1. Algorithmic Analysis of Elections: Voting Rules and Manipulability (minicourse)
Piotr Faliszewski
AGH University Kraków, Poland

2. Why Are Elections Important?
Politics
Electing leaders
Deciding policies, laws …
Business settings
Stock holders make decisions in companies
University Senates, hiring decisions
Competitions (e.g., racing, Eurovision)
Multiagent systems
Meta-search engines
Planning
Recommendation systems
Elections aggregate the opinions of the voters

3. Agenda Computational Social Choice Why study elections? Formal model
What are elections?
How to model votes?
What voting rules to use?
Manipulating Elections
Example
Gibbard-Satterthwaite Theorem
Median Voter Theorem
Computational Social Choice
Winner Determination
Exercises

4. How to Study Elections? We need to answer several questions:
What are elections?
How can we model voters’ preferences?
What election rules are we interested in?
A mathematical object, a pair E = (C,V). C – set of candidates, V – set of voters
A single top guy? A ranking of candidates? Numerical values, “utility” a voter derives from having a particular candidate elected?
An election rule is a function that given an election returns the set of winners.

5. Formal Model of Elections
L(C) – linear orders over C
W(C) – weak orders over C
E = (C, V) – an election
C = {c1, c2, …, cm} – set of candidates
V = (v1, v2, …, vn) – collection of voters
Each vote is a ranking of the candidates:
vi : c1 > c7 > c2 > …  L(C)
Finding a result:
Social choice function: f: L(C)n  C
Social choice correspondence: f: L(C)n  2C
Social welfare function: f: L(C)n  W(C)
Social welfare correspondence: f: L(C)n  2W(C)

6. Who Wins?
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b

7. Who Wins?
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b
Plurality
p = 3 c = 2
b = 1 d = 1 others = 0

8. Who Wins?
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b
Borda
a = 12 b = 22 c = 21 d = 20 v = 13 p = 17

9. Who Wins?
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b
Veto
a = 5 b = 6 c = 6 d = 6 v = 7 p = 5

10. Who Wins?
Condorcet
4 c : a 3
4 c : b 3
4 c : d 3
5 c : v 2
4 c : p 3
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b b a c p d v

11. Who Wins?
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b
Approval
a = 7 b = 6 c = 4 d = 6 v = 3 p = 4

12. Who Wins?
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b
Approval
a = 7 b = 6 c = 4 d = 6 v = 3 p = 4

13. Who Wins?
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b
Dictatorship!

14. Which Election Rule Is Best?
Nondictatorship
There is no voter vi such that f(vi) = vi
Unrestricted domain
Every vote is legal (i.e., every linear order can be cast as a vote, we can have as many candidates as we like)
Unanimity
If all voters prefer a to b, then a is socially preferred to b
Independence of irrelevant alternatives
Whether a is socially preferred to b is independent of how other candidates are ranked
Arrow’s Theorem: There is no non-dictatorial social welfare function with unrestricred domain that satisfies unanimity and independence of irrelevant alternatives.

15. Agenda Computational Social Choice Why study elections? Formal model
What are elections?
How to model votes?
What voting rules to use?
Manipulating Elections
Example
Gibbard-Satterthwaite Theorem
Median Voter Theorem
Computational Social Choice
Winner Determination
Exercises

16. Puzzle: Why would Mac win this elections? 
Who Wins?
Voting for an operating system to use in the lab.
Rule: Borda
Linux : 21
Windows : 20
Mac : 5
Votes:
7: Linux > Windows > Mac 6: Windows > Linux > Mac 2: Mac > Windows > Linux 1: Mac > Linux > Windows
Puzzle: Why would Mac win this elections? 

17. Which Election Rule Is Best?
Nondictatorship
There is no voter vi such that f(vi) = top(vi)
Unrestricted domain
Every vote is legal (i.e., every linear order can be cast as a vote, we can have as many candidates as we like)
Unanimity
If all voters rank c first, then f(V) = c
Gibbard-Satterthwaite Theorem: Consider a set C of at least three candidates. Dictatorship is the only social choice function
f: L(C)n  C
that accepts unrestricted domain, is unanimous, and reporting truthful preferences is the dominant strategy for the votgers.

18. Gibbard-Satterthwaite Theorem
Gibbard--Satterthwaite Theorem: Consider a set C of at least 3 candidates. Dictatorship is the only social choice function f: L(C)n  C that accepts unrestricted domains, is unanimous, and reporting truthful preferences is the dominant strategy for the votgers.
History of the proofs
1973 – Gibbard
1975 – Satterthwaite
1978 – Schmeider/Sonnenschein
1983 – Barbera
1990 – Barbera/Peleg
2000 – Benoit
2001 – Sen
2001 – Remy
2009 – Cato
2012 – Ninjbat

19. Gibbard-Satterthwaite Theorem
Gibbard--Satterthwaite Theorem: Consider a set C of at least 3 candidates. Dictatorship is the only social choice function f: L(C)n  C that accepts unrestricted domains, is unanimous, and reporting truthful preferences is the dominant strategy for the votgers.
History of the proofs
1973 – Gibbard
1975 – Satterthwaite
1978 – Schmeider/Sonnenschein
1983 – Barbera
1990 – Barbera/Peleg
2000 – Benoit
2001 – Sen
2001 – Remy
2009 – Cato
2012 – Ninjbat

20. Proof of the G-S Theorem: Start
C = {a, b, c, …} – set of at least 3 candidates
f: L(C)n  C – social choice function
(unanimous, nonmanipulable)
v1: a > b > c > …
v2: a > c > b > … …
vn: b > c > a > … V
Lemma (shifting): If f(V) = a and we form V’ by shifting b forward by one position in some vote, then f(V’)  {a, b}
v1: a > b > c > …
v2: a > b > c > … …
vn: b > c > a > … V’
Proof
If f( V’ ) = c (c ≠ a, c ≠ b) then
v2: a > c – v2 would prefer to lie in V’
v2: c > a – v2 would prefer to lie in V

21. Proof of the G-S Theorem: Step 1
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b
f(V0) ≠ b V0
What if f(V0) = b?
v1: > ... > … > b
v2: > ... > … > b …
vt-1: > ... > … > b
vt: > ... > … > b
vn: > ... > … > b
f(V’0) = b
V’0

22. Proof of the G-S Theorem: Step 1
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b
f(V0) ≠ b V0
What if f(V0) = b?
v1: a > ... > … > b
v2: > ... > … > b …
vt-1: > ... > … > b
vt: > ... > … > b
vn: > ... > … > b
f(V’0) = b
V’0

23. Proof of the G-S Theorem: Step 1
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b
f(V0) ≠ b V0
What if f(V0) = b?
v1: a > ... > … > b
v2: a > ... > … > b …
vt-1: > ... > … > b
vt: > ... > … > b
vn: > ... > … > b
f(V’0) = b
V’0

24. Proof of the G-S Theorem: Step 1
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b
f(V0) ≠ b V0
What if f(V0) = b?
v1: a > ... > … > b
v2: a > ... > … > b …
vt-1: a > ... > … > b
vt: > ... > … > b
vn: > ... > … > b
f(V’0) = b
V’0

25. Proof of the G-S Theorem: Step 1
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b
f(V0) ≠ b V0
What if f(V0) = b?
v1: a > ... > … > b
v2: a > ... > … > b …
vt-1: a > ... > … > b
vt: a > ... > … > b
vn: > ... > … > b
f(V’0) = a
V’0
A step before vt also preferred a to b, so vt would have preferred to lie then
This had to happen due to unanimity

26. Proof of the G-S Theorem: Step 1
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b
f(V0) ≠ b V0
shift b to top position as long as it does not become a winner
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: k > ... > … > b
Vr+1: a1 > ... > … > b
vn: awhatever> … > b V1
f(V1) ≠ b

27. Proof of the G-S Theorem: Step 1
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b
f(V0) ≠ b V0
shift b to top position as long as it does not become a winner
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: k > ... > … > b
Vr+1: a1 > ... > … > b
vn: awhatever> … > b
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: b > k > … > …
Vr+1: a1 > ... > … > b
vn: awhatever> … > b V1 V2
shift b in vr
f(V1) ≠ b
f(V2) = b

28. Proof of the G-S Theorem: Step 1
If vi with i > r changes a vote to another one, then b is still winning
(otherwise, this voter would lie in V2)
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b V0
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: k > ... > … > b
Vr+1: a1 > ... > … > b
vn: awhatever> … > b
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: b > k > … > …
Vr+1: a1 > ... > … > b
vn: awhatever> … > b V1 V2
f(V1) ≠ b
f(V2) = b

29. Proof of the G-S Theorem: Step 1
If vi with i > r changes a vote to another one, then b is still winning
(otherwise, this voter would lie in V2)
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b V0
If vi with i < r changes a vote to another one, where b is first, b still is winning
(otherwise, this voter would lie in the new profile)
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: k > ... > … > b
Vr+1: a1 > ... > … > b
vn: awhatever> … > b
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: b > k > … > …
Vr+1: a1 > ... > … > b
vn: awhatever> … > b V1 V2
f(V1) ≠ b
f(V2) = b

30. Proof of the G-S Theorem: Step 1
Conclusion
If v1 … vr rank b first, then b is winning
If vr … vn rank b last, then b is losing
v1: > ... > … > b
v2: > ... > … > b …
vn: > ... > … > b V0
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: k > ... > … > b
Vr+1: a1 > ... > … > b
vn: awhatever> … > b
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: b > k > … > …
Vr+1: a1 > ... > … > b
vn: awhatever> … > b V1 V2
f(V1) ≠ b
f(V2) = b

31. Proof of the G-S Theorem: Step 2
If v1 … vr rank b first, then b is winning
If vr … vn rank b last, then b is losing
We want to show f(V3) = k
f(V4) = k
shift b to the top in v1 … vr-1
v1: > ... > … > b
v2: > ... > … > b …
vr-1: > ... > … > b
vr: k > ... > … > b
Vr+1: > ... > … > b
vn: > ... > … > b
v1: b > k > … > …
v2: b > k > … > … …
vr-1: b > k > … > …
vr: k > ... > … > b
vr+1: k > ... > … > b
vn: k > ... > … > b
f(V3’) = k V3 V4
v1: k > ... > … > b
v2: k > ... > … > b …
vr-1: k > ... > … > b
vr: k > ... > … > b
Vr+1: k > ... > … > b
vn: k > ... > … > b
V3’
shift k to the top position everywhere
unanimity

32. Proof of the G-S Theorem: Step 2
If v1 … vr rank b first, then b is winning
If vr … vn rank b last, then b is losing
We want to show f(V3) = k
f(V4) = k
f(V5) = k
v1: > ... > … > b
v2: > ... > … > b …
vr-1: > ... > … > b
vr: k > ... > … > b
Vr+1: > ... > … > b
vn: > ... > … > b
v1: b > k > … > …
v2: b > k > … > … …
vr-1: b > k > … > …
vr: k > ... > … > b
vr+1: k > ... > … > b
vn: k > ... > … > b
v1: b > k > … > …
v2: b > k > … > … …
vr-1: b > k > … > …
vr: k > b > … > …
vr+1: k > ... > … > b
vn: k > ... > … > b V3 V4 V5
Shift b – no winner change or vr would want to lie

33. Proof of the G-S Theorem: Step 2
If v1 … vr rank b first, then b is winning
If vr … vn rank b last, then b is losing
Assume f(V3) = g, g ≠ k
f(V4) = k
f(V5) = k
v1: > ... > … > b
v2: > ... > … > b …
vr-1: > ... > … > b
vr: k > ... > … > b
Vr+1: > ... > … > b
vn: > ... > … > b
v1: b > k > … > …
v2: b > k > … > … …
vr-1: b > k > … > …
vr: k > ... > … > b
vr+1: k > ... > … > b
vn: k > ... > … > b
v1: b > k > … > …
v2: b > k > … > … …
vr-1: b > k > … > …
vr: k > b > … > …
vr+1: k > ... > … > b
vn: k > ... > … > b V3 V4 V5
Shift b – no winner change or vr would want to lie

34. Proof of the G-S Theorem: Step 2
If v1 … vr rank b first, then b is winning
If vr … vn rank b last, then b is losing
Assume f(V3) = g, g ≠ k
By shifting lemma
f(V6)  {b, g}
If g were winning in V6, then vr could shift b to top, and b would win
vr prefers b to g (or b = g)
f(V5) = k
v1: > ... > … > b
v2: > ... > … > b …
vr-1: > ... > … > b
vr: k > ... > … > b
Vr+1: > ... > … > b
vn: > ... > … > b
v1: b > k > … > …
v2: b > k > … > … …
vr-1: b > k > … > …
vr: k > b > … > …
vr+1: k > ... > … > b
vn: k > ... > … > b V3 V5
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: k > b > …
Vr+1: > ... > … > b
vn: > ... > … > b V6
shift b to top in v1 … vr-1 and second to last in vr

35. Proof of the G-S Theorem: Step 2
If v1 … vr rank b first, then b is winning
If vr … vn rank b last, then b is losing
3) So… b = f(V5) = k contradiction!
Assume f(V3) = g, g ≠ k
f(V5) = k
2) shifting k in vr+1 … vn cannot change the result from b
v1: > ... > … > b
v2: > ... > … > b …
vr-1: > ... > … > b
vr: k > ... > … > b
Vr+1: > ... > … > b
vn: > ... > … > b
v1: b > k > … > …
v2: b > k > … > … …
vr-1: b > k > … > …
vr: k > b > … > …
vr+1: k > ... > … > b
vn: k > ... > … > b
1) shifting k in v1 … vr-1 cannot change the result from b V3 V5
f(V6) = b
v1: b > ... > …
v2: b > ... > … …
vr-1: b > ... > …
vr: k > b > …
Vr+1: > ... > … > b
vn: > ... > … > b V6
shift b to top in v1 … vr-1 and second to last in vr
shift k to 2nd best in v1 … vr-1
shift k to top in vr+1 … vn

36. Proof of the G-S Theorem: Step 3
If v1 … vr rank b first, then b is winning
If vr … vn rank b last, then b is losing
v1: > ... > …
v2: > ... > … …
vr-1: > ... > …
vr: k > ... > …
Vr+1: > ... > …
vn: > ... > …
v1: > ... > … > b
v2: > ... > … > b …
vr-1: > ... > … > b
vr: k > ... > … > b
Vr+1: > ... > … > b
vn: > ... > … > b
If v1 … vn rank b last, but vr ranks k first, k wins
Proof is complete!
v1: b > ... > c
v2: b > ... > c …
vr-1: b > ... > c
vr: b > … > c
Vr+1: a > ... > c
vn: a > … > c V7 V8
shift b back V9
f(V7) {k, b} (by shifting lemma)
f(V8) = k (by step 2)
If k = b then we’re good!
If k ≠ b then… rerun the whole proof for c in the place of b
We discover pivotal voter vm for c.
b must win in V9 (step 1)
Top choice of vm must win in V9
Thus m ≤ r
Symmetrically, r ≤ m
f(V7) {k, c}

37. Solution: Choose median vote
How to Circumvent G-S?
Consider a mayor of a city (which happens to be a line) asking the citizens where the train station should be located.
Each citizen suggests the position for the train station. How to aggregate these votes?
Solution: Choose median vote

38. Single-Peaked Preferences
> > > >
> > > >
> > > >
Def. Preferences are single-peaked if there is an order S over the candidates such that for every vote v with preference order > it holds that:
If (c S d S e) or (e S d S c) then
If c > d then d > e
left
right

39. Median Voter Theorem > > > > > > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
left
right
Theorem
If election E = (C,V) is single-peaked and the numer of voters is odd, then
The top-ranked candidate of the median voter is a Condorcet winner
Choosing the Condorcet winner is strategy-proof
median voter

40. Agenda Computational Social Choice Why study elections? Formal model
What are elections?
How to model votes?
What voting rules to use?
Manipulating Elections
Example
Gibbard-Satterthwaite Theorem
Median Voter Theorem
Computational Social Choice
Winner Determination
Exercises

41. Computational issues in elections
Winner determination

42. Who Wins?
Condorcet
4 c : a 3
4 c : b 3
4 c : d 3
5 c : v 2
4 c : p 3
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b b a c p d v

43. Condorcet-Consistent Rules
C = { a, b, c, d, p, v } v1: p > b > a > v > c > d v2: p > b > d > v > a > c v3: p > c > b > d > v > a v4: b > d > a > c > v > p v5: d > c > b > a > v > p v6: c > v > b > d > p > a v7: c > d > a > v > p > b
Def. A voting rule is Condorcet-consistent if it always elects the Condorcet winner whenever it exists.
Example. Condorcet winner does not always exist (a group of rational people can be irrational; surprise? ) b a c p d v

44. Condorcet-Consistent Rules
C = { a, b, c } v1: a > b > c v2: b > c > a v3: c > a > b
Def. A voting rule is Condorcet-consistent if it always elects the Condorcet winner whenever it exists.
Example. Condorcet winner does not always exist (a group of rational people can be irrational; surprise? )
Rules
Copeland
Maximin
Dodgson …
Pairwise comparisons
2 a : b 1
2 b : c 1
2 c : a 1
a > b > c > a

45. Condorcet-Consistent Rules
Def. A voting rule is Condorcet-consistent if it always elects the Condorcet winner whenever it exists.
Notation
E = (C, V) – an election
NE(c,d) =|{v  V | v: c > d}|
c defeats d if NE(c,d) > NE(d,c)
c ties d if NE(c,d) = NE(d,c)
A Condorcet winner defeats everyone else
Rules
Copeland
Maximin
Dodgson
Copelandα Score:
scoreE(c) = |{d  C| c defeats d}| + α |{d  C| c ties d}|
Maximin Score:
scoreE(c) = mind  C NE(c,d)
Dodgson Score:
scoreE(c) = smallest number of swaps of adjacent candidates that make c a Condorcet winner

46. Dodgson Score is NP-Complete
Reduction from X3C
Election
C = {1,2, …, 3k, 1’, 2’, … 3k’, S1, …, Sn, p}
X3C votes:
1 > 3 > 4 > S1 > p > …
1 > 3 > 6 > S2 > p > …
2 > 4 > 5 > S3 > p > … … 3 1
Equalizing votes:
1 > 1’ > p > …
4 > 4’ > p > … … 4 6 2 5
ensure that every set element is preferred to p in the same numer of votes

47. Dodgson Score is NP-Complete
Reduction from X3C
Election
C = {1,2, …, 3k, 1’, 2’, … 3k’, S1, …, Sn, p}
X3C votes:
1 > 3 > 4 > S1 > p > …
1 > 3 > 6 > S2 > p > …
2 > 4 > 5 > S3 > p > … … 3 1
Equalizing votes:
1 > 1’ > p > …
4 > 4’ > p > … … 4 6 2 5
Defeating votes:
1 > 2 > … > 3k > 1’ > 2’ > … > 3k’ > p > … …
enough to ensure that for each set element i we have
NE(i,p) – NE(p,i) = 1

48. Dodgson Score is NP-Complete
Reduction from X3C
Election
C = {1,2, …, 3k, 1’, 2’, … 3k’, S1, …, Sn, p}
X3C votes:
1 > 3 > 4 > S1 > p > …
1 > 3 > 6 > S2 > p > …
2 > 4 > 5 > S3 > p > … … 3 1
Equalizing votes:
1 > 1’ > p > …
4 > 4’ > p > … … 4 6 2 5
Defeating votes:
1 > 2 > … > 3k > 1’ > 2’ > … > 3k’ > p > … …
p defeats everyone except for the set elements
p has Dodgson score 4k if and only if there is an exact cover

49. Dodgson in Single-Peaked Elections
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
left
right
Theorem
If election E = (C,V) is single-peaked and the numer of voters is odd, then the top-ranked candidate of the median voter is a Condorcet winner
median voter
Theorem
Computing Dodgson winners under single-peaked elections is in P

50. Computational issues in elections
Cheating in
elections
Winner determination
Manipulation
control
bribery
Gibbard-Satterthwaite Theorem

51. Complexity Barrier Approach
Model: Represent each cheating strategy as a computational decision problem.
Complexity barrier approach: If manipulating elections is hard, then we can ignore the fact that it is in principle possible.
Approach initiated by Bartholdi, Tovey, and Trick in the late 80s and the early 90s
Outdated!

52. Complexity Barrier: Results
Effects of complexity barrier research
Dozens of computational problems identified
Multiple standard election systems analyzed
Quite thorough understanding of worst case complexity of elections
Complications…
We would like some of the problems to be efficiently computable
Determining winners
Organizing a campaign
Worst-case analysis seems problematic…

53. Computational issues in elections
Cheating in
elections
Winner determination
Manipulation
control
Analyzing elections
bribery
Margin of victory
Measures of success
Gibbard-Satterthwaite Theorem

54. How to measure candidate performance?15 10 11 9 15
C = { , , , , }
V = (v1, … , v6)
V1:
V2:
V3:
V4:
V5:
V6:
Borda

55. How to measure candidate performance?21 16 17 15 21
Copeland
C = { , , , , }
V = (v1, … , v6)
win = 1, tie = 0.5, lose = 0
V1:
:1.5
V2: 1:
V3: :1
V4: 3:
V5:
:3.5
V6:
Borda

56. How to measure candidate performance?21 16 17 15 21
Copeland
C = { , , , , }
V = (v1, … , v6)
win = 1, tie = 0, lose = -1
V1:
:-1
V2:
-2:
V3:
:-2
V4: 2:
V5: :3
V6:
Borda

57. Thank you!