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Regular Expression, Set of strings, Concatenating the strings, Defining Languages, Possible strings, Remark, Unique language, Language of strings of even length are the key points and terms you can learn in this lecture. Other related concepts are also explained in this course.
Typology: Slides
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Kleene Star Closure, Plus operation, recursive definition of languages, INTEGER, EVEN, factorial, PALINDROME, { anbn }, languages of strings (i) ending in a, (ii) beginning and ending in same letters, (iii) containing aa or bb (iv)containing exactly aa,
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Q)
Let S={ab, bb} and T={ab, bb, bbbb} Show that S^ = T ^ [Hint S^ ⊆ T^ and T ^ ⊆ S^ ]
Let S={ab, bb} and T={ab, bb, bbb} Show that S^ ≠ T^ But S^ ⊂ T Solution: Since S ⊂ T , so every string belonging to S^ , also belongs to T^ but bbb is a string belongs to T^ but does not belong to S^.
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Q1)Is there any case when S +^ contains Λ? If yes then justify your answer.
S +^ ={Λ, a, aa, aaa, …} Here Λ is in S +^ as member of S. Thus Λ will be in S +^ , in this case.
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Q2) Prove that for any set of strings S
i. (S +)*^ =(S ^ )
Solution : In general Λ is not in S +^ , while Λ does belong to S ^. Obviously Λ will now be in (S +)^ , while (S ^ )^ and S ^ generate the same set of strings. Hence (S +)^ =(S ^ )^.
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iii) Is (S ^ )+=(S +)
Solution : since Λ belongs to S *^ ,so Λ will belong to (S *^ )+^ as member of S ^ .Moreover Λ may not belong to S +, in general, while Λ will automatically belong to (S +)^. Hence (S ^ )+=(S +)
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As discussed earlier that a*^ generates
Λ, a, aa, aaa, … and a+^ generates a, aa, aaa, aaaa, …, so the language L 1 = {Λ, a, aa, aaa, …} and L 2 = {a, aa, aaa, aaaa, …} can simply be expressed by a^ and a+, respectively. a^ and a+^ are called the regular expressions (RE) for L 1 and L 2 respectively.
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Method 3 (Regular Expressions) Consider the language L={Λ, x, xx, xxx,…} of strings, defined over Σ = {x}. We can write this language as the Kleene star closure of alphabet Σ or L=Σ^ ={x} this language can also be expressed by the regular expression x*^. Similarly the language L={x, xx, xxx,…}, defined over Σ = {x}, can be expressed by the regular expression x+.
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Now consider another language L, consisting of all possible strings, defined over Σ = {a, b}. This language can also be expressed by the regular expression (a + b)^. Now consider another language L, of strings having exactly double a, defined over Σ = {a, b}, then it’s regular expression may be b^ aab*
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It may be noted that a language may be expressed by more than one regular expressions, while given a regular expression there exist a unique language generated by that regular expression.
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Example:
Consider the language, defined over Σ={a , b} of words having at least one a, may be expressed by a regular expression (a+b)^ a(a+b)^. Consider the language, defined over Σ = {a, b} of words having at least one a and one b, may be expressed by a regular expression (a+b) *^ a(a+b) *^ b(a+b) *^ + (a+b) *^ b(a+b) *^ a(a+b) *^. Docsity.com
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Consider the language, defined over
Σ={a, b} of words beginning with a , then its regular expression may be a(a+b)*
Consider the language, defined over
Σ={a, b} of words beginning and ending in same letter , then its regular expression may be (a+b)+a(a+b) *^ a+b(a+b) *^ b
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Consider the language, defined over
Σ={a, b} of words ending in b , then its regular expression may be (a+b) *^ b.
Consider the language, defined over
Σ={a, b} of words not ending in a , then its regular expression may be (a+b) *^ b + Λ. It is to be noted that this language may also be expressed by ((a+b) *^ b) *^.
