Memoryless Determinacy of Parity Games - PowerPoint Presentation

Slide1

Memoryless Determinacy of Parity Games

Itay

HarelSlide2

Table of Contents

Quick recap

Complexity results

Definitions and lemmassub-games-trapsAttractors-paradiseDeterminacy3 important lemmasnon-constructive proof

 Slide3

Quick recap – parity games

Game :

A – arena :

– coloring function:

Acc

– winning condition for inf. In parity games:

0-player wins if is evenStrategy: , a partial function

 Slide4

The main eventSlide5

Determinacy

Theorem

:

The set of vertices of a parity game is partitioned into a 0-paradise and a 1-paradise

M.W.S

M.W.S Slide6

Complexity ResultSlide7

Complexity class of finite parity games

Theorem:

 Slide8

A non-deterministic algorithm:

Given G and v, guess a memoryless strategy w

Check whether w is a M.W.SWhy is it O.K. to only guess memoryless?

How can we check a strategy quickly?Slide9

Creating

A memoryless strategy w can be represented using a sub-graph of G:

 Slide10

Checking a strategy (1)

Check whether there exists a vertex

such that:

is reachable from

is odd

lies on a cycle in

containing only vertices of priority less or equal Claim:w is a winning strategy iff such a vertex doesn’t exist Slide11

Let’s assume such a vertex

exists

Observation: In

, if a vertex is reachable from v, player-1 can force the token into it (formal proof with induction)Observation: In , if the token is inside a cycle of vertices, player-1 can force the token to go over the entire cycleA winning play for player-1:Force the token into

Go over the cycle forever

If

exists, w is not a M.W.SOther direction – very similar Checking a strategy (2)Slide12

We saw that:

For

, we need to show that for:

Exactly the same (switch odd with even)!

 Slide13

Definitions and LemmasSlide14

Sub-games

Definition

:

Let U ⊆V be a subset of V. Denote:

The graph G[U] is a sub-game of G if every dead-end in G[U] is also a dead-end in G

 Slide15

Sub-game example

is the only dead-end

Let’s look at

G[U] is a sub-game of G

Let’s look at

G[W] not a sub-game

 Slide16

Sub-games lemma

Claim

:

Let and be subsets of V s.t.

If

is a sub-game of and is a sub-game of then is a sub-game of GProof:Notice that If

is a dead-end in

, it is a dead-end in

Since

is a sub-game of

, v is a dead-end in

Using the same argument, v is a dead-end in

 Slide17

In English:

can’t force the token out of U,

can always stay inside U.

is ‘trapped’.

Definition

:A set will be called a if:

 Slide18

example

 Slide19

lemmas (1)

Claim 1

: For every

– trap U in G ,

is a sub-game of

Proof:Let be a dead-end in . If then , which means v was a dead-end in .If then

, which means it can’t be a dead-end in

 Slide20

lemmas (2)

Claim 2

: For every family

of

– traps the union

is a σ-trap as well. Proof:Trivial… Slide21

lemmas (3)

Claim 3

: Let X be a σ-trap in G and Y is a subset of XY is a σ-trap in G iff

Y is a σ-trap in G[X]

Proof

:Doodle timeSlide22

Attractor Sets

Mathematical definition was given in chapter 2

For a game G and a set

, denote

as:

The set of vertices from which Player σ has a strategy to attract the token to X or a dead-end in in a finite (possibly 0) number of stepsClaim:Said strategy can be memoryless  Slide23

Attractor Sets example

X

 Slide24

Attractor sets lemmas (1)

Claim 1

:

The set

is a σ-trap in G.

Proof

:Let us look at If and then: There is a move σ can do to take the token to some From , there is a σ-strategy that reaches a member of X in a finite set of movesThis means that there is a σ-strategy that reaches X from in a finite set of moves

in contradiction

If

then

 Slide25

Attractor sets lemmas (2)

Claim 1

:

The set

is a σ-trap in G.

Proof (cont.)

:If and then:Notice that v can’t be a dead-end. must move the token to some

From

, there is a σ-strategy that reaches a member of X in a finite set of moves

This means that there is a σ-strategy that reaches X from

in a finite set of moves

in contradiction

If

then

 Slide26

Attractor sets lemmas (3)

Claim 2

:

If X is a σ-trap in G, then so is

Proof

:

Trivial… Do a doodle proof!Claim 3:X is a σ-trap in G iff Proof:Another doodle proof  Slide27

Attractor sets lemmas (4)

Claim 4

:

where

is the greatest (w.r.t. set inclusion) σ-trap contained in Proof:Is U well defined? is a σ-trap so there is at least one σ-trap. Further more, union of σ-traps is a σ-trap, which means U is well defined! Now… doodle away! Slide28

In English:

A region from which:

cannot escape

σ wins from all vertices of this region using a memoryless strategy

Definition

:A set will be called a if:U is a

 Slide29

(1)

Claim 1

:

If U is a σ-paradise, then so is

Proof

:U is a , which means that is also a a Also, we know that player has a memoryless strategy from that either:Brings the token to UBrings the token to a dead-end for

Combine said strategy with

and you get a M.W.S for

 Slide30

(2)

Claim 2

:

For every family

of

– paradises the union is a σ-paradise as wellProof:A union of is a , which means that U is a We will find a M.W.S using Rom’s trick!Denote the M.W.S on for player Define a well-ordering relation < on

.

Define the strategy:

where

i

is the least element of I

s.t.

It’s easy to see that using said construction, a play p is either finite and ends with a dead-end to

, or infinite and its suffix conforms with some

 Slide31

Another quick recap!

Sub-game: a sub graph with no new dead-ends

: a sub set of the arena from which:

can’t leave can always stay

– all vertices from which

can:

Force to a dead-endForce the game into X paradise: can’t leave has a M.W.S Slide32

DeterminacySlide33

Determinacy

Theorem

:

The set of vertices of a parity game is partitioned into a 0-paradise and a 1-paradiseWe will prove this using an induction over

 Slide34

Base case: n = 0

Lemma 0

:

If the maximum parity of G is 0, then V is partitioned into a 0 and a 1-paradise Proof:Observation: player 1 can only win by taking player 0 to a dead-endThe winning region of player 1 is

From above lemmas,

is a 1-paradise (why?)

Let’s look at From above lemmas, we know that it is a 1-trap. Also, since the maximum parity is 0, and there are no dead-ends for 0 in , 0 always has a winning strategy! is a 0-paradise and

is a 1-paradise

 Slide35

The construction (1)

Induction step: assume the theorem holds for every parity game with maximum parity less than n

Let us mark

Let

be a

s

.t. Define: Define

 Slide36

The construction (2)

A few observations:

is a trap

is a sub-game of

Z is a σ-trap in

since it is the complement of an Attr setFrom the above is a sub-game of From a previous lemma: is a sub-game of  Slide37

The construction (3)

A few more observations:

The induction hypothesis applies to

!

We can partition Z to

and

- 0 and 1 paradises in  Slide38

Important lemma

Lemma 1:

Proof:

Let’s look at the sketch

 Slide39

Last lemma before we win!

Lemma 2:

Proof:

Let’s look at the sketch

 Slide40

So… what’s left?

We need to find

such that:

is a

) is a

is the empty setReminder: we saw that Maybe, if is the maximal

, we can finish…?

 Slide41

Creating

s

:

since paradises are close under union,

is the largest

We have seen before that the attractor set of a paradise is still a paradise

Since

is the largest paradise:

We have seen before that the complement of an attractor set is a trap:

Just like we wanted!

 Slide42

Proving determinacy

Theorem

:

The set of vertices of a parity game is partitioned into a 0-paradise and a 1-paradise.Outline:Do an induction over

We have proved the base case

We shall define

as the union of all and as its complementUse lemma 1 to show that is a Use lemma 2 to show that is a  Slide43

Complexity ResultSlide44

Complexity class of finite parity games

Theorem:

 Slide45

A non-deterministic algorithm:

Given G and v, guess a memoryless strategy w

Check whether w is a M.W.SWhy is it O.K. to only guess memoryless?

How can we check a strategy quickly?Slide46

Creating

A memoryless strategy w can be represented using a sub-graph of G:

 Slide47

Checking a strategy (1)

Check whether there exists a vertex

such that:

is reachable from

is odd

lies on a cycle in

containing only vertices of priority less or equal Claim:w is a winning strategy iff such a vertex doesn’t exist Slide48

Let’s assume such a vertex

exists

Observation: In

, if a vertex is reachable from v, player-1 can force the token into it (formal proof with induction)Observation: In , if the token is inside a cycle of vertices, player-1 can force the token to go over the entire cycleA winning play for player-1:Force the token into

Go over the cycle forever

If

exists, w is not a M.W.SOther direction – very similar Checking a strategy (2)Slide49

We saw that:

For

, we need to show that for:

Exactly the same (switch odd with even)!