This is 2nd Part of Regular expression in theory of computation solved examples. You can also read Regular expression in theory of computation solved examples Part – 1.
6. Construct a regular expression for all strings over {0, 1} of length 6 or less.
Solution:
We have the input alphabets ∑ = {0, 1}
Here, the resultant regular expression will denote the set of all strings over the given input alphabets ∑ of length 6 or less. The set of all strings of length zero or one can be given by the following regular expression-
(λ + (0 + 1))
Thus, the resultant regular expression which denote the set of all strings over the given ∑ of length 6 or less can be given by-
(λ + (0 + 1)) (λ + (0 + 1)) (λ + (0 + 1)) (λ + (0 + 1)) (λ + (0 + 1)) (λ + (0 + 1))
As per the mathematically representation we can write it as-
(λ + (0 + 1))6
It can only handle consecutive a’s! If we have a string like
aabcbca or abcaba
it can not handle them!