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4 votes
4 votes

The regular expression for the complement of the language $L=\{a^nb^m \mid n \geq 4, m \leq 3\}$ is:

  1. $(\lambda +a+aa+aaa)b^*+a^*bbbb^*+(a+b)^*ba(a+b)^*$
  2. $(\lambda +a+aa+aaa)b^*+a^*bbbbb^*+(a+b)^*ab(a+b)^*$
  3. $(\lambda +a+aa+aaa)+a^*bbbbb^*+(a+b)^*ab(a+b)^*$
  4. $(\lambda +a+aa+aaa)b^*+a^*bbbbb^*+(a+b)^*ba(a+b)^*$

3 Answers

Best answer
7 votes
7 votes

(1) using the term a*bbbb* it can accept a4b3 hence it is not complement

(2) Using (a+b)* ab (a+b)*  it can accept a4b3 hence it is not complement

(3) Using (a+b)* ab (a+b)*  it can accept a4b3 hence it is not complement

(4) (a+b)* ba (a+b)* cannot produce strings in a n b m  format and other two terms cannot produce any string a n b m such that n>=4 and m<=3. 

Hence I think 4 is the answer

Correct me if I am wrong

selected by
6 votes
6 votes
(D) is correct

as option (B) and (C) are eliminated easily because complement surely will contain term (a + b)*ba(a + b)*

[ From B and C, we can get $a^5b$ from (a+b)*ab(a+b)* part, that should not belong to complement of L]

so now to chose between A and D option will eliminate like this :
a*bbbbb*  -----> it contain atleast 4 b's so this is correct as compared to (A) option

[From A, we can get $a^5b^3$ from a*bbbb* part, that should not belong to complement of L]
0 votes
0 votes

Language L contain string aaaabbb. then complement of L never contain string aaaabbb.

but in option a,b and c all are genrate string aaaabbb. so it is false . option d is right.

Answer:

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