FORMAL LANUGAGE

7/28/2019 FORMAL LANUGAGE

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CSCI 3130: Formal languages

and automata theory

Tutorial 7

Chin


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Reminder

Homework 4 is due at 23:59, today!


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Turing Machines

a b a q1 q2

a/b L

Replace a with b, and move head left

current state

a b b q1 q2

a/b L

new state


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Turing Machines: Low-level Design

Useful tools

1. Move to left/right until pointing to x

S = {a, b, x} G= {a, b, x, }

q1

a/a R

b/b R

a b a x b

a b a x b


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Useful tools

2. Go to first symbol of the tape S = {a, b, x}

G= {a, b, x, a, b, x, }

At the beginning

Now, use tool 1.

q0

a b a x b

a b a x b

q1b/b R

a/a R

x/x R


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Useful tools

3. Go to last symbol of the tape S = {a, b, x}

Use component 1, then go 1 step to the left.

G= {a, b, x, }

q1

a/a R

b/b R

a b a x b

a b a x b

x/x R

/ La b a x b


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Turing Machines

Useful tips

1. Check if the input is in the right form to

reduce the possible inputs you have to deal with later


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Turing Machines

Exercise

1. L1 = {0n1n: n 0}


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Turing Machines

L1 = {0n1n: n 0}

High-level idea:

If input is e, accept

Check if the string is of the form 00*11*

For each 0

replace with x

find the first 1 on the right and cross it

Check if everything on the tape is x


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Turing Machines

L1 = {0n1n: n 0}

Implementation:

If input is e, accept

q0/ R qacc


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Turing Machines

L1 = {0n1n: n 0}

Implementation:


q0 q10/0 R

0/0 R

q21/1 R

1/1 R

/ L


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Turing Machines

L1 = {0n1n: n 0}

Implementation:


(Go back to the first position)

For each 0

replace with x




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Turing Machines

L1 = {0n1n: n 0}

Implementation:

Go back to the first position

q0 q10/0 R

q1

0/0 L

1/1 L


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Turing Machines

L1 = {0n1n: n 0}

Implementation:

For each 0 (resp. 0)

replace with x (resp. x)



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Turing Machines

L1 = {0n1n: n 0}

Possible situations:

x 0 0 1 1 1

x 0 0 x 1 1

x x 0 x x

x x x x 1 1


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Turing Machines

L1 = {0n1n: n 0}

Implementation:


replace with x (resp. x)


q1 q20/x R

0/0 Rx/x R

q31/x L

x 0 0 x 1 1

x x 0 x x 1


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Turing Machines

L1 = {0n1n: n 0}

Implementation:


replace with x (resp. x)find the first 1 on the right and cross it

How to go back to the left after each iteration?

q1 q20/x R

0/0 Rx/x R

q31/x L

x/x L

x x 0 x x 1

x x 0 x x 1

0/0 L q4

0/0 L

x/x R

x/x R


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Turing Machines

L1 = {0n1n: n 0}

Implementation:


replace with x (resp. x)find the first 1 on the right and cross it

What will happen after the last iteration?

q1 q20/x R

0/0 Rx/x R

q31/x L

x/x L

x x x x x x

x x x x x x

0/0 L q4

0/0 L

x/x R

x/x R


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Turing Machines

L1 = {0n1n: n 0}

Implementation:


q1 q20/x R

0/0 Rx/x R

q31/x L

x/x L

0/0 L q4

0/0 L

x/x R

x/x R

x x x x x x

x/x R

q2

x/x R

/ R qacc

x x x x x x


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Turing Machines

L1 = {0n1n: n 0}

Implementation:

Putting everything together

q1 q20/x R

0/0 Rx/x R

q31/x L

x/x L

0/0 L q4

0/0 L

x/x R

x/x R

x/x R

q2

x/x R

/ R qacc

If input is e, acceptCheck if the string is of the form 00*11*


For each 0

replace with x



q0/ R qacc

q0 q10/0 R

0/0 R

q21/1 R

1/1 R

/

L

q0 q10/0 R

q1

0/0 L

1/1 L


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Turing Machines

L1 = {0n1n: n 0}

Implementation:

Putting everything together

q4 q50/x R

0/0 Rx/x R

q61/x L

x/x L

0/0 L q7

0/0 L

x/x Rx/x R

x/x R

q8

x/x R

/

R qacc

q1

0/0 R

q21/1 R

1/1 R

/ Lq00/0 R

q3

0/0 L

1/1 L

If input is e, acceptCheck if the string is of the form 00*11*


For each 0

replace with x



0/x R

/ R


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Variants of Turing Machines

Turning machine with left reset

The machines head can move one symbol to the right, or

jump to the leftest position of the tape

It cannot move one symbol to the left

Show that it is equivalent to a Turing Machine.


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Show that TM with left resetis equivalent to a TM

Two steps

1. Show how to simulate a TM with left reseton a TM

2. Show how to simulate a TM on a TM with left reset


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How to go to first symbol of the tape? e.g. a/b reset

S = {a, b, x} G= {a, b, x, a, b, x,}

At the beginning

To reset

q0 q1b/b R

a/a R

x/x R

q1

a/a L

b/b L

x/x L

a b a x b

a b a x b

a b a x b

a b a x b


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Every time you change the symbol in the first position

e.g. a / bR

Add dots to the symbols e.g. a / bR

a b a x b

b b a x b

a b a x b

b b a x b


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2. Show how to simulate a TM on a TM with left resetIdea:

First, mark the current position with a dot

Then resetand mark the first position with a *

Reset

While (the dot is not one symbol to the right of the * position)

Move the star to one position to the right

Now the star is at the position you want to reach

Remove the dot, reset, and go to the star position

.

* .

* .


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While (the dot is not one symbol to the right of the * position)

Move the star to one position to the right

How?

1. reset

2. Go to the * position

3. Move one position to the right

4. If it is a dot, done

Else mark it with a *

5. reset

6. Find the first * and remove it

* .

* .

* .

* * .

* * .

* * .

* .


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e.g. a/x L

a b a x b

a b x b b


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e.g. a/x L

1. a/x reset

2. a/a reset3. a/a R

4. b/b reset

a b a x b

a b x x b

*

a b x x b *

*

a b x x b *

*

a b x x b * *


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e.g. a/x L

5. a/a R

6. b/b R7. x/x reset

8. a/aR

Done.

*

*

a b x x b *

*

a b x x b *

a b x x b *


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End

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