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On Problem Kernels for Possible Winner Determination Under the k - - PowerPoint PPT Presentation

Introduction Kernelization results Conclusion On Problem Kernels for Possible Winner Determination Under the k -Approval Protocol Nadja Betzler Friedrich-Schiller-Universit at Jena, Germany 3rd Workshop on Computational Social Choice


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Introduction Kernelization results Conclusion

On Problem Kernels for Possible Winner Determination Under the k-Approval Protocol

Nadja Betzler

Friedrich-Schiller-Universit¨ at Jena, Germany

3rd Workshop on Computational Social Choice September 2010

Nadja Betzler (Universit¨ at Jena) 1/17

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Introduction Kernelization results Conclusion

Motivation

Typical voting scenario for joint decision making: Voters give preferences over a set of candidates as linear orders. Example: candidates: C = {a, b, c, d} profile: vote 1: a > b > c > d vote 2: a > d > c > b vote 3: b > d > c > a Aggregate preferences according to a voting rule

Nadja Betzler (Universit¨ at Jena) 2/17

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Introduction Kernelization results Conclusion

Partial information

Realistic settings: voters might only provide partial information. For example: not all voters have given their preferences yet new candidates are introduced a voter cannot compare several candidates because of lack of information

How to deal with partial information?

We consider whether a distinguished candidate can still win.

Nadja Betzler (Universit¨ at Jena) 3/17

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Introduction Kernelization results Conclusion

Partial vote

A partial vote is a transitive and antisymmetric relation. Example: C = {a, b, c, d} partial vote: a ≻ b ≻ c, a ≻ d a c b d possible extensions:

1

a > d > b > c

2

a > b > d > c

3

a > b > c > d An extension of a profile of partial votes extends every partial vote.

Nadja Betzler (Universit¨ at Jena) 4/17

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Introduction Kernelization results Conclusion

Computational Problem

Possible Winner

Input: A voting rule r, a set of candidates C, a profile of partial votes, and a distinguished candidate c. Question: Is there an extension profile where c wins according to r? Considered voting rule:

k-approval

In every vote, the best k candidates get one point each. A candidate with most points in total wins.

Nadja Betzler (Universit¨ at Jena) 5/17

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Introduction Kernelization results Conclusion

Known results for Possible Winner

Results for several voting systems, [Konczak and Lang, 2005], [Pini et

al., IJCAI 2007], [Walsh, AAAI 2007], [Xia and Conitzer, AAAI 2008], ...

Results for k-approval Possible Winner is NP-hard for two (or more) partial votes

[Betzler, Hemmann, and Niedermeier, IJCAI 2009]

Possible Winner is NP-hard for any fixed k ∈ {2, . . . , m − 2} for m candidates

[Betzler and Dorn, JCSS 2010]

Nadja Betzler (Universit¨ at Jena) 6/17

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Introduction Kernelization results Conclusion

Known results for Possible Winner

Results for several voting systems, [Konczak and Lang, 2005], [Pini et

al., IJCAI 2007], [Walsh, AAAI 2007], [Xia and Conitzer, AAAI 2008], ...

Results for k-approval Possible Winner is NP-hard for two (or more) partial votes

[Betzler, Hemmann, and Niedermeier, IJCAI 2009]

Possible Winner is NP-hard for any fixed k ∈ {2, . . . , m − 2} for m candidates

[Betzler and Dorn, JCSS 2010]

This work

Is the Possible Winner problem easy to compute when the number k of “one-positions” and the number of votes is small?

Nadja Betzler (Universit¨ at Jena) 6/17

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Introduction Kernelization results Conclusion

Combined parameters

2 scenarios: “number of partial votes” and “number of one-positions” k “number of partial votes” and “number of zero-positions” k0

11..1 000 .... 00 votes 1111 ... 11110...0 partial

k k0 . . . . . . Motivation: Small committee selects few winners/losers out of a large set of candidates (grants, graduate students, ...)

Nadja Betzler (Universit¨ at Jena) 7/17

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Introduction Kernelization results Conclusion

Parameterized Complexity

Given an NP-hard problem with input size n and a parameter p Basic idea: Confine the combinatorial explosion to p p n instead of p n

Definition

A problem of size n is called fixed-parameter tractable with respect to a parameter p if it can be solved in f (p) · nO(1) time. Parameters: pairs of integers

Nadja Betzler (Universit¨ at Jena) 8/17

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Introduction Kernelization results Conclusion

Problem kernel

Let L ⊆ Σ∗ × Σ∗ be a parameterized problem. An instance of L is denoted by (I, p).

Kernelization

data reduction rules

p p′ I I ′ poly(|I|, k) (I, p) ∈ L ⇐ ⇒ (I ′, p′) ∈ L |p′| ≤ |p| |I ′| ≤ g(|p|) If g is a polynomial, we say L admits a polynomial problem kernel.

Nadja Betzler (Universit¨ at Jena) 9/17

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Introduction Kernelization results Conclusion

Main results

partial linear votes: votes: 11..1 000 .... 00 1111 ... 1111 0...0

k . . . . . . . . . . . . t t k0

Nadja Betzler (Universit¨ at Jena) 10/17

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Introduction Kernelization results Conclusion

Main results

partial linear votes: votes: 11..1 000 .... 00 1111 ... 1111 0...0

k . . . . . . . . . . . . t t k0 Parameter (t, k) (t, k0) FPT FPT “Tower of k’s”-kernel no polynomial kernel polynomial kernel

Nadja Betzler (Universit¨ at Jena) 10/17

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Introduction Kernelization results Conclusion

Polynomial kernel for (t, k0)

  • 1. Fix the distinguished candidate c as good as possible
  • 2. z(c′) := # zero-positions such that c′ is beaten by c
  • 3. If

c′∈C z(c′) > t · k0, then return “no”

else replace “irrelevant” candidates by a bounded number.

Example: C := {a, b, c, d1, . . . , ds}, k0 = 2

candidate points in linear votes di ≤ 10 b 11 a 12 c 12 partial votes: v1 : a ≻ c, b ≻ d1, d2 ≻ d3 v2 : d1 ≻ d2 ≻ · · · ≻ ds ≻ b ≻ c v3 : ds ≻ c, a ≻ d2

Nadja Betzler (Universit¨ at Jena) 11/17

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Introduction Kernelization results Conclusion

Polynomial kernel for (t, k0)

  • 1. Fix the distinguished candidate c as good as possible
  • 2. z(c′) := # zero-positions such that c′ is beaten by c
  • 3. If

c′∈C z(c′) > t · k0, then return “no”

else replace “irrelevant” candidates by a bounded number.

Example: C := {a, b, c, d1, d2, . . . , ds}, k0 = 2

candidate points in linear votes di ≤ 10 b 11 a 12 c 12 partial votes: v1 : a ≻ c, b ≻ d1, d2 ≻ d3, c ≻ C \ {c, a} ⇒ a > c > . . . v2 : d1 ≻ d2 ≻ · · · ≻ ds ≻ b ≻ c v3 : ds ≻ c, a ≻ d2, c ≻ C \ {c, ds} ⇒ ds > c > . . .

Nadja Betzler (Universit¨ at Jena) 12/17

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Introduction Kernelization results Conclusion

Polynomial kernel for (t, k0)

  • 1. Fix the distinguished candidate c as good as possible
  • 2. z(c′) := # zero-positions such that c′ is beaten by c
  • 3. If

c′∈C z(c′) > t · k0, then return “no”

else replace “irrelevant” candidates by a bounded number.

Example: C := {a, b, c, d1, d2, . . . , ds}, k0 = 2

candidate points in linear votes # zero-positions di ≤ 10 b 11 1 a 12 2 c 12 — partial votes: v1 : a ≻ c, b ≻ d1, d2 ≻ d3, c ≻ C \ {c, a} v2 : d1 ≻ d2 ≻ · · · ≻ ds ≻ b ≻ c v3 : ds ≻ c, a ≻ d2, c ≻ C \ {c, ds}

Nadja Betzler (Universit¨ at Jena) 13/17

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Introduction Kernelization results Conclusion

Polynomial kernel for (t, k0)

  • 1. Fix the distinguished candidate c as good as possible
  • 2. z(c′) := # zero-positions such that c′ is beaten by c
  • 3. If

c′∈C z(c′) > t · k0, then return “no”

else replace “irrelevant” candidates by a bounded number.

Example: C := {a, b, c, d, d1, d2 . . . , ds}, k0 = 2 ————-

candidate points in linear votes # zero-positions di ≤ 10 b 11 1 a 12 2 c 12 — partial votes: v1 : a ≻ c, b ≻ d1, d2 ≻ d3, c ≻ C \ {c, a} ⇒ a ≻ c ≻ b ≻ d v2 : d1 ≻ d2 ≻ · · · ≻ ds ≻ b ≻ c ⇒ d ≻ b ≻ c v3 : ds ≻ c, a ≻ d2, c ≻ C \ {c, ds} ⇒ c ≻ a ≻ d

Nadja Betzler (Universit¨ at Jena) 14/17

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Introduction Kernelization results Conclusion

Polynomial kernels

Theorem

For k-approval, Possible Winner with t partial votes and k0 zero-positions admits a polynomial kernel with O(t · k2

0)

candidates. Crucial idea: Number of candidates that have to take zero-positions is bounded in a yes-instance. Why does this not work for (t, k)?

Nadja Betzler (Universit¨ at Jena) 15/17

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Introduction Kernelization results Conclusion

Polynomial kernels

Theorem

For k-approval, Possible Winner with t partial votes and k0 zero-positions admits a polynomial kernel with O(t · k2

0)

candidates. Crucial idea: Number of candidates that have to take zero-positions is bounded in a yes-instance. Why does this not work for (t, k)? In a yes-instance, there might be an unbounded number of candidates that may take a one-position.

Theorem

For k-approval, Possible Winner with t partial votes does not admit a polynomial kernel wrt. (t, k) unless coNP ⊆ NP/poly.

Nadja Betzler (Universit¨ at Jena) 15/17

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Introduction Kernelization results Conclusion

Overview of kernelization results

Possible Winner for k-approval with t partial votes k0 denotes the number of zero-positions (t, k0) (t, k) polynomial kernel (FPT) superexponential kernel (FPT) no polynomial kernel 2-approval : polynomial kernel with O(t2) candidates

Nadja Betzler (Universit¨ at Jena) 16/17

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Introduction Kernelization results Conclusion

Future work

Counting variant: In how many extensions does a distinguished candidate win? Some first results in

[Bachrach, Betzler, and Faliszewski, AAAI 2010]

Can the results from this work be transferred to other voting rules. kernelization for related problems

Nadja Betzler (Universit¨ at Jena) 17/17

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Introduction Kernelization results Conclusion

Future work

Counting variant: In how many extensions does a distinguished candidate win? Some first results in

[Bachrach, Betzler, and Faliszewski, AAAI 2010]

Can the results from this work be transferred to other voting rules. kernelization for related problems

Thank you

Nadja Betzler (Universit¨ at Jena) 17/17