User:BetterVotingAdvocacy/Negative vote-counting approach for pairwise counting: Difference between revisions
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Revision as of 23:08, 14 August 2020
, 3 years ago→Formula for counting the required number of marks to be made
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* 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 numbers can be derived as such: the voter's N/2 (rounded down to nearest integer) highest-ranked candidates are counted with negative counting's worst-case, and the remaining candidates are counted with regular pairwise counting separately. So if a voter ranks 5 out of 7 candidates, half of the candidates (3.5) rounded down (=3) are counted with negative counting (negative counting for 3 candidates requires at most 6 marks), and the remaining 2 ranked candidates are counted with regular pairwise counting (that means having to do regular pairwise counting for R=2 and N=4, because out of the 7 candidates, you ignore the 3 candidates already counted by negative counting; so plugging that into 2N-3 yields: 2(4)-3= 5 marks). So that is 6+5=11 marks for R=5 and N=7 with semi-negative counting.
** 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: ▼
***The reasoning for this can be derived as follows: separate the voter's ranked candidates into their N/2 most-preferred candidates (their "top-ranked half") and their N/2 least-preferred candidates (the "bottom-ranked half"); if there are an odd number of candidates, consider the median candidate to be in the bottom-ranked half. Any candidate in the top-ranked half will be ranked above more candidates than they are ranked below, so it will be easier or equally easy to use negative counting to count them. Any candidate in the bottom-ranked half will be in the opposite situation, so regular counting is easier for them.
*** 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). ▼
▲** Semi-negative counting's performance is always better than or equal to negative counting's performance
*** 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. ▼
▲*** When
▲*** 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.
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