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CSC 4170 Theory of Computation. Turing Machines. Section 3.1. 3.1.a. Components of a Turing machine (TM). a a b a b b - - - - - -. (Q,  ,  ,  ,start,accept,reject). Q is the set of states

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turing machines
CSC 4170

Theory of Computation

Turing Machines

Section 3.1

components of a turing machine tm
3.1.aComponents of a Turing machine (TM)

a a b a b b - - - - - -

(Q,,,,start,accept,reject)

  • Q is the set of states
  •  is the input alphabet not containing the blank symbol -
  •  is the tape alphabet, where - and 
  •  is the transition function of the type Q Q{L,R}
  • start,accept,rejectQ, where rejectaccept; the states
  • accept and reject are called halting states.
  • There are no transitions from the halting states, and they immediately take effect!

xy,R

q1

q2

If the current tape symbol is x, replace it with y, move the head right and go to q2. The label xx,R is simply written as xR.

If we here have L instead of R, then the head is moved left, unless it was on the first cell, in which case it remains where it was.

how a turing machine works
3.1.b

0  L

x  L

How a Turing machine works

q5

xR

-  R

-  L

xR

start

q2

q3

0 -,R

0x,R

  •  R
  • x  R

-  R

0 x,R

0R

xR

q4

reject

accept

  • Configuration:
  • Current state;
  • Tape contents;
  • Head position

-  R

-

x

0

0

-

-

-

-

-

-

-

definitions
3.1.c1Definitions

A TM accepts an input string iff, for this input, sooner or later it enters

the accept state.

Otherwise the string is considered rejected.

  • Thus, the input is rejected in two cases:
  • The machine enters the reject state at some point, or
  • The machine never halts (never enters a halting state).

A Turing machine is said to be a decider iff it halts for every input.

The language recognized by a TM --- the set the strings that TM accepts;

If this machine is a decider, then we say that it not only recognizes,

but also decides that language.

A language is said to be Turing-recognizable iff some TM recognizes it.

A language is said to be Turing-decidable iff some TM decides it.

recognizing vs deciding
3.1.c2Recognizing vs deciding

accept

0  R

-  R

-  R

  •  R
  • 0 R

0  R

reject

What language does the above machine recognize?

Does it decide that language?

accept

0  R

-  R

0  R

-  R

reject

What language does the above machine recognize?

Does it decide that language?

diagrams without the reject state
3.1.dDiagrams without the reject state

0  L

x  L

xR

q5

-  R

-  L

start

xR

q2

q3

0 -,R

0x,R

-  R

0R

0 x,R

accept

xR

q4

The reject state can be safely removed.

It will be understood that all the missing transitions lead to the reject state.

designing tm example 1
3.1.e1Designing TM: Example 1

Design a TM that recognizes (better – decides) {#n =m | n=m}

1. Sweep left to right across the tape, testing if the input has the form

#*=*; if not, reject; if yes, go back to the beginning of the tape

and go to step 2 (state q3).

L

= L

x L

- L

R

R

= R

- L

# R

start

q1

q2

q9

#  L

q3

designing tm example 11
3.1.e2Designing TM: Example 1

2. Keep going to the right, replace the first  you see before = with

x andgo to step 3 (state q5);

if you reach = without seeing a , go to step 4 (state q6).

xR

x,R

q4

q3

q5

# R

= R

q6

designing tm example 12
3.1.e3Designing TM: Example 1

3. Keep going to the right, pass = and then replace the first  you see

with x andgo to the beginning of the tape, step 2 (this can be

done by going to state q9);

if you reach a blank without seeing a , reject.

q9

x,R

R

= R

xR

q7

q5

designing tm example 13
3.1.e4Designing TM: Example 1

4. Keep going to the right as long as you see x. If you see a  before

reaching a blank, reject; otherwise accept.

xR

- R

accept

q6

designing tm example 14
3.1.e5Designing TM: Example 1

L

= L

x L

- L

R

R

= R

- L

# R

start

q1

q2

q9

#  L

x,R

xR

R

x,R

= R

xR

q4

q7

q3

q5

# R

xR

= R

- R

accept

q6

-

-

-

#

=

designing tm example 2
3.1.fDesigning TM: Example 2

Design a TM that decides {#n <m | n

L

< L

x L

L

< L

x L

- L

R

R

< R

- L

# R

start

q1

q2

q9

#  L

x,R

xR

R

x,R

< R

xR

q4

q7

q3

q5

# R

xR

< R

R

accept

q6

designing tm example 3
3.1.gDesigning TM: Example 3

Design a TM that decides {#n +m = k | n+m=k}

+ L

L

= L

x L

- L

R

R

R

= R

- L

+ R

# R

start

q1

q2

q9

q0

#  L

x,R

xR

+R

R

+R

x,R

= R

xR

q4

q7

q3

q5

# R

xR

= R

- R

accept

q6

designing tm example 4
3.1.hDesigning TM: Example 4

Design a TM that decides {#n m = k | nm=k}

Step 1: Check if the string has the form #* * = * . If not, reject;

If yes, go back to the beginning of the tape, step 2.

Step 2: Find the first  between # and, delete it and go to step 3;

If no suchwas found, go to step 5.

Step 3: Find the first  between and =,delete it and go to step 4;

If no suchwas found, go to the beginning of the tape, restoring on

the way back all the deletedbetween and=,and go to step 2.

Step 4: Find the first  after=, delete it, go back to ,and go to step 3;

If no suchwas found before seeing a blank, reject.

Step 5: Go right past = ; if no  is found there before reaching a blank,

Accept; otherwise reject.

=

-

#

testing whether the head is at the beginning of the tape
3.1.i

Design a fragment of a TM that, from a state “Beg?”, goes to a state “Yes” or “No”,

depending on whether you are at the beginning of the tape or not,

without corrupting the contents of the tape.

Testing whether the head is at the beginning of the tape

Tape alphabet: {x1,…,xn}

Add a new tape symbol: $

  • Read the current symbol, remember it, type $, and move left;
  • Read the current symbol.
  • If it is $, restore the remembered symbol and go to “Yes”.
  • If it is not $, move right, restore the remembered symbol and go to “No”.

x1,…,xnR

a1

b1

$ x1,L

x1,R

x1$,L

L

Beg?

Yes

Temp

No

xn$,L

$ xn,L

xn,R

an

bn

x1,…,xnR

implementing the go to the beginning of the tape operation
3.1.jImplementing the “go to the beginning of the tape” operation

Design a fragment of a TM that, from a state “Go to the beginning”,

goes to the beginning of the tape and state “Done”.

Move left and test if you are at the beginning of the tape.

If yes, go to “Done”.

If not, repeat the step.

No

Done

Go to the beginning

Beg?

Yes

shifting tape contents
3.1.kShifting tape contents

Design a fragment of a TM that types 0 in the current cell and shifts the

(old) contents of the remaining tape one cell to the right.

Assume the tape alphabet is {0,1,-}.

  • Read the current symbol, remember it, type 0 and move right.
  • While the current symbol is not -, remember it, type the
  • previously remembered symbol and move right.
  • Once you see a blank, type
  • the remembered symbol
  • and you are done.

Shift

Done

a tm for the element distinctness problem
3.1.lA TM for the element distinctness problem

Design a TM that decides the language

E={#x1#x2#…#xn | each xi{0,1} and xixj for each ij}

See page 135 for a description of such a TM

# x1 # x2 # x3 # x4 #x5

x1 x2 x3 x4

#

0

1

1

#

0

1

0

#

0

#

1

1

0

-

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