1. with NFA M = ({0,1,2,3}, {a, b}, f, 0, {3}), where f (0, a) = {0,1} f (0, b) = {0} f (1, b) = {2} f (2, b) = {3}
State transition matrix shown, state transition diagrams, identification and description of the NFA is what kind of language.
| a |
b |
|
| 0 |
0，1 |
0 |
| 1 |
|
2 |
| 2 |
|
3 |
State transition diagram
Recognition of language is (a | b) * abb
2.NFA determine into DFA
1. Solution multifunctions: Method subset
NFA 1). Above the Exercise 1
2). P64 page Exercise 3
2. Empty arc solution: seeking ε- closure of all the initial state and a new state
1). Figure 2 distributed to you
2) .P50 3.6 FIG.
Child collection method:
Subset f (q, a) = {q1, q2, ..., qn}, the state set
The {q1, q2, ..., qn} regarded as a state A, to record the set of all possible states achieved after NFA reads input symbol.
step:
1. The conversion matrix configuration DFA state NFA
① determined alphabet DFA, the initial state (initial state sets all the NFA)
② From the initial state, the state set alphabet arrived as a new state
③ adding a new state to the current DFA state
④ Repeat step 23 until no new DFA state
2 .. Draw DFA
3. NFA and DFA to see if the identified symbol string consistent.
1.(1)
|
|
a |
b |
|
| A |
0 |
01 |
0 |
| B |
01 |
01 |
02 |
| C |
02 |
01 |
03 |
| D |
03 |
01 |
0 |
2 .. Draw DFA
3. NFA and DFA to see if the identified symbol string consistent.
Not the same, because:
DFA identified symbol string is: (a | b) * abb
NFA is identified symbol string: b * aa * (ba) * bb
1.(2)
|
|
0 |
1 |
|
| S |
S |
VQ |
QU |
| A |
VQ |
VZ |
QU |
| B |
QU |
V |
quz |
| C |
VZ |
WITH |
WITH |
| D |
V |
WITH |
|
| E |
quz |
VZ |
quz |
| F |
WITH |
WITH |
WITH |
Symbol string NFA identification are: (00 ((100) | (0 | 1)) (0 | 1) *) | (1 ((00) | (11 * 0 (0 | 1) *)) (0 | 1)*)
2.(1)
|
|
0 |
1 |
2 |
|
| X |
ε{A}={ABC} |
ε{A}={ABC} |
ε{B}={BC} |
ε{C}={C} |
| Y |
ε{BC}={BC} |
|
ε{B}={BC} |
ε{C}={C} |
| WITH |
ε{C}={C} |
|
|
ε{C}={C} |
Recognition of language: 0 * (11 * 2 | 2) 2 *
2.(2)
|
|
a |
b |
|
| 0 |
e {0} = {01247} |
e {38} = {1234678} |
e {5} = {124567} |
| 1 |
{1234678} |
e {38} = {1234678} |
e {59} = {1245679} |
| 2 |
{124567} |
e {38} = {1234678} |
e {5} = {124567} |
| 3 |
{1245679} |
e {38} = {1234678} |
e {5,10} = {12456710} |
| 4 |
{12456710} |
e {38} = {1234678} |
e {5} = {124567} |
Recognition of language: (a | bb * a) a * (ba) * bb ((bb * aa * (ba) * bb) * | (aa * (ba) * bb) *)