meljun cortes automata lecture turing machines 1

8/21/2019 MELJUN CORTES Automata Lecture Turing Machines 1

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Theory of Computation (With Automata Theory)

* Property of STI

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Turing Machines

TURING MACHINES

Turing Machines

Introduction

• A finite automaton is a

machine with a finite number

of states.

It is used to model computers

with a limited amount of

memory or programs that

process only a small amountof data.

The disadvantage of a DFA

and an NFA is its limited

number of states.


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Turing Machines

• A pushdown automaton is

similar to a finite automaton

except that it has access to a

stack.

It is used to model machineswith unlimited memory.

The problem with PDAs is that

access to the data in the stack

is limited on a last-in, first-outbasis.

• In general, finite automata and

pushdown automata are

limited to performing onlysimple tasks.

• Is there a model that can truly

represent a general-purpose

computer?


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Turing Machines

Turing Machines

• A Turing m ach ine (TM) is

similar to a finite automaton

and a pushdown automatonexcept that it has unlimited

and unrestricted memory.

• It is similar to a PDA except

that its storage device is called

a tape .

• A tape can be viewed as an

infinite, one-dimensional array

of cells. Each cell can store

one symbol.

• The Turing machine has a

read/write head that can

access any cell on the tape.


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Turing Machines

• Representation of a TM:

• The read/write head can move

left or right on the tape.

• It can read or write one symbol

on each move.

TM

1 0

Read/Write

Head

. . .0 o

Tape

o


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Turing Machines

• Initially, the first few cells

contain the symbols of the

input string while the rest are

blank.

• During the computation

process, the TM considers the

current state it is in and the

symbol on the tape beingpointed to by the read/write

head.

The TM can then replace the

symbol on the tape, move to

another state, and move the

read/write head to the left or to

the right.


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* Property of STI

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Turing Machines

• Case Study

Consider the language

L1 = {w #w w {0, 1}*}.

TM Operation:

1. Read the first symbol of

the input string.Remember the symbol

read and replace it with an

x .

2. Move the head to the firstsymbol to the right of #. If

no # is found, reject the

string.


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Turing Machines

3. If the first symbol to the

right of the # matches the

symbol read earlier,

replace this symbol with an x . If not, reject the string.

4. Repeat the actions of steps

1 to 3 for each symbol on

the left side of #.

5. When all symbols on the

left side of # have been

replaced with x ’s, check if

there are remaining

symbols to the right of #. If there are, reject the string.

Otherwise, accept the

string.


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Turing Machines

Formal Definition of TuringMachines

• A Turing machine is a 7-tuple

(Q, , , , q0, qaccept, qreject),

where Q, , and are all finite

sets, and

1. Q is the set of states,

2. is the input alphabet,

3. is the tape alphabet,

4. is the transition function,

5. q0 Q and is the start

state,

6. qaccept Q and is the acceptstate, and

7. qreject Q and is the reject

state, where qaccept qreject.


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Turing Machines

• Some points:

1. The input alphabet does

not include the blank

symbol .

2. The tape alphabet is

composed of the input

alphabet , the blank

symbol , and some other

symbols that may be

stored on the tape.

3. The read/write head starts

at the leftmost cell. The

head cannot go to the left

of the leftmost cell.

4. The first blank symbol

encountered by the

read/write indicates the

end of the input string.


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Turing Machines

5. A TM has one accept stateand one reject state.

If the TM goes in the

accept state, it accepts the

string being processed,and stops or halts the

computation immediately.

If the TM goes in the reject

state, it rejects the stringbeing processed, and

stops or halts the

computation immediately.

If the TM ends up in any

other state, the TM is said

to be in a loop. A loop

simply states that the TM

will go on forever, never

halting.


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Turing Machines

• In a TM, transitions are

represented by

(qi , a) = (q j , b, R)

This equation indicates that if

the current state is state qi and

the symbol below the

read/write head is a, the TM

goes to state q j , writes thesymbol b on the tape, and

moves the head to the right.

Transitions can also be

(qi , a) = (q j , b, L)

where the head moves to the

left instead of right.


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Turing Machines

• Configuration of a Turing

Machine

The conf igurat ion of a TM is

defined by the current state itis in, the contents of the tape,

and the position of its

read/write head.

It is written as

xqy

where q is the current state of

the TM, x is the string on the

left side of the current head

position, and y is the string

position on the right side of the

head.


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Turing Machines

Example:

Configuration = 101q30011

q 3

TM

. . .

Tape

o1 0 1 0 0 1 1 o


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Turing Machines

Configuration C 1 is said to

yield configuration C 2 if the TM

can reach C 2 from C 1 in one

step. This is represented as

C 1 C 2

If the current configuration is

x 0qi1y and the transition

function is (qi

, 1) = (q j

, 0, L),

then

x 0qi1y xq j00y

If the current configuration is

x 0qi1y and the transitionfunction is (qi , 1) = (q j , 0, R),

then:

x 0qi1y x 00q jy


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Turing Machines

• Sample Transitions in TM

State Diagrams

The label 0 → x , R means thatif the symbol under the

read/write head is a 0, then the

TM moves from state qi to

state q j , writes the symbol x on

the tape, and moves the headto the right.

q j

0 → x, R

q i


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Turing Machines

The label 0 → L means that if the symbol under the

read/write head is a 0, then the

TM moves from state qi tostate q j , and moves the head

to the left.

The label 0 → 0, L willaccomplish the same thing.

q j

0 → L

q i


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Turing Machines

The label → L means that if

the symbol under theread/write head is , then the

TM moves from state qi to

state q j , and moves the head

to the left.

The label 0, 1 → x , L means

that if the symbol under theread/write head is a 0 or 1,

then the TM moves from state

qi to state q j , writes the symbol

x on the tape, and moves the

head to the left.

q j

→ L

q i

q j

0, 1 → x, L

q i


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Turing Machines

• Language of a Turing Machine

If a language is recognized by

a Turing machine, then it is

said to be Tur ing- recognizable .

A TM recognizes a language if

it accepts all the strings of the

language.

If a TM accepts all the strings

of a language and rejects all

strings not in the language,

then the language is said to be

decided by the TM.

If a language is decided by a

Turing machine, then it is said

to be Turing-decidable .


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Turing Machines

• Continuation of the Case Study

The TM T 1 that can decide

language L1 = {w #w w {0,

1}*}:

q 0

# → R

q accept

q 7

→ R

x → R q 2q 1

q 3 q 4

q 5

q 6

1 → x , R

0 →

x , R

0 →

x , L 1

→ x , L

0,1 → R

x → R

0,1 → R

x → R

# → R# → R

0, 1, x → L

# → L

0, 1 → Lx → R


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Turing Machines

Simulating how M 1 accepts thestring 110#110, the following

are the series of configurations

M 1 enters:

q0110#110 xxq60#xx0xq210#110 xq6x0#xx0

x1q20#110 xxq00#xx0

x10q2#110 xxxq1#xx0

x10#q4110 xxx#q3xx0

x10q5#x10 xxx#xq3x0x1q60#x10 xxx#xxq30

xq610#x10 xxx#xq5xx

q6x10#x10 xxx#q5xxx

xq010#x10 xxxq5#xxx

xxq20#x10 xxq6x#xxxxx0q2#x10 xxxq0#xxx

xx0#q4x10 xxx#q7xxx

xx0#xq410 xxx#xq7xx

xx0#q5xx0 xxx#xxq7x

xx0q5#xx0 xxx#xxxq

7


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Turing Machines

• Example 2:

Consider the language L2 =

{02n

n ≥ 0}.

Algorithm:

1. Move the read/write from

left to right across the tape.

As this is done, replace

every other 0 that is

encountered with an x .

2. If there is only one

remaining 0 after step 1,

accept.

3. If the number of remaining

0s is an odd number greater than 1, reject the

string.

4. If the number of remaining

0s is an even number,

repeat step 1.


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Turing Machines

The TM M 2 that can decide the

language L2 = {02n n ≥ 0} is:

→ L

0, x → L

0 → ,Rq 0

q reject

q 1 q 2

q 4

q 3q accept

→ R

0 → x, R

x → Rx → R

→ R

→ R

0 → R 0 → x, R

x → R

X, → R


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Turing Machines

Simulating how M 2 accepts thestring 00000000, the following

are the series of configurations

M 2 enters:

q000000000 q1x0x0x0x

q10000000 xq10x0x0x

xq2000000 xxq2x0x0x

x0q300000 xxxq20x0x

x0xq20000 xxx0q3x0x

x0x0q3000 xxx0xq30x

x0x0xq200 xxx0xxq2xx0x0x0q30 xxx0xxxq2x0x0x0xq2 xxx0xxq4x

x0x0x0q4x xxx0xq4xx

x0x0xq40x xxx0q4xxx

x0x0q4x0x xxxq40xxxx0xq40x0x xxq4x0xxx

x0q4x0x0x xq4xx0xxx

xq40x0x0x q4xxx0xxx

q4x0x0x0x q4xxx0xxx

q4x0x0x0x q1xxx0xxx


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Turing Machines

xq1

xx0xxx xq4

xxxxxx

xxq1x0xxx q4xxxxxxx

xxxq10xxx q4xxxxxxx

xxxxq2xxx q1xxxxxxx

xxxxxq2xx xq1xxxxxx

xxxxxxq2x xxq1xxxxx

xxxxxxxq2 xxxq1xxxx

xxxxxxq4x xxxxq1xxx

xxxxxq4xx xxxxxq1xx

xxxxq4xxx xxxxxxq1x

xxxq4xxxx xxxxxxxq1xxq4xxxxx


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Turing Machines

• Example 3:

Consider the language L3 =

{ai b j c k i j = k and i , j , k ≥ 0}.

Algorithm:

1. Scan the tape from left to

right to determine if the

input string is composed of consecutives a’s, followed

by consecutive b’s, and

then followed by

consecutive c ’s. If not,

reject the string.

2. Move the read/write head

back to the first cell (the

leftmost cell) of the tape.


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3. Replace an a with a y .Scan to the right (skipping

a’s) until a b is reached.

Sweep between b’s and

c ’s, replacing each c with

an x until all b’s (eachreplaced by z along the

way) are processed. If

there are not enough c’s,

reject the string.

4. Restore the b’s (previously

replaced by z ’s). If there

are a’s remaining, then

repeat step 3.

5. If all the a’s are replaced

by y ’s, determine if all the

c ’s are replaced by x ’s. If

yes, then accept the string,

else reject the string.

meljun cortes automata lecture turing machines 1

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