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The assignment describes a parity-based encoding machine that accepts binary messages. The machine starts in an initial state and transitions based on the parity of each bit in the message. The parity determines whether to move to states 1, 2, 3, or 4, which then determine the next state transition. The final state is accepting if the machine reaches it, indicating that the encoded message is valid.

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Solution:

: Final state

: Initial state

: Self loop state

: Final state

: Initial state

: Self loop state

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STATE 0 TRANSITION 1 TRANSITION ACCEPTING

0 1 2

1 3 4

2 5 6

3 7 3

4 8 4

5 5 9

6 6 10

7 3 7

8 4 8

9 9 5

10 10 6

Here we are assuming that 0 is the Initial state then based on parity we will proceed further

suppose if the parity is 0 then we move to the 1 state but if the parity bit is 1 then we move to

the 2 state

suppose if the parity is odd 0 then we move to the state 3 from the state 1 but if the parity is

odd 1 then we move to the state 4 from state 1

suppose if the parity is even 0 then we move to the state 5 from the state 2 but if the parity is

even 1 then we move to the state 6 from the state 2

once the parity is completed then we can proceed for the message and if it reaches the final

state then we need to append 1 otherwise append 0

0 1 2

1 3 4

2 5 6

3 7 3

4 8 4

5 5 9

6 6 10

7 3 7

8 4 8

9 9 5

10 10 6

Here we are assuming that 0 is the Initial state then based on parity we will proceed further

suppose if the parity is 0 then we move to the 1 state but if the parity bit is 1 then we move to

the 2 state

suppose if the parity is odd 0 then we move to the state 3 from the state 1 but if the parity is

odd 1 then we move to the state 4 from state 1

suppose if the parity is even 0 then we move to the state 5 from the state 2 but if the parity is

even 1 then we move to the state 6 from the state 2

once the parity is completed then we can proceed for the message and if it reaches the final

state then we need to append 1 otherwise append 0

Machine Testing:

A B C Parity

0010 1101 001 Odd0

Case 1: ABCB is the message whose binary value is 001011010011101

So initially we are in state 0, so first we look for the parity right now the parity is 0, so we move to the

state 1 further the parity is odd, so we move to the state 3 now we start our message so initial bit is 0 so

it move to the next state which is 7 now the next bit is 0 so it again move to the state number 3, next bit

is 1, and it is self-loop so it remains on same state i.e. 3 now next bit is 0 so it again move to the state 7

now again it receive next bit as 1 so again it is self-loop for two times then next bit is 1 so it move to the

state 3, next bit is 0 so move to state 7 again next bit is 0 move to the state 3 now there is three

continuous but it is self-loop so it does not affect and remains in the same state 3 now next bit is 0 so

move to the state 7 then final bit is 1 so remains in 7 states which is accepting state

So the encoded message is 0010110100111011

Case 2: BCB is the message whose binary value is 11010011101

So initially we are in state 0, so first we look for the parity right now the parity is 0 so we move to the

state 1 further the parity is odd, so we move to the state 3 now we start our message so initial bit is 1 so

we remains in the same state 3 next bit is also 1 so no need to move now the next bit is 0 so we need to

move the state 7 next bit is again 0 so move to the state 3 now there is three continuous 1 bit, so we

remain in the same state and further next is 0 so move to the state 3 and final bit is 1 so remains in the

same state, our string is over and right now we are in non-final state, so it is non-accepting

So the encoded message is 110100111010

A B C Parity

0010 1101 001 Odd0

Case 1: ABCB is the message whose binary value is 001011010011101

So initially we are in state 0, so first we look for the parity right now the parity is 0, so we move to the

state 1 further the parity is odd, so we move to the state 3 now we start our message so initial bit is 0 so

it move to the next state which is 7 now the next bit is 0 so it again move to the state number 3, next bit

is 1, and it is self-loop so it remains on same state i.e. 3 now next bit is 0 so it again move to the state 7

now again it receive next bit as 1 so again it is self-loop for two times then next bit is 1 so it move to the

state 3, next bit is 0 so move to state 7 again next bit is 0 move to the state 3 now there is three

continuous but it is self-loop so it does not affect and remains in the same state 3 now next bit is 0 so

move to the state 7 then final bit is 1 so remains in 7 states which is accepting state

So the encoded message is 0010110100111011

Case 2: BCB is the message whose binary value is 11010011101

So initially we are in state 0, so first we look for the parity right now the parity is 0 so we move to the

state 1 further the parity is odd, so we move to the state 3 now we start our message so initial bit is 1 so

we remains in the same state 3 next bit is also 1 so no need to move now the next bit is 0 so we need to

move the state 7 next bit is again 0 so move to the state 3 now there is three continuous 1 bit, so we

remain in the same state and further next is 0 so move to the state 3 and final bit is 1 so remains in the

same state, our string is over and right now we are in non-final state, so it is non-accepting

So the encoded message is 110100111010

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