Distributed Score Voting: Difference between revisions

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Distributed Score Voting (DSV) is a [[Single Member system|Single-Winner]] and [[Multi-Member System|Multi-Winner]] [[Cardinal voting systems| Cardinal voting system]].
 
In the [[Single Member system|Single-Winner]] part, it's similar to [[Smith//Score]]. In the [[Multi-Member System|Multi-Winner]] part, [[Distributed Multi-Voting]], the more preferred the winning candidate is in a vote, the more the weight of that vote is decreased in the choice of the next winner.
 
[[Category:Multi-winner voting methods]]
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==Procedure==
 
===Voting===
Each voter has 100 points to distribute among the candidates according to his preferences (it's also possible to write the vote even in a simpler form, with range from 0 to 5 points for each candidate).
[[File:DVS procedure.jpg|thumb|DSV counting]]
Each voter has 100 points to distribute among the candidates according to his preferences.
 
All candidates in the vote have 0 points by default.
[[File:DSV procedure.jpg|alt=|thumb|DSV counting]]
 
===Counting the votes===
W = sum of all the points in the original vote (100 for all votersvotes, at the beginning).
 
#1) All head-to-head matches are conducted between candidates. In head-to-head, the candidate who has more points in a vote than his opponent receives W points from the vote. The candidate who gets the most points wins the head-to-head. Graphically, each candidate is a node; the head-to-head is represented by an arrow, leaving the winning candidate, entering the losing candidate. The tie is represented as a double arrow entering, that is both candidates are considered losers.
 
# Find the smallest set X of nodes that don’t have incoming arrows, coming from outside the set. Then remove all candidates not in X from the votes.
Graphically, each candidate is a node; the head-to-head is represented by an arrow, leaving the winning candidate, entering the losing candidate. The tie is represented as a double arrow entering, that is both candidates are considered losers.
# Convert the marks into a range form, assigning 0 points to the candidates with the lowest score and normalizing* the remaining candidates, using the following formula: M = candidate with the highest score, before normalization. v0 = current value of candidate C, to be normalized. v1 = value of candidate C, after normalization.
 
2) Find the smallest set X ([[Smith set]]) of nodes that don’t have incoming arrows, coming from outside the set.
 
3) Convert the votes using the following formula:
 
M = highest score among the candidates in the vote, before normalization.
 
v0 = current value of candidate C, to be normalized.
 
v1 = value of candidate C, after normalization.
 
<math>
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v1=\frac{v0}{M} \cdot W
\end{equation}</math>
 
# Add up the points for each candidate of the range votes, and the candidate who has the highest sum, wins. The choice of the single winner ends here.
Then remove all candidates not in X from the votes.
# If you want to have more winners, then remove the single-winner from all original votes, repeating the whole procedure from point 1. The value W of each original vote is reduced by the points assigned to the removed candidate. By repeating this process several times, I can get as many winners as I like, which will be those removed in point 5.
 
# If you want to know the % of victory of the winning candidates then, in each original vote, you must remove all the candidates who haven’t won, and normalize* the vote with the formula used in point 3 (with W=100 fixed). The sum of points for each candidate will indicate the % of victory.
4) Add up the points for each candidate of the range votes, and the candidate who has the highest sum, wins.
 
The choice of the single winner ends here.
 
5) If you want to have more winners, then remove the single-winner from all original votes, repeating the whole procedure from point 1.
 
The value W of each original vote changes according to the following formula:
 
M = highest score among the candidates in the vote (before removing the candidate).
 
e = candidate's score eliminated.
 
W0 = previous value of W
 
W1 = new value of W
 
<math>
\begin{equation}
W1=\frac{W0}{\left( 1+\frac{e}{M}\right)}
\end{equation}</math>
 
By repeating this process several times, you can get as many winners as you like, which will be those removed in point 5.
 
6) If you want to know the % of victory of the winning candidates then, in each original vote, you must remove all the candidates who haven’t won, and normalize each vote with the following formula:
 
S = sum of the points left in the vote.
 
v0 = current value of candidate C, to be normalized.
 
v1 = value of candidate C, after normalization.
 
<math>
\begin{equation}
v1=\frac{v0}{S} \cdot 100
\end{equation}</math>
 
The sum of points for each candidate will indicate the % of victory.
 
===Head-to-head===
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This form avoids all the problems mentioned above.
 
===Simplified vote writing===
 
To make the writing of the vote more comprehensible and simple, the voter can be left with almost complete freedom in the use of numerical values or only X.
 
Before the counting process, the votes will be normalized to 100-point votes, where the Xs are considered as equal weight values.
 
Examples of how a vote can be written by the voter and subsequently, before the counting, converted into 100 points:
 
X,0,0,0,0 → 100,0,0,0,0
 
X,X,X,X,0 → 25,25,25,25,0
 
4,3,2,1,0 → 40,30,20,10,0
 
40,6,3,1,0 → 80,12,6,2,0
 
101,0,0,0,0 → 100,0,0,0,0
 
The complexity in writing the vote adapts to the voter, and it’s also noted that, if 101 or 99 points are mistakenly distributed, the vote will still be valid.
 
==Criteria==
 
{| class="wikitable" style="text-align:center"
<!-- criteria headers -->
|- style="font-size:80%;"
! rowspan=1 |
! rowspan=1 style="border-left: 2px solid #a0a0a0;" | [[Majority criterion|Majority]]
! rowspan=1 | [[Majority loser criterion|Maj. loser]]
! rowspan=1 | [[Mutual majority criterion|Mutual maj.]]
! rowspan=1 | [[Condorcet criterion|Condorcet]]
! rowspan=1 | [[Condorcet loser criterion|Cond. loser]]
! rowspan=1 | [[Smith criterion|Smith]]
! rowspan=1 | [[Pareto criterion|Pareto]]
! rowspan=1 | IIA*
! rowspan=1 | [[Independence of irrelevant alternatives|IIA]]
! rowspan=1 | [[w:Independence of clones criterion|Clone proof]]
! rowspan=1 style="border-left:2px solid #a0a0a0;" | [[Monotonicity criterion|Monotone]]
! rowspan=1 | [[Consistency criterion|Consistency]]
! rowspan=1 | [[Participation criterion|Participation]]
! rowspan=1 | [[w:Reversal symmetry|Reversal<br>symmetry]]
! rowspan=1 style="border-left:2px solid #a0a0a0;" | [[Later-no-help criterion|Later-no<br>Help]]
! rowspan=1 | [[Favorite betrayal criterion|Favorite<br>betrayal]]
|- style="font-size:80%;"
 
<!-- Methods -->
|-
! [[Distributed_Score_Voting|DSV<br>single-winner]]
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #fd8787; font-weight: inherit;" | No
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #fd8787; font-weight: inherit;" | No
! style="background: #fd8787; font-weight: inherit;" | No
! style="background: #98ff98; font-weight: inherit;" | Yes
! style="background: #fd8787; font-weight: inherit;" | No
! style="background: #fd8787; font-weight: inherit;" | No
|}
 
<b>IIA*</b>: X is a set containing all the preferred candidates over B. If I add C a less appreciated candidate (in head-to-head) than the candidates in X, then all candidates in X continue to be preferred over B.
 
This method also passes [[ISDA]].
 
All the criteria not met are linked to the fact that, through tactical votes, it's possible add / remove a candidate from the [[Smith set]].
 
- add one more candidate into the [[Smith set]] isn't a big problem because that additional candidate must then beat all the other candidates in point 4 of the procedure (and if he manages to beat them all it makes sense that he wins).
 
- removing a candidate from the [[Smith set]] is only possible when that candidate lose all the head-to-head with the candidates contained in the [[Smith set]]. This actually becomes a problem only if the excluded candidate is the one who really should have won.
 
Below is an example in which, through tactical votes, it's possible to bring out a candidate, who should have won, from the [[Smith set]] (making him lose).
 
===Tactical votes===
 
In an election, the results of the head-to-head are the following: A>B , B>C , C>D , D>A , A>C , D>B and in the end wins B.
 
A voter who in this case supported the candidates as follows: A>D>B>C he could change his vote as follows: A>D>C>B to favor C more than B (without disadvantaging A and D).
 
This tactical vote could cause B to lose head-to-head between B and C and in this case B would be the candidate who loses all head-to-head, being eliminated immediately. The winner would no longer be B.
 
This type of tactical vote works only if:
* there is a [[Condorcet paradox]] which includes at least 4 candidates.
* through the tactical vote, the candidate who should have been the winner can be taken out of the [[Smith set]].
* the new winner is actually a better candidate than the previous one (the new winner in the example could also be C).
* the voter has a fairly precise knowledge of the likely ballots result, without which this tactical vote would turn against him.
 
[[Category:Smith-efficient Condorcet methods]]