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 :
\((1 + 01)^{*} (1 + 01)\)
A binary string with no 00 and ending in 1 can be cut into blocks where every 0 is immediately followed by a 1, i.e. blocks 1 and 01.
So the language is \((1 + 01)^{+} = (1 + 01)^{*}(1 + 01)\), which forces at least one block and makes the last symbol a 1.
\((1 + 01)^{*}(10 + 01)\) allows strings ending in 0, \((1 + 01)^{*}01\) misses strings such as 1 and 11, and \((10 + 01)^{*}01\) can create 00 as in 1001.
Check: 1, 101, 0111 are generated, while 100 and 10 are not.
Hence, \((1 + 01)^{*} (1 + 01)\).
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 ?
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? 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)?
