# parity games

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Parity Games. Games, logic and Automata Seminar 2014 Tal Zelmanovich. Overview. The parity game problem. The parity game problem Uses Terms and definitions Exponential algorithm Sub-exponential algorithm. Uses. Notation, & terms. Exponential algorithm. Sub-exponential algorithm. - PowerPoint PPT PresentationTRANSCRIPT

Parity Games

Parity Games

Games, logic and Automata Seminar 2014

Tal Zelmanovich

Lecture about stuff

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Overview

The parity game problem

Uses

Terms and definitions

Exponential algorithm

Sub-exponential algorithm

The parity game problem

Uses

Notation, & terms

Exponential algorithm

Sub-exponential algorithm

Exponential actually 2^n, exponential exists due it being in NP

Consider also talking more about:

Different attempts to solve the problem

Proof it is in NP\co-NP (mean payoff reduction?)

Mew calculus

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The Parity Game

The parity game problem:

A two-headed llama wonders infinitely through the forest.

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The Parity Game

Each head remembers only one part of the forest (to conserve precious llama memory) and so only he can navigate through these places.

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The Parity Game

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Goal: Each head tries to force his favorite treat to be the maximal treat infinitely visited

(no ties)

Each head also have a favorite snack: blue flowers for the right head, and bright olive-green apples for the left one.

But more than they like their treats, they hate each other, so each llama wants her treat to appear more than the others treat. And since theyre having an infinite walk, each head wants the maximal treat appearing in a single place to be his favorite.

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The Parity Game

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+ priorities

Each vertex has at least one outgoing edge

Starting vertex

Player 0: Green

Player 1: Blue

U = vertices visited infinitely

Winner: (Max(priority in U))%2

Mathematical definition

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Properties

Is it a full information game

Is it a zero sum game

Does the prefix of a play matters

Yes!

Yep!

No, winner stays the same

Is it a determined

Indeed!

Winning strategies are memory-less

Third level of Borel hierarchy

Strategy given by algorithm

First ones are trivial (ask the audience)

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Why is it interesting?

Verification and logics:- equivalent to some cases of -automata emptiness problem (polynomial reduction)

- Equivalent to model checking problem of the modal -calculus (modal fixpoint logic)

Complexity:

- Known to be in and even in

same status as integer factorization problem

The problem of solving parity games is polynomial time equivalent to the non-emptiness problem of -automata on infinite trees with Rabin-chain acceptance conditions

(consider talking more about mew-calculus)

UP like NP, but exactly one witness string for Yes (as opposed to at least one)not known to be in P just like factorization problem

also helps solving churchs problem!

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Notation

: The subset of vertices from which player i has a wining strategy

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In general: =V

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Notation

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: The subset of vertices from which player I has a wining strategy

Observation:

Controlling the choices does not guarantee victory

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Notation

: all vertices from which player i may force a visit to one of the vertices of A (same as attractor)

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Notation

: all vertices from which player i may force a visit to one of the vertices of A (same as attractor)

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A

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Notation

: all vertices from which player i may force a visit to one of the vertices of A (same as attractor)

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A

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Notation

: a set of vertices s.t. player i has a strategy that doesnt allow the game to go outside the set

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Alternative definition:

a set is called i-closed iff:

:

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Notation (Summary)

: The subset of vertices from which player I has a wining strategy

: all vertices from which player i may force a visit to one of the vertices of A (same as attractor)

: a set of vertices s.t. player i has a strategy that doesnt allow the game to go outside the set

All of these terms depends on the player

Reach & Closed ignores graph priorities (llama treats)

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The exponential algorithm

By R. McNaughton & W. Zielonka

Using a two-steps recursion to separate the problem into smaller (possibly overlapping) instances

The algorithm is simple, the reason it works isnt

Some observations (and of course proofs) are needed

Lets start by finding a way to detect closed areas

Mathematical definition

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The exponential algorithm (proofs)

Is always i-closed?

Lemma 1: is always i-closed

Proof: Assume is not i-closedAccording to closed-i definition, there must be a vertex to which one of the following apply:

, and :

:

But since we can win from , we must be able to win from one of these s ContradictionReminder: emitting play prefix does affect the result

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The exponential algorithm (proofs)

Lemma 2: is i-closed

Proof: Like the last proof, assume it is not i-closed.If it wasnt closed, then while following a reach-A strategy for i, on some vertex along the way one of the following will happen:

player j may force the play to leave the region.

Either player i will have to leave it

but since the rest of the strategy must reach A eventually, this cannot be.

WRONG!

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A

not-closed

It doesnt work since A may forceus to leave the area:

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The exponential algorithm (proofs)

Lemma 2 (fixed): is j-closed

Proof: assume it is not j-closed.Then there exists some so either:

But that means player i has a strategy forcing a play from to pass to and eventually to A, meaning .

Contradiction since

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The exponential algorithm (proofs)

For the algorithm, well need to compute multiple timesSo how fast can we do it?

Trivial idea expanding back:start with

Repeatedly add to B vertices answering one of the following conditions:

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The exponential algorithm (proofs)

How can we speed it up?

Count for each vertex how many outgoing edges we already checked (maintain an adjacency list)

While there is still an edge entering B that we didnt visit, do the following:

- If , add to B

- If , substruct s edge count by 1 On count=0, add to B

Total time:

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The exponential algorithm (proofs)

How can we speed it up?

Count for each vertex how many outgoing edges we already checked (maintain an adjacency list)

While there is still an edge entering B that we didnt visit, do the following:

- If , add to B

- If , substruct s edge count by 1 On count=0, add to B

Total time:

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The exponential algorithm (proofs)

So up to now we got:

Lemma 1: is always i-closed

Lemma 2: is j-closed

Algorithm: compute with time complexity of O(m)

We can now easily compute closed areasNext: breaking down a game to smaller (faster to solve) subgames

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Subgames

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B

G\B

must stay a valid game having at least 1 outgoing edge

If was i\j-closed in G, then were promised G\B is valid

Defining a subset of a game by removing set B of vertices (to get G\B)

Remaining vertices must form a game (must have at least one outgoing edge)

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The exponential algorithm (proofs)

Can a subgame tell us anything about the original game?

Lemma 3: if , the vertices of a subgame , were i-closed in the original game , then any winning state of is also a wining state of

Proof: since i has a wining strategy from the same strategy can be used to win the game . i follows the strategy and stays in , while j cannot escape it since is i-closed in the original game.We get as required

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G=G\B

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not in

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G=G\B

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Purple arrows are is winning strategy choices that must leave him inside G.

(in the drawing it is a positional strategy, but the proof also work if it is an historic strategy)

The upper left one is outside of win G, so it may (or may not) be in G

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The ex

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