MELJUN CORTES Automata Theory 16

8/13/2019 MELJUN CORTES Automata Theory 16

1/23

CSC 3130: Automata theory and formal languages

Variants of Turing Machines

Fall 2008MELJUN P. CORTES MBA MPA BSCS ACS

MELJUN CORTES


2/23

Recognizing versus deciding

Looping is an undesirable behavior so we

say

Such TM are called deciders

The language recognized by a TM is theset ofall inputs that make it reach qacc

A TM decides language Lif it recognizes L

and

does not loop on any input

accept reject loop


3/23

Turing Machines

state control

infinite tape

0 1 0

input

Why is this a universal model of computation?


4/23

The Church-Turing Thesis

On Computable Numbers, with an

Application to the Entscheidungsproblem

1936:

Section 9. The extent of the computable numbers

All arguments [for the CT Thesis] which can be given arebound to be, fundamentally, appeals to intuition, and for this

reason rather unsatisfactory mathematically.

The arguments which I shall use are of three kinds:

1. A direct appeal to intuition

2. A proof of the equivalence of two definitions

(In case the new definition has greater intuitive appeal)

3. Giving examples of large classes of numbers which arecomputable.


5/23

Variant 1: The multitape Turing

Machine

The transition may depend on the contents

of

all the cells

Different tape heads can be moved

independently

state control tape 1 0 1 0

tape 2 0 1

tape 3 1 0 0


6/23

Variant I: The multitape Turing

Machine

Multitape Turing Machines recognize thesame

languages as single-tape Turing Machines

M

0 1 0

0 1

1 0 0

G

= {0, 1,

}

S 0 1 0 10 # # 0 #1 0

G= {0, 1, , 0, 1, , #}

#


7/23

Variant I: The multitape Turing

MachineFor every transition inM, do following in S:

Until you find the last #,

Find the next dotted symbolUpdate this symbol according to transition inM

Move the dot according to transition inMIf there is no space to move the dot,

Move everything to the right of the dot

to make space firstMove head of Sto beginning of tape

S 0 1 0 10 # # 0 #1 0

G= {0, 1, , 0, 1, , #}

#


8/23

Variant 2: The random access

machine

It has registersthat can store arbitrary

values, a

program counter, and a random-accessmemor

load -7 R0:= -7write R3 M[R3] := R0store R5 R5 := R0add R5 R0:= R0+ R5jpos 3 if R

0> 0 thenPC := 3

accept

instruction meaning012345

0PC

0R0

0R1

0R2

2M 1 2 2 0

0 1 2 3 4

program

counter

registers

memory


9/23


machineload -7 R0:= -7write R2 M[R2] := R0save R1 R1 := R0add R1 R0:= R0+ R1jzero 3 if R

0

= 0 thenPC := 3accept

instruction meaning012345

0PC

0R0

0R1

0R2

0M 0 0 0 0

0 1 2 3 4

-7

-7

-7

-14

12345

The instructions are indexed by the

program counter


10/23


machine

Simulating a Turing Machine on a RAM:

Random assess machines recognize thesame

languages as Turing Machines

PC

2R0

1M 2 1 0 0

M 2 11


11/23

save R1 handle for state q00

Simulating a TM on a RAM

M

q0

q1

qacc

save R1 save head position

read R1 read tape contents

xadd -1

jzero 6 if x= 1goto line 6

load 2 new value of cell

write R1 write in memory

load R1 recall head position

add 1 move head to right

jump 30 go to state q1

program0123

6789

10

2 11 save R1 handle for state q1

accept handle for state qacc

30

200


12/23

Simulating a RAM on a Turing

MachineThe configuration of a RAM consists of

Program counter

Contents of registers

Indices and contents of all nonempty memory

cells14PC

3R0

-7R1

5R2

2M 0 1 2 0

1 2 3 4

configuration =(14, 3, -7, 5, (0, 2), (2, 1), (3, 2))

0


13/23


MachineThe TM has a simulation tape and a

scratch tape

The simulation tape stores RAM

configuration

The TM has a set of states corresponding

to each

(14,3,-7,5,(0,2),(2,1),(3,2))

M


14/23


MachineExample: load R1

(14,3,-7,5,(0,2),(2,1),(3,2))

2. Write R1to conf tape

c

1. Copy R1to scratch tape -7s

(14,-,-7,5,(0,2),(2,1),(3,2))

-7

c

s

(14,-,-7,5,(0,2),(2,1),(3,2))c

(14,-,-7,5,(0,2),(2,1),(3,2) )c

.

(14,- ,-7,5,(0,2),(2,1),(3,2))c

.

.

Make more space

as needed

(14,-7,-7,5,(0,2),(2,1),(3,2))c3. Erase scratch tape


15/23

Variant 3: Nondeterministic

Turing Machine

The transition function is nondeterministic:

The language recognized byNare those

strings that

can lead N to end in qacc

N0 1 0

d: (Q{qacc, qrej}) G subsets of (Q G {L, R})


16/23

Equivalence of NTM and TM

Let us look more deeply into NTMs

Nondeterministic Turing Machinesrecognize

the same languages as Turing Machines

q3

q5

q6

Nondeterministic choices can

be numbered

The first msteps of the computation

are then fully specified by a sequenceof mnumbers


17/23

Simulating a nondeterministic

TMN

1 00

k= maximum number ofnondeterministic

choices in any state

M 1 00

1 00

3 21

input tape

2

simulation tape

address tape

Address tape specifies nondeterministic choices ofN

First, input is copied from input tape to simulation tapeThen,M simulatesNusing address tape data

G = G{1,..., k}


18/23

How to use the address tape

SupposeNaccepts xwhen nondeterminism =11221321

Then we want to make sure the address tape contains

the string 11221321 at some pointin its execution

N

1 00

M 1 00

1 00

3 21

input tape

2

simulation tape

address tape

To ensure this we try all possibilities for the address tape


19/23

Simulating a nondeterministic

TM1 00

1 00

3 21

input tape x

2

simulation tape z

address tape a

Initially:

x= input ofN

ais empty

For all possible strings a:Copy xto z.

SimulateNon input zusing a as the nondeterminism

If aspecifies an invalid choice or

simulation runs out of choices, abandon simulation.IfNenters its accept state, accept and halt.

IfNenters its reject state, abandon simulation.

Description ofM:


20/23

Correctness of simulation

IfNaccepts x:

For all possible strings a:

Copy xto z.SimulateNon input zusing a as the nondeterminism


simulation runs out of choices, abandon simulation.

IfNenters its accept state, accept and halt.IfNenters its reject state, abandon simulation.

Eventually,Mwill

encounterthe correct a

Description ofM:

SoMwill

accept

Provided that all previoussimulations halt!


21/23


Claim: Simulation step always halts (never

loops)For all possible strings a:





Description ofM:

Since ais finite, the simulation will

eventually

run out of choices


22/23


IfNdoes not accept x, then for every a:

For all possible strings a:





Description ofM:

EitherNwill loop, so

simulationruns out of choices

OrNrejects, so simulation is

abandoned anyway

In either case, simulation fails, so M loops forever!


23/23

Context-free languages are

recognizableCorollary

In fact

If Lis context-free, then it is recognizable by a TM

If Lis context-free, then it is decidedby some TM

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MELJUN CORTES Automata Theory 16