1. Write the following description are regular grammar and formal languages formula:
L1={abna|n≥0}
Regular grammar:
S->aA
A->Ba
B->ε|bB
Regular Expression: ab * a
L2={ambn|n≥1,m ≥1}
Regular grammar:
S->AB
A->aA|a
B->bB|b
Regular Expression: aa * bb *
L3={（ab）n|n≥1}
Regular grammar:
S->aA
A->bB
B->aA|ε
Regular Expression: (ab) * (ab)
2. Convert the following regular grammar to the regular formula
0A → Z
A → 0A | 0B
B → 1A | e
answer:
A=0A+0(1A+ε)=0A+01A+0=(0+01)A+0=(0+01)*0=(0|01)*0
Z = 0 (0 | 01) 0
Z→U0|V1
U→Z1|1
V→Z0|0
answer:
V = (V0 + V1) 1 + 1
V = (V0 + V1) 0 + 0
Z = (Z1 + 1) 0+ (Z0 + 0) 1
Z = Z10 + Z01 + 10 + 01
Z = (10 + 01) * (01 + 10)
Z = (10 | 01) * (01 | 10)
S → aA
A → bA | aB | b
B → aA
answer:
S=a(bA＋aaA＋b)
S = longer two AAAA +
S=(ab＋aaa)*ab
S=(ab|aaa)*ab
I → L | IL | A
answer:
I = L + IL + Id = L + I (L + d) = (L + d) * L = (L | d) * L
L1={abna|n≥0}
Regular grammar:
S->aA
A->Ba
B->ε|bB
Regular Expression: ab * a
L2={ambn|n≥1,m ≥1}
Regular grammar:
S->AB
A->aA|a
B->bB|b
Regular Expression: aa * bb *
L3={（ab）n|n≥1}
Regular grammar:
S->aA
A->bB
B->aA|ε
Regular Expression: (ab) * (ab)
2. Convert the following regular grammar to the regular formula
0A → Z
A → 0A | 0B
B → 1A | e
answer:
A=0A+0(1A+ε)=0A+01A+0=(0+01)A+0=(0+01)*0=(0|01)*0
Z = 0 (0 | 01) 0
Z→U0|V1
U→Z1|1
V→Z0|0
answer:
V = (V0 + V1) 1 + 1
V = (V0 + V1) 0 + 0
Z = (Z1 + 1) 0+ (Z0 + 0) 1
Z = Z10 + Z01 + 10 + 01
Z = (10 + 01) * (01 + 10)
Z = (10 | 01) * (01 | 10)
S → aA
A → bA | aB | b
B → aA
answer:
S=a(bA＋aaA＋b)
S = longer two AAAA +
S=(ab＋aaa)*ab
S=(ab|aaa)*ab
I → L | IL | A
answer:
I = L + IL + Id = L + I (L + d) = (L + d) * L = (L | d) * L