st, while disapproving a candidate can be thought of as ranking them last. Given that connection, and that in Approval voting, the vote-counting is done by counting the number of ballots that approve/mark a candidate, rather than by harder pairwise counting, it is clear that a similar counting procedure could be applied to pairwise counting itself. The only complexity is that when voters rank candidates, they are allowed to express that they don't prefer certain candidates in certain matchups; thus, negative votes are necessary to communicate that lack of preference for specific matchups.

This has the advantage of, when every voter does bullet voting, being counted exactly like an FPTP election (one mark per ballot for the candidate it marked), which also shows that FPTP can be thought of as a constrained form of Approval.

Comprehensive example

Here is a more comprehensive picture of how the negative approach works with an example: for a voter who voted A>B=C>D with 2 unranked candidates E and F, their ballot would be processed into this matrix:

	A	B	C	D	E	F
A	(Ranked on) 1 ballot
B	-1 vote	1 ballot
C	-1 vote		1 ballot
D	-1 vote	-1 vote	-1 vote	1 ballot
E
F

A second voter who voted C>D=E>A would have their ballot processed into a matrix like this:

	A	B	C	D	E	F
A	(Ranked on) 1 ballot		-1 vote	-1 vote	-1 vote
B
C			1 ballot
D			-1 vote	1 ballot
E			-1 vote		1 ballot
F

The two ballots can be combined into this matrix:

	A	B	C	D	E	F
A	(Ranked on) 2 ballots		-1 vote	-1 vote	-1 vote
B	-1 vote	1 ballot
C	-1 vote		2 ballots
D	-1 vote	-1 vote	-2 votes	2 ballots
E			-1 vote		1 ballot
F

Supposing that these two voters were the only voters in the precinct, this matrix would be the matrix containing all of the necessary information to get the final vote totals for this precinct, so the precinct could send it along. Each different precinct's matrix could be summed in the way shown above for the two voters. Once all precinct matrices have been summed, the following is the math done to get the final vote totals:

	A	B	C	D	E	F
A	2 ballots	2 (votes)	1	1	1	2
B	0	1 ballot	1	1	1	1
C	1	2	2 ballots	2	2	2
D	1	1	0	2 ballots	2	2
E	1	1	0	1	1 ballot	1
F	0	0	0	0	0	0

For example, look at the row for candidate A. Because A was ranked on 2 ballots, they received 2 votes in every head-to-head matchup against another candidate. But because they were ranked below C, D, and E on 1 ballot, their votes against C, D, and E were reduced by 1 vote, so that they only had 1 vote in favor of them against those 3 candidates in the respective head-to-head matchups.

Note: In the B vs C matchup, B has 1 vote and C has 2 votes; yet in reality, 1 of the 2 voters equally ranked B and C (giving 0 votes to either of them), and the other ranked C over B (giving 1 vote to C and 0 to B); this discrepancy can be explained as being because the equal-ranking voter was treated as voting for both B and C, rather than neither of them.

Dealing with truncation

When voters aren't allowed to do truncation (and Write-in candidates aren't allowed), then it can be useful to skip the part of the negative counting procedure where the vote-counters mark how many ballots a candidate is ranked on, and instead assume a candidate is preferred in every matchup on every voter's ballot. The negative votes would then be applied as usual.

• This works because every candidate will actually be marked on every voter's ballot.

• This would obviously be impractical when voters are allowed to truncate, however, because it could mean negative votes would have to be applied to each and every ballot for a candidate no voters ranked.

Independence of unranked candidates

The negative approach doesn't require additional marks to be made for a given ballot when candidates are added to the election that that ballot doesn't vote for. For example, at most 3 marks need be made for a voter whose ballot is A>B, regardless of whether there are 2 candidates in the election or 100.