Cumulative voting: Difference between revisions
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Fix typo and add zero-info strategy for quadratic voting.
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A variant of cumulative voting which gained popularity in 2018 is "[[quadratic voting]]". Quadratic voting was conducted in an experiment by the Democratic caucus of the Colorado House of Representatives in April 2019. It differs from cumulative voting by altering "the cost" and "the vote" relation from linear to quadratic.
Quadratic voting is characterized by the optimal zero-information strategic vote (using Myerson-Weber strategy) being to vote honestly -- that is, to vote according to any affine scaling of one's honest utilities.
=== Criticism ===
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The optimal tactical vote depends heavily on <math>p</math>. For instance, vote-splitting is a problem in standard cumulative voting but not in Range voting. On the other hand, the Burr dilemma is a problem in Approval voting (and thus in tactical Range voting) but not in standard cumulative voting.
=== Clone independence ===
Every form of <math>\ell_p</math>-norm cumulative voting except <math>p=\infty</math> (plain [[Range voting]]) is vulnerable to vote-splitting.
For continuous cumulative voting, suppose each candidate can be given any score between 0 and 1. Then let <math>x = \frac{\sqrt[p]{2} - 1}{2\big(\sqrt[p]{2} + 1\big)}</math> and consider the election
* <math>\frac{1}{2}+x-\epsilon</math>: A (1) B (0)
* <math>\frac{1}{2}-x+\epsilon</math>: A (0) B (1)
for some <math>\epsilon</math> where <math>x > \epsilon > 0</math>. A wins by majority rule no matter the value of p.
Clone A into A1 and A2 so that everybody rates the two clones equal. The <math>\ell_p</math> normalization then leads to the first faction giving each A candidate <math>y = \frac{1}{\sqrt[p]{2}}</math> points each. As a result, there's a three-way tie when <math>\epsilon=0</math>, and B wins for any <math>\epsilon > 0</math>, which demonstrates the clone failure.
==== Example ====
Suppose we want to find a clone failure for <math>p=2</math> (quadratic voting). Then <math>x = \frac{\sqrt{2}-1}{2+2\sqrt{2}} \approx 0.086</math> and <math>y = \frac{1}{\sqrt{2}} \approx 0.707</math>. Let <math>\epsilon=0.001</math> to compensate for the roundoff error. Before cloning, the ballots are
* 0.585: A (1) B (0)
* 0.415: A (0) B (1)
where A wins 0.585 points to 0.415.
After cloning:
* 0.585: A1 (0.707) A2 (0.707) B (0)
* 0.415: A1 (0) A2 (0) B (1)
A1 gets 0.413 points, A2 the same, and B gets 0.415 points, thus making B the winner.
== Notes ==
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