Regular expression for the complement of language \(L = \{a^n b^m \mid n \ge 4,\ m \le 3\}\) is
(λ + a + aa + aaa)b* + a*bbbbb* + (a + b)* ba(a + b)*
L contains exactly the strings of the form \(a^n b^m\) (all a's before all b's) with at least four a's and at most three b's. Its complement over {a, b} is every other string, which falls into three groups.
Group 1 – strings not of the form a*b*: some b is followed by an a, i.e. the string contains ba: (a + b)* ba(a + b)*.
Group 2 – strings a*b* with fewer than four a's (n ≤ 3, any m): (λ + a + aa + aaa)b*.
Group 3 – strings a*b* with four or more b's (m ≥ 4, any n): a*bbbbb* (that is, a* followed by bbbb and then b*).
Every string of the complement lies in at least one group and no string of L lies in any, so the union (λ + a + aa + aaa)b* + a*bbbbb* + (a + b)* ba(a + b)* is the complement. Group 1 alone misses ab; a*bbbbb* alone misses ba; and since a complete expression exists, “None of the above” fails.
Hence, the answer is (λ + a + aa + aaa)b* + a*bbbbb* + (a + b)* ba(a + b)*.
Similar Questions
Given two languages :
\(L_1 = \{(ab)^n a^k \mid n \gt k,\ k \ge 0\}\)
\(L_2 = \{a^n b^m \mid n \ne m\}\)
Using pumping lemma for regular language, it can be shown that
Which of the following statements is false ?
The regular expression corresponding to the language L where
\(L = \{ x \in \{0, 1\}^{*} \mid x\) ends with 1 and does not contain substring 00 \(\}\) is :
Let L be any language. Define even (W) as the strings obtained by extracting from W the letters in the even-numbered positions and even(L) = {even (W) | W ∈ L}. We define another language Chop (L) by removing the two leftmost symbols of every string in L given by \(\text{Chop}(L) = \{W \mid \nu W \in L, \text{ with } |\nu| = 2\}\).
If L is regular language then
For the above specified passage, which of the following represents the regular expression?
Which of the following properties correctly describe a Regular Grammar ?
(A) All production rules are of the form A→xB or A→x, where A and B are non terminal symbols and x is a terminal symbol.
(B) Regular grammars are more powerful than context-free grammars and can express any type of language.
(C) There is a direct correspondence between regular grammar and finite automata.
(D) Regular grammars can generate languages that are not recognised by any type of automata.
Choose the correct answer from the options given below :
The precedence of regular expression operators from lowest to highest precedence is:
Which of the following is TRUE about the Pumping Lemma for regular language?
Consider the following two languages :
L1 = {0i1j| gcd (i, j) = 1}
L2 is any subset of 0*.
Which of the following is correct ?
The regular expression for the complement of the language L = {anbm|n ≥ 4, m ≤ 3} is :
Important Questions from Regular Languages
Which of the following are correct on regular expressions?
A. φ + L = L + φ = L
B. εL = Lε = L
C. φL = Lφ = φ
D. φL = Lφ = L
Choose the correct answer from the options given below:
- Let $L_1$ and $L_2$ be two languages over a finite alphabet, such that $L_1 \cap L_2$ and $L_2$ are regular languages.
Which of the following statements is/are always true? Given two languages :
\(L_1 = \{(ab)^n a^k \mid n \gt k,\ k \ge 0\}\)
\(L_2 = \{a^n b^m \mid n \ne m\}\)
Using pumping lemma for regular language, it can be shown that
- Let $\Sigma = \{a, b, c\}$. For $x \in \Sigma^*$, and $a \in \Sigma$, let $\#_a(x)$ denote the number of occurrences of $a$ in $x$.
Which one or more of the following option(s) define(s) regular language(s)? Which of the following statements is false ?
