Regular grammar formal ceremony finite automata

1. regular conversion to the regular grammar

For any regular formula R selecting a nonterminal Z → Z generation rule R

1. rules of the form A → ab converted into A → aB, B → b

2. the form A → a | b rules, be converted into A → a, A → b (A → a | b)

3. The form A → a * b rules, be converted into A → aA, A → b

   The form A → ba * As a rule, be converted into A → Aa, A → b

Continue using the conversion rules, each rule comprising at most until a terminator far.

1(0|1)*101

S->A1

A->B0

B->C1

C->1(0|1)*

 ->1|C0|C1

 

(a|b)*(aa|bb)(a|b)*

S->(a|b)S

S->(aa|bb)(a|b)*

S->S(a|b)

S->(aa|bb)

S->aA

A->a

S->bB

B->b

S-> ThE | P | In | SB | AA | BB

 

((0|1)*|(11))*

S->((0|1)*|(11))S|E

S->(0|1)*S|11S|E

S->(0|1)*S

S->(0|1)S

S->0S|1S

S->1A

A->1S

S->0S|1S|1A|E

 

(0|110)

 S->0

S->1A

A->1B

B->0

S->0|1A

 

 

2. automaton M = ({q0, q1, q2, q3}, {0,1}, f, q0, {q3})

Where f:

(q0,0)=q1

(q1,0)=q2

(Q2,0) = q3

(q0,1)=q0

(q1,1)=q0

(Q2,1) = q0

(Q3,0) = q3

(Q3,1) = q3

Now draw the state transition matrix and state transition diagram, identify what language.

Transformation matrix:

  0 1
q0 q1 q0
q1 q2 q0
q2 q3 q0
q3 q3 q3

State transition diagram:

 

 

 

Recognition of language: (1 | 01 | 001) * 000 (0 | 1) *


3. configured by regular automatic machine NFA formula R 

(A | b) * abb

 

 

(a|b)*(aa|bb)(a|b)*

 

 

1(1010*|1(010)*1)*0

 

Guess you like

Origin www.cnblogs.com/zhff/p/11730018.html