The precedence of regular expression operators from lowest to highest precedence is:
Union, Concatenation, Star
Regular expressions use three operators — union (+ or |), concatenation and Kleene star (*) — and, as in arithmetic, a precedence convention lets parentheses be omitted (Hopcroft–Ullman).
Star has the highest precedence: it applies to the smallest well-formed expression on its left, so ab* means a(b*).
Concatenation comes next: it binds tighter than union, so ab + c means (ab) + c.
Union has the lowest precedence, so from lowest to highest the order is union, concatenation, star.
The orders that start with star invert the convention, and the order that ranks star below concatenation would make ab* mean (ab)*, which is not how regular expressions are read.
Union, Concatenation, Star option is correct.
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
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 :
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 :
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:
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