Question

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Consider the language L={All strings of odd length} over the input alphabet {a, b}.
Which of the following is the correct regular expression for the language L?
a) ((a+b)a(a+b)) * +((a+b)b(a+b))
b) a(a+b)* (a+b)*+b(a+b)*(a+b)*
c) a((a+b)(a +b))* +b((a+b)(a+b)) *
d) None of the above
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Answer 1

Explanation:

Regular Expression can be recursively defined as follows −

ε is a Regular Expression indicates the language containing an empty string. (L (ε) = {ε})

φ is a Regular Expression denoting an empty language. (L (φ) = { })

x is a Regular Expression where L = {x}

If X is a Regular Expression denoting the language L(X) and Y is a Regular Expression denoting the language L(Y), then

X + Y is a Regular Expression corresponding to the language L(X) ∪ L(Y) where L(X+Y) = L(X) ∪ L(Y).

X . Y is a Regular Expression corresponding to the language L(X) . L(Y) where L(X.Y) = L(X) . L(Y)

R* is a Regular Expression corresponding to the language L(R*)where L(R*) = (L(R))*

If we apply any of the rules several times from 1 to 5, they are Regular Expressions.

Some RE Examples

Regular Expressions Regular Set

(0 + 10*) L = { 0, 1, 10, 100, 1000, 10000, … }

(0*10*) L = {1, 01, 10, 010, 0010, …}

(0 + ε)(1 + ε) L = {ε, 0, 1, 01}

(a+b)* Set of strings of a’s and b’s of any length including the null string. So L = { ε, a, b, aa , ab , bb , ba, aaa…….}

(a+b)*abb Set of strings of a’s and b’s ending with the string abb. So L = {abb, aabb, babb, aaabb, ababb, …………..}

(11)* Set consisting of even number of 1’s including empty string, So L= {ε, 11, 1111, 111111, ……….}

(aa)*(bb)*b Set of strings consisting of even number of a’s followed by odd number of b’s , so L = {b, aab, aabbb, aabbbbb, aaaab, aaaabbb, …………..}

(aa + ab + ba + bb)* String of a’s and b’s of even length can be obtained by concatenating any combination of the strings aa, ab, ba and bb including null, so L = {aa, ab, ba, bb, aaab, aaba, …………..}

Answer 2

Answer:

Option (d) is not correct as the correct answer is (c).

Explanation:

The correct regular expression for the language L={All strings of odd length} over the input alphabet {a, b} is:

c) a((a+b)(a+b))* + b((a+b)(a+b))*

This regular expression describes a string that starts with either an "a" or a "b", followed by any number of pairs of "a" and "b" (represented as (a+b)), and then ends with a star (denoting zero or more occurrences) of (a+b)(a+b), which ensures that the length of the string is odd.

Option (a) is incorrect because it allows for strings of even length, as it includes the possibility of ending with "b" (represented by (a+b)b(a+b)).

Option (b) is also incorrect because it allows for strings of even length, as it includes the possibility of starting with "b" (represented by b(a+b)(a+b)).

Option (c) is the correct regular expression, as explained above.

Option (d) is not correct as the correct answer is (c).

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