Theory of Automata

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(CS402)
Theory of Automata
Lecture N0. 21
Introduction to Computer Theory
Chapter 8
Summary
Example of Moore machine, Mealy machine, Examples, complementing machine, Incrementing machine.
Example
To identify the relation between the input strings and the corresponding output strings in the following Moore
machine,
a
a
b
a
b
q0/0
q2/0
q1/0
q3/1
a
b
b
if the string bbbabaabbaa is run, the output string will be 000010000010, as shown below
Input
b
b
b
a
b
a
a
b
b
a
a
State
q0
q1
q2
q2
q3
q1
q0
q0
q1
q2
q3
q0
output
0
0
0
0
1
0
0
0
0
0
1
0
It can be observed from the given Moore machine that q3 is the only state which prints out the character 1 which
shows that the moment the state q3 is entered, the machine will print out 1. To enter the state q3, starting from q0
the string must contain bba. It can also be observed that to enter the state q3 once more the string must contain
another substirng bba. In general the input string will visit the state q3 as many times as the number of substring
bba occurs in the input string. Thus the number of 1's in an output string will be same as the number of
substring bba occurs in the corresponding input string.
Mealy machine
A Mealy machine consists of the following
A finite set of states q0, q1, q2, ... where q0 is the initial state.
An alphabet of letters Σ = {a,b,c,...} from which the input strings are formed.
An alphabet G={x,y,z,...} of output characters from which output strings are generated.
A pictorial representation with states and directed edges labeled by an input letter along with an output
character. The directed edges also show how to go from one state to another corresponding to every possible
input letter.
(It is not possible to give transition table in this case.)
Note
It is to be noted that since, similar to Moore machine, in Mealy machine no state is designated to be a final state,
so there is no question of accepting any language by Mealy machine. However in some cases the relation
between an input string and the corresponding output string may be identified by the Mealy machine. Moreover,
the state to be initial is not important as if the machine is used
b/1
q2
q1
several times and is restarted after some time, the machine
will be started from the state where it was left
b/1 a/0
a/0  a/1
off. Following are the examples
b/0
q3
q0
a/1
Example
b/1
Consider the Mealy machine shown aside, having the states q0, q1, q2, q3 , where q0 is the start state and
62 Theory of Automata
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Σ = {a,b},
G={0,1}
Running the string abbabbba over the above machine, the corresponding output string will be 11011010, which
can be determined by the following table as well
Input
a
b
b
a
b
b
b
a
States
q1
q0
q1
q2
q3
q3
q0
q3
q0
output
0
1
1
1
1
0
1
0
It may be noted that in Mealy machine, the length of output string is equal to that of input string.
Example
Consider the following Mealy machine having the states q0, q1, q2 , where q0 is the start state and
Σ = {a,b},
b/0
G={0,1}
q2
a/1
q1
b/1
a/0
a/0
b/0
q0
It is observed that in the above Mealy machine, if in the output string the nth character is 1, it shows that the nth
letter in the input string is the second in the pair of double letter.
For babaababba as input string the machine will print 0000100010.
Example
Consider the following Mealy machine having the only state q0 as the start state and
Σ = {0,1},
0/1, 1/0
G= {0,1}
q0
If 0011010 is run on this machine then the corresponding output string will be 1100101.
This machine is called Complementing machine.
Constructing the incrementing machine
In the previous example of complementing machine, it has been observed that the input string and the
corresponding output string are 1's complement of each other. There is a question whether the Mealy machine
can be constructed, so that the output string is increased, in magnitude, by 1 than the corresponding input string?
This machine is called the incrementing machine. Following is how to construct the incrementing machine.
Before the incrementing machine is constructed, consider how 1 is added to a binary number.
Since, if two numbers are added, the addition is performed from right to left, so while increasing the binary
number by 1, the string (binary number) must be read by the corresponding Mealy machine from right to left,
and hence the output string (binary number) will also be generated from right to left.
a)
100101110
b) 1001100111
+1
+1
100101111
1001101000
It may be observed from the above that
If the right most bit of binary number, to be incremented, is 0, the output binary number can be obtained by
converting the right most bit to 1 and remaining bits unchanged.
If the right most bit of binary number is 1 then the output can be obtained, converting that 1 along with all its
concatenated 1's to 0's, then converting the next 0 to 1 and remaining bits unchanged.
The observations (a) and (b) help to construct the following Incrementing (Mealy) machine.
The Mealy machine have the states q0, q1, q2 , where q0 is the start state and
63 Theory of Automata
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Σ = {0,1},
G={0,1}
0/0,1/1
0/1
q2
q1
1/0
0/1
1/0
q0
It may be observed that, in the incrementing machine, if 0 is read at initial state q0, that 0 is converted to 1 and a
no change state q1 (no carry state) is entered where all 0's and all 1's remain unchanged. If 1 is read at initial
state, that 1 is converted to 0 and the state q2(owe carry state) is entered, where all 1's are converted to 0's and
at that state if 0 is read that 0 is converted to 1 and the machine goes to no change state.
If the strings 100101110 and 1001100111 are run over this machine, the corresponding output strings will be
100101111 and 1001101000 respectively.
Note
It is to be noted that if the string 111111 is run over the incrementing machine, the machine will print out
000000, which is not increased in magnitude by 1. Such a situation is called an overflow situation, as the length
of output string will be same as that of input string.
It may also be noted that there exists another incrementing machine with two states.
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