Solving parity games Definition (Parity game) G = V E , V A , R , - - PowerPoint PPT Presentation

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Solving parity games

Definition (Parity game)

G = VE, VA, R, Ω : V → N where

• VE is the set of vertices for Eve,
• VA is the set of vertices for Adam,
• R ⊆ V × V is the move relation,
• Ω defines the parity winning condition.

Definition (Winning region)

A winning region for Eve is the set of vertices from which she has a winning strategy.

Problem

Given a parity game, find winning regions for Eve and for Adam.

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Some definitions

Definition (Attractor, Trap)

Given a set S of nodes, let AttrE(S) be the set of nodes from which Eve has a strategy to reach S. A set S traps Eve if Adam has a strategy to keep Eve in S.

Remarks

• The winning region of Eve traps Adam.
• G − AttrE(S) traps Eve.

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Basic algorithm

AttrE (Ω−1(0))

G1 Ω−1(0)

Algorithm

• Solve G1. Let (W 1

E, W 1 A) be the result.

• If W 1

A = ∅ then we are done.

• If W 1

A = ∅ then solve G2. Let (W 2 E, W 2 A)

• Return (W 2

E, AttrA(W 1 A) ∪ W 2 A).

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Basic algorithm

AttrE (Ω−1(0))

G1 Ω−1(0)

Algorithm

• Solve G1. Let (W 1

E, W 1 A) be the result.

• If W 1

A = ∅ then we are done.

• If W 1

A = ∅ then solve G2. Let (W 2 E, W 2 A)

• Return (W 2

E, AttrA(W 1 A) ∪ W 2 A).

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Basic algorithm

AttrA(W1 A)

G2 W 1

A

Algorithm

• Solve G1. Let (W 1

E, W 1 A) be the result.

• If W 1

A = ∅ then we are done.

• If W 1

A = ∅ then solve G2. Let (W 2 E, W 2 A)

• Return (W 2

E, AttrA(W 1 A) ∪ W 2 A).

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Basic algorithm

Algorithm

• Solve G1. Let (W 1

E, W 1 A) be the result.

• If W 1

A = ∅ then we are done.

• If W 1

A = ∅ then solve G2. Let (W 2 E, W 2 A)

• Return (W 2

E, AttrA(W 1 A) ∪ W 2 A).

Complexity: O(2n) T(n) = T(n − 1) + T(n − 1) Complexity: O(2d) where d is the number of priorities used

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Finding small winning sets

Definition (Dominion)

E-dominion is a set S of vertices such that Eve has a winning strategy from all

• f them without leaving S. Similarly for A-dominion.

Dominion is either A-dominion or E-dominion.

Lemma

One can find (if it exists) a dominion of size l < n/3 in time O(2ln

l

• ).

Proof

• There are at most

n

l

• sets of size l.
• To check that a given set is a dominion takes O(2l) time

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Faster algorithm

D

G

Faster algorithm [Jurdzinski, Peterson, Zwick]

• Given G find if there is a dominion of size ≤ l = √n.
• If there is one, call it D, then solve G − AttrP(D).
• If not then solve G − AttrE(Ω−1(0)). Let (W 1

A, W 1 E) be the result

• If W 1

A = ∅ then W 1 A > √n.

• Apply the algorithm recursively to G − AttrA(W 1

A).

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Faster algorithm

AttrE (Ω−1(0))

G1 Ω−1(0)

Faster algorithm [Jurdzinski, Peterson, Zwick]

• Given G find if there is a dominion of size ≤ l = √n.
• If there is one, call it D, then solve G − AttrP(D).
• If not then solve G − AttrE(Ω−1(0)). Let (W 1

A, W 1 E) be the result

• If W 1

A = ∅ then W 1 A > √n.

• Apply the algorithm recursively to G − AttrA(W 1

A).

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Faster algorithm

AttrE (W1 A)

G2 W 1

A

Faster algorithm [Jurdzinski, Peterson, Zwick]

• Given G find if there is a dominion of size ≤ l = √n.
• If there is one, call it D, then solve G − AttrP(D).
• If not then solve G − AttrE(Ω−1(0)). Let (W 1

A, W 1 E) be the result

• If W 1

A = ∅ then W 1 A > √n.

• Apply the algorithm recursively to G − AttrA(W 1

A).

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Faster algorithm

G

Faster algorithm [Jurdzinski, Peterson, Zwick]

• Given G find if there is a dominion of size ≤ l = √n.
• If there is one, call it D, then solve G − AttrP(D).
• If not then solve G − AttrE(Ω−1(0)). Let (W 1

A, W 1 E) be the result

• If W 1

A = ∅ then W 1 A > √n.

• Apply the algorithm recursively to G − AttrA(W 1

A).

Complexity: n

√n