Read the following and answer the questions that follow. A machine is represented by states Q, input alphabet Σ, transition function δ. Initial state q0 and final state F. The machine accepts all the strings over Σ = {a, b}, which start and ended with any combination of all alphabet and abb works /lies as substring in all the strings to be accepted.
For the above specified passage, which of the following represents the regular expression?
(a+b)*abb(a+b)*
A language of the form “all strings containing a fixed word w” is written as \(\Sigma^{*}\,w\,\Sigma^{*}\): any string before w, then w itself, then any string after it. Over the alphabet {a, b}, \(\Sigma^{*} = (a+b)^{*}\).
Step 1 — The passage fixes w = abb, with any combination of a and b at the start and at the end.
Step 2 — Substituting gives \((a+b)^{*}\,abb\,(a+b)^{*}\), which matches abb, aabba, babbb and every other string containing abb, and nothing else.
The expression (a+b)*aab describes strings ending in aab, so it misses abb itself. The expression aba(a+b)* describes strings starting with aba, again excluding abb. The expression b(a+b)*b(a+b)*a(a+b)* requires a leading b and a later a, so it rejects abb and accepts bba, which does not contain abb.
(a+b)*abb(a+b)* option is correct.
Similar Questions
Regular expression for the complement of language \(L = \{a^n b^m \mid n \ge 4,\ m \le 3\}\) is
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
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? Regular expression for the complement of language \(L = \{a^n b^m \mid n \ge 4,\ m \le 3\}\) is
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)?
