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CS2303-THEORY OF COMPUTATION

•Chapter:

Minimization of Automata

1

Minimized DFA

• Myhill-Nerode Theorem• DFA with minimum number of states that will

solve the problem• There may be states which are equivalent and

not necessary

Equivalent States

Two states p and q are said to be equivalent if and only if

for all input strings w *, both *(p, w) and *(q, w)

result in accepting states or both *(p, w) and *(q, w)

result in non accepting states.

Distinguishable States

Two states p and q are said to be distinguished when

there exist a string s * such that either *(p, w)

result in accepting state and *(q, w) results in a

non accepting state or vice versa.

A D B

E F G H

C 0 1

0 1

0

1

0

0 1

1

0 1 1

0

0

1

• By default final states of a DFA are distinguishable from non final states

• Hence in the given DFA, state C is distinguishable from other states

• To determine whether two states are equivalent we have to analyze the states reachable from the states for various strings from length 0 to some ‘n’

• Difficult hence we go for table filling algorithm

Table filling algorithm• Build a (n - 1) x (n - 1) table where n is the

number of states in the given DFA• Each column is named by the name of the

states starting with the first state till n-1 states• Each row is named with name of the states

starting from the second state• Duplicate cells are eliminated (i.e.) only one

cell is used to determine whether two states are equal

Table filling algorithm• Each cell in the table represents a pair of

states in the DFA. • Cells are marked with a “x” as soon as the two

states represented by it are decided to be distinguishable

• At the end of the process, the pairs of states corresponding to the cells of the table without the mark are concluded as equivalent states

Basis

If state p is an accepting and state q is a non

accepting then the pair of states {p, q} is

distinguishable.

Recursive part

Given two distinguished states r and s, if p and q are

states such that for some input symbol a, (p, a) = r

and (q, a) = s then one may conclude that the pair

of states {p, q} is also distinguishable.

x x

x

x

x

x

x

B

C

D

E

F

G

H

A B C D E F G

Table after the base step in the table filling algorithm

Distinguishable states are analyzed, by analyzing the transition of the

states on the input symbol 0

*(A, 0) = B

*(B, 0) = G

*(D, 0) = C*

*(E, 0) = H

*(F, 0) = C*

*(G, 0) = G

*(H, 0) = G

The two conditions to decide that the pair of states is

distinguishable are repeated below

(i) When one of the states moves to a non final

state on reading an input symbol and the other

one move to final state on the same symbol.

(ii) When the states in the pair moves to the states

already distinguished

{A, D}, {A, F}, {B, D}, {B, F}, {D, E}, {D, G}, {D, H}, {E, F}, {F, G} and {F, H} are distinguishable as per the first condition

It may also be concluded from the above transitions, that the pair

states {D, F}, {B, G}, {B, H} and {G, H} cannot be distinguished by

strings that begin with 0, since both of the states in the pair move to

the same state on reading 0.

x x

x x x

x x

x x x x

x x x

x x x

B

C

D

E

F

G

H

A B C D E F G Table after the using recursive step once in the table filling algorithm

(A, 1) = F

(B, 1) = C*

(D, 1) = G

(E, 1) = F

(F, 1) = G

(G, 1) = E

(H, 1) = C*

As per the condition 1 of distinguishability, the pair of states {A, B},

{B, D}, {B, E}, {B, F}, {B, G}, {A, H}, {D, H}, {E, H}, {F, H}, {G,

H} are distinguished

Next distinguishability condition 2 is applied.

Using the known fact that the pair of states {E, F} and {E, G} are

distinguished, some more distinguishable pairs {A, G}, {E, G} are

obtained x

x x

x x x

x x x

x x x x

x x x x x x

x x x x x x

{AE}

{BH}

{DF} G

C 1

0

0

1 0

0

1

1 0

1

Minimized DFA

Testing equivalence of two FA• To test whether the given two FAs M1 and M2

are equivalent it is imagined as if the two FAs are combined.

• The imaginary FA will contain m + n states in it where m is the number of states in M1 and n is the number of states in M2.

• The imaginary FA contains two initial states, initial state of M1 and M2 which is not allowed in real case.

Testing equivalence of two FA

• Table filling algorithm is used for the imaginary DFA. A table of size (m + n-1) x (m + n-1) is used and the two initial states are tested for equivalence.

• If they are equivalent then it may be concluded that both the DFAs represent same language.

A B

0 1

1

0

DFA M1

E

C D

0 0

0 1

1

1

DFA M2

x

x x

x x x

B C D E

A B C D

Then transition from all the states on the symbol 0 is determined as below

(A,0) = A

(B,0) = A

(C,0) = D

(D,0) = D

(E,0) = C

No pair of states may be distinguished by looking at the

above transition since the entire set of states move to

non distinguished states on reading the symbol 0.

The transition of the states on the symbol 1 are analyzed as given below

(A,1) = B

(B,1) = B

(C,1) = E

(D,1) = E

(E,1) = E

transition on the symbol 1 also does not distinguish any state

Hence the pair of states {A, C}, {A, D}, {B, E}, {C, D} are equivalent.

The initial states A and C of the two given automata are

non distinguishable hence the automata M1 is

equivalent to M2