Arjun Suresh (talk | contribs) |
Arjun Suresh (talk | contribs) |
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L₁ - L₃ = L₁, hence CFL | L₁ - L₃ = L₁, hence CFL | ||
| − | + | Proof | |
L₁ - L₃ = L₁ ∩ L₃ʻ | L₁ - L₃ = L₁ ∩ L₃ʻ | ||
= L₁ ∩ {∊,a,b,ab,aab,....}ʻ | = L₁ ∩ {∊,a,b,ab,aab,....}ʻ | ||
Consider $L_1 = \{a^nb^nc^md^m | m,n \ge 1\}$
$L_2 = \{a^nb^n | n \ge1\}$
$L_3 = \{(a+b)^*\}$
(1) Intersection of $L_1$ and $L_2$ is
(A) Regular (B) CFL but not regular (C) CSL but not CFL (D) None of these
(2) $L_1$ - $L_3$ is
(A) Regular (B) CFL but not regular (C) CSL but not CFL (D) None of these
(1) Regular.
GATECSEL₁ ∩ L₂
= {abcd,aabbcd,aaabbbccdd,.....} ∩ {ab, aabb, aaabbb,....}
= ϕ
(2) CFL
L₁ - L₃ = L₁, hence CFL
Proof
L₁ - L₃ = L₁ ∩ L₃ʻ
= L₁ ∩ {∊,a,b,ab,aab,....}ʻ
= L₁ ∩ {(a+b)* (c+d)⁺}
= L₁
Consider $L_1 = \{a^nb^nc^md^m | m,n \ge 1\}$
$L_2 = \{a^nb^n | n \ge1\}$
$L_3 = \{(a+b)^*\}$
(1) Intersection of $L_1$ and $L_2$ is
(A) Regular (B) CFL but not regular (C) CSL but not CFL (D) None of these
(2) $L_1$ - $L_3$ is
(A) Regular (B) CFL but not regular (C) CSL but not CFL (D) None of these
(1) Regular.
L₁ ∩ L₂
= {abcd,aabbcd,aaabbbccdd,.....} ∩ {ab, aabb, aaabbb,....}
= ϕ
(2) CFL
L₁ - L₃ = L₁, hence CFL
Proof
L₁ - L₃ = L₁ ∩ L₃ʻ
= L₁ ∩ {∊,a,b,ab,aab,....}ʻ
= L₁ ∩ {(a+b)* (c+d)⁺}
= L₁