turing machines for dummies why representations do matter peter van emde boas illc-fnwi-univ. of...
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Turing Machines for DummiesWhy representations do matter
Peter van Emde BoasILLC-FNWI-Univ. Of Amsterdam
Bronstee.com Software & Services B.V.
SOFSEM 2012 – Jan 25 2012Špindlerúv MlýnCzech Republic
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Turing Machine
Finite Program : P
Tape
Read/Writehead
P (K ) (K {L,0,R}) :(q,s,q’,s’,m) P denotes the instruction:When reading s in state q print s’, performmove m and proceed to state q’ . Nondeterminism!
K: States: tape symbols
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Transitions
Configuration c : finite string in *(K) * $ A B A A C <q,A> C B B A $
Transition c --> c’ obtained by performing instruction in P E.G., the instruction <q,A,r,B,R>
$ A B A A C <q,A> C B B A $ |-- $ A B A A C B <r,C> B B A $
Computation: sequence of transitions
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Configurations
Three ingredients are required for describing a Configuration:
The Machine State : qThe contents of the tape (preferably with endmarkers) : $ A B A A C B C B B A $The position of the reading head : i
Available options
Mathematical Representation: < q , xj xj+1 …. xi …. xl-1 xl , i >
Intrinsic Representation: $ xj xj+1 …. q xi …. xl-1 xl $ or $ xj xj+1 …. <q xi > …. xl-1 xl $
Semi Intrinsic:< q , $ xj xj+1 …. xi …. xl-1 xl $ >
↓
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Theme of this presentation
• The Convenience of the Intrinsic Representation
– Its History : who invented it, who saw its usefulness ?– Applications which are hard, if not impossible when the
Mathematical representation is used• Chomsky Hierarchy and Automata models
• Master reductions for NP: Cook-Levin and Tilings
• Stockmeyer on Regular Expressions
• Parallel Computation Thesis : the Second Machine Class
– Is there a real problem ?
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HISTORY
Turing Machine and their Use
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The teachings of our Master
Our textbooks present Turing Machine programs in the format of quintuples or quadruples.
What format did Turing use himself ?
Some fragments of the 1936 paper
Configuration means state in our terminology
Looks like quintuples….
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For Turing Composite transitions are allowed
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This is an example of the Intrinsic Representation
Complete Configuration means Configuration on our terminology
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A Macro language for Turing Machine programs
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This Macro Language supportsRecursion !
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The format of TM programs which today is conventional arises as a simplification introduced for the purpose of constructing the Universal Turing Machine
Turing operates as an Engineer(Programmer) rather than aMathematician / Logician
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Nondeterminism
• Our concept of Nondeterminism (the applicable instruction is not necessarily unique) is for Turing a serious programming error
• Nondeterminism became accepted in the late 1950-ies as a consequence of the needs of Automata Theory
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Using the model
• Initial configuration on some input• Final configuration
– No available instruction– By final state (accept , reject)– Evaporation of the state
• Complete Computation– From initial to final configuration (or infinity)
• Result of computation– Language recognition (always halting, condition on final
configuration)– Language acceptance (accept by termination)– Function evaluation (partial function/relation, requires termination)
• Non Terminating Computations – Stream Computing– Interactive computation
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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;
Increments a number in binary._ represents the empty halting state. 11 + 1 = 12
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Variants
Semi Infinite Tape
B B BBBB BA AAA OO OOO
BBB BAAO OO
BBB AA OO
q
q
Tape folding; remember which track you are on….
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Multiple Tapes
B B BBBB BA AAA OO OOO
Z U XZYU XX YZX ZZ ZYU
q
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Multiple Tapes
B B BBBB BA AAA OO OOO
Z U XZYU XX YZX ZZ ZYU
q
Tapes become tracks on a single tapeMarkers used for maintaining head positions on the tracks
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Invariance Thesis
• Other variants– Multi dimensional tapes– Multi heads on a single tape– Jumps to other head positions
• All models of Turing Machines are equivalent– up to polynomial overhead in time and
constant factor overhead in space• First Machine Class– Includes RAM / RASP model as well
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How Turing Machines are used
• Marvelous TM algorithms do exist in the literature– Hennie Stearns: oblivious k-tapes on two tapes– Slisenko: Real-time Palindromes regognition and string
matching– Vitányi: Real-time Oblivious multi-counter simulation
• The main use of TM’s is for proving negative results (using reductions)– Undecidability– NP-hardness– Other hardness results
• Requires direct encoding of TM-computations in target formalisms: Master Reductions
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Time-Space Diagram
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
Master Reductions use this Time-Space Diagram as representation of the computation subject to the Reduction
The Intrinsic Representation is far more useful, if not required, for constructing these Master Reductions
WHY ??
Because validity of the Diagram can be checked Locally
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Time for a new hero
Larry StockmeyerFOCS 1978, Ann Arbor
© Peter van Emde Boas ; 19781016 Thesis MIT 1974Rep. MAC-TR-133
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Stockmeyer on representations
This is a Mathematical Representation
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Stockmeyer on representations
For the Single Tape model the Intrinsic Representation is used
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Stockmeyer’s Lemma
Validiy of transition becomes a local check on a 2 by 3 window in the Time-Space Diagram
NB: for Stockmeyer Functions are partial and multi-valued, I.E., Relations
Is this the first time the convenience of the Intrinsic Representation is mentioned explicitly ?
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Applications
• Automata Theory
• Master reductions for NP– Cook/Levin reduction to SAT– Bounded Tiling
• Stockmeyer on Regular Expressions
• Parallel Computation Thesis
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Automata Theory
• The Machine based characterization of the Chomsky Hierarchy– Regular grammars Finite Automata– Context Free grammars Push Down
Automata– Context Sensitive grammars Linear
Bounded Automata– Unrestricted grammars Turing Machines
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A Side Remark
Traditional Textbooks present Automata Theory in the order:REG , CF, CSL, Type 0 resp. FA, PDA, LBA, TM
Alternative: start with Turing Machines, treating the alternative models as restricted models
Advantage: the concepts involving Configurations and Computations don’t need a separate presentation for each model
The desired characterizations are obtained by correlating production steps in the grammar world and computation segments in the Machine world, observing the required restrictions on both sides
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A trivial Observation
• The production proces in the Grammar world can be simulated by a single Tape Turing Machine
• Turing Machines are after all perfect symbol manipulators
• Remains to show that restricted grammar classes can be simulated by restricted Machine models
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The Converse Direction
• Using the Intrinsic Representation, the transitions of a Turing Machine are described by Context Sensitive Rules:
<q,a> <p,b,R> corresponds to qaX bpX<q,a> <p,b,0> corresponds to qa pb<q,a> <p,b,L> corresponds to Xqa pXb
etcetera
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The context
• In the grammar world a production starts with the start symbol S, and terminates in a string of terminals
• In the machine world a computation starts with an initial ID with the terminal string on the input tape, and ends in an accepting configuration
• Hence: Mutual Simulations require some adaptations……
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TM simulation of a type 0 grammar
In the initial Configuration the TM writes the start symbol S in a second track of the tape.
The productions in the grammar are stepwise simulated in this second track (shifting the symbols left/right of the rewritten ones over the required distance)
When the production is completed the TM checks whether the two tracks contain the same string, and accepts accordingly
Hence: the language generated by a type 0 grammar can be recognized by a Turing Machine
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A type 0 grammar simulates a TM
From S generate the Initial Configuration in two tracks (this can be done using Regular productions only)
Simulate the TM computation using the CS rules in the first track, leaving the symbols in the second track invariant
If the machine accepts, erase the entire first track(this may require lenght reducing rules, hence type 0….)
Hence: The language accepted by a Turing Machine can be produced by a Type 0 grammar.
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CS grammars and LBA
• The same proof idea works• LBA simulates CS grammar: no
intermediate string exceeds the given input in lenght
• CS grammar simulates LBA: in previous proof erasing rules are only needed to remove extra workspace on the tape, and the LBA doesn’t use extra workspace
• Beware for the endmarkers: better have them printed as markers on the first and last input symbol….
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CF grammars and PDA
• Snag: the PDA is a two tape device• Solution: code configurations as
<processed input segment><state><reversed stack>
• Yields correspondence between leftmost derivation and PDA computations
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S
A B
A CAB
C
z
y
zx
CF Rules:S AB , A AC , B BA, A C,A x , B y, C z
Left Derivation:
*S *AB *ACB x*CB xz*B xz*BA xzy*A xzy*C xzyz*
PDA Instructions:
*,λ,S *,AB*,λ,A *,AC*,λ,B *,BA*,λ,A *,C*,x,A *, λ*,y,B *, λ*,z,S *, λ
Syntax tree
The Left derivation is equal to the time-space diagram of the PDA computation
Hence: a single state PDA can accept what the CF grammar produces
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PDA and CFG
• A single state PDA can simulate a CFG
• The PDA accepts by empty stack
• If the PDA has several states the CFG rules must encode these states
PDA Instructions:
q,λ,S r,ABr,λ,A s,ACq,λ,A s,Cr,x,A s, λ
CF Rules:
[qSα] [rAβ][βBα][rAα] [sAβ][βCα] [qAα] [sCα] [qAs] x
α , β , range over the state symbols
[rAr] means “in state r, with A on top of the stack, a computation starts after which in state s the symbol below A is exposed”
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Regular Grammars and Finite Automata
The standard translation: q,a r q ar
instruction production rule
Machine configuration and partial derivation
abbaqacbac abbaq
For the computation the input already processed is irrelevant; all information resides in the state, and the unread input determines the computation.In the grammar the past symbols are produced and the future symbols are invisible.A matter of perspective: past vs. future
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Master reductions for NP
• Cook-Levin reduction to SAT– Based on Mathematical representation– Based on Intrinsic representation
• What is the difference ?
• Tiling based reduction– Does it require an Intrinsic
representation ?
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Time-Space Diagram
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
0 1 0 1 1 B q 0 0 1 0 1 1 B q 1 0 1 0 1 1 B q 2 0 1 0 1 1 B q 3 0 1 0 1 1 B q 4 0 1 0 1 1 B q 5 0 1 0 1 1 B r 4 0 1 0 1 0 B r 3 0 1 0 0 0 B r 2 0 1 1 0 0 B - 2
Intrinsic Mathematical
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Reduction to SAT (intrinsic)
The key idea is to introduce a family of propositional variables:
P[i,j,a] expressing at row i (time i) on position j (space j) the symbol a is written in the diagram
Conditions:I at every position some symbol is writtenII at no position more than one symbol is writtenIII the diagram starts with the initial configuration on the inputIV the diagram terminates with an accepting configurationV the transitions follow the Turing Machine program
Va : expressed using implications (beware for Nondeterminism)Vb : expressed by exclusion of illegal transitions
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Reduction to SAT (mathematical)The key idea is to introduce a family of propositional variables:
P[i,j,a] expressing at row i (time i) on position j (space j) the symbol a is written in the diagramQ[i,q] expressing at time i the machine is in state qM[i,j] expressing at time i the head is in position j
Extra Conditions:VI at every time the machine is in some stateVII at no time the machine is in more than one stateVIII at every time the head is in some positionIX at no time the head is in more than one position
The correctness conditions III , IV and V are rephrased, somewhat easier to understand….
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What’s the difference ??
Assume that a computation of T steps is described, hence the height (but also the width) of the diagram is O(T)Assume that the number of symbols used is K K = O( # states * # tape symbols )
Investigate the size of the required propositional formulas.
Investigate whether these formulas are expressed as clauses.
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Expressing the ConditionsConditions:I at every position some symbol is written O(T2K)II at no position more than one symbol is written O(T2K2)III the diagram starts with the initial configuration O(T)
on the inputIV the diagram terminates with an accepting O(1)
configurationV the transitions follow the Turing Machine program Va : expressed using implications O(T2K2) Vb : expressed by exclusion of illegal transitions O(T2K6)VI at every time the machine is in some state O(TK)VII at no time the machine is in more than one state O(TK2)VIII at every time the head is in some position O(T2)IX at no time the head is in more than one position O(T3)
All conditions (except Va) are easily expressed by clauses
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Conclusion
• The standard proof (EG., Garey & Johnson) uses the Mathematical representation, yielding a cubic formula size blow-up
• However, a quadratic formula size blow-up is achievable when using the intrinsic representation
• Same overhead is obtained when taking the detour by the tiling based reduction (next)
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Tiling based ReductionTile 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.
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Turing Machines and Tilings
Idea: 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....
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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
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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
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Implementation in Hardware
© Peter van Emde Boas ; 19950310 © Peter van Emde Boas ; 19950310 © Peter van Emde Boas ; 19921031
htpp://www.squaringthecircles.com/turingtiles
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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
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Tiling Problems
Square Tiling: Tiling a given square withboundary condition: Complete for NP.
Corridor Tiling: Tiling a rectangle withboundary 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!
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Detour to SAT
A reduction from Bounded Tiling to SAT requires propositional variables t[i,j,s] expressing at position (i,j) a tile of type s is placed
Conditions:I Everywhere some tile is placed O(T2K)II Nowhere more than one tile is placed O(T2K2)III Boundary conditions are observed O(TK)IV Adjacency conditions are observed O(T2K2)
T height and width of the tiled regionK number of tile types
All conditions are expressed as clauses
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Is the intrinsic representation needed ?
• A tiling reduction is posible for the semi-intrinsic representation (state information can be transmitted through rows…)
• Translating numeric information into geometric info (without introducing a semi-intrinsic representation) seems hard if not impossible…
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Stockmeyer on Regular Expressions
Thesis MIT 1974Rep. MAC-TR-133
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Regular ExpressionsS finite alphabet (in our applications Σ U (K ˟ Σ) U { $ } )
REG(S) :0 REG(S) M(0) = empty language1 REG(S) M(1) = {λ} singleton empty worda REG(S) for a S M(a) = {a} singleton letter a word
If f, g REG(S) then f + g REG(S) M(f + g) = M(f) U M(g) union f.g REG(S) M(f.g) = M(f).M(g) concatenation f* REG(S) M(f*) = M(f)* Kleene star
f* = 1 + f + f.f + f.f.f. + ….
Extra operations:
f2 = f.f squaring f ∩ g M(f ∩ g) = M(f) ∩ M(g) intersection~ f M(~ f) = S* \ M(f) complementation
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Regular Expressions
• Describe the Regular languages over S• Transformation expression Finite automaton is easy (construction
where the parentheses in the expression become the states in the Finite Automaton)
• Converse transformation more difficult but standard textbook material (induction over number of states)
• Other interpretations exist and are useful: Regular Algebra’s, EG., in programming logics (PDL)
• Complete axiomatizations exist• No direct algebraic expressions for intersection and complementation • Regular languages being closed under intersection and
complementation implies that these operations are expressible in all individual instances
• Extra operators yield succinctness
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Stockmeyer’s Decision Problems
NEC(f,S) is M(f) a proper subset of S* ?
EQ(f,g) is M(f) = M(g) ?INEQ(f,g) is M(f) ≠ M(g) ?
NEC(f,S) is equivalent to INEQ(f,S*)EQ and INEQ are complementary problems
Stockmeyer (1974) characterizes the complexity of these problems, depending on the set of available operators
Considering complementary problems was meaninful: the Immerman-Szelepsényi result was discovered only 13 years later….
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Stockmeyer’s Master Reduction
Given a TM program P and some input string ω, there doesn’t exist a regular expression denoting the (linearizations of) accepting time-space diagrams.
But Violations against representing such a diagram can be described by regular expressions
Syllabus Errorum approach: construct a Regular expression which enumerates all possible violations, and test whether there remains a string not covered by this expression (NEC problem)
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Syllabus Errorum
A correct time-space diagram consists of configurations, all of equal lenght, separated by $ symbols
The first configuration must be the intitial configuration on the given input
The last configuration must be accepting (the unique accepting) configuration
The diagram may contain no illegal transitions
This condition is captured by the absense of forbidden 2 by 3 windows in the diagram, as expressed by Stockmeyer’s lemma;For this method to work the Intrinsic representation seems essential.
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Yardstick expressions
Alphabet S used: Σ U (K ˟ Σ) U { $ } ; The width of the time space diagram (space consumed by the computation) is denoted M .Let V = Σ U (K ˟ Σ) , W = Σ
Given alphabet Z and number N we construct an regular expressionYa(Z,N) representing strings of length N of symbols from Z
Note that Ya(1+Z,N) now represents strings of lenght ≤ N
Using these yardstick expressions the various sources of errors can be described
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Error Descriptions
There is a substring inbetween two $ symbols which is to short:
S*.$. Ya(1+V,M-1) .$.S*
There is a substring inbetween two $ symbols which is to long:
S*.$. Ya(V,M+1) .S*.$.S*
There is an incorrect transition in the diagram:
S*.xyz. Ya($+V,M-2) .uvw.S* where is a forbidden 2 by 3 window in the diagram
Similar (even more simple) expressions for the properties “starts wrong” and “ends wrong”
x y z u v w
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The reduction
• The regular expression which is the sum of all these error types represents the exact complement of the set of time space diagrams of accepting computations by P on input ω in space M
• Denote this expression by ER(P, ω, M)• Input is accepted iff NEC( ER(P, ω, M) , S)• Remains to estimate the lenght of this expression• Remember that we consider Nondeterministic
space bounded computations
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The size of the yardstick expressions
• Without extra operators Ya(Z,N) is of size O(N) yielding NPSPACE hardness for NEC
• With squaring 2 Ya(Z,N) is of size O(log(N)) yielding NEXPSPACE hardness for NEC
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Expressions without *
• The same method also works without using * , but now the height of the diagram (time) must be restricted
• Yields reductions showing NP hardness and NEXPTIME hardness for the INEQ problem (without or with squaring)
• Nonelementary hardness if complementation is added
• Matching upper bounds are also obtained
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Parallel Computation Thesis
// PTIME = // NPTIME = PSPACE
True for Computational Models which combineExponential Growth potential withUniform Behavior.
The Second Machine Class
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Representative examplesSequential devices operating on huge objects
Vector machine Pratt & Stockmeyer 74,76MRAM Hartmanis & Simon 74MRAM without bit-logic Bertoni, Mauri, Sabadini 81EDITRAM Stegwee, Torenvliet, VEB 85ASMM Tromp, VEB 90,93
Alternating TM (RAM) Chandra, Stockmeyer & Kozen 81
Parallel DevicesParallel TM Savitch 77PRAM Savitch & Stimson 76,79SIMDAG Goldschlager 78,82Aggregate Goldschlager 78,82Array Proc. Machine v Leeuwen & Wiedermann 85
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How to prove it
• Inclusion //NPTIME PSPACE :– Guess computation trace– Verify that it accepts by means of
recursive procedure– Validate that the parameters are
polynomially bounded in size– Uniformity of behavior is essential
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How to prove it
• Inclusion PSPACE //PTIME :– Today’s authors show that
QBF //PTIME– Original proofs give direct simulations
of PSPACE computations, based on techniques originating from the proof of Savitch’ Theorem PSPACE = NPSPACE
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Walter Savitch
Amsterdam; CWI, Aug 1976 San Diego, Oct 1983
© Peter van Emde Boas © Peter van Emde Boas© Peter van Emde Boas
Proved in 1970 PSPACE = NPSPACE
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Understanding PSPACE
Acceptance = Reachability in Computation Graph
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
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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
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Polynomial Space Configuration Graph
• 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)) )
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Polynomial Space Configuration Graph
• Uniqueness Initial & Final Accepting Configuration: – Before Accepting Erase Everything– Return Focus to Starting Positions– Halt in Unique Accepting State
Start Goal
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Path Finding in Configuration Graph
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Path Finding in Configuration Graph
Cycles in accepting path are irrelevant
Trash Nodes: Unreachable: or Useless78
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Unreasonable Algorithm
• Step 1: generate this 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 work is wasted.
• :‘• Which is just what the Parallel Models do.....
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Unreasonable Algorithm
Transitive Closure of Adjacency Matrix byIterated squaring ==> // Models
Recursive approaches ==> // Models,Savitch' Theorem & Hardness QBF and Games
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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 :=
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Adjacency Matrix
1 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 =
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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)
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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
• More in Second Machine Class paper and/or chapter in Handbook of TCS; (both publications from the 1980-ies)
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How to obtain this Matrix ?
The row/column index is a binary numberwhich codes a configuration.
This code must be efficient in order that itis easy to recognize whether two configurationsare connected by a transition
"Locality" of the transitions is key: configurationonly changes at focus; everywhere else it remains the same.
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How to obtain this Matrix ?The intrinsic representation for Turing Machine has all desired properties.
We need some routine to extract from a binary bitstring a group of bits coding a single symbol.
On a RAM model your values are numbers - notbitstrings. Extracting these symbol codes requires number-to-binary conversion, whichpresupposes the availability of some "multiplicative" instruction which lacks in thestandard model (but which is always granted).
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The Problem ??!Is there a dragon out there ??
© Games Workshop
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What do actual authors Use ?
• Whenever TM computations are used in master reductions, the author will almost always chose either an intrinsic or a semi intrinsic representation
• This holds even if the formal definition for configurations is based on the mathematical representation
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Representative authors
Author Title Mathematical Intrinsic Semi-Intrinsic
Reidel Ency of Math X
RI Soare RE sets and degrees 87
X
SC Kleene Intro to Metamath 52
X
Boolos & Jeffrey Computability & Logic 74
X
Börger Computability 89
X
Cohen Computability & Logic 87
X
M Davis Computability & Unsolv. 58
X
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Representative authors
Author Title Mathematical Intrinsic Semi-Intrinsic
Hopcroft & Ullman
Formal Lang & Autom 69
X X
illustration
Hopcroft & Ullman
Formal Lang & Autom 79
X
F Hennie Intro to Comput 77
XWithout states
Harrison Intro Formal
Languages 78
X
Sudkamp Intro TCS
06
Xearlier eds as well
Lewis & Papadimitriou
Elts th of comp 81
X Xfor reductions
Mehlhorn EATCS mon 2 84
X
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Representative authors
Author Title Mathematical Intrinsic Semi-Intrinsic
Rudick & Wigderson
Comp Compl Theory 04
X
J Savage Models of Computation 98
X
For k tapes
X
Balcazar Diaz & Gabarro
Structural Complexity 88
X
H Rogers Th Recursive Functions 67
X
Odifreddi Classical Rec Theory 89
X
WJ Savitch Abstr Mach & Grammars 82
X X
J Martin Intr Lang & th of comp 97
X
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We are Safe
• In practice, all authors on the basis of their intuition use the intrinsic representation
• Why then it seems that Stockmeyer is the unique author who makes the advantages explicit ??
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