(I) State the pumping lemma with all its conditions and indicate how you go abou proving that a language is not regular. (II) If the languages are regular, give an automaton. Otherwise, show it is not regular by using pumping lemma. L= {b²anbma³ | m, n ≥ 0}. L = {b²anbma³ | m = n}. L = {ww|w is any string over {a.b}} L = {a³b4} L= {anbn+1} L = {ambnamn | m,n≥ 0 }

(I) State the pumping lemma with all its conditions and indicate how you go abou proving that a language is not regular. (II) If the languages are regular, give an automaton. Otherwise, show it is not regular by using pumping lemma. L= {b²anbma³ | m, n ≥ 0}. L = {b²anbma³ | m = n}. L = {ww|w is any string over {a.b}} L = {a³b4} L= {anbn+1} L = {ambnamn | m,n≥ 0 }

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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dont post any existing answers directly dislike Let regular language L1 recognized by DFA (Q1,Σ,δ1,s1,F1) and regular language L2 recognized by DFA (Q2,Σ,δ2,s2,F2). We will construct a product DFA as the quintuple (Q,Σ,δ,s,F) where: •Q=Q1×Q2•For allx∈Σ and (q1,q2)∈Q1×Q2,δ((q1,q2),x) = (δ1(q1,x),δ2(q2,x))•s= (s1,s2) Suppose we have a DFA that recognizes concatenation of two languages L1 L2. Also suppose we have a DFA that recognizes L2. Using these two DFAs, can we construct a new DFA to recognize L1? If yes, construct it.If no, explain why. Let regular language L1 recognized by DFA (Q1,Σ,δ1,s1,F1) and regular language L2 recognized by DFA (Q2,Σ,δ2,s2,F2). We will construct a product DFA as the quintuple (Q,Σ,δ,s,F) where: •Q=Q1×Q2•For allx∈Σ and (q1,q2)∈Q1×Q2,δ((q1,q2),x) = (δ1(q1,x),δ2(q2,x))•s= (s1,s2) Suppose we have a DFA that recognizes concatenation of two languages L1 L2. Also suppose we have a DFA that recognizes L2. Using these two DFAs, can we construct a new DFA to recognize…
Dont copy the answer otherwise will surely dislike Let regular language L1 recognized by DFA (Q1,Σ,δ1,s1,F1) and regular language L2 recognized by DFA (Q2,Σ,δ2,s2,F2). We will construct a product DFA as the quintuple (Q,Σ,δ,s,F) where: •Q=Q1×Q2•For allx∈Σ and (q1,q2)∈Q1×Q2,δ((q1,q2),x) = (δ1(q1,x),δ2(q2,x))•s= (s1,s2) Suppose we have a DFA that recognizes concatenation of two languages L1 L2. Also suppose we have a DFA that recognizes L2. Using these two DFAs, can we construct a new DFA to recognize L1? If yes, construct it.If no, explain why. Let regular language L1 recognized by DFA (Q1,Σ,δ1,s1,F1) and regular language L2 recognized by DFA (Q2,Σ,δ2,s2,F2). We will construct a product DFA as the quintuple (Q,Σ,δ,s,F) where: •Q=Q1×Q2•For allx∈Σ and (q1,q2)∈Q1×Q2,δ((q1,q2),x) = (δ1(q1,x),δ2(q2,x))•s= (s1,s2) Suppose we have a DFA that recognizes concatenation of two languages L1 L2. Also suppose we have a DFA that recognizes L2. Using these two DFAs, can we construct a new DFA to…
(I) State the pumping lemma with all its conditions and indicate how you go about proving that a language is not regular. (II) If the languages are regular, give an automaton. Otherwise, show it is not regular by using pumping lemma.
3. show L = {0^n1^n0^n1^n | n >=0} is not context free language. Prove by stating that even though L follows condition 3 of pumping lemma , it will not follow condition 1, hence not context free?
Consider the following languages over the alphabet Σ = {a, b}: 1. L1 = {xwx : x, w ∈ Σ∗}2. L2 = {xwx : x, w ∈ Σ∗, |w| = 1}3. L3 = {xwx : x, w ∈ Σ∗, |x| = 1} For each language, decide whether it is regular or not regular. Prove each of your claims. If you construct a FA or a regular expression, you do not need to prove the correctness of your construction.
Question 1 a) Based of your study for automata and theory of computer science, explain push down automata and give examples for each. b) State the Pumping Lemma for regular languages. Is every language that satisfies the pumping lemma property a regular language?
Please prove or disprove: If a language L ⊆ Σ∗ is recognized by a FA, then there is an NFA M = (K,Σ,δ,s0,F) with |F|= 1 such that L = L(M).
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