The document Turing Machine (TM) Computer Science Engineering (CSE) Notes | EduRev is a part of the Computer Science Engineering (CSE) Course Theory of Computation.

All you need of Computer Science Engineering (CSE) at this link: Computer Science Engineering (CSE)

**Turing Machines**

Consider the following figure:

From the diagram, unrestricted grammars produce unrestricted languages (Type-0 languages). These Type-0 languages are recognised using Turing machines.

Following is an example for unrestricted grammar:

S → ACaB

CaBc → aaC

CB → DB

aD → Da

aEC → Ea

AE → ε

Here LHS and RHS can contain any set of terminals and non terminals.

**Part I. Turing Machine (TM)**

A Turing machine is a computing device that can be represented as,

From the above diagram, a TM consists of a,**1. Tape**

It is divided into a number of cells. Tape is infinite. A cell can hold a tape symbol or a blank symbol (#).**2. Finite control**

At a particular instant, finite control is in a state. States are divided into,

Start state (q_{0}),

Halt state (h),

Intermediate states (q_{1}, q_{2}, .....).**3. Tape head**

Tape head points to one of the tape cells. It communicates between finite control and tape. It can move left ,right or remain stationary.

**1 Definition of a TM**

A Turing machine is,

TM = (Q,∑, Γ, δ, q_{0}, #, h)

where

Q is a set of states,

∑ is a set of input symbols,

Γ is a set of tape symbols, including #,

δ is the next move function from

(Q × Γ) → (Q × Γ × {L, R, N}),

q_{0} is the start state,

# is the blank symbol,

h is the halt state.

**Example:**

Consider the following TM,

Here TM is in state, q and head points to tape symbol, c.

Let a move is defined as,

δ(q, c) = (q_{1}, b, R)

This means, if TM is in state q and points to input symbol, c, then it enters state, q_{1}, replaces symbol, c with b and moves one position towards right (R). Thus the TM now is,

Now TM is in state q1,and head points to the tape symbol, a.

Let a move of the TM is defined as,

δ(q_{1}, a) = (q_{2}, d, L)

This means, if TM is in state q1 and points to input symbol, a, then it enters state, q_{2}, replaces symbol, a with d and moves one position towards left (L). Thus the TM now is,

Now TM is in state q_{2},and head points to the tape symbol, b.

Let a move of the TM is defined as,

δ(q_{2}, b) = (q_{1}, y, N)

This means, if TM is in state q_{2} and points to input symbol, b, then it enters state, q_{1}, replaces symbol, b with y and head

remains in the same position (N). Thus the TM now is,

Now TM is in state q_{1},and head points to the tape symbol, y.

**Example:**

Consider a Turing machine,

TM = (Q,∑, Γ, δ, q_{0}, #, h)

where

Q = {q_{0}, h}

∑= {a, b}

Γ = {a, b, #}

q_{0} = q0

# = #

h = h

δ is defined as,

δ(q_{0}, a) = (q_{0}, #, L)

δ(q_{0}, b) = (q_{0}, #, L)

δ(q_{0}, #) = (h, #, N)

δ(h, #) = ACCEP T

**Language Acceptability by TM**

A string, w is said to be accepted by a Turing machine, if TM halts in an accepting state. That is,

A string, w is not accepted by a TM, if TM halts in a non-accepting state or does not halt.

**Example 1:**

Consider a Turing machine,

TM = (Q,∑, Γ, δ, q_{0}, #, h)

where

Q = {q_{1}, q_{2}, q_{3}, q_{4}, q_{5}, q_{6}}

∑ = {0, 1, x, y}

Γ = {0, 1, x, y, #}

q_{0 }= q1

# = #

h = q6

δ is defined as,

Check whether the string 011 is accepted by the above TM?

Initially, TM is in start state, q_{1} and the tape contains the given string 011. Initially, head points to symbol 0 in the input string.

A given transition is, δ(q_{1}, 0) = (q_{2}, x, R)

Then, TM becomes,

A given transition is, δ(q_{2}, 1) = (q_{3}, y, L)

Then, TM becomes,

A given transition is, δ(q_{3}, x) = (q_{5}, x, R)

Then, TM becomes,

A given transition is, δ(q_{5}, y) = (q_{5}, x, R)

Then, TM becomes,

δ(q_{5}, 1) is not defined for the TM. So the string 011 is not accepted by the above TM.

**Example 2:**

Consider a Turing machine,

TM = (Q,∑, Γ, δ, q_{0}, #, h)

where

Q = {q_{1}, q_{2}, q_{3}, q_{4}, q_{5}, q_{6}}

∑= {0, 1, x, y}

Γ = {0, 1, x, y, #}

q_{0} = q_{1}

# = #

h = q_{6}

δ is defined as,

Check whether the string 0011 is accepted by the above TM?

Initially, TM is in start state, q_{1} and the tape contains the given string 0011. Initially, head points to symbol 0 in the input string.

A given transition is, δ(q_{1}, 0) = (q_{2}, x, R)

Then, TM becomes,

A given transition is, δ(q_{2}, 0) = (q_{2}, 0, R)

Then, TM becomes,

A given transition is, δ(q_{2}, 1) = (q_{3}, y, L)

Then, TM becomes,

A given transition is, δ(q_{3}, 0) = (q_{4}, 0, L)

Then, TM becomes,

A given transition is, δ(q_{4}, x) = (q_{1}, x, R)

Then, TM becomes,

A given transition is, δ(q_{1}, 0) = (q_{2}, x, R)

Then, TM becomes,

A given transition is, δ(q_{2}, y) = (q_{2}, y, R)

Then, TM becomes,

A given transition is, δ(q_{2}, 1) = (q_{3}, y, L)

Then, TM becomes,

A given transition is, δ(q_{3}, y) = (q_{3}, y, L)

Then, TM becomes,

A given transition is, δ(q_{3}, x) = (q_{5}, x, R)

Then, TM becomes,

A given transition is, δ(q_{5}, y) = (q_{5}, x, R)

Then, TM becomes,

A given transition is, δ(q_{5}, y) = (q_{5}, x, R)

Then, TM becomes,

A given transition is, δ(q_{5}, #) = (q_{6}, #, R)

Then, TM becomes,

q_{6 }is the halt state or accepting state of the TM. So the string 0011 is accepted by the above TM.

**Example 3:**

Consider a Turing machine,

TM = (Q,∑, Γ, δ, q_{0}, #, h)

where

Q = {q_{1}, q_{2}, q_{3}, q_{4}, q_{5}, q_{6}}

∑= {0, 1, x, y}

Γ = {0, 1, x, y, #}

q_{0} = q_{1}

# = #

h = q_{6}

δ is defined as,

δ(q_{1}, 0) = (q_{2}, x, R) δ(q_{1}, #) = (q_{5}, #, R)

δ(q_{2}, 0) = (q_{2}, 0, R) δ(q_{2}, 1) = (q_{3}, y, L)

δ(q_{2}, y) = (q2, y, R) δ(q_{3}, 0) = (q_{4}, 0, L)

δ(q_{3}, x) = (q_{5}, x, R) δ(q_{3}, y) = (_{q3}, y, L)

δ(q_{4}, 0) = (q_{4}, 0, L) δ(q_{4}, x) = (q_{1}, x, R)

δ(q_{5}, y) = (q_{5}, x, R) δ(q_{5}, #) = (q_{6}, #, R)

Check whether the string 001 is accepted by the above TM?

Initially, TM is in start state, q_{1} and the tape contains the given string 001. Initially, head points to symbol 0 in the inputstring.

A given transition is, δ(q_{1}, 0) = (q_{2}, x, R)

Then, TM becomes,

A given transition is, δ(q_{2}, 0) = (q_{2}, 0, R)

Then, TM becomes,

A given transition is, δ(q_{2}, 1) = (q_{3}, y, L)

Then, TM becomes,

A given transition is, δ(q_{3}, 0) = (q_{4}, 0, L)

Then, TM becomes,

A given transition is, δ(q4, x) = (q1, x, R)

Then, TM becomes,

A given transition is, δ(q_{1}, 0) = (q_{2}, x, R)

Then, TM becomes,

A given transition is, δ(q_{2}, y) = (q_{2}, y, R)

Then, TM becomes,

Now TM halts because no δ is defined for q_{2} and #. But q_{2} is not the halt state of TM. So the string 001 is not accepted

by the above TM.

**2 Representation of Turing Machines**

We can represent a TM in different ways. They are,

Instantaneous descriptions,

Transition tables,

Transition diagrams.

**Representation of TM by Instantaneous Descriptions**

In all the above examples, Turing machines shown in the pictures were instaneous descriptions. This means an instant of a TM is shown.

Following shows an instantaneous description of a TM,

Currently, symbol under the head is 1.

Current state is q_{2}.

Symbols on the left of 1 are x and 0.

Symbnols on the right of 1 are 1, #, #, .....

This instant of TM can be written as,

x 0 q_{2 }1 1 # # #.

Let a transition is defines as, δ(q_{2}, 1) = (q_{3}, y, L)

Then TM moves to,

That is,

x q_{2} 0 y 1 # # #.

This transition from one state to other can be written as,

**Representation of TM by Transition Table**

Let a Turing machine is defined as,

TM = (Q,∑, Γ, δ, q_{0}, #, h)

where

Q = {q_{1}, q_{2}, q_{3}, q_{4}, q_{5}, q_{6}}

∑ = {0, 1, x, y}

Γ = {0, 1, x, y, #}

q_{0} = q_{1}

# = #

h = q_{6}

δ is defined as,

This TM can be represented as a Transition table.

Transition table corresponding to the above TM is,

Above is the transition table representation of a TM.

In the above transition table, the entry, (q_{2}, y, R) corresponding to row _{q2} and column y denotes the transition,

δ(q_{2}, y) = (q_{2}, y, R)

**Representation of TM by Transition Diagram**

Let a Turing machine is defined as,

TM = (Q,∑, Γ, δ, q0, #, h)

where

Q = {q_{1}, q_{2}, q_{3}, q_{4}, q_{5}, q_{6}}

∑ = {0, 1, x, y}

Γ = {0, 1, x, y, #}

q_{0} = q_{1}

# = #

h = q_{6}

δ is defined as,

δ(q_{1}, 0) = (q_{2}, x, R) δ(q_{1}, #) = (q_{5}, #, R)

δ(q_{2}, 0) = (q_{2}, 0, R) δ(q_{2}, 1) = (q_{3}, y, L)

δ(q_{2}, y) = (q_{2}, y, R) δ(q_{3}, 0) = (q_{4}, 0, L)

δ(q_{3}, x) = (q5, x, R) δ(q_{3}, y) = (q_{3}, y, L)

δ(q_{4}, 0) = (q_{4}, 0, L) δ(q_{4}, x) = (q_{1}, x, R)

δ(q_{5}, y) = (q_{5}, x, R) δ(q_{5}, #) = (q_{6}, #, R)

This TM can be represented as a Transition diagram.

Transition diagram corresponding to the above TM is,

In the above transition diagram, the part of th picture,

denotes the transition,

δ(q_{1}, 0) = (q_{2}, x, R)

Also the part of the picture,

denotes the transition,

δ(q_{5}, y) = (q_{5}, x, R)

**Exercises:**

1. Consider the TM given in the following transition table:

Check whether the string 00 is accepted by the above TM.

2. Consider the TM given in the following transition table:

Check whether the string 0011 is accepted by the above TM.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

18 videos|43 docs|39 tests