Theory of computing pdf
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This document discusses how to prove that a language is not regular using the pumping lemma for regular languages. It provides two examples: 1) It shows that the language Leq consisting of strings with an equal number of 0s and 1s is not regular by pumping a string w in Leq to get a string with an unequal number of 0s and 1s, violating the definition of Leq. 2) It shows that the language Lpal consisting of palindrome strings is not regular by pumping a string w in Lpal to get a string that is not a palindrome, again violating the definition of the language.
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Pumping lemma
1) The pumping lemma can be used to prove that a language is not regular or context-free.
2) For regular languages, the pumping lemma states that for any string x longer than n, x can be broken into uvw such that pumping v any number of times results in strings still in the language.
3) For context-free languages, the pumping lemma similarly states that any string x longer than n can be broken into five parts such that pumping the second and fourth parts any number of times results in strings still in the language.
Pumping lemma (1)
The pumping lemma states that for any regular language L, there exists an integer n such that any string x in L with length at least n can be divided into uvw such that: (1) x=uvw, (2) the length of uv is at most n, (3) v is not empty, and (4) for all i>=0, uviw is in L. This allows proving that a language is non-regular by finding a string that cannot be pumped.
1.10. pumping lemma for regular sets
The document discusses the pumping lemma for regular sets. It states that for any regular language L, there exists a constant n such that any string w in L of length at least n can be broken down into sections xyz such that y is not empty, xy is less than or equal to n, and xykz is in L for all k. The pumping lemma can be used to show a language is not regular by finding a string that does not satisfy the lemma conditions. Examples are provided to demonstrate how to use the pumping lemma to prove languages are not regular.
Pumping lemma for regular set h1
This document summarizes a lecture on automata theory, specifically discussing non-regular languages, the pumping lemma, and regular expressions. It introduces the language B={0n1n | n ≥ 0} as a non-regular language that cannot be recognized by a DFA. It then proves the pumping lemma theorem and uses it to show that B and the language of strings with equal numbers of 0s and 1s are non-regular. Finally, it defines regular expressions as a way to describe languages and provides examples of regular expressions and their meanings.
3.6 & 7. pumping lemma for cfl & problems based on pl
The document discusses the pumping lemma for context-free grammars (CFGs). It defines the pumping lemma, which states that for any CFG there is a pumping length n such that any string in the language of length >= n can be broken into five parts that satisfy certain properties related to pumping. The document gives examples of using the pumping lemma to determine whether languages are context-free or not, such as the language L={0n1n2n | n >= 1}. It concludes by listing additional problems involving using the pumping lemma to prove languages are not context-free.
Flat
This document discusses deterministic finite automata (DFAs) and regular languages. It defines DFAs formally and explains their components. It also proves several theorems about regular languages, such as how they are closed under operations like union and intersection. Additionally, it shows how to build DFAs that recognize specific languages and provides examples. Finally, it proves that the language of strings with an equal number of a's and b's is not regular using the pigeonhole principle.
Graph representation of DFA’s Da
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The document outlines 6 rules of inference used to derive logical conclusions: 1) Conjunctive simplification, 2) Disjunctive amplification, 3) Hypothetical syllogism, 4) Disjunctive syllogism, 5) Modus ponens, and 6) Modus tollens. Each rule is symbolically represented and includes an explanation of its application using truth tables to prove the validity of the logical inferences.
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1) The pumping lemma can be used to prove that a language is not regular or context-free.
2) For regular languages, the pumping lemma states that for any string x longer than n, x can be broken into uvw such that pumping v any number of times results in strings still in the language.
3) For context-free languages, the pumping lemma similarly states that any string x longer than n can be broken into five parts such that pumping the second and fourth parts any number of times results in strings still in the language.
Pumping lemma (1)
The pumping lemma states that for any regular language L, there exists an integer n such that any string x in L with length at least n can be divided into uvw such that: (1) x=uvw, (2) the length of uv is at most n, (3) v is not empty, and (4) for all i>=0, uviw is in L. This allows proving that a language is non-regular by finding a string that cannot be pumped.
1.10. pumping lemma for regular sets
The document discusses the pumping lemma for regular sets. It states that for any regular language L, there exists a constant n such that any string w in L of length at least n can be broken down into sections xyz such that y is not empty, xy is less than or equal to n, and xykz is in L for all k. The pumping lemma can be used to show a language is not regular by finding a string that does not satisfy the lemma conditions. Examples are provided to demonstrate how to use the pumping lemma to prove languages are not regular.
Pumping lemma for regular set h1
This document summarizes a lecture on automata theory, specifically discussing non-regular languages, the pumping lemma, and regular expressions. It introduces the language B={0n1n | n ≥ 0} as a non-regular language that cannot be recognized by a DFA. It then proves the pumping lemma theorem and uses it to show that B and the language of strings with equal numbers of 0s and 1s are non-regular. Finally, it defines regular expressions as a way to describe languages and provides examples of regular expressions and their meanings.
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The document discusses the pumping lemma for context-free grammars (CFGs). It defines the pumping lemma, which states that for any CFG there is a pumping length n such that any string in the language of length >= n can be broken into five parts that satisfy certain properties related to pumping. The document gives examples of using the pumping lemma to determine whether languages are context-free or not, such as the language L={0n1n2n | n >= 1}. It concludes by listing additional problems involving using the pumping lemma to prove languages are not context-free.
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This document discusses deterministic finite automata (DFAs) and regular languages. It defines DFAs formally and explains their components. It also proves several theorems about regular languages, such as how they are closed under operations like union and intersection. Additionally, it shows how to build DFAs that recognize specific languages and provides examples. Finally, it proves that the language of strings with an equal number of a's and b's is not regular using the pigeonhole principle.
Graph representation of DFA’s Da
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The document outlines 6 rules of inference used to derive logical conclusions: 1) Conjunctive simplification, 2) Disjunctive amplification, 3) Hypothetical syllogism, 4) Disjunctive syllogism, 5) Modus ponens, and 6) Modus tollens. Each rule is symbolically represented and includes an explanation of its application using truth tables to prove the validity of the logical inferences.
Regular expressions and languages pdf
Also since {a} is regular, {a}* is a regular language which is the set of strings consisting of a's such as , a, aa, aaa, aaaa etc. Note also that *, which is the set of strings consisting of a's and b's, is a regular language because {a, b} is regular. Regular expressions are used to denote regular languages.
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Regular expressions and languages pdf
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1 introduction
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- The history of theory of computation from Turing machines to complexity theory and applications in computer science.
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- 1. Topic - 4: Regular LanguagesTopic - 4: Regular Languages Theory of ComputingTheory of Computing Dilouar Hossain dilouarbd@gmail.com
- 2. Proving Languages not to be Regular The Pumping Lemma for Regular Languages Applications of the Pumping Lemma Closure Properties of Regular Languages Topic ContentsTopic Contents
- 3. 3
- 4. 4
- 5. 5 Problem (Example 4.2 of Text Book): Show that the language Leq consisting of all strings with an equal number of 0s and 1s (not in any particular order) is not a regular language. Soln : Suppose that Leq were regular. Then there would be a constant n satisfying the conditions of the pumping lemma. Let By the pumping lemma, we can break, w = xyz such that y ≠ ε and |xy| ≤ n. .10 nn w =
- 6. 6 Now, |w| = |0n 1n | = n + n = 2n. Now by the pumping lemma, the string xyn z is in Leq. Here, |xyn | = |x| + n|y| ≥ np > n Therefore the string xyn z has n 1s and more than n 0s. So xyn z cannot be in Leq, which contradicts our assumption. Thus the language Lpr is not a regular language. py npy ynxy p p ==∴ ≤<=∴ ≠≤ 0 .0where,0 and ε
- 7. 7 Problem (Page No. 190 of Text Book): Show that the language Lpal consisting of all strings of 0s and 1s which is a palindrome is not a regular language . Soln : Suppose Lpal is regular. Then there will be a constant n satisfying the conditions of the pumping lemma. Let By the pumping lemma, we can break, w = xyz such that y ≠ ε and |xy| ≤ n. .100 nn w =
- 8. 8 Now, |w| = |0n 10n | = n + 1 + n = 2n + 1. Now by the pumping lemma, the string xyn z is in Lpal. Here, |xyn | = |x| + n|y| ≥ np > n py npy ynxy p p ==∴ ≤<=∴ ≠≤ 0 .0where,0 and ε
- 9. 9 Therefore the string xyn z has more than n 0s to the left of the single 1 and n 0’s to the right of that 1. Hence xyn z is not a palindrome. So xyn z cannot be in Lpal, which contradicts our assumption. Thus the language Lpal is not a regular language. [Proved]
- 10. L − M 10
- 11. 11
- 12. Complement language construction: 1. Convert the regular expression to an є-NFA 2. Convert the є-NFA to a DFA by the subset construction. 3. Complement the accepting states of that DFA. 4. Turn the complement DFA back into a regular expression. 12 Closure Properties of Regular Languages: Complement
- 13. 13 Closure Properties of Regular Languages: Complement
- 14. 14
- 15. 15
- 16. 16
- 17. 17