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$L_3 = \{(a+b)^*\}$
 
$L_3 = \{(a+b)^*\}$
  
'''(1) Intersection of $L_1$ and $L_2$ is'''
+
(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
 
'''(A) Regular''' (B) CFL but not regular (C) CSL but not CFL (D) None of these
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'''(2) $L_1$ - $L_3$ is'''
+
(2) $L_1$ - $L_3$ is
  
 
(A) Regular '''(B) CFL but not regular''' (C) CSL but not CFL (D) None of these
 
(A) Regular '''(B) CFL but not regular''' (C) CSL but not CFL (D) None of these

Revision as of 23:07, 20 February 2014

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

Solution

(1) Regular.

L₁ ∩ L₂ 
= {abcd,aabbcd,aaabbbccdd,.....} ∩ {ab, aabb, aaabbb,....} 
= ϕ

(2) CFL

L₁ - L₃ = L₁, hence CFL
Proof,
L₁ - L₃ = {abcd,aabbcd,aaabbbccdd,.....} - { ∊,a,b,ab,aab,.....} 
= {abcd,aabbcd,aaabbbccdd,.....} 
= L₁




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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

Solution[edit]

(1) Regular.

L₁ ∩ L₂ 
= {abcd,aabbcd,aaabbbccdd,.....} ∩ {ab, aabb, aaabbb,....} 
= ϕ

(2) CFL

L₁ - L₃ = L₁, hence CFL
Proof,
L₁ - L₃ = {abcd,aabbcd,aaabbbccdd,.....} - { ∊,a,b,ab,aab,.....} 
= {abcd,aabbcd,aaabbbccdd,.....} 
= L₁




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