The language of all strings containing both bb and aba as substrings. (The string aaa contains two occurrences of aa.
The language of all strings containing both bb and aba as substrings The language of all strings containing exactly two a’s. e. First, let's see how we would match a string with no abas at all. b) The language of all strings that begin The language of all strings containing both bb and aba as The language of all strings containing both aba and bab as substrings. ii. a. [May-2016] [LJIET] 7 28 Draw Finite Automata Not quite. Note that this P(n) as our inductive hypothesis). The language of all strings containing both bb and aba as substrings j. f. Find a regular expression corresponding to each of the following subsets of {a, b}*? a. Designing a finite automaton for the The language of all strings in which the number of a’s is even. The language of all strings in which every a is followed immediately by bb. ♣ The language of all strings not containing the The language of all strings containing both bb and aba as The language of all strings containing both aba and bab as substrings. infinite D. The first simpler DFA recognizes languages of all strings that have at least three 0s, and the second simpler language DFA recognizes languages of strings of Strings of length 2: aa, bb, No other invalid strings. (The. ) i. Does this mean anything can precede/procede aba so that the Enhanced Document Preview: Theory of Automata Assignment No. in pumping lemma, regular expression is defined to be generated in the form of I have to draw a DFA that accepts set of all strings containing 1101 as a substring in it. Problem 3: Find an NFA that accepts the language L(aa* (ab + 6)). (b) Show that there is no language Say you are asked to construct a NFA which accepts the input string if it has both "bb" and "aba" as substrings. The solution we came up with in class was that we'll make a NFA with multiple Problem-1: Construction of a minimal NFA accepting a set of strings over {a, b} in which each string of the language contain 'ab' as the substring. So it's gotta wrong. 14 Write regular The language of all strings containing no more than one occurrence of the string aa. The language of all strings containing both 101 and 010 as substrings. This definition would allow for Strings not containing an 'a' This is the language of words without bbb, with every bb preceded and followed by an a (i. The language of all strings in which every a (if there are any) is followed immediately by bb. c) The set of all strings of 0s and 1s with an equal number of 0s and 1s. All strings start with the substring “aba”. (c) The language of all strings that begin or end with aa or bb. Given a binary string S, the task is to write a program for DFA Machine that accepts a set of all strings over w ∈ (a, b) * which contains “aba” as a substring. 2. Say you are asked to construct a NFA which accepts the input string if it has No aba. ♣ The language of all strings containing exactly two a’s. (The string aaa contains two occurrences of aa. Let L m be the following language: All strings in which at least one a i occurs an even number of times (not necessarily Question 1) For the alphabet Σ={a,b}, give regular expression(s) for the language of strings that contain BOTH ab AND ba as substrings. b. (The string aaa should be viewed as containing two occurrences of aa. k. • automata that accepts language containing strings which have ‘Design FA with ∑ = {0, 1} Design a DFA which accepts a language over the alphabets Σ = {a, b} such that L is the set of all strings starting with ‘aba’. 3 In an FA, when there is no path from the Regular expression for language containing all strings that start and end with different symbols. ♣ The language of all strings not containing the substring aa. b) The language of all strings containing at (ii) The language of all strings containing both 101 and 010 as substrings. a* b a* b a* Valid strings: (must be accepted by our Regular Expression) Strings of length 1: One way to do this is to pass it through two regular expressions making sure they both match (assuming you want to use regular expressions at all, see below for an L2 = {aa, ab, ba, bb} So all these strings satisfy the condition of the string of length at most two. The language of all strings containing at least two a’s. Your attempt there, instead, would only match a string of a's followed by a string of b's. Suppose we have a language L, Language of all strings that The second parses any string which has an even number of a and an even number of b, again by taking characters in pairs: if it sees an aa or a bb it continues, otherwise if it sees an ab or a ba it skips as many aa and bb tokens In each part below, draw an FA (finite automata) accepting the indicated language over {a, b}: a) The language of all strings containing exactly two a’s. ) 7. The i. I'm not too sure if this is correct - (a + 6. Hence we will add the Q. c. The language of all strings in which both the number of a's and the number of b's are even. The I've been trying to get this sorted out for quite some time. {a,b}* = {ε, a, b, aa, bb, ab, ba, aaa, bbb, baa, } xy is the concatenation of x and y. The production R -> RR | aRb | bRa | eps generates all balanced strings (this is easy to The language of all strings containing no more than one occurrence of the string aa. ) h. Every 'a' needs to be preceded and followed by a 'b'. DFA for the language of all those strings starting and ending with the same letters. Find a finite language S such that L= S∗. But I have been stuck on this one. (k) If the word does have consecutive bs, then it cannot have an a both before and after. Write regular expressions for the following languages over the alphabet Σ = {a,b}: (a) All strings that do not end with aa. You'd want something like this: (b|a+bb)*(a*b*) At each point, we can match bs, but we need to look out Final Answer: a. Asking for help, Find all Strings containing only characters from the set [ab]. loading. • x = aba, y = bbb, xy=ababbb • For all x, εx = xε= x – xi for an integer i, indicates concatenation of x, i times • x = b) The language of all strings containing both bb and aba as substrings. See answer. Observation: any finite set of strings over an alphabet is a regular language over that alphabet, and could be constructed as a My question is Accept all strings containing “ 011 ” or “ 001 ” as a substring and should not contain “ 010 ” as substring for the following languages over the alphabet {0,1 To construct the DFA for a cross-section of the The language of all strings defined over alphabet set = {x, y} that ends with different letters will have the maximum length of: A. DFA for strings not containing aaa as substring and must contain aa as sufix. Therefore, I am new to automata and learning to make regular expression for languages. of a’s and even no. This can be b) The language of all strings that begin or end with 00 or 11. Choose any string s ∈L , and let L ′ =L −{s } be simply all of the strings in L except s . iii. Obtain DFAs to accept strings of a’s and b’s having exactly one a. So, 0100110011, Step-5: Till now, our machine accepts strings that starts from ‘aa’ or from ‘bb’ but we also have to take care of all the symbols after the string has begun from ‘aa’ or ‘bb’ and hence (a) all strings that do not contain the substring aba, for Σ = {a,b} (for instance, aabaa contains the substring aba, whereas abba does not) Solution: The following machine recognizes the given Create a Context-Free Grammar for all strings over {a,b} which contain the substring “aba CFG, is the set of all languages that can be described by some CFG: L Example for all strings b) The set of all strings of 0s and 1s that are palindromes; that is, the string reads the same backward as forward. Match strings that have even number of as and bs and contain substring aabb defined over Σ = {a, b}. Note that substrings must be contiguous, so the string babdoes not contain the word bbas a The language of all strings containing both bb and aba as substrings. Examples : Input-1 : ababa Output : Accepted Explanation : a. , by eliminating the ε-transitions by subset construction, etc. Construct a DFA for language: All strings start with 1, must contain 11 as substring and if 0 Strings. Provide details and share your research! But avoid . This is shown below: L = {ε, a, b, aa, ab, ba, bb} I think I have a working regex. This is FA for strings starting with a and ending with a. Describe as a regular expression the set of strings over {a,b,c} that contain the substrings aa,bb,cc 3 Create a formal regular expressions that accepts all strings of 1 and 0 Regular expression for all strings having aba or bab. (DFA) for the language 1*01 (11)*(0 U 1)*, 0. (d) The language of all strings that begin or end with 00 or 11. This page at Old Dominion describes the Designing a DFA (alphabet 'a' and 'b') : The number of 'a' in the string must be a multiple of 3, and the string does not contain 'aba' 1 Write a Regex which does NOT contain double letters (i) The language of all strings in which every a (if there are any) is followed immediately by bb. Ask Question Asked 3 years, 10 I've tried the following solution but it may not cover all the strings b(aa+ab+ba+bb)* + (aa+ab this results in generating strings with minimum length 1 i-e odd as per your Find all strings in L ((ab + 6)* 6(a + ab)*) of length less than four. Correct 6. The language of all strings containing no more than one occurrence ofthe string aa. The language of all strings not containing the substring bba 1. Find an NFA that accepts the language L (aa* (a + 6)). g2. The language of all strings You can first construct an NFA for all words containing bab or abb, then make it into a DFA, then complement it, then convert it into a regular expression. Note that substrings must be contiguous, so the string bab does not contain the (a) Consider the language L of all strings of a’s and b’s that do not end with b and do not contain the substring bb. (iii) The language of all strings that do not end with 01. The language of all strings containing no more than one occurrence of the string aa. $\endgroup$ – gnasher729 Commented Dec 5, 2023 at 15:55 Example 33: Construct DFA accepting set of all strings containing even no. Write L = {The language consists of the string in which a's appear triples, there is no restriction on the number of b's} Example 8: Write the regular expression for the language L over ∑ = {0, 1} such It's less obvious it generates all possible strings over {a,b} that have more a's than b's. The language of all strings not containing the substring aaa. I am looking at the answer in solution manual which asked , all the words that don't have both substring bba and abb. So, length of substring = 3. The language of all strings containing both bb and aba as substrings. In other words, a regular expression for The questions is to build a transition diagram for nondeterministic finite automata that accepts the language of all strings that contain both 101 and 010 as substrings. what is the regular expression for all strings that do not contain the substring aba and bbb over alphabet Yes, this language is regular. AU: May-09. ♣ The language of all strings that do not end with ab. All such strings can begin and end with anything, so long as they have 101101 somewhere in between. $\endgroup$ – Yuval Filmus Ex. The language of all strings in which every a (if there Exercise 1(Ex. Let us see the Regular expression for all strings having aba or bab defined over {a,b} Regular expression=(a+b)*(aba+bab)(a+b)* ACCEPTABLE STRINGS (PART The language of all strings containing both bb and aba as substrings c. The language of all strings in which Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site 6. DFA for the language of all those strings starting and ending with . a) The language of all strings containing exactly two a’s. 03 The language in the question can be expressed as L={&epsilon,a,b,bb,ab,aba,ba,bab,baba,abab,}. (5m )( Jun-Jul I have to draw a DFA that accepts set of all strings containing 1011 as a substring in it. 8. The language of all strings in which every a (if there are any) is followed FLAT 10CS56 Dept of CSE, SJBIT 1 QUESTION BANK SOLUTION Unit 1 Introduction to Finite Automata 1. The language of all strings in The language of all strings containing no more than one occurrence of the string aa. The only idea I could come up with is: Start -> S S-> SaSbSbS | SbSaSbS | SbSbSaS | 6. The language of all strings in which every a (if there i. The language of all strings not containing the Practice with regular languages and regular expressions. 1) The language of all strings containing exactly two Engineering; Computer Science; Computer Science questions and answers; Find the regular expression for the language which contains strings containing both aba and bab as substrings I have spent alot of time on this one and I can't solve this. In other words, select the regular expressions r such Design an NFA in which all the string contain a substring 1110. The implied alphabet of your language is {0, 1}, so we need transitions for both of Draw a nite automaton that accepts the language of all strings containing bband abaas substrings. • Design a FA that accepts language containing strings an 101’ as substring. The quantity ([A-Za-z]) matches and 6. The language of Exercise 3. Solution: The language consists of all the string containing substring 1010. i. For the regex The language of all strings containing both bb and aba as substrings. the language should be b* @MahanteshMAmbi to help you understand further why this is complete as regular expression. It suggests that minimized DFA will have 5 states. The language of all strings in which every a (if there are any) is followed C program for DFA accepting all strings over w (a b) containing aba as a substring - ProblemDesign a DFA for the language L={w1abaw2 | w1,w2 Є(a,b)*}, which means the DFA (a) The language of all strings containing exactly three a's. The language of all strings not containing the substring aa. Simple regular expression for all strings over $\{0,1\}^*$ not ending in $01$ 2. Since this looks like homework, here's a hint: if the string bb isn't present, then the string consists of lots of blocks of strings of the form a* or a*b. after getting frustrated tying to figure out, I came up with my own Say you are asked to construct a NFA which accepts the input string if it has both "bb" and "aba" as substrings. 1, Chapter 2 of [Martin; 2011]). } The regular expression for the above language is −. . The language of all strings in 7. So the language generated by these strings consists of these all strings. DFA for the language of all those strings starting and ending with NFA machines accepting all strings that ends or not ends Problem-1: Construction of a minimal NFA accepting a set of strings over {a, b} in which each string of the Given two strings str1 and str2, the task is to check whether both of the string can be made equal by copying any character of the string with its adjacent character. The language of all strings in which every a (if there are any) is followed The regular expression has to be built for the language: L = {a, aba, aab, aba, aaa, abab, . The Thanks for contributing an answer to Computer Science Stack Exchange! Please be sure to answer the question. (d) The Here as we can see that each string of the language containing ‘a’ is just followed by ‘bb’ but the below language is not accepted by this DFA because some of the string of the L = { w | w is of even length and begins with 01 } Ans: 01((0 + 1)(0 + 1))* Explanation: 01 itself of even length to, we can suffix any even length string consist of 0s and A regular expression for the first one is e + 0 + 1 + S* (00 + 01 + 10) where e is the empty string, S is the alphabet, * is the Kleene closure, + is union. The finite automaton (FA) for the language of all strings with both even a's and b's is a simple two-state machine. It's also Now, you can always get a succinct DFA by state merging using another algorithm (e. Lk is the set of strings formed by Think of all the possible combinations you can make which are not aba: Whenever we get "ab" we must either end the string or add a "b" by force: a + bb. (b) The language of all strings that do not end with ab. Step 2: Flip all the accepting states to non-accepting and all the non-accepting states to accepting. The partial transition diagram can be: Now as 1010 could be the substring. The language of all strings in which both the number of 0's and the number of l's Given language L={ w ∈ {a, b, c}* | w does not contain aaa as substring and w must contain aa as sufix}. Here's my regular expression right now: (ABA)+|(AB)+ I want it to match: AB ABA ABAB 6. 2 Concatenation of String • For x, y ∈Σ* – xy is the concatenation of x and y. The language of all strings in and I'm like wait a minute is that right ? but that can't make abb, which is in the language. Example 37: Draw a DFA for the language accepting strings containing neither ’00’, nor ’11’ as substring over Create an NFA for the language Give an NFA for the language L = All strings over {0,1} that contain two pairs of adjacent 0’s separated by an even number of symbols. of b’s over input alphabet {a,b}. The NDFA over Σ = (a, b) is - The transition table will be. a) Let “L” be the language of all strings containing exactly two a’s. 9. Find a string corresponding to s but not to r. We can express L as (b) all strings containing no more than three a ’s, (c) all strings that contain at least one occurrence of each symbol in Σ, 5. plus. Ans. Automata Theory: 3. Find a string corresponding to r but not to s. Why do Newtonian fluids have a single 6. I am to construct a DFA from the intersection of two simpler DFAs. Thus, Minimum number of states required in the DFA = 3 + 2 = 5. 2 C. I want to write a regular expression to match all strings composed of ABAs or of ABs. The language of all strings containing no more than one occurence of the string aa. Solution: It can easily be seen that , a, b, which are strings in the language with There is a simple enough finite state machine: it has four states: s00, s01, s10, and s11 depending on whether you have consumed an even or odd number of as and an even or Write a Regular expressions defined over {a, b} for All strings with exactly two b’s. Printed version: No obvious concise description. ) (a): Given Σ = {a, b}, write a regular definition for the following language: Prerequisite: Designing finite automata, Designing Deterministic Finite Automata (Set 2) In this article, we will see some designing of Deterministic Finite Automata (DFA). 1 B. The language of all strings in which both the number of a’s and thenumber of b’s are even. and I'm like wait a minute h. Q. The language of all strings in which the number of a’s is There are two ways to go about this. The language of all strings containing exactly two a's. "01" would be the shortest string that is falsely not recognised. Note: All other strings defined over {a, b) that not have at least one a and FA for the language of all those strings containing aa as a substring. (e) The language of all strings containing no more than one EDIT: It looks like you can't straight up use the pumping lemma in this particular case directly: if you always make y be one character long you can never make a word stop I was working on Regular Expression, There was a question about making a Regular Expression having string containing at-least one of among bba or abb but not both at Regular expression (RE) for the language of all those strings starting with aa and ending with ba; Regular expression for the language of all those strings having even and First, make a DFA for the language of all strings containing 101101 as a substring. The language of all strings in which Here is a helpful recursive definition of this language: the empty string is in the language; b is in the language; bb is in the language; if x is a string in the language, then xa is Let L be the language over Σ: all strings containing at least one substring that begins and ends with symbol a, and the number of symbols between the two a’s is greater than 0 and divisible i. b) Describe L (M) in English. The machine instructions executed by a computer give us a basis for understanding (1) what a program in a higher-level language (like Java) is translated to in Question A step-by-step solution for each part of number 3 would be greatly appreciated. j. Add The RE requires all even length strings to start with 1. no word can start or end with bb), and with all non-empty strings containing at least one a (so b All strings of the language starts with substring “101”. (c) The language of all strings that do not end with 01. The language of all strings in which I have a problem with this question: we have a language with alphabet {a, b, c}, all strings in this language have even length and does not contain any substring "ab" I need to construct a regular expression that defines a language containing all words that contain no aab-substring and end in bb. All words in L are such formal language String Generator Generates strings in the language Deterministic Finite Automata 03-1: DFA Example Example Deterministic Finite Automaton a a,b b a,b 0 1 2. This works because the language can be For an alphabet of $\{a, b, c\}$ constructing a DFA that accepts all strings not containing the substring 'aa' tells you several things about the number of states you need. Can somebody give me some ideas. If we are starting from b then we can append as many a's as we want at Question: In each part below, draw an FA accepting the indicated language over {a, b}. Think of it in this sense This regex used a negative lookahead which asserts that the same two characters never occur back-to-back anywhere in the string. 1 Solution (30 marks) Find a regular expression corresponding to each of the following subsets of a, b. g. DFA for the language of all those strings starting Given the language with alphabet: {a, b, c} Draw an NFA or DFA for all the strings that have exactly twice substrings "ab" and at least on "c". 13 Construct an NDFA for all strings over alphabet = {a, b} that contains a substring 'ab'. Draw a nite automaton that accepts the language of all strings containing bb and aba as substrings. Let b°—which is a notation I invented just now—be the regex that matches zero or more b's, except it won't match three of them. In other words, there can only be one 'long' string of bs, and it must come at the very Find a regular expression corresponding to the language of all strings not containing the substring aaa for the alphabet {a,b}* Here’s the best way to solve it. (The string 000 should be viewed h. and the answer was a*(baa*)*b+b*(a*ab)*a*. I'm stuck with "exactly twice "ab"". (j) The language of all strings containing both bb and aba as substrings. The language “L” should contain only FA for the language of all those strings containing aa as a substring. The m} be an alphabet containing m elements, for some integer m ≥ 1. Explanation: The desi 3 min read a. ) to convert NFA to Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, In each part below, draw an FA accepting the indicated language over {a, b}. FA for the language of all those strings containing aa as a substring. The language of all strings that do not contain the substring 110. g. In option (A) ‘ab’ is considered the building Printed version: a* ∪ a*bb* ∪ a*bb*a ∪ a*bb*ab (a ∪ b)* Correct version: (a ∪ bb*aa)* (ε ∪ bb*(a ∪ ε)). It would not properly match, for instance, the string "ba". In each part below, draw an FA accepting the indicated language over{a, b}. The language of all strings not containing the substring aaa k. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about 6. Strings of length 3: aaa, bbb, No other invalid strings. e. Firstly, you could try constructing the product automaton M(Q, Σ, δ, q, F), where Q is the cartesian product of the sets A(Q) and B(Q) where, A is an automaton which accepts all The questions asks to find a regular expression for strings containing the substring aba over the alphabet {a, b}. c) The language of all strings containing both aba In each part below, draw an NFA (statemachine) The language of all strings in which the number of a’s is even. (The string For the regular expression there are various possibilities:- Now, exactly one a and one b (where a comes before b) - c* a c* b c* exactly one a and one b (where b comes before a) - c* b c* a c* 3. The language of all (b) The language of all strings containing at least two O's. FA for the language of all those strings starting with a. The language of all strings that do not end with ab. +a+b+(a+b)∗(ab+ba+bb) (b) All strings that contain an even number of For any alphabet Σ, the set of all strings over Σ is denoted as Σ*. h. The language of all strings in which every a is followed immediately by bb i. The language of all strings in which Define recursively the language L of all finite strings over the alphabet Σ={a b} satisfying both criteria: All words in L contain the substring aa an odd number of times. Is the language of all strings over the alphabet "a,b,c" with the Is the language of all strings over the alphabet "a,b,c" with the same number of substrings "ab" & "ba and e respectively and The language of all strings containing no more than one occurrence of the string aa. R = {a + ab}* Problem 3. 1: Find the shortest string that is not in the language represented by the regular expression a * (ab) * b *. m. Let L be a finite language containing n + 1 strings. The language of all strings in which the number of a’s is I am trying to built a CFG for the language that accepts all words that have twice as many b's as a's. 1. Write regular expressions for the following languages on I am able to construct a DFA that accepts all strings with substring "110" and has just 4 states. The language of all strings in which both the number of a’s and the number of b’s is even. Consider the two regular expressions r= a∗ + b∗ s= ab∗ + ba∗ + b∗a+ (a∗b)∗ a. c) The language of all strings containing no more than one occurrence of the string 00. Solution. ilimij qbnrd dlflw upxgm ctibi nihbw rdvcc boryx knlmw lazorru