1. DFA minimization: Textbook P65 Question 9
solution:
| I |
{1,2,3,4,5} |
{6,7} |
|
|
{1,2}b->{2} {3,4}b->{6,7} {5}b->ε
|
{6,7} |
| II |
{1,2} {3,4}{5} |
{6,7} |
|
|
{3}c->{3},{4}c->{3} {3}d->{5},{4}d->{5} {3}b->{6},{4}b->{7} Therefore indistinguishable {3,4}
{1}b->{2},{2}b->{2} {1}a->{3}.{2}a->{4} Therefore indistinguishable {1,2} |
{6,7} indistinguishable |
Therefore, the conversion matrix is minimized
|
|
|
a |
b |
c |
d |
| 1 |
{1,2} |
{3,4} |
{1,2} |
|
|
| 2 |
{3,4} |
|
{6,7} |
{3,4} |
{5} |
| 3 |
{5} |
{3,4} |
|
|
|
| 4 |
{6,7} |
|
{6,7} |
|
|
Therefore, DFA minimization
2. The minimum configuration of the DFA corresponding grammar
S→ 0A|1B
A→ 1S|1
B→0S|0
solution:
S->0(1S|1)｜1(0S|0)
S->01S|01|10S|10
S->01S|10S|01|10
S->(01|10)S|(01|10)
S->(01|10)*(01|10)
It is NFA
Into DFA, which is a transformation matrix:
|
|
|
0 |
1 |
| 1 |
ε{S}={SAD} |
{THIS} |
{BG} |
| 2 |
{THIS} |
|
ε {by} = {ADF} |
| 3 |
{BG} |
ε {by} = {ADF} |
|
| 4 |
{ADF} |
{THIS} |
{BG} |
By converting DFA graph matrix obtained
| I |
{1,2,3} |
{4} |
|
|
{1}0->{2} {2}0->ε {3}0->{4}
|
{4} |
| II |
{1}{2}{3} |
{4} |
Found at the very beginning has been most of the DFA, so as to minimize DFA
3. Given the following grammar G [ S ]:
S →AB
A → aA | ɛ
B → b | bB
给出句子aaab 的一个自顶向下语法分析过程，并说明回溯产生的原因是什么？
解：
S -> AB
S -> aAB
S -> aaAB
S -> aaaAB
S -> aaaɛB
S -> aaaɛb
S -> aaab
回溯产生的原因：反复提取公共左因子
4.P100 练习4，反复提取公共左因子，对文法进行改写
解：
A->a( ɛ |C) | bAA
A->aA' | bAA
B->b( ɛ |C) | aBB
B->bB' | aBB
C->bA | aB
C->baA' | aB
C->a(bA' | B)
C->bAA | bB'