User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting: Difference between revisions
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[[File:Adding ballot matrices in negative pairwise counting approach.png|thumb|1088x1088px|[[File:Pairwise counting negative counting with ranked ballot GIF.gif|thumb|454x454px|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 ==
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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:
{| class="wikitable"
|+Negative counting values ''italicized'', regular pairwise counting values normal, and semi-negative counting values bolded and <big>big</big>
!Voter's ranking
!
!A
!B
!C
!D
!E
|-
|1st (choice)
|A
|'''''1 ballot'''''
|<small>1</small>
|<small>1</small>
|<small>1</small>
|<small>1</small>
|-
|2nd
|B
|'''''-1'''''
|'''''1 ballot'''''
|<small>1</small>
|<small>1</small>
|<small>1</small>
|-
|3rd
|C
|''<small>-1</small>''
|''<small>-1</small>''
|''<small>1 ballot</small>''
|'''<big>1</big>'''
|'''<big>1</big>'''
|-
|4th
|D
|''<small>-1</small>''
|''<small>-1</small>''
|''<small>-1</small>''
|''<small>1 ballot</small>''
|'''<big>1</big>'''
|-
|5th
|E
|''<small>-1</small>''
|''<small>-1</small>''
|''<small>-1</small>''
|''<small>-1</small>''
|''<small>1 ballot</small>''
|}
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 ==
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** 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.
** This is if miscounting write-ins' matchups, which is the usual way to treat them.
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|}
* Negative counting approach requires at least '''
** Calculation: (48 + 3*35 + 6*22 + 10*39 + 15*25 + 21*16 + 28*10 + 36*6 + 45*4 + 55*4 + 55*40)
*
**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.
** Calculation: (10*48 + 19*35 + 27*22 + 34*39 + 40*25 + 45*16 + 49*10 + 52*6 + 54*4 + 55*4 + 55*40)
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