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Moore and Mealy machines

Finite state Machines with output

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Moore Machine output is associated with states

A Moore machine is a six tuple (Q , , , , , q0) : Is the output alphabet : I s a mapping from Q to giving output

associated with each state.Output in response to input a1 a2 …an is(q0) (q1) (q2)… (qn)

Where q0, q1,…qn is the sequence of states such that (qi-1,ai) = qi

STRING LENGTH N+1

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Note

Any Moore machine gives output

(q0) in response to input - empty string

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DFA may be viewed as a special case of Moore machine where

output for accepting state is 1

output for nonaccepting state is 0

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Example

To find the residue mod 3 for binary string treated as a binary integer

if a number i written in binary is followed by 0, resulting string has value 2i

If I is followed by 1 the resulting string has value 2i +1

If a string x(when treated as binary integer) has remainder p(when divided by 3) then string ‘x0’ will have remainder 2p mod 3

And string ‘X1’ will have remainder 2p+1 mod 3

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q0 q1 q2

0

input 1010 value 10 so 10 mod 3 = 1

q0 1 q1 0 q2 1 q2 0 q1

output 0 1 2 2 1

1 20

1

1

1

0

0

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• 25 = 1 1 0 0 1• Q0- 1 q1 – 1- q0 0- q0- 0 - q0 – 1 – q1• 0 1 0 0 0 1

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Mealy Machine out put is associated with transition

A Mealy machine is a six tuple (Q , , , , , q0) : Is the output alphabet : I s a mapping from QX to giving

output (q,a) associated with the transitions from state q and input each state.

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Output in response to input a1 a2 …an is

(q0, a1) (q1,a2) (q2,a3)… (qn-1,an)

Where q0, q1,…qn is the sequence of states such that (qi-1,ai) = qi

Note that this sequence has length n rather that n+1 as in Moore machine

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Mealy machine to add two integers using their binary expansions

Inputs xn xn-1 …x0

yn yn-1…y0

We assume that both xn and yn are zero. a state q0 represents that previous carry is

zeroA state q1 represents that previous carry was

1Inputs to the machine are pairs of bits: there

are four possible inputs(00,01,10,11)

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Input 0 1 0 0 1 0 18

0 1 1 1 0 1 29

47

q0 q1

11, 0

01, 1 01, 0

00, 100, 0

10, 1

11, 1

10, 0

q0 - 01- q0 - 10 - q0 - 01 - q0- 01 - q0- 11 - q1 - 00- q0

1 1 1 1 0 1

out put is : 1 0 1 1 1 1 value 47

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Equivalence of Moore and Mealy Machines

A Moore machine M can never be correctly equivalent to a Mealy machine M’ because the length of output string from a Moore machine M is one greater than that from the Mealy machine M’ on the same input.

However we may neglect the response of Moore machine to input and say

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Moore machine M and a Mealy machine M’ are equivalent if for all inputs w

b TM’ (w) = TM(w)

where b is the output of Moore machine M for its initial state

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• Theorem: I If M1= (Q , , , , , q0)is a Moore machine then there is a Mealy machineM2 equivalent to M1

• Proof: Construct Mealy machine M2 as

(Q , , , , ’, q0) ’ is defined as:

’(q,a) = ((q,a)) for all states q and input symbols a

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q0 q1 q2

0

input 1010 value 10 so 10 mod 3 = 1

q0 1 q1 0 q2 1 q2 0 q1

output of Moore 0 1 2 2 1

1 20/2

1/2

1/1

1/0

0/0

0/1

Out put for Mealy machine on the same input

q0 - 1 (1) q1 - 0 (2) q2- 1 (2) q2 - 0 (1) q1

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• on input 1010 out by Moore machine is 0 1 2 2 1

While output from the constructed Mealy machine is

1 2 2 1By the condition of equivalence we should add out put of q0 0 of Moore

machine in the beginning of out put of Mealy machine

Hence the machine made is equivalent to the give Moore machine.

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• Theorem: If M1= (Q , , , , , q0)is a Mealy machine then there is a Moore machine M2 equivalent to M1.

• Proof:We can do the reverse of what we did in the construction of Mealy machine.

As the different edges terminating to a node may have different outputs.

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We look into each state and see the edges with different outputs are terminating to it.

Say state qi is having all the terminating edges with 0,1, 2 as outs on them.

State qi is split into

Qi0, qi1, qi2 ie. In the Moore machine in place of qi state we take three state

qi0, qi1, qi2.

With these states we associate outputs as 0,1,2 respectively in the Moore machine

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In q0 four edges are terminating and in different outputs on them are only two 0 and 1

So q0 is split into two states q00 and q01

q0 q1

11, 0

01, 1 01, 0

00, 100, 0

10, 1

11, 1

10, 0

Similarly on state q1 four edges are terminating with two different output vales 0 and 1 so it is also split into two new states as q10 and q11.

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q00

q01

q10

q11

00

01

10

11

01,10

1111

10, 01

01,10

00

11

0 1 0 0 1 0 18 q00 -01-q01-10-q01-01-q01-01-q01-11-q10-00-q01

0 1 1 1 0 1 29 0 1 1 1 1 0 1 ie 101111 = 47

0

1

0

1

00

00

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• 0 1 0 0 1 0 18

• 0 1 1 1 0 1 29 47

q00 - 01- q01 -10 - q01 - 01- q01 - 01- q01-11- q10 – 00 - q01

• 0 1 1 1 1 0 1 ie

• 101111 = 47