1. DFA minimization: Textbook P65 Question 9
| I |
{1,2,3,4,5} |
{6,7} |
|
|
{1,2}b->{2} {3,4}b->{6,7} {1,2} a -> {3,4} indistinguishable {3,4} c -> {3} indistinguishable |
|
| II |
{1,2}{3,4}{5} |
{6,7} |
2. The minimum configuration of the DFA corresponding grammar
S→ 0A|1B
A→ 1S|1
B→0S|0
answer:
S → 0 (1 S + 1) | 1 (0S | 0)
S→01S+01|10S+10
S→(01+10)S|(01+10)
S→(01|10)*(01|10)
DFA：
|
|
0 |
1 |
|
| 0 |
e {S} = {AD} |
{BE} |
{CF} |
| 1 |
{BE} |
|
{ADG} |
| 2 |
{CF} |
{ADG} |
|
| 3 |
{ADG} |
{BE} |
{CF} |
| I |
{0,1,2} |
{3} |
|
|
{0}->{1,2} {1}1->{3} {2}0->{3} |
|
| II |
{0}{1}{2} |
{3} |
Simplification DFA:
3. Given the following grammar G [ S ]:
S →AB
A → aA | ɛ
B → b | bB
Given sentences aaab of a top-down parsing process, and explain the reasons for backtracking produce what is?
answer:
S →AB
S →aAB
S →aaAB
S →aaaɛB
S →aaabɛ
S →aaab
The reason backtracking produced: Left repeatedly extracted Public factor
4.P100 练习4，反复提取公共左因子，对文法进行改写。
答：
S -> C$
C -> bA | aB
A -> aC' | bAA
B -> bC' | aBB
C' -> C | ɛ