regular expression tutorial in automata TOC
Regular expression tutorial Theory of Automata (TOA) Theory of Computation (TOC)
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Summary of the Lecture:
Regular Expressions Theory of computations or automata
Why the theory of computations and automata?
We use this theory to process any textual data (natural languages and programming languages.) It is the pre-requisite for the compiler construction.
Today’s Topic is Regular Expressions.
We use Regular Expressions to represent a set of words (language)
Let’s dive into it.
Superscript * Means zero to n
Superscript + Means 1 to n (at least one)
Or (we have to pick one letter from the given list of letters)
Examples
A+={A, AA,AAA,AAAA,AAAAA,AAAAAA, ……… so on}
A*={null, A, AA,AAA,AAAA,AAAAA,AAAAAA, ……… so on} // when * is zero, it will generate null.
Here we can say that
A+ = AA*
(a+b) = {a,b}
(1+2)(a+b) = {1a, 1b,2a,2b}
(a+b)(a+b) = {aa,ab,ba,bb}
(a+b)(a+b)(a+b) = now how many words can we generate here?
We have 2 options at first place, 2 options at second place and 2 options at third place. So 2 x 2 x 2 = 8 this language contains 8 words.
(a+b)(a+b)(a+b) = { aaa,aba,baa,bba, aab,abb,bab,bbb }
Now we are going to write a very regular expression that will be used in subsequent lectures as well.
(a+b)*
Here as discussed earlier, the superscript plus + means (1 to n)
Lets see, how many words can be generated from this regular expression
(a+b)* = {(a+b)null, , (a+b)1 , (a+b)2 , (a+b)3 , (a+b)4 , (a+b)5 , (a+b)6 , (a+b)7 , (a+b)8 , …………………………………… }
Here
• (a+b)null = NULL,
• (a+b)1 = (a+b) = {a,b}
• (a+b)2 = (a+b)(a+b) = {aa,ab,ba,bb}
• (a+b)3 = (a+b)(a+b)(a+b) = { aaa,aba,baa,bba, aab,abb,bab,bbb }
So
(a+b)* can generate literally any word from the letter a and b including null.
(a+b)+ = (a+b)(a+b)*
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