Post on 17-Dec-2015
But, how??
Explaining all possible positional, pairwise voting paradoxes & prop.
Don SaariInstitute for Math Behavioral Sciences
University of California, Irvine CA 92697-5100
3 A>B>C4 A>C>B6 C>B>A2 B>C>A2 B>A>C
Vote for one (1, 0, 0):A
Vote for two (1, 1, 0):BBorda (2, 1, 0):C
Paired comparisons?
B>A, A>C, C>B, a cycle
How would you explain, create, this example?
Approval voting? Cumulative voting?Any of the 13 possible rankings is a
sincere outcomeStandard approach since Borda, Condorcet,
Arrow:“Proof” by example
Put forth “desirable properties that a voting rule “must” satisfy
Example: “Will voting rule always elect Condorcet winner?”
What have we accomplished?Many impressive results (and difficult), but …
Answer above challenges? Why paradoxes? Arrow? Consensus -- even among those in voting theory?
My approach was influenced by:Fishburn -- paradoxes, which are properties, n=7, 1050
Nurmi -- Nothing goes rightFor a price …..
Need new approach
Rather than “select rule and find supporting properties”Science approach is to find all properties and identify
appropriate rules.
New approachWell, not so new (1999, 2000) Symmetry
groups, new way to achieve objective for any number
of candidatesFind appropriate coordinate system
that will identify all properties
Polar coordinates
y = A xEigenvectors
My goal: find “profile coordinate system” so that
we can explain all differences in paired comparisons,
in positional outcomes, in everything
Coordinate system for morning coffeeWaterCoffeeCreamSugar
A B
C
Tallying ballots
3 A>B>C4 A>C>B6 C>B>A2 B>C>A2 B>A>C
34
0 62
2
7 10
7
108
9
Positional voting(w1, w2, 0)
Normalize; i.e., divide each weight by w1
So, weights are (1, s, 0)s=0, plurality, s=½,
Borda, s=1, vote-for-two7+2s 4 + 9s
6+6s
Coordinates:Water, nothing happens
A B
C
x
xx
xx
x
More alternatives, more complicated
and more dimensions
A B
C
Paired comparisonsA>B
-1-1
1
11
-1
A B
C
Paired comparisonsB>C
11
-1
-11
-1
A B
C
Paired comparisonsA>C
1-1
-1
11
-1
Basis for all pairwise votesIs there a better one?
n=6, Kernel has dimension 720- 130= 590
Better (by giving more information) pairwise basis
A B
C
Paired comparisonsA>B
-1-1
1
11
-1
A B
C
Paired comparisonsA>C
1-1
-1
11
-1+ = (well, after taking ½)
A B
C
-1-11
1
2 -2
-2
2
0
0
ABasic
2 + 0s
-1+0s
-1+0s
A B
C
11
-1-1
BBasic
Complete coordinate system for pairs;
source of all properties of any rule depending on paired
comparisons such as Dodgsen, Borda, Condorcet, Kemeny,
Arrow, Sen, etc.Nothing else is needed
A B
C
-11
1
1 -1-1
Condorcet
1
1 -1
1
-1
-1
A>B>C, B>C>A, C>A>BCriticism: Borda need not elect Condorcet winner
Nothing goes wrong; complete agreement
Source of all problemsAnalysis depends on how rule treats
this feature
More accurate criticism: Condorcet winner need not be Borda
winnerKelly: n=3, Ostrogorski paradox implies no Condorcet winner: Now trivial. Conjecture true for
odd n>3
(Add A>B, B>C, C>A)
A B
C
-1-11
1
ABasicA B
C
11
-1-1
BBasic
A B
C
-11
1
1 -1-1
Condorcet
1
1 -1
1
-1
-1
For n candidatesSame idea--start with “n choose 2” X>Y vectors
XBasic; n-1 everywhere X is top
ranked, n-3 everywhere X is
second ranked, ….
n-(2j-1) everywhere X is jth ranked
Basis for n-1 dimensional space
Condorcet, (n-1)!/2 of themA>B>...>Y>Z,B>...>Y>Z>A,
…
-Z>Y>...>B>A-A>Z>Y>...>B
Source of all possible
properties or rules that are based on
paired comparisons;
this is all that is needed
Arrow, Sen, likelihood of cycles, agendas, Kemeny, Borda, Condorcet,
etc.
Example of how answer is easily found for typical
problem (problem, from earlier
analysis, is flawed): What voting rules will always elect a Condorcet winner?
Kelly Conjecture: Ostrogorski paradox implies no Condorcet winner for odd n>3
True for all n >2.
n=6, dimension is 60
Back to n=3 and finding coordinate system
A B
C
-1-11
1
ABasicA B
C
11
-1-1
BBasicA B
C
-11
1
1 -1-1
Condorcet
1
1 -1
1
-1
-1
Positional (1, s, 0)What does not affect pairs and is not included?A>B>C, C>B>A
A B
C
1
111
-2
-2
AReversal
2-4s
-1+2s
-1+2sA B
C
11
11
-2
-2
BReversal
That is all there is;all three alternative properties follow from these coordinates
Includes runoffs, etc.
All positional outcome problems and differences are caused by how rule
(Plurality, AV, Cumulative, etc.) react to reversal
(Only Borda is not affected by Reversal)
Example: Which positional rules have rankings that coincide (in any way) with
paired comparisions?Borda
Source of paradoxes
A B
C
Creating initial example
3 A>B>C4 A>C>B6 C>B>A2 B>C>A2 B>A>C
1
This C>B>A preference will be Borda Outcome
Next, to get A to be Plurality winner and B vote-for-two winner
add reversal components
xxy
y
So, solvex+y>1+y>x to get plurality A>C>B;
e.g., x=2, y=3
A B
C
223
4
To obtain cyclic effect, add Condorcet term
z
zzSolve:
5+z < 6+2z5+z<6+2z6+z< 5+2ze.g., z=2
6
224
3
Cambridge University press
Coordinates for n=4 (same idea holds for all n>3)
A B
C
11
11
-2
-2
BReversal
-2
-2A B
C
1
111
AReversal
Need this structure for triplets, but also need to ensure that it
does not affect paired comparisons (already done) and
set of all 4A>B>C>D, D>C>B>AB>A>D>C, C>D>A>B
Same approach: Sum all of the A-reversals to find coordinate over all triplets (as with the coordinate over all pairs)
then find “cyclic part” of “A>B>C, B>C>D, C>D>A, D>A>C” kind.
Finally, behavior for set of all four candidates
that cannot for triplets, and pairs
This coordinate system, then, can be used to identify all possible properties of four candidate election rules
where positional and paired comparisons are involvedSame approach continues for n>4