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Computational Democracy: Algorithms, Game Theory, and Elections

Computational Democracy: Algorithms, Game Theory, and Elections. Steven Wolfman 2011/10/27. A voting system is software. . 2001 BC General Election. Some other oddities: - 1926 MB Federal - 1992 and 2000 US Presidential - 2008 Vancouver Municipal. MANY Other Algorithms.

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Computational Democracy: Algorithms, Game Theory, and Elections

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  1. Computational Democracy:Algorithms, Game Theory, and Elections Steven Wolfman 2011/10/27

  2. A voting system is software.

  3. 2001 BC General Election

  4. Some other oddities:- 1926 MB Federal- 1992 and 2000 US Presidential- 2008 Vancouver Municipal

  5. MANY Other Algorithms Approval (hand-raising in class, often), Range (IMDB), Party List Proportional (Germany), Cumulative (??), Block (Vancouver municipal), Condorcet (UBC AMS, “ranked pairs” flavour), ... But.. what do we really want out of a voting system?

  6. We have apple and blueberry. I’ll take the apple. Oh, wait! We have cherry, too! In that case, I’ll take the blueberry “Independence of Irrelevant Alternatives” (Intuition. Thanks to Sidney Morgenbesser.)

  7. Formally: Independence of Irrelevant Alternatives If under one set of votes, A beats B, then... A still beats B under another set of votes with the same relative rankings of A and B. 1CADEB 2BADCE 3ACDEB 4AEBCD 5DECBA SA...B 1EACBD 2CEBAD 3ABCDE 4CAEBD 5DBECA S.A..B

  8. General Definition: “Pareto (In)Efficient” If a change in the solution can make everyone better off, then the solution is “Pareto inefficient”. Question: Which of these is Pareto Inefficient? Key: Candidate (“option”) Voter A B C D E F G

  9. Formal: Pareto Efficient For any two candidates A and B, if all voters prefer A to B, A must beat B. Key: Candidate (“option”) Voter A B

  10. Formal: Dictator d is a dictator iff for any set of votes, the outcome precisely matches d’s vote. 1CADEB 2BADCE 3ACDEB 4AEBCD 5DECBA SAEBCD 1EACBD 2CEBAD 3ABCDE 4CAEBD 5DBECA SCAEBD All hail 4!

  11. Arrow’s Impossibility Theorem(Kenneth Arrow, 1951  Nobel Prize) IIA Dictatorship + PE Let’s prove it. We assume: no ties in votes or outcome.This assumption is unnecessary. We assume: finite #voters, at least three candidates, election is a function.All three are necessary.

  12. Let’s play with votes. Run the election. Go back in time and change the votes. Rerun the election. (AKA: explore the result of the election function on various inputs.) Scenario: move all B votes to the top 1 CADEB 2 BADCE 3 ACDEB 4 AEBCD 5 DECBA S ACDBE 1 BCADE 2 BADCE 3 BACDE 4 BAECD 5 BDECA S BACDE PE and all B at top  B wins top

  13. Scenario: all B at bottom Run election Change votes Rerun election 1 CADEB 2 BADCE 3 ACDEB 4 AEBCD 5 DECBA S ACDBE 1 CADEB 2 ADCEB 3 ACDEB 4 AECDB 5 DECAB S ACDEB PE and all B at bottom  B “wins” bottom

  14. Scenario: B at top or bottom Here, A > Band B > C. Generally, if B is in the middle, something beats B and some-thing loses to B. (Key results will always apply to the general case.) 1 CADEB 2 BADCE 3 ACDEB 4 AEBCD 5 DECBA S ACDBE 1 CADEB 2 BADCE 3 ACDEB 4 AECDB 5 BDECA S ABCDE Can B end up in the middle like this? Let’s play!

  15. Scenario: C just above A PE  C beats A IIA  A (still) beats B IIA  B (still) beats C CONTRADICTION Remember: A beats B B beats C 1 CADEB 2 BADCE 3 ACDEB 4 AECDB 5 BDECA S ABCDE 1 CADEB 2 BCADE 3 CADEB 4 CAEDB 5 BDECA S ABCDE B at top or bottom in all votes  B wins top or bottom

  16. Scenario: find who moves B up Reminder: if every voter puts B at top or bottom, B wins top or bottom. 1 CADEB 2 ADCEB 3 ACDEB 4 AECDB 5 DECAB S ACDEB 1 BCADE 2 BADCE 3 BACDE 4 BAECD 5 BDECA S BACDE PE  left side has B at bottom, right side has B at top

  17. Scenario: find who moves B up Reminder: if every voter puts B at top or bottom, B wins top or bottom. 1 CADEB 2 ADCEB 3 ACDEB 4 AECDB 5 DECAB S ACDEB 1 BCADE 2 ADCEB 3 ACDEB 4 AECDB 5 DECAB S ACDEB At some point as we flip B up, it will move to the top.

  18. Scenario: find who moves B up Reminder: if every voter puts B at top or bottom, B wins top or bottom. 1 BCADE 2 ADCEB 3 ACDEB 4 AECDB 5 DECAB S ACDEB 1 BCADE 2 BADCE 3 ACDEB 4 AECDB 5 DECAB S ACDEB At some point as we flip B up, it will move to the top.

  19. Scenario: find who moves B up Reminder: if every voter puts B at top or bottom, B wins top or bottom. 1 BCADE 2 BADCE 3 ACDEB 4 AECDB 5 DECAB S ACDEB 1 BCADE 2 BADCE 3 BACDE 4 AECDB 5 DECAB S BACDE At some point as we flip B up, it will move to the top.

  20. Scenario: find who moves B up Reminder: if every voter puts B at top or bottom, B wins top or bottom. Let’s focus on the voter who “controls” B. (3 in this case, but someone in general.) Starting with the right side. 1 BCADE 2 BADCE 3 ACDEB 4 AECDB 5 DECAB S ACDEB 1 BCADE 2 BADCE 3 BACDE 4 AECDB 5 DECAB S BACDE At some point as we flip B up, it will move to the top.

  21. Scenario: 3 moves A above B 1 BCADE 2 BADCE 3 ABCDE 4 AECDB 5 DECAB S ABCDE 1 BCADE 2 BADCE 3 BACDE 4 AECDB 5 DECAB S BACDE Why is A at the top? Compare to when we discovered 3...

  22. Scenario: find who moves B up Reminder: if every voter puts B at top or bottom, B wins top or bottom. Let’s focus on the voter who “controls” B. (3 in this case, but someone in general.) This time we want the left side. 1 BCADE 2 BADCE 3 ACDEB 4 AECDB 5 DECAB S ACDEB 1 BCADE 2 BADCE 3 BACDE 4 AECDB 5 DECAB S BACDE At some point as we flip B up, it will move to the top.

  23. Look at the A/B rankings here versus... our “3 puts A above B” scenario. IIA, left says A > B  A > B 1 BCADE 2 BADCE 3 ACDEB 4 AECDB 5 DECAB S ACDEB 1 BCADE 2 BADCE 3 ABCDE 4 AECDB 5 DECAB S ABCDE 1 BCADE 2 BADCE 3 ACDEB 4 AECDB 5 DECAB S ACDEB 1 BCADE 2 BADCE 3 ABCDE 4 AECDB 5 DECAB S ABCDE In fact, all but 3 can rearrange their votes, if they don’t move B...

  24. Scenario: free up all but B and 3’s A As long as we keep voters’ A/B ordering the same, A > B, by IIA. 1 BCADE 2 BADCE 3 ABCDE 4 AECDB 5 DECAB S ABCDE 1 BDEAC 2 BCDEA 3 ABEDC 4 ACEDB 5 DACEB S ABDEC B beats C. Must it? Compare to when we discovered 3...

  25. Scenario: find who moves B up Reminder: if every voter puts B at top or bottom, B wins top or bottom. Let’s focus on the voter who “controls” B. (3 in this case, but someone in general.) Back to the right side. 1 BCADE 2 BADCE 3 ACDEB 4 AECDB 5 DECAB S ACDEB 1 BCADE 2 BADCE 3 BACDE 4 AECDB 5 DECAB S BACDE At some point as we flip B up, it will move to the top.

  26. Scenario: free up all but B and 3’s A 1 BDEAC 2 BCDEA 3 ABEDC 4 ACEDB 5 DACEB S ABDEC 1 BDEAC 2 BCDEA 3 ABEDC 4 ACEDB 5 DACEB S ABDEC 1 BCADE 2 BADCE 3 BACDE 4 AECDB 5 DECAB S BACDE 1 BCADE 2 BADCE 3 BACDE 4 AECDB 5 DECAB S BACDE But, must B > C? Yes, by IIA, since B/C rankings match.

  27. Scenario: free up the B’s Reminder: As long as we keep voters’ A/B ordering the same, A > B and B > C by IIA. Now, everyone moves their Bs around, and 3 moves its A around but keeps it above C. 1 BDEAC 2 BCDEA 3 ABEDC 4 ACEDB 5 DACEB S ABDEC 1 DEBAC 2 CDEAB 3 BAEDC 4 ACEDB 5 BDACE S BDACE IIA A > C, but the only fixed ranking is 3’s A > C!

  28. Brief pause for formality: In general, A and C are arbitrary options other than B. So, 3 is a dictator with respect to all relative orderings except those involving B. All (but B) hail 3! 1 BDEAC 2 BCDEA 3 ABEDC 4 ACEDB 5 DACEB S ABDEC 1 DEBAC 2 CDEAB 3 BAEDC 4 ACEDB 5 BDACE S BDACE Rest of proof is a repeat with different candidate for B; skipping...

  29. Arrow’s Impossibility Theorem(Kenneth Arrow, 1951  Nobel Prize) IIA Dictatorship + PE

  30. More algorithms/game theory rangevoting.org Ka-Ping Yee zesty.ca

  31. Some Other Contexts of Use • Usability and interface design • Security and voter verifiability • District creation and member allocation (and gerrymandering) • Design specifications and standards • Election auditing and fraud detection • Voter files/databases • Voter registration systems

  32. Extra Slides(just in case)

  33. Scenario: what about B Repeat proof, except with C at top/bottom rather than B. We find a dictator over B vs A (and B vs all but C), but remember... 1 BADEC 2 ADBEC 3 ABDEC 4 AEBDC 5 DEBAC S ABDEC 1 CBADE 2 CADBE 3 CABDE 4 CAEBD 5 CDEBA S CABDE

  34. Scenario: 3 moves A above B Because 3 moves A to the top, A > B. So, 3 is that dictator as well. And any candidate but B can “play” A. So, 3 is the dictator. All hail 3! 1 BCADE 2 BADCE 3 ABCDE 4 AECDB 5 DECAB S ABCDE 1 BCADE 2 BADCE 3 BACDE 4 AECDB 5 DECAB S BACDE

  35. Canada’s Algorithm, Roughly • Start counts for each candidate and for “rejected” at 0. • For each ballot: • If the ballot does not show a single candidate choice or has stray marks that might identify the voter, add one to the “rejected” total. • Else, add one to the marked candidate’s total. • Report the results of all counts. After recounts, ties are do-overs: “deemed by-election” for the “deemed vacant” seat. Based on the Canada Elections Act.(Ties specified in the Parliament of Canada Act.)

  36. A Harder Case: STV In Single-Transferable Voting, voters rank the candidates, a single election results in multiple candidates elected, and votes are transferred among candidates during the count based on preference order...

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