Peter van Emde Boas: Games in Computation and Complexity Theory 20060925-26 Games in Computation and...

Peter van Emde Boas: Games in Computation and Complexity Theory 20060925-26

Games in Computation and Complexity Theory

Peter van Emde Boas

ILLC-FNWI-Univ. of AmsterdamReferences and slides available at: http://staff.science.uva.nl/~peter/teaching/gtis07.html

Workshop Logic & GamesKazimierz DolnySep 24-30, 2006

© Games Workshop© Games Workshop


Topics

• Relations Games and Computations

• Algorithmic Problems on Games and Game Models

• Complexity of problems on Games

• Games and PSPACE

• Other connections


Non Topics

• Computer programs for game playing– Chess, go, bridge, poker, robot-soccer

• Computer Games– Pong, Doom, Civilization , .....

• Games, Logic and Language– That’s for the other tutorials ……


Games and Computer Science



Are Games interesting at all ???

Past (1970/80) position of Games in Mathematics & CS:

Study object for a marginal part of AI(Chess playing programs)

Recreational Mathematics (cf. Conway, Guy &Berlekamp Theory)

Game Theory (Economy): von Neumann, Morgenstern, Aumann,

Games in Logic: Determinacy in set theory, Ehrenfeucht-Fraise Games, ....


A previous development

Epistemic Logic used to be a strange topic inPhilosophy up-to the early 1980-ies.

But then at IBM its use for Distributed Computingwas recognized and it became well fundable ComputerScience:

Fagin, Halpern, Moses, Vardi: Reasoning about Knowledge

Biannual TARK conference series since 1986


Noam NisanAlgorithms for Selfish AgentsMechanism Design for Distributed ComputingProc. STACS’99, Springer LNCS 1563, pp 1--15Invited paper.

Elias Koutsoupias & Christos PapadimitriouWorst-Case Equilibria, Proc. STACS’99, Springer LNCS 1563, pp 404--413

STACS’99 IN TRIER: Games !!!

© Peter van Emde Boas



Computer Science Theory• Computation Theory

• Complexity Theory

• Machine Models

• Algorithms

• Knowledge Theory

• Information Theory

• Semantics


Games in Computer Science• Evasive Graph properties (1972-74)• Information & Uncertainty (Traub ea. - 1980+)• Pebble Game (Register Allocation, Theory 1970+)• Tiling Game (Reduction Theory - 1973+)• Alternating Computation Model (1977-81)• Interactive Proofs /Arthur Merlin Games (1983+)• Zero Knowledge Protocols (1984+)• Creating Cooperation on the Internet (1999+)• E-commerce (1999+)• Logic and Games (1950+)• Language Games, Argumentation (500 BC)


Games and Computation

Computation: Set of Configurations, connectedby transitions, with initial one and final ones;

entire computation graph determines outcome:Accept, Reject, obtained value.....

Game: Set of positions connected by moves, with initial one and final ones; entire Game graphdetermines outcome: Win, Loose, Utility Payoff

What is the Difference ?????


COMPUTATION

• Deterministic

• Nondeterministic

• Probabilistic

• Alternating

• Interactive protocols

• Concurrency


COMPUTATION TREES• Notion of Configurations: Nodes• Notion of Transitions: Edges• Non-uniqueness of transition:

Out-degree > 1 - Nondeterminism• Initial Configuration : Root • Terminal Configuration : Leaf• Computation : Branch Tree • Acceptance Condition:

Property of trees


Types of Games

• Single player - no choices : Routine :

• Single player - choices : Solitaire :

• Two players – choices : Finite Combinatorial Games :

• Single player - random moves : Gambling :

• Two players - choices, randomness, information hiding:• Several players - concurrent moves :


Types of Games (and Computations)

• Single player - no choices : Routine : Deterministic Computation

• Single player - choices : Solitaire : Nondeterministic Computation

• Two players – choices : Finite Combinatorial Games : Alternating Computation

• Single player - random moves : Gambling : Probabilistic Algorithms

• Two players - choices, randomness, information hiding: Interactive Proof Systems

• Several players - concurrent moves : Multi Prover Systems


The Questions



A Game….

X

X

O

Questions: Termination ?

Is there a winner ?

If so, who ?

And how does he do it ?

Trivial …..

No ….


A Game….

X

X

O


Is there a winner ?

If so, who ?


Trivial …..

Yes ….

O

X ….

X


A Game

from H.W. Lenstra: Aeternitatem Cogita


Is there a winner ?

If so, who ?


Not evident: 50 moves rule…

Unknown in general…..


The game of HEX


Position is winning for Green


Green wins the game


Questions about HEX: Termination ?

Is there a Winner ?

If so, Who ?

And how does he achieve to win ….?

trivial

Topological argument => Draw is impossible !

Zermelo principle => There must be a winner !

Strategy-copy argument: It can’t be the second player who wins the game…..

Unknown


Why is a Draw impossible?


A Mathematical Game

Position: finite list of non negative numbers:

5, 5, 3, 4, 0, 1, 12

Move: replace a single number by a list of smaller numbers ( 0 only can be removed….. )

5, 3, 3, 2, 0, 3, 4, 0, 1, 12

Player unable to move loses the game.


Questions on the Mathematical Game: Termination ?

Is there a winner ?

If so, Who ?

And how to win ….?

Yes, based on set theory (Königs Lemma)

Yes, but not grace to the Zermelo principle ….

Parity principle:

First player wins iff some number occurs with odd parity

Establish position with even parities only…..


de Bono’s L-Game

Move: 1) remove your L-piece, and place it somewhere else2) may move one of the neutral stones

Board: 4 by 4 square

Pieces: Red’s L-piece Greens’s L-piece 2 Neutral stones


de Bono’s L-Game





de Bono’s L-Game





de Bono’s L-Game





de Bono’s L-GameBoard: 4 by 4 square




Move / Lost position

Complete move for Red

Lost Position for Green Initial position


Questions on L-Game: Termination ?

Is there a winner ?

If so, Who ?

And how to do that ….?

No – the game can run in cycles….

Some positiosn are determined, but not the intitial one

Initial position is a draw due to infinite play…..

Complicated; there exists a Database for this game…..

Source: van Gijlswijk, J.J.O.O Wiegerinck et.al.: Computer Analysis of E. de Bono’s L-GameRep. MI-UvA-76-18


Time for Computer Science…

In order to answer such questions the games must become input for some Computer Program

How to represent a Game on the Computer ?


Game Models



© Games Workshop © Games Workshop

URGATOrc Big Boss

THORGRIMDwarf High King

Introducing the Opponents

Games involve strategic interaction ......Games involve strategic interaction ......


CONWAY’S PERSPECTIVE

© Peter van Emde Boas 19761208 @ SMC, Amsterdam

© Peter van Emde Boas 19761208 @ SMC, Amsterdam © Peter van Emde Boas 19761208 @ SMC, Amsterdam

A Game isa Set.


A GAME IS A SET

• Pick an element, this is again a set

• Give it to the other player, it is now his turn

• Player looses if there is no element remaining: the empty set represents the (unique) lost game

• Let’s play REALS !


Another way of playing multisets

• A position is a finite collection of sets, repetitions allowed (multiset)

• A move consists of– Select one of the sets– Replace this set by a finite collection of

its elements, repetitions allowed

• The player unable to move looses• Why does this game terminate?


Playing Multisets

{ A, A, B, B, B, C , D , E }

( B = { p, p, q, r, r, s, t } )

{ A, A, B, B, p, q, q, s, s, t, C , D , E }


Playing Multisets

Game tree of the multiset game becomes infinite

Tree representing an individual play of the multisetgame is both Finitely Branching (due to the rules)and anti-well founded (due to well foundedness of sets)and therefore Finite on behalf of Königs Lemma

Therefore the multiset game terminates.


Game Trees

Root

Terminal node:Thorgrim looses

Thorgrim’s turn

Urgat’s turn

Terminal node:Urgat looses

Standard Interpretation:Player unable to move looses the game


Game Trees

Root

Terminal node:

Thorgrim’s turn

Urgat’s turn

Terminal node:

Free Interpretation:Winner explicitly designated at terminal node

T

TU

U

T

UT


Game Trees

Root

Terminal node:

Thorgrim’s turn

Urgat’s turn

Terminal node:

Non Zero-Sum Game:Payoffs explicitly designated at terminal node

2 / 0

5 / -71 / 4

-1 / 4

3 / 1

-3 / 21 / -1


Game Trees

Root

Terminal node:

Thorgrim’s turn

Urgat’s turn

Terminal node:


T

TU

U

T

UT

SUB-GAME


Game Graphs

start

U

T

An extension of theGame Tree model

May contain cycles

Acyclic Game Graphsas an intermediate model


Backward Induction

Terminal node:

Thorgrim’s turn

Urgat’s turn


Root

Terminal node: T

TU

U

T

UT

T

T

UU

U


Backward Induction

Root

Terminal node:

Thorgrim’s turn

Urgat’s turn

Terminal node:

Non Zero-Sum Game:Pay-offs computed for allgame nodes including the Root.

2 / 0

5 / -71 / 4

-1 / 4

3 / 1

-3 / 21 / -1

2 / 0

3 / 1

1 / 4-3 / 2

1 / 4


Backward Induction

At terminal nodes: Pay-off as explicitly given

At Thorgrim’s nodes: Pay-off inherited from Thorgrim’s optimal choice

At Urgat’s nodes: Pay-off inherited from Urgat’s optimal choice

For strictly competetive games this is the Max-Min rule

Snag ! How to chose between equal utilities ….???!

2 / 0

5 / -71 / 4

-1 / 4

3 / 1

-3 / 21 / -1

2 / 0

3 / 1

1 / 4-3 / 2

1 / 4

T

TU

U

T

UTT

T

UU

U


Zermelo Principle

Zermelo Principle presents a sufficient condition for Backward Induction to be possible and correct :

In a finite two person game tree where draws are impossible, one of the players must have a winningStrategy

Original intended application: Chess.

Extends to the case of Finite Acyclic Game Graphs


Why Finitely Branching ?Game of Finity:

Thorgrim: calls a random integerUrgat: decrements the number until it becomes zero and claims victory

Any mathematician knows that this game is won by Urgat

Backward Induction fails to prove so at everyfinite stage (there remain undecided successorsof the initial position)

BI has to be extended over ordinal time in this example at stage + 1 the game isdeclared to be a win for Urgat

U

UU

U

.....

.....


Backward induction on Game Graphs

T

Initial labeling:only sinks are labeled. Player unable tomove looses

start

TU

Final labeling:iterative apply BI rulesuntil no new nodes arelabeled. Remaining nodes are Draw D

D

T

U

U

U

T

T

TT

D D

start

U


Bi-Matrix Games

© Games Workshop © Games Workshop© Games Workshop© Games Workshop


Runesmith Dragon SquiggOgre

R

D

O S

1/-1

1/-1

-1/1

-1/1

A Game specified by describing A Game specified by describing the Pay-off Matrix ....the Pay-off Matrix ....


Von Neumann’s Theorem

( )/2 :+ ( )/2+© Games Workshop © Games Workshop© Games Workshop© Games Workshop


R D SO

R

D

O S

1/-1

1/-1

-1/1

-1/1

Mixed Strategy Mixed Strategy Nash EquilibriumNash Equilibrium; ; no player can improve his pay-off by deviation.no player can improve his pay-off by deviation.


2 / 0

5 / -71 / 4

-1 / 4

3 / 1

-3 / 21 / -1

L R

RL

l

r

r

rl lm

lll llr lml lmr lrl lrr rll rlr rml rmr rrl rrrLL 1/-1 1/-1 1/-1 1/-1 1/-1 1/-1 -3/2 -3/2 -3/2 -3/2 -3/2 -3/ 2LR 1/-1 1/-1 1/-1 1/-1 1/-1 1/-1 -3/2 -3/2 -3/2 -3/2 -3/2 -3/2RL 2/ 0 3/ 1 1/ 4 1/ 4 5/-7 5/-7 2/ 0 3/ 1 1/ 4 1/ 4 5/-7 5/-7 RR -1/ 4 3/ 1 1/ 4 1/ 4 5/-7 5/-7 -1/ 4 3/ 1 1/ 4 1/ 4 5/-7 5/-7

Extensive and Strategic Form


2 / 0

5 / -71 / 4

-1 / 4

3 / 1

-3 / 21 / -1

L R

RL

l

r

r

rl lm

lll llr lml lmr lrl lrr rll rlr rml rmr rrl rrrLL 1/-1 1/-1 1/-1 1/-1 1/-1 1/-1 -3/2 -3/2 -3/2 -3/2 -3/2 -3/ 2LR 1/-1 1/-1 1/-1 1/-1 1/-1 1/-1 -3/2 -3/2 -3/2 -3/2 -3/2 -3/2RL 2/ 0 3/ 1 1/ 4 1/ 4 5/-7 5/-7 2/ 0 3/ 1 1/ 4 1/ 4 5/-7 5/-7 RR -1/ 4 3/ 1 1/ 4 1/ 4 5/-7 5/-7 -1/ 4 3/ 1 1/ 4 1/ 4 5/-7 5/-7

Winning Strategy and Equilibrium


Extensive vs. Strategic Form

Game tree ===> Strategic Form• Exponential Blow-Up in size• Strategies deciding unreached positions• Strategy is non-adaptive• Representation is correct (for trees;• but what about arbitrary game graphs)


Extensive vs. Strategic Form

Strategic Form ===> Game Tree• Leads to games with incomplete information• Strategy is non-adaptive• Equilibria don’t always exist, • Translation is doomed to fail…..


Urgat doesn’t know the position he is in !

INCOMPLETE INFORMATION

Thorgrim’s Choice of strategy

Urgat’s Choice of strategy

Pay-off phase


A Real Life Game

© Donald Duck 1999 # 35

Starting with 15 matchesplayers alternatively take1, 2 or 3 matches away untilnone remain. The playerending up with an oddnumber of matches winsthe game

A Game specified by describing A Game specified by describing the rules of the game ....the rules of the game ....


Complexity Theory



COMPLEXITY ENDGAME ANALYSIS

Input Data:Game G , Position p in G

Question: Is position p a winning position for Thorgrim ?for Urgat ?a Draw ?

Relevant Issues: Game presentation,Game structure (tree, graph, description)Determinacy (Incomplete Information!)


The ProblemThinking about simple games like Tic-Tac-Toeone considers the size of the game to be indicated by measures like:-- size configuration ( 9 cells possibly with marks)-- depth (duration) game (at most 9 moves)

The full game tree is much larger : 986410 nodes(disregarding early terminated plays)

And what about the size of the Strategic Form ???!

What size measure should we use for complexityestimates ??


The ProblemThe gap between the experienced sizes (configuration size & depth) and the sizeof the game tree / graph is exponential !

Another exponential gap with the strategic form !

Here: use configuration size and depth assize measures for input games. Estimatecomplexity of endgame analysis in termsof these sizes: the Wood Measure of the Game


Why Worry About Representations?

Algorithmic problem

Instances

Solutions

InstanceFormat

Question

Instance Size

Algorithm Space/TimeComplexity

The rules of the game called “Complexity Theory”

MachineModel


Games as Acceptors

Input X is mapped to some game G(X)

The mapping X G(X) is easy to compute(computable in Polynomial Time or Logarithmic Space)

Consequence: G(X) has a Polynomial Size Description. (Leaving open what the Proper Descriptions are. Works only when using Wood Measure ....)

LG := { X | G(X) has a winning strategy forthe first player }

Which Languages L can be characterized in this way ?


The Folklore Belief

Reasonable Games Capture PSPACE

Many Conterexamples:

-- Games to easy-- Solitaire games which are to hard-- Two person perfect information games to hard

So what is the base for this belief ???


REASONABLE GAMESAssumptions for the sequel:

Finite Complete Information Zero Sum GamesStructure: tree given by description, where deciding properties likeis p a position ?, is p final ? is p starting position ?, who has to move in p ?, and thegeneration of successors of p are all trivialproblems ..... The tree can be generatedAnd traversed in time proportional to its size.....Moreover the duration of a play is polynomialwith respect to the Wood Measure.


Backward Induction in PSPACE?

The Standard Dynamic Programming Algorithm forBackward Induction uses the entire ConfigurationGraph as a Data Structure: Exponential Space !

Instead we can Use Recursion over Sequences of Moves:

This Recursion proceeds in the game tree from theLeaves to the Root.



Recursive Scheme in case of absense of Draws:

Let denote a sequence of movesc denote a configuration denote the empty sequence of movespos() denotes the position obtained by playing play(c) denotes the player who has tomove in position cmov(c) the finite set of legal moves in cfin(c) denotes whether c is finalwinf(c) denotes the winner in position cin case c is final.


Backward Induction in PSPACE?win( ) :=if fin( pos( ) )then winf( pos( ) )else if play( pos ( ) ) = Thorgrim

then if mov( pos ( ) )[ win( .) = Thorgrim ]

then Thorgrimelse Urgatfi

else if mov( pos ( ) )[ win( .) = Urgat ]

then Urgatelse Thorgrimfi

fifi



This Recursive scheme combines recursion(over move sequence) with iteration (over locallylegal moves). Correct only for determinated games!

Space Consumption =O( | Stackframe | . Recursion Depth )

| Stackframe | = O( | Move sequence | + | Configuration| )

Recursion Depth = | Move sequence | =O( Duration Game )

So the game duration should be polynomial!


Backward induction on Game Graphsstart

TU

Initial labeling:only sinks are labeled. Player unable tomove looses

start

TU

Final labeling:iterative apply BI rulesuntil no new nodes arelabeled. Remaining nodes are Draw D

D

T

U

U

U

T

T

TT

D D


Analyzing a Graph

If Thorgrim has no move the position is lost (in 0 moves) for Thorgrim

If Urgat has no move the position is lost (in 0 moves) for Urgat

If Thorgrim can move to a position which is lost (in k moves) for Urgat the position is won (in k+1 moves) for Thorgrim

If Thorgrim only can move to positions which are won (in ≤ k moves) for Urgat the position is lost (in ≤ k+1 moves) for Thorgrim


Analyzing a GraphIf Urgat can move to a position which is lost (in k moves) for Thorgrim the position is won (in k+1 moves) for Urgat

If Urgat only can move to positions which are won (in ≤ k moves) for Thorgrim the position is lost (in ≤ k+1 moves) for Urgat

Apply the above rules until no new nodes are declared loss or won for any of the two players. The remaining nodes areDraws (for both players).


Best Neighbour AlgorithmInitially: assign to every position a safe neighbour (if possible); otherwiseposition is lost (in 0 moves).Repeat (round k+1) as long as changes occur: Look for wins: if some position in the last round was declared lost (in k moves), declare its parents won (in k+1 moves) (unless they were declared won previously): Reset safe neighbour: if your safe neighbour in the last round was declared won, try to find a new safe neighbour; if no more neighbours are available declare position lost (in k+1 moves)End Repeat;All remaining positions are declared Draw


Best Neighbour Algorithm

In this algorithm each edge is investigated atmost once in every direction. Generating edges on the fly in stead of precomputingthe entire game graph becomes advantageous.

Original Result obtained by this method: E. de Bono’s L-game is a draw if playedby a perfect player. The longest decisive play from a won position requires 9 moves.


Incomplete Information Games



Runesmith Dragon SquiggOgre

R

D

O S

1/-1

1/-1

-1/1

-1/1

O S

RD

R D

-1/1

-1/11/-1

1/-1


Incomplete Information Games

Indications of trouble….:-- Simple games no longer are determinated-- Information sets capture uncertainty-- Uniform strategies are required-- Earlier algorithms become incorrect

WANTED: a complexity theory for IncompleteInformation Games......

WARNING: such a complexity theory is unlikelyto exist…..


Games and PSPACE



THE HOLY QUADRINITY

FINITE COMBINATORIAL

GAMES

QUANTIFIEDPROPOSITIONAL

LOGIC: ALTERNATION

PSPACE

UNRESTRICTEDUNIFORM

PARALLELISM


Walter Savitch

ICSOR; CWI, Aug 1976 San Diego, Oct 1983

© Peter van Emde Boas © Peter van Emde Boas© Peter van Emde Boas


The Invariance Thesis

Reasonable Sequential MachinesSimulate Each Other with

Polynomial overhead in Time andConstant Factor Overhead in Space

M’ simulates M : …. ; no definition given

TM’( x ) ≤ k.TM( x )k

SM’( x ) ≤ k.SM( x )


Parallel Computation Thesis

// PTIME = // NPTIME = PSPACE

True for Computational Models which combineExponential Growth potential withUniform Behavior.

The Second Machine Class


UNDERSTANDING PSPACE

• The most Robust Complexity Class• Solitaire Problem: finding an Accepting

path in an Exponentially large, but highly Regular Graph

• Matrix Powering Algorithm: Parallelism• Recursive Procedure: Savitch Theorem• Logic: QBF, Alternation, Games


Polynomial Space Configuration Graph

• Configurations & Transitions: – (finite) State, Focus of Interaction &

Memory Contents– Transitions are Local (involving State

and Memory locations in Focus only; Focus may shift). Only a Finite number of Transitions in a Configuration

– Input Space doesn´t count for Space Measure



• Exponential Size Configuration Graph:– input length: |x| = k ; Space bound: S(k)– Number of States: q (constant)– Number of Focus Locations: k.S(k)t

(where t denotes the number of “heads”)– Number of Memory Contents: CS(k)

– Together: q.k.S(k)t. CS(k) = 2O(S(k))

(assuming S(k) = (log(k)) )



• Uniqueness Initial & Final Accepting Configuration: – Before Accepting Erase Everything– Return Focus to Starting Positions– Halt in Unique Accepting State

Start Goal


Path Finding in Configuration Graph


Path Finding in Configuration Graph

Cycles in accepting path are irrelevant

Trash Nodes: Unreachable: or Useless


Unreasonable Algorithm• Step 1: generate Exponentially large

structure• Step 2: Perform Exponentially long heavy

computation on this structure• Step 3: Extract a single bit of information

from the result - the rest of the efforts are wasted.

• :‘• And this is just what the Parallel Models

do.....


Unreasonable Algorithm

Transitive Closure of Adjacency Matrix byIterated squaring ==> // Models

Recursive approaches ==> // Models,Savitch' Theorem & Hardness QBFand Games

Playing Savitch: a direct connection betweenPSPACE and Games


Adjacency Matrix

1 0 0 1 10 1 1 0 01 0 1 0 00 0 0 1 11 0 0 0 1

1

5

4

3

2

Matrix describes Presence of Edges in Graph;1 on diagonal: length zero paths

M :=


Adjacency Matrix1 0 0 1 11 1 1 0 01 0 1 1 11 0 0 1 11 0 0 1 1

1

5

4

3

2

In Boolean Matrix AlgebraM2 : Paths up to length 2M4 : paths up to length 4

M2 =

1 0 0 1 11 1 1 1 11 0 1 1 11 0 0 1 11 0 0 1 1

1

5

4

3

2

M4 =


Matrix Squaring

M[i,j] := ( M[i,k] M[k,j] )k

On an N node graph, a single squaring requires O(N3) operations

Log(N) squarings are required to compute N-th Power of the Matrix

Remember that N = 2O(S)


Think Parallel• O( N3 ) processors can compute these

squarings in time– O( log(N)) if unbounded fan-in is allowed– O( log(N)2 ) if fan-in is bounded

• This is the basis for recognizing PSPACE in polynomial time on PRAM models

• Don’t forget that the Matrix must be generated first …..

• See Second Machine Class paper* and/or chapter in Handbook of TCS

* Published in the proc’s of the 1985/86 Banach Center Workshop (Rasiowa)


Recursive Matrix Squaring

M[i,j,p+1] := ( M[i,k,p] M[k,j,p] )k

Log(N) recursion depth is required to compute N-th Power of the MatrixLog(N) recursion depth is required to replace N fold Iteration by RecursionOverall Recursion depth: Log(N)2

M[i,j,0] is the given Adjacency Matrix


Recursive Path FindingReach(x,y,0) := x = y trans(x,y)

Reach(x,y,p+1) := z [ Reach(x,z,p) Reach(z,y,p) ]

Reach(x,y,p+1) :=z [u,v [ (u=x v=z) (u=z v=y) ==> Reach(u,v,p) ]]

Rabin, Meyer & Stockmeyer trick! The Exponential Growth of the formula is prevented! Size ≈ O( S ) . O(Rec-depth)

Cook/Levin Formula


Recursive Path Finding

• Space Consumption of Recursive Procedure: O( | stackframe |.depth )

• In this case: | stackframe | and depth are both O(S)

• For path finding Nondeterminism of the original machine is irrelevant!

• Savitch Theorem: PSPACE = NPSPACE


Quantified Boolean Formulas (QBF)

INSTANCE: Formula of the form:F = Qx[Qy[Qz[….. [ P(x,y,z, …… ) ]]] … ]Where P is propositional and Q is or

QUESTION: is F true ?

THEOREM: QBF is PSPACE-Complete


QBF as a LOGIC GAME

• Game is a Propositional Formula (x1,x2, .... xn, y1,y2, .... yn)

• THORGRIM and URGAT select values for the xi and yi in a specified order

• THORGRIM wins if the formula eventually evaluates to true: otherwise URGAT wins the game.


QBF is PSPACE HardReach(x,y,0) := x = y trans(x,y)

Reach(x,y,p+1) := z [ Reach(x,z,p) Reach(z,y,p) ]

Reach(x,y,p+1) :=z [u,v [ (u=x v=z) (u=z v=y) ==> Reach(u,v,p) ]] The resulting formula is polynomial sizeand reduces an arbitrary PSPACE computation to QBF .Brute force Evaluation runs in PSPACE.

Cook/Levin Formula


QBF in PSPACE?

There exists an evident Recursive Algorithm for QBF

x[ P(x,….)] --> ( P(0, …. ) P(1, …. ) ) x[ P(x,….)] --> ( P(0, …. ) P(1, …. ) )

Space consumption: O( Depth . | Stackframe | ) = O(L2) where L = size input


Some Direct Connections



THE HOLY QUADRINITY

FINITE COMBINATORIAL

GAMES

QUANTIFIEDPROPOSITIONAL

LOGIC: ALTERNATION

PSPACE

UNRESTRICTEDUNIFORM

PARALLELISM


PLAYING SAVITCH

Usually the connection between PSPACE andGames is established indirectly, using eitherAlternation or QBF (or both) as intermediate.

In the Literature occasionally the same holds for the reverse connection: (this game can beanalyzed in PSPACE because it can beanalyzed in Polynomial Time on a ATM ....)

A direct approach is possible !


The Savitch Game

Given: some input x for a PSPACE acceptor M(M can be nondeterministic)

To Construct: a 2 person Complete Informationreasonable Game G(M,x) such thatx is accepted by M iff the first playerhas a winning strategy in G(M,x)

WLOG: time accepting computation ≤ 2 (S(|x|))


The Savitch GameAethis Thorgrim


Typical Position: Configurations C1 , C2 and Time Interval t1 < t2

| C1 | , | C2 | ≤ (S(|x|)) , 0 ≤ t1 < t2 ≤ 2 (S(|x|))

ROUND of the Game :Thorgrim chooses t3 such that t1 < t3 < t2

Aethis chooses C3 at t3

Thorgrim decides to continue with either C1 , C3 and t1 < t3 or C3 , C2 and t3 < t2


The Savitch Game

Initial Position: C1 is the startting position and C2 the (unique)

accepting Configuration. 0 = t1 and t2 = 2 (S(|x|))

Final Position: t2 - t1 = 1

Aethis wins if C1 ---> C2 is a legal transition; otherwiseThorgrim wins the game

Polynomial duration enforced by requiring (t2 - t1).e ≤ (t3 - t1) ≤ (t2 - t1).(1-e) for some fixede satisfying 0 < e ≤ 1/2


The Savitch GameWinning Strategies:

If x is accepted Aethis can win the game bybeing truthful (always play the true configuration insome Accepting Computation...)

If x is not accepted the assertion entailed by theinitial position is false. Regardless the configurationC3 chosen by Aethis he must make a false assertioneither on the first or on the second interval (or both). Thorgim wins by always attacking the false interval....


The Savitch GameThe Punchline:

Endgame Analysis of the Savitch Game is in Deterministic PSPACE, even if the original acceptor was Nondeterministic:

NPSPACE = PSPACE !

an Alternative (direct) proof of the SavitchTheorem....


The Savitch GameFinal remarks:Aethis can play his winning strategy if heknows the accepting computation.Thorgrim can play his winning strategy if hecan locate errors. Utterly unfeasible....And the new proof isn’t that much different.....

Reach(x,y,p+1) :=z [u,v [ (u=x v=z) (u=z v=y) ==> Reach(u,v,p) ]]

Move Aethis Move Thorgrim


The Power of your EditorTypical sequential models with the power ofparallelism:Vector Machines (Pratt, Rabin & Stockmeyer 74)MRAM (Hartmanis & Simon 74, Bertoni et. al 81)ASSM (Tromp & vEB 93)Vector Machines (Iwama & Iwamoto 98)

But also a simple model of a Text Editor (EDITRAM) solves QBF in polynomial time.

So have patience when your word processor makes you wait .....


EDITRAM

Program

RAMMemoryIntegers

Text files

Cursors


EDITRAM InstructionsThe standard RAM instructions on the main memory (store, load, store-I, load-I, ....)

Read at cursor positionWrite at cursor positionMove cursor (one position, to address, to end,..)Copy segment of text (marked by pair of cursors)Systematic string replacement (literal stringsonly) : C / aab / aacba /


EDITRAM Program for QBF1. Eliminate Quantifiers and variables: Starting at innermost quantifier replace xi F(...xi ...) by ( F(... 0 ....) F(... 1 ... ) ) xi F(...xi ...) by ( F(... 0 ....) F(... 1 ... ) )2. Eliminate connectives Working inside out evaluate logical connectives: C / (0 1) / 1 / etc.3. The result is a literal 0 or 1 . Read the answer .

step 1 requires a subroutine for identifyingand marking variables (due to literal only condition)


Alternation



Alternating Computation

+ -

+ +- - - ++

- Computation Tree

Configuration Type

Existential

Universal

Negating

Accepting

Rejecting

Assumption: no infinite branches


Alternating Computation

+ -

+ +- - - +

+

-Evaluation Full Computation Tree

This Tree AcceptsThis is just backward induction on a game tree ;

But what is the Game ??

Configuration Type: Game meaning

Existential: Thorgrim moves

Universal: Urgat moves

Negating: Role Switch

Accepting: win (for Thorgrim)

Rejecting: Loose

+

+

+

+

+

+

-

- -

- -

-

--

++

+

+


Negating Nodes ?• Create for every node its dual node which

yields the “same” transitions• Dual of accepting node is rejecting • Dual of rejecting node is accepting• Dual of universal node is existential• Dual of existential node is universal• Dual of Dual is identity• Replace every negating node by an

existential one, dualizing the entire subtree below it (think de Morgan!)


Eliminating Negating Nodes

+ -

+ +- - - +

+

+

+

+

+

+

-

- -

- -

-

--

++

+

+

+ +

+ +- - -

+

+

+

+

+

+

-

-

- -

-

--

+

+

+

-

-

-

-

+

-

-

Dualized nodes


Alternating Computation as a Game

Negating states are unnecessary - by dualizingparts of the computation tree they can be removed.

What remains is a Computation Game where bothThorgrim and Urgat control nondeterministicchoices in the computation. Thorgrim wants thecomputation to accept. Urgat wants to prevent this from happening......

Infinite branches (if present) don’t contribute to the game value (non-trivial to prove)


The Alternation Theorems

THM 1: APTIME PSPACE:

Recursive evaluation of the quality of the root of theAlternating Computation Tree:Depth = O(time) , | Stackframe | = O(time)

Resulting Overhead: Square

Can be improved to linear by storing only a constantamount of information in a stack-frame



THM 2: PSPACE APTIME:

QBF trivially can be solved by an ATMQBF is PSPACE-hardWhat Further Evidence do we Need?



THM 3: ALOGSPACE PTIME:

Iterative evaluation of the quality of the all nodes in theAlternating Computation Tree:

#Nodes = Exponential in S; # Stages = O(# Nodes )

In terms of Games: This is BackwardInduction on a Game Graph.



THM 4: PTIME ALOGSPACE:

Guess a Space-Time diagram of AcceptingComputation, (thinking in terms of a correct tiling)starting from the “accepting” tile and moving Backwards in time:

At cell X guess contents of three upper neighbors;Universally validate these three upper neighbors.

THIS IS CORRECT DUE TO DETERMINISM!


Guess and Validate

“accepting Configuration”Coherence of validated Guesses is enforced fromTop to bottom

“Start Configuration”


Tiling Games



Tiling GamesTile Type: square divided in 4coloured triangles.Infinite stock availableNo rotations or reflections allowed

Tiling: Covering of region of the plane such that adjacent tiles havematching colours

Boundary condition: colours given along (part of) edge of region, or some giventile at some given position.


Turing Machine

Finite ControlProgram : P

Tape

Read/WriteHead

P (Q ) (Q {L,0,R}) :(q,s,q’,s’,m) P denotes the instruction:when reading s in state q print s’ performmove m and go to state q’ . Nondterminism!

Q: states: tape symbols


ComputationsConfiguration c : finite string in *(Q) *Computation Step c --> c’ obtained by performing an instruction in PComputation: sequence of stepsFinal Configuration: no instruction applicableInitial Configuration: start state & leftmost

symbol scannedComplete Computation: computation starting

in initial configuration and terminatingin final one

Accepting / Rejecting computation ....


Turing Machines and TilingsIdea: tile a region and let successive color sequences along rows correspond tosuccessive configurations.....

s

s

symbol passing

tile

s

qs

state accepting

tilesq

s

qsq

s’

qs instruction steptiles

q’s’

qsq’

q’s’

qs

(q,s,q’,s’,0) (q,s,q’,s’,R) (q,s,q’,s’,L)

SNAG: Pairs of phantom heads appearing out of nowhere...Solution: Right and Left Moving States....


Example Turing Machine

K = {q,r,_}S = {0,1,B}P = { (q,0,q,0,R),

(q,1,q,1,R),(q,B,r,B,L),(r,0,_,1,0),(r,1,r,0,L),(r,B,_,1,0) }

q0 1 0 1 1 B 0 q1 0 1 1 B 0 1 q0 1 1 B 0 1 0 q1 1 B 0 1 0 1 q1 B 0 1 0 1 1 qB 0 1 0 1 r1 B 0 1 0 r1 0 B 0 1 r0 0 0 B 0 1 1 0 0 BSuccessor Machine;

adds 1 to a binary integer._ denotes empty halt state. 11 + 1 = 12


Reduction to Tilings

q0 1 0 1 1 B 0 q1 0 1 1 B 0 1 q0 1 1 B 0 1 0 q1 1 B 0 1 0 1 q1 B 0 1 0 1 1 qB 0 1 0 1 r1 B 0 1 0 r1 0 B 0 1 r0 0 0 B 0 1 1 0 0 B

© Peter van Emde Boas ; 19921029


Implementation in Hardware

© Peter van Emde Boas ; 19950310 © Peter van Emde Boas ; 19950310 © Peter van Emde Boas ; 19921031


Tiling reductions

initial configuration

accepting configuration/by construction unique

blank border

space

blank border

time

Program : Tile TypesInput: Boundary

condition

Space: Width regionTime: Height region


Tiling ProblemsSquare Tiling: Tiling a given square with

boundary condition: Complete for NP.Corridor Tiling: Tiling a rectangle with

boundary conditions on entrance and exit(length is undetermined): Complete for PSPACE .

Origin Constrained Tiling: Tiling the entire planewith a given Tile at the Origin.Complete for co-RE hence Undecidable

Tiling: Tiling the entire plain without constraints. Still Complete for co-RE(Wang/Berger’s Theorem). Hard to Prove!


Two Person Tiling GameDesired property of the construction:

A winning strategy for Thorgrim for the given instance corresponds to an Accepting Alternating Computation.

Thorgrim wins by simulating a winning strategy on the Alternating Turing Machine (understood as game).Urgat, however has additional possibilities: he couldattack the legality of the encoding of a Turing Machine computation.....Urgat therefore must be forced to stay within theconstraints allowed by the encoding.

Solved by Bogdan Chlebus.


Conventions on ATM

• Tape uses even number of cells

• Universal and Existential States Alternate

• Left or Right moving States only

• All Instructions Move

• Unique Accepting Configuration


Chlebus´ Bag of Tricks

Thorgrim and Urgat, each obtain their own set oftile types; this is enforced by introducing two flavorsof the vertical colors, indicated by a pink shade.

Base Shading for Thorgrim

Base Shading for Urgat

Both borders of Rectangle to be tiled are shaded white.


Chlebus´ Bag of Tricks

Tiles types representing the Turing Machine instructionsare replaced by pairs: Decide & Move

s’

q’

qs

q’s’

qs

q’

q’s’

s’

Decide

MoveThe blue shadingindicates a rightmoving state q’for left moving statesuse grey shading


THORGRIM’S TILES

s

s

s

s

q

pass symbol &prevent introfrom right

pass symbol &force intro from right for Urgat q’

qs

q’s’

q’s’

s’

q’

qs

q’s’

q’s’

s’

Decide &move right

Decide &move leftqs

s

qs

s

qq

accept from left

accept from right


URGAT’S TILES

s

s

pass symbol;accept fromright is prevented q’

qs

q’s’

q’s’

s’

q’

qs

q’s’

q’s’

s’

Decide &move right

Decide &move leftqs

s

qs

s

qq

accept from left

accept from rightis forced

q


Other Games



GEOGRAPHY

Objects Selected: Directed Edges

Constraint: Edge is connected to the previously selected edge, and has notbeen selected before

Winning: Player unable to move looses

Morale: Both players build a maximalEulerian Path.


GEOGRAPHYThe original Childrens game:

Objects Selected: City names

Constraint: Name starts with final letterof the name of the previous city

Winning: Player unable to invent a newcity name looses

Question: Explain the graph formulation


GEOGRAPHY is PSPACE HARD

Proof: Reduction from QBF

Special Contraints on QBF:Propositional Formula in CNF

Qx1 Qx2 .... Qxk [ C1 C2 ... Cm ]

Q = or , Cj are clauses


GEOGRAPHY is PSPACE HARD

Component Design: Vars

selection component

for xi followed

by xi+1

xixi

ci

ci+1

di

xixi

ci

ci+1

di

selection component

for xi followed

by xi+1

if no alternation occurs thenodes di and ci+1 are identified


GEOGRAPHY is PSPACE HARDComponent Design: Clauses

cn

yky1 ym

towards thosexi and xi which occur inC1

Here it is crucial that Urgat chooses for a clauseand Thorgrim chooses a literal in the clause


GEOGRAPHY is PSPACE HARDc1 is the startnode

First Player depends on the Type of First Quantifier: if it is Existential, Thorgrim willstart the game as usual.

Play: each play will first traverse the varscomponents and select a truth assignment.Subsequently Urgat selects a clause, Thorgrimone of its literals. Only if this literal node isunvisited in the first stage Urgat has a move left.


STRATEGY MAPPINGPlayers select truth values for their variables as in the QBF logic game.

If now some clause is false Urgat can select it;and all literals chosen by Thorgrim will be unvisited, so Urgat still has a free move toa visited out-degree 1 node. Urgat wins.If all clauses are true Urgat must select atrue clause, so Thorgrim can select a visitedliteral. Urgat can’t go anywhere. Thorgrim wins.

NB! Urgat is to move, even if at this node Thorgrim had to move previously!


Known Examples of Games used in Complexity Theory (1980+)

• Tiling Games (NP, PSPACE, NEXPTIME,....)

• Pebbling Game (Solitaire game !) (PSPACE)• Geography (PSPACE)• HEX (generalized or pure) (PSPACE)• Checkers, Go (PSPACE)• Block Moving Problems (PSPACE)• Chess (EXPTIME)

The Common View is that Games Characterize PSPACE


Hard Solitaire Games

Solitaire Games are in NP when a winning playcan be guessed and validated......

OOPS! The game can require to many steps:repetition of moves and/or positions

Typical Examples:

Pebble GameBlock Moving Games


The Combinatorial Game perspective

26th MFCS 2001Marianske Lazne

Aug 30


Erik D. Demaine, LCS MITbirthdate: Feb 28 1981

Playing Games with Algorithms:Algorithmic combinatorial game theory

URL : http://theory.lcs.mit.edu/~edemaine/games/


GAME OVER

© Morris & Goscinny

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