Arjun Suresh (talk | contribs) (Created page with "Consider $L_1 = {a^nb^nc^md^m, m,n \ge 1}$ $L_2 = {a^nb^n, n \ge1}$ $L_3 = {(a+b)^*}") |
Arjun Suresh (talk | contribs) |
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− | Consider $L_1 = {a^nb^nc^md^m, m,n \ge 1}$ | + | Consider $L_1 = \{a^nb^nc^md^m, m,n \ge 1\}$ |
− | $L_2 = {a^nb^n, n \ge1}$ | + | $L_2 = \{a^nb^n, n \ge1\}$ |
− | $L_3 = {(a+b)^*} | + | |
+ | $L_3 = \{(a+b)^*\}$ | ||
+ | |||
+ | '''(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 | ||
+ | |||
+ | ===Solution=== | ||
+ | (1) Regular. | ||
+ | L₁ ∩ L₂ | ||
+ | = {abcd,aabbcd,aaabbbccdd,.....} ∩ {ab, aabb, aaabbb,....} | ||
+ | = ϕ | ||
+ | |||
+ | (2) CFL | ||
+ | L₁ - L₂ = L₁, hence CFL | ||
+ | |||
+ | Alternatively, | ||
+ | L₁ - L₃ = L₁ ∩ L₃' | ||
+ | = L₁ ∩ {∊,a,b,ab,aab,....}' | ||
+ | = L₁ ∩ (a+b)* (c+d)⁺ | ||
+ | = L₁ | ||
+ | |||
+ | {{Template:FBD}} | ||
+ | |||
+ | [[Category:Automata Theory]] | ||
+ | [[Category: Questions]] |
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 - L₂ = L₁, hence CFL
Alternatively,
L₁ - L₃ = L₁ ∩ L₃' = L₁ ∩ {∊,a,b,ab,aab,....}' = 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 - L₂ = L₁, hence CFL
Alternatively,
L₁ - L₃ = L₁ ∩ L₃' = L₁ ∩ {∊,a,b,ab,aab,....}' = L₁ ∩ (a+b)* (c+d)⁺ = L₁