Automata Theory Questions And Answers.docx 5j142p

Automata Theory Questions and Answers – Finite Automata This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “Regular Language & Expression”. 1. There are ________ tuples in finite state machine. a) 4 b) 5 c) 6 d) unlimited View Answer Answer:b Explanation: states,input symbols,initial state,accepting state and transition function. 2. Transition function maps. a) Σ * Q -> Σ b) Q * Q -> Σ c) Σ * Σ -> Q d) Q * Σ -> Q View Answer Answer:d Explanation: Inputs are state and input string output is states. 3. Number of states require to accept string ends with 10. a) 3 b) 2 c) 1 d) can’t be represented. View Answer Answer:a Explanation: This is minimal finite automata. 4. Extended transition function is . a) Q * Σ* -> Q b) Q * Σ -> Q c) Q* * Σ* -> Σ d) Q * Σ -> Σ View Answer Answer:a Explanation: This takes single state and string of input to produce a state.

5. δ*(q,ya) is equivalent to . a) δ((q,y),a) b) δ(δ*(q,y),a) c) δ(q,ya) d) independent from δ notation View Answer Answer:b Explanation: First it parse y string after that it parse a. 6. String X is accepted by finite automata if . a) δ*(q,x) E A b) δ(q,x) E A c) δ*(Q0,x) E A d) δ(Q0,x) E A View Answer Answer:c Explanation: If automata starts with starting state and after finite moves if reaches to final step then it called accepted. 7. Languages of a automata is a) If it is accepted by automata b) If it halts c) If automata touch final state in its life time d) All language are language of automata View Answer Answer:a Explanation: If a string accepted by automata it is called language of automata. 8. Language of finite automata is. a) Type 0 b) Type 1 c) Type 2 d) Type 3 View Answer Answer:d Explanation: According to Chomsky classification. 9. Finite automata requires minimum _______ number of stacks. a) 1 b) 0

c) 2 d) None of the mentioned View Answer Answer:b Explanation: Finite automata doesn’t require any stack operation . 10. Number of final state require to accept Φ in minimal finite automata. a) 1 b) 2 c) 3 d) None of the mentioned View Answer Answer:d Explanation: No final state requires. 11. Regular expression for all strings starts with ab and ends with bba is. a) aba*b*bba b) ab(ab)*bba c) ab(a+b)*bba d) All of the mentioned View Answer Answer:c Explanation: Starts with ab then any number of a or b and ends with bba. 12. How many DFA’s exits with two states over input alphabet {0,1} ? a) 16 b) 26 c) 32 d) 64 View Answer Answer:d Explanation: Number of DFA’s = 2^n * n^(2*n). 13. The basic limitation of finite automata is that a) It can’t arbitrary large amount of information. b) It sometimes recognize grammar that are not regular. c) It sometimes fails to recognize regular grammar. d) All of the mentioned View Answer

Answer:a Explanation:Because there is no memory associated with automata. 14. Number of states require to simulate a computer with memory capable of storing ‘3’ words each of length ‘8’. a) 3 * 2^8 b) 2^(3*8) c) 2^(3+8) d) None of the mentioned View Answer Answer:b Explanation: 2^(m*n) states requires . 15. FSM with output capability can be used to add two given integer in binary representation. This is a) True b) False c) May be true d) None of the mentioned View Answer Answer:a Explanation: Use them as a flip flop output

Automata Theory Questions and Answers – The Language of DFA This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “The Language of DFA” 1. How many languages are over the alphabet R? a) countably infinite b) countably finite c) uncountable finite d) uncountable infinite View Answer Answer: d Explanation: A language over an alphabet R is a set of strings over A which is uncountable and infinite. 2. According to the 5-tuple representation i.e. FA= {Q, ∑, δ, q, F} Statement 1: q ϵ Q’; Statement 2: FϵQ a) Statement 1 is true, Statement 2 is false b) Statement 1 is false, Statement 2 is true

c) Statement 1 is false, Statement 2 may be true d) Statement 1 may be true, Statement 2 is false View Answer Answer: b Explanation: Q is the Finite set of states, whose elements i.e. the states constitute the finite automata. 3. δˆ tells us the best: a) how the DFA S behaves on a word u b) the state is the dumping state c) the final state has been reached d) Kleene operation is performed on the set View Answer Answer: a Explanation: δ or the Transition function describes the best, how a DFA behaves on a string where to transit next, which direction to take. 4. Which of the following option is correct? A= {{abc, aaba}. {ε, a, bb}} a) abcbb ₵ A b) ε₵A c) ε may not belong to A d) abca ₵ A View Answer Answer: b Explanation: As the question has dot operation, ε will not be a part of the concatenated set. Had it been a union operation, ε would be a part of the operated set. 5. For a DFA accepting binary numbers whose decimal equivalent is divisible by 4, what are all the possible remainders? a) 0 b) 0,2 c) 0,2,4 d) 0,1,2,3 View Answer Answer: d Explanation: All the decimal numbers on division would lead to only 4 remainders i.e. 0,1,2,3 (Property of Decimal division).

6. Which of the following x is accepted by the given DFA (x is a binary string ∑= {0,1})?

a) divisible by 3 b) divisible by 2 c) divisible by 2 and 3 d) divisible by 3 and 2 View Answer Answer: d Explanation: The given DFA accepts all the binary strings such that they are divisible by 3 and 2.Thus, it can be said that it also accepts all the strings which is divisible by 6. 7. Given: L1= {xϵ ∑*|x contains even no’s of 0’s} L2= {xϵ ∑*|x contains odd no’s of 1’s} No of final states in Language L1 U L2? a) 1 b) 2 c) 3 d) 4 View Answer

Answer: c Explanation:

8. The maximum number of transition which can be performed over a state in a DFA? ∑= {a, b, c} a) 1 b) 2 c) 3 d) 4 View Answer Answer: c Explanation: The maximum number of transitions which a DFA allows for a language is the number of elements the transitions constitute. 9. The maximum sum of in degree and out degree over a state in a DFA can be determined as: ∑= {a, b, c, d} a) 4+4 b) 4+16 c) 4+0 d) depends on the Language View Answer Answer: d Explanation: The out degree for a DFA I fixed while the in degree depends on the number of states in the DFA and that cannot be determined without the dependence over the Language. 10. The sum of minimum and maximum number of final states for a DFA n states is equal to: a) n+1 b) n

c) n-1 d) n+2 View Answer Answer: a Explanation: The maximum number of final states for a DFA can be total number of states itself and minimum would always be 1, as no DFA exits without a final state. Therefore, the solution is n+1

Automata Theory Questions and Answers – Non Deterministic Finite Automata-Introduction This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “Non Deterministic Finite Automata-Introduction” 1. Which of the following options is correct? Statement 1: Initial State of NFA is Initial State of DFA. Statement 2: The final state of DFA will be every combination of final state of NFA. a) Statement 1 is true and Statement 2 is true b) Statement 1 is true and Statement 2 is false c) Statement 1 can be true and Statement 2 is true d) Statement 1 is false and Statement 2 is also false View Answer Answer: a Explanation: Statement 1 and 2 always true for a given Language. 2. Given Language: L= {ab U aba}* If X is the minimum number of states for a DFA and Y is the number of states to construct the NFA, |X-Y|=? a) 2 b) 3 c) 4 d) 1 View Answer Answer: a Explanation: Construct the DFA and NFA individually, and the attain the difference of states. 3. An automaton that presents output based on previous state or current input: a) Acceptor b) Classifier c) Transducer d) None of the mentioned. View Answer

Answer: c Explanation: A transducer is an automaton that produces an output on the basis of what input has been given currently or previous state. 4. If NFA of 6 states excluding the initial state is converted into DFA, maximum possible number of states for the DFA is ? a) 64 b) 32 c) 128 d) 127 View Answer Answer: c Explanation: The maximum number of sets for DFA converted from NFA would be not greater than 2n. 5. NFA, in its name has ’non-deterministic’ because of : a) The result is undetermined b) The choice of path is non-deterministic c) The state to be transited next is non-deterministic d) All of the mentioned View Answer Answer: b Explanation: Non deterministic or deterministic depends upon the definite path defined for the transition from one state to another or undefined(multiple paths). 6. Which of the following is correct proposition? Statement 1: Non determinism is a generalization of Determinism. Statement 2: Every DFA is automatically an NFA a) Statement 1 is correct because Statement 2 is correct b) Statement 2 is correct because Statement 2 is correct c) Statement 2 is false and Statement 1 is false d) Statement 1 is false because Statement 2 is false View Answer Answer: b Explanation: DFA is a specific case of NFA. 7. Given Language L= {xϵ {a, b}*|x contains aba as its substring} Find the difference of transitions made in constructing a DFA and an equivalent NFA? a) 2 b) 3 c) 4 d) Cannot be determined. View Answer

Answer: a Explanation: The individual Transition graphs can be made and the difference of transitions can be determined. 8. The construction time for DFA from an equivalent NFA (m number of node)is: a) O(m2) b) O(2m) c) O(m) d) O(log m) View Answer Answer: b Explanation: From the coded NFA-DFA conversion. 9. If n is the length of Input string and m is the number of nodes, the running time of DFA is x that of NFA.Find x? a) 1/m2 b) 2m c) 1/m d) log m View Answer Answer: a Explanation: Running time of DFA: O(n) and Running time of NFA =O(m2n). 10. Which of the following option is correct? a) NFA is slower to process and its representation uses more memory than DFA b) DFA is faster to process and its representation uses less memory than NFA c) NFA is slower to process and its representation uses less memory than DFA d) DFA is slower to process and its representation uses less memory than NFA View Answer Answer: c Explanation: NFA, while computing strings, take parallel paths, make different copies of input and goes along different paths in order to search for the result. This creates the difference in processing speed of DFA and NFA

Automata Theory Questions and Answers – Simpler Notations This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “Simpler Notations”. 1.Given Language: L= {xϵ∑= {a, b} |x has a substring ‘aa’ in the production}. Which of the corresponding representation notate the same?

a)

b)

c)

d)

View Answer Answer: a Explanation: The states transited has been written corresponding to the transitions as per the row and column. The row represents the transitions made and the ultimate.

2.Let u=’1101’, v=’0001’, then uv=11010001 and vu= 00011101.Using the given information what is the identity element for the string? a) u b) v c) u v d) ε -1

-1 -1

-1

View Answer Answer: d Explanation: Identity relation: εw = wε = w, thus the one satisfying the given relation will be the identity element.

3.Which of the following substring will the following notation result?

a) 0101011 b) 0101010 c) 010100 d) 100001 View Answer Answer: c Explanation: The given DFA notation accepts the string of even length and prefix ‘01’.

4.Predict the following step in the given bunch of steps which accepts a strings which is of even length and has a prefix=’01’ δ (q0, ε) =q0 < δ(q0,0) =δ (δ (q0, ε),0) =δ(q0,0) =q1 < _______________ a) δ (q0, 011) =δ (δ (q0,1), 1) =δ (q2, 1) =q3 b) δ (q0, 01) =δ (δ (q0, 0), 1) = δ (q1, 1) =q2 c) δ (q0, 011) =δ (δ (q01, 1), 1) =δ (q2, 0) =q3 d) δ (q0, 0111) =δ (δ (q0, 011), 0) = δ (q3, 1) =q2 View Answer Answer: b Explanation: Here, δ refers to transition function and results into new state or function when an transition is performed over its state.

5. Fill the missing blank in the given Transition Table: Language L= {xϵ∑= {0,1} |x accepts all the binary strings not divisible by 3}

a) Q0 b) Q1 c) Q2 d) No Transition View Answer

Answer: Q1 Explanation: The tabular representation of DFA is quite readable and can be used to some ore complex problems. Here, we need to form the transition graph and fill up the given blank.

6.Which among the following is the missing transition in the given DFA? L= {xϵ∑= {a, b} | x starts with a and ends with b}

a) δ (q0, a) =q0 b) δ (F, a) =q1 c) δ (F, a) =D d) δ (q1, a) =D View Answer Answer: b Explanation: For the given Language, the transition missing is δ (F, a) =q1.

7.The complement of a language will only be defined when and only when the __________ over the language is defined. a) String b) Word c) Alphabet d) Grammar View Answer Answer: c Explanation: It is not possible to define the complement of a language without defining the input

alphabets. Example: A language which does not consist of substring ‘ab’ while the complement would be the language which does contain a substring ‘ab’.

8.Which among the following is not notated as infinite language? a) Palindrome b) Reverse c) Factorial d) L={ab}* View Answer Answer: Factorial Explanation: Factorial, here is the most appropriate non-infinite domain. Otherwise, palindrome and reverse have infinite domains.

9.Which among the following states would be notated as the final state/acceptance state? L= {xϵ∑= {a, b} | length of x is 2}

a) q1 b) q2 c) q1, q2 d) q3 View Answer Answer: b Explanation: According to the given language, q2 Is to become the final/acceptance state in order to satisfy.

10.Which of the following are the final states in the given DFA according to the Language given.? L= {xϵ∑= {a, b} |length of x is at most 2}

a) q0, q1 b) q0, q2 c) q1, q2 d) q0, q1, q2 View Answer Answer: d Explanation: According to the given language, the length is at most 2, thus the answer is found accordingly.

Automata Theory Questions and Answers – Extended Transition Function This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “Extended Transition Function”. 1. The number of tuples in an extended Non Deterministic Finite Automaton: a) 5 b) 6 c) 7

d) 4 View Answer Answer: a Explanation: For NFA or extended transition function on NFA, the tuple elements remains same i.e. 5.

2. Choose the correct option for the given statement: Statement: The DFA shown represents all strings which has 1 at second last position.

a) Correct b) Incorrect, Incomplete DFA c) Wrong proposition d) May be correct View Answer Answer: c Explanation: The given figure is an NFA. The statement contradicts itself.

3. What is wrong in the given definition? Def: ({q0, q1, q2}, {0,1}, δ, q3, {q3}) a) The definition does not satisfy 5 Tuple definition of NFA. b) There are no transition definition. c) Initial and Final states do not belong to the Graph. d) Initial and final states can’t be same. View Answer Answer: c Explanation: q3 does not belong to Q where Q= set of finite states.

4. If δ is the transition function for a given NFA, then we define the δ’ for the DFA accepting the same language would be: Note: S is a subset of Q and a is a symbol. a) δ’ (S, a) =U δ (p, a) b) δ’ (S, a) =U δ (p, a) c) δ’ (S, a) =U δ(p) d) δ’ (S) =U δ(p) View Answer pϵs

p≠s

pϵs

p≠s

Answer: a Explanation: According to subset construction, equation 1 holds true.

5. What is the relation between DFA and NFA on the basis of computational power? a) DFA > NFA b) NFA > DFA c) Equal d) Can’t be said View Answer Answer: c Explanation: DFA is said to be a specific case of NFA and for every NFA that exists for a given language, an equivalent DFA also exists.

6. If a string S is accepted by a finite state automaton, S=s s s ……s where s ϵ∑ and there exists a sequence of states r0, r1, r2…… rn such that δ(r(i), s ) =r for each 0, 1, …n-1, then r(n) is: a) initial state b) transition symbol c) accepting state d) intermediate state View Answer 1

2

3

i+1

n

i

i+1

Answer: c Explanation: r(n) is the final state and accepts the string S after the string being traversed through r(i) other states where I ϵ 01,2…(n-2).

7. According to the given table, compute the number of transitions with 1 as its symbol but not 0:

a) 4 b) 3 c) 2 d) 1 View Answer Answer: d Explanation: The transition graph is made and thus the answer can be found.

8. From the given table, δ*(q0, 011) =?

a) {q0} b) {q1} U {q0, q1, q2} c) {q2, q1} d) {q3, q1, q2, q0} View Answer Answer: b Explanation: δ*(q0,011) = Urϵδ*(q0,01) δ (r, 1) = {q0, q1, q2}.

9. Number of times the state q3 or q2 is being a part of extended 6 transition state is

a) 6 b) 5 c) 4

d) 7 View Answer Answer: a Explanation: According to the question, presence of q2 or q1 would count so it does and the answer according to the diagram is 6.

10. Predict the missing procedure:

1.Δ(Q0, ε) ={Q0}, 2.Δ(Q0, 01) = {Q0, Q1} 3.δ(Q0, 010) =? a) {Q0, Q1, Q2} b) {Q0, Q1} c) {Q0, Q2} d) {Q1, Q2} View Answer Answer: c Explanation: According to given table and extended transition state implementation, we can find the state at which it rests

Automata Short Question & Answers QNo1.What is the difference between the strings and the words of a language? Answer:A string is any combination of the letters of an alphabet where as the words of a language are the strings that are always made according to certain rules used to define that language.For example if we take Alphabet Σ = { a , b } Here a , b are the letters of this alphabet. As you can see we can make a lot of strings from these letters a and b. For example a,b,aa,ab,ba,bb,aaa,aab,aba,baa,…………………………………………………… and so on. But when we define a language over this alphabet having no a’s and only odd number ofb’s. Then the words of this language would have only those strings that have only odd number of b’s and no a’s.some example words of our defined language are

b , bbb , bbbbb , bbbbbbb ,……………………………..and so on. So we can say that all the words are strings but all the strings may not be the words of a language.Hence strings are any combination of letters of an alphabet and the words of a language are strings made according to some rule. QNo.2 What is the difference between an Alphabet and an element of a set. Whether Alphabet is an element of a set or it is a set itself? Answer:An Alphabet is a set in itself. The elements of an Alphabet are called letters . For example Binary Alphabet Σ = {0,1} Here 0,1 are the letters of binary alphabet. Binary Alphabet is very important because it the Alphabet used by the computer. Set of Natural Numbers N={1,2,3,4,5,…………………………………..} Here 1,2,3……………………………………. are the elements of set of Natural Numbers. QNo.3 What is Null String (Λ) ? Answer:The string with zero occurrences of symbols (letters) from ∑. It is denoted by (Small Greek letter Lambda) λ or (Capital Greek letter Lambda) Λ, is called an empty string or null string. The capital lambda will mostly be used to denote the empty string, in further discussion. QNo.4 What is PALINDROME ? Answer:The language consisting of Λ (Null String) and the strings s defined over an Alphabet Σ such that Rev(s)=s. Some example words of this language are aa As Rev(aa) = aa aba As Rev(aba) = aba bbb

As Rev(bbb) = bbb aabaa As Rev(aabaa) = aabaa bbbaaabbb As Rev( bbbaaabbb ) = bbbaaabbb It is to be noted that the words of PALINDROME are called palindromes. QNo5.What is the concept of valid and invalid alphabets ? Answer:While defining an alphabet of letters consisting of more than one symbols, no letter should be started with any other the letter of the same alphabet i.e. one letter should not be the prefix of another. However, a letter may be ended in the letter of same alphabet i.e. one letter may be the suffix of another. Σ= { a , b } ( Valid Alphabet) Σ= { a , b , cd } ( Valid Alphabet) Σ= { a , b , ac } ( Invalid Alphabet) QNo 6. What is ALGOL ? Answer:ALGOL (ALGOrithmic Language) is one of several high level languages designed specifically for programming scientific computations. It started out in the late 1950’s, first formalized in a report titled ALGOL 58, and then progressed through reports ALGOL 60, and ALGOL 68. It was designed by an international committee to be a universal language. Their original conference, which took place in Zurich, was one of the first formal attempts to address the issue of software portability. ALGOL’s machine independence permitted the designers to be more creative, but it made implementation much more difficult. Although ALGOL never reached the level of commercial popularity of FORTRAN and COBOL, it is considered the most important language of its era in of its influence on later language development. ALGOL’s lexical and syntactic structures became so popular that virtually all languages designed since have been referred to as “ALGOL – like”; that is they have been hierarchical in structure with nesting of both environments and control structures. QNo7. What are the Sequential Operators? Answer:Sequencing Operators: Sequencing operators a >> b

Sequence

Match a and b in sequence

a && b

Sequential-and

Sequential-and. Same as above, match a and b in sequence

a || b

Sequential-or

Match a or b in sequence

The sequencing operator >> can alternatively be thought of as the sequential-and operator. The expression a && b reads as match a and b in sequence. Continuing this logic, we can also have a sequential-or operator where the expression a || b reads as match a or b and in sequence. That is, if both a and b match, it must be in sequence; this is equivalent to a >> ! b | b. QNo 8.What is Non-Determinism and Determinism and what is the difference between them ? Answer:Determinism means that our computational model (machine) knows what to do for every possible inputs. Non determinism our machine may or may not know what it has to do on all possible inputs. As you can conclude from above definition that Non-Deterministic machine can not be implemented ( used ) on computer unless it is converted in Deterministic machine. QNo 9. What is meant by equivalent FA’s ? Answer:FA’s that accept the same set of languages are called Equivalent FA’s. QNo 10. What is the difference between Palindrome and Reverse function? Answer:It is to be denoted that the words of PALINDROME are called palindromes. Reverse

=w

Example: Σ={a,b}, PALINDROME={Λ , a, b, aa, bb, aaa, aba, bab, bbb, …} If a is a word in some language L, then reverse (a) is the same string of letters spelled backwards, called the reverse of a. e.g reverse (xxx) = xxx reverse (623) = 326 reverse (140) = 041 QNo11.Define Kleene Star? Answer:Given Σ, then the Kleene Star Closure of the alphabet Σ, denoted by Σ*, is the collection of all strings defined over Σ, including Λ It is to be noted that Kleene Star Closure can be defined over any set of strings. Examples If Σ = {x}

Then Σ* = {Λ, x, xx, xxx, xxxx, ….} If Σ = {0,1} Then Σ* = {Λ, 0, 1, 00, 01, 10, 11, ….} If Σ = {aaB, c} Then Σ* = {Λ, aaB, c, aaBaaB, aaBc, caaB, cc, ….} Note: Languages generated by Kleene Star Closure of set of strings, are infinite languages. (By infinite language, it is supposed that the language contains infinite many words, each of finite length) QNo12.Valid/In-Valid alphabets? Answer:Any alphabet is valid if any of its letter does not appear in the start of any other letter otherwise it is invalid. QNo13.What is Reverse of a string? Answer:Alphabet provides only a set of symbols. A string is a concatenation of these symbols. Reverse of the string means to write the string in reverse order. It has no effect on alphabet. Alphabet will remain same. QNo14.Differentiate Kleene Star Closure and PLUS? Answer:Given Σ, then the Kleene Star Closure of the alphabet Σ, denoted by Σ*, is the collection of all strings defined over Σ, including Λ. Plus Operation is same as Kleene Star Closure except that it does not generate Λ (null string), automatically. You can use other symbol for alphabet but we are mostly use sigma symbol. QNo15.Define Regular Expression? Answer:Regular Expression is the generalized form of any regular language through which you can construct any string related to that language. Take an example from your handouts L1 = {Λ, a, aa, aaa, …} and L2 = {a, aa, aaa, aaaa, …} can simply be expressed by a* and a+, respectively. so a* and a+ are the generalized form of Languages L1, L2. And a* and a+ are called the regular expressions (RE) for L1 and L2 respectively.

FAQ’s about Lectures 6 to 10

Automata Theory FAQ’s about Lectures 6 to 10Q No.1 What is the concept of FA also known as FSM ( Finite State Machine) ? FA (Finite Automaton) is a finite state machine that recognizes a regular language. In computer science, a finite-state machine (FSM) or finite-state automaton (FSA) is an abstract machine that has only a finite, constant amount of memory. The internal states of the machine carry no further structure. This kind of model is very widely used in the study of computation and languages. Q No.2 What is the difference between FA , TG , GTG. ? In every FA, we mark transitions with single letter of the given alphabet but in TG transitions can be marked with letters or strings (combination of letters). In every FA, every state shows transition for all letters of given alphabet but in any TG it is not necessary to show all transition for all letters of given alphabet. In TG, we may or may not show all letter transitions according to requirement. We can also show transitions on reading any strings in TGs but it is not possible in FA’s. In GTG Directed edges connecting some pair of states are labeled with regular expressions . It may be noted that in GTG, the labels of transition edges are corresponding regular expressions. In TG we write strings and in GTG we are bound to write RE. Every FA is also a TG but not every TG is FA. Q No.3 What is the difference between FA’s and TG’s .Why we need TG’s when we have FA’s? The Transition Graphs (TG) differ from FA in the following areas TG’s are generalizations of FA’s. TG’s can change state without an input ( Null transition). Can read more than one letter (words of the language they are accepting) along the transition edges at a time. Can have a regular expression as a edge label. Can have more then one start state. We have been given more freedom in TG’s. But this freedom is on the cost of more memory and processing power it means that if we implement TG’s on computer using some programming language it will need more memory and processing power of computer than used in the implementation of FA’s. Q No.4 What is the concept of the Union of FA’s ? When we take Union of two FA’s it means that resultant FA’s should accept all the words that were accepted by the two FA’s individually. It is like taking union of two sets, the resultant set contain of both sets.

For example Let A ={1,3,5,7,9} and B = {0,2,4,6,8,10} then, A U B = { 0,1,2,3,4,5,6,7,8,9,10 } you can see that A U B contain elements of both sets similar is the case with FA’s. Q No.5 What is the difference between is TG and GTG ? In TG, there are letter transitions for the strings. While in GTG, one can write whole RE as a transition from one state to another one. Q No.6 How one can create RE of a particular language? First thing about RE and FA is that there is no hard and fast formula or method to generate these. One can generate them by its mental approach. And this mental approach can be acquired through only PRACTICE. Here are some useful tips to write RE’s, · Let our language consist of the words of length three exactly over alphabet Σ= {a,b} then it consists of the words L = {aaa, aab, aba,abb,baa,bab,bba,bbb}. Its RE can be simply written as RE = aaa + aab + aba + abb + baa + bab + bba + bbb which simply means that our language consists of only these words. So we can make RE for a finite language by writing its all words with + operator between them. · We should also keep the null string in our mind. If our language generates null string than our RE should also generate it) For example language having all the words of even length has null string in it as well so we can write its RE as follows RE = ((a+b)(a+b))* This RE also generates null string.

If a language generates all strings starting with a. then strings will be of type a , aa, ab, aab, aaa, aba, abb,…. Here RE should start with ‘a’ and then all strings including null. So this will be (a + b)* and complete RE is a (a+ b)*. Similarly languages of strings ending in b will have RE (a + b)*b. Q No.7 What is the diagrammatically difference between FA’s and TG’s? The main differences between FA’s and TG’s are as follows · · · Q No.8 What is the corresponding FA for RE =aa((a+b)(a+b))* RE is aa((a + b)(a + b))*. Its corresponding FA is as follows. Q No.9 What is difference between FA’s and NFA’s. Are they opposite to each other ? FA stands for finite automata while NFA stands for non-deterministic finite automata In FA there must be a transition for each letter of the alphabet from each state. So in FA number of transitions must be equal to (number of states * number of letter in alphabet). While in NFA there may be more than one transition for a letter from a state. And finally every FA is an NFA while every NFA may be an FA or not. Q No.10 Differentiate between (a,b) and (a+b)? (a, b) = Represents a and b. (a + b) = Represents either a or b. FAQ’s about Lectures 11 to 15 Q No.1 What is the difference between how’s FA and TG .Why we need TG’s when we have FA’s? The Transition Graphs (TG) differ from FA in the following areas

· TG’s can change state without an input ( Null transition).

We have been given more freedom in TG’s. But this freedom is on the cost of more memory and processing power it means that if we implement TG’s on computer using some programming language it will need more memory and processing power of computer than used in the implementation of FA’s. Q No.2 What is the concept of the Union of FA’s ? When we take Union of two FA’s it means that resultant FA’s should accept all the words that were accepted by the two FA’s individually. It is like taking union of two sets the resultant set contain of both sets. For example Let A ={1,3,5,7,9} and B = {0,2,4,6,8,10} then, A U B = { 0,1,2,3,4,5,6,7,8,9,10 } you can see that A U B contain elements of both sets similar is the case with FA’s. Q No.3 What is the difference between GT and GTG ? In TG, there are transitions for the strings. While in GTG, one can write whole RE as a transition from one state to another one. Q No.4 How to create a RE of a particular Language? Regular expression is used to express the infinite or finite language, these RE are made in such a way that these can generate the strings of that unique language also for the cross check that the defined RE is of a specified language that RE should accept all the string of that language and all language strings should be accepted by that RE. Q No.5 How diagrams of FA’s are created ? It depends upon the question how many states involve in a FA. There is not any formal procedure to design FA for a language. This ability just improves with time and practice. Every FA is also a TG but not every TG is FA. In every FA, every state shows transition of all letters of given alphabet but in any TG it is not must. In TG, we may or may not show all letters transition according to requirement. We can also show transitions on reading any strings in TGs but it is not possible in FAs. Q No.6 How one can create RE of a particular language? First thing about RE and FA is that there is no hard and fast formula or method to generate these. One can generate them by their mental approach. And this mental approach can be acquired through only PRACTICE.

I am giving you few tips. I hope those will help you. If we have a finite language then it will always be regular and will not have * in RE. e.g. L={aaa, aba, bb}. L language generates given three strings then its RE will be (aaa + aba + bb}. So in finite language + of all strings can be it’s RE. If we have an infinite language, then there will be * in it’s RE. We should also keep the null string in our mind. For practice just try to create RE of simple languages. Don’t try to confuse yourself with complex languages. For example if a language generates all strings starting with a. then strings will be of type a , aa, ab, aab, aaa, aba, abb,…. Here RE should start with ‘a’ and then all strings including null. So this will be (a + b)* and complete RE is a (a+ b)*. Similarly languages of strings ending in b will have RE (a + b)*b. I hope now you will be able to generate the RE of simpler languages. Gradually, increase the complexity of languages to become a perfect in RE’s. Now as similar to RE, FA of finite language will not have any loop in it. If language is infinite then there will always be at least one loop in its FA. From RE, if you want to generate its FA, then first get the smallest strings and generate their FA and then gradually get the strings of bigger length and keep amending the created FA. After some practice, you will be able to generate the FA’s. And the last thing nobody can do the new task accurately for the first time. Practice is the key to success. In the start you will have lot of mistakes but after practice you will be able to clear all of them. Q No.7 What is the difference between FA’s ,and TG’s ? There are two or three big differences between FA’s and TG’s. In FA there can be maximum one initial or starting state while in TG there may be more than one initial state. In FA there can be transition for letters only while in TG transitions from a state to another one can be for strings. In FA there must be transition from each state for each letter (deterministic) while in TG there may be no transition for specific letter from a state and there may be more than

one path for a string or letter from a state. Q No.8 What is the exact definition of FA ? Definition: A Finite automaton (FA), is a collection of the followings Finite number of states, having one initial and some (maybe none) final states. Finite set of input letters (Ó) from which input strings are formed. Finite set of transitions i.e. for each state and for each input letter there is a transition showing how to move from one state to another. Q No.9 What is the difference between TG and GTG ? In TG, there are transitions for the strings. While in GTG, one can write whole RE as a transition from one state to another one. For RE =aa((a+b)(a+b))* what will be its corresponding FA ? RE is aa((a + b)(a + b))*. Its corresponding FA is as follows. Q No.10 What is the difference between FA and NFA ? FA stands for finite automata while NFA stands for non-deterministic finite automata In FA there must be a transition for each letter of the alphabet from each state. So in FA number of transitions must be equal to (number of states * number of letter in alphabet). While in NFA there may be a transition for a letter from a state. In NFA there may be more than one transition for a letter from a state. And finally every FA is an NFA while every NFA may be an FA. FA: NFA: Q No.11 What is the method to understanding FA’s and NFA’s Firstly we know that an FA is used to describe a language. Now a language consists of strings. FA will describe the specific language only if it accepts all the strings of that particular language and all the strings generated by the FA are in the language. So confirmation is of two ways. Now, how to traverse the FA. It is very easy. Every FA has one initial state (state with -sign). From every state of FA there is one transition for every letter of the alphabet. Read the string letter by letter and move according to transitions from state to state. If the string ends in the final state (state with a + sign), that particular string will be accepted otherwise rejected. So, every string ending in final state will be accepted by FA and will be a word of the

language. For NFA, there may be no path or more than one path for a letter from a specific state. As similar to FA just start traversing from the initial state and if the string ends in the final state, it will be accepted. , as there may be more than one path for a letter from a state. So any path can be used. Goal is to reach the final state. Remaining theory is same to the FA. Practice is the key to success. Just try simple FA’s and NFA’s in the start.

FAQ’s about Lectures 16 to 20 Q No 1.What is the concept of Nondeterministic Finite Automaton (NFA) ? Nondeterminism plays a key role in the theory of computing. A nondeterministic finite state automaton is one in which the current state of the machine and the current input do not uniquely determine the next state. This just means that a number of subsequent states (zero or more) are possible next states of the automaton at every step of a computation.Of course, nondeterminism is not realistic, because in real life, computers must be deterministic. Still, we can simulate nondeterminism with deterministic programs. Furthermore, as a mathematical tool for understanding computability, nondeterminism is invaluable. As with deterministic finite state automata, a nondeterministic finite state automaton has five components. · a set of states · a finite input alphabet from which input strings can be constructed · a transition function that describes how the automaton changes states as it processes an input string · a single designated starting state · a set of accepting states The only difference lies in the transition function, which can now target subsets of the states of the automaton rather than a single next state for each state, input pair. Q No 2. If a language can be expressed in the form of FA than why it is needed to use NFA ? NFA stands for non-deterministic FA and this sort of structure has relaxation compared with FA. So it is rather more easy to represent a language using NFA. We have methods to convert NFA into FA’s so sometimes it is easier to build NFA of a given language and than convert its NFA into FA using these methods rather than directly building an FA for a language which may be very difficult.

Q No 3.How to made NFA corresponding to the closure of an FA ? While generating NFA corresponding to closure of an FA one should take care of the null string. Simple way to accept null string is declare initial state, final as well. But in this way a lot of other strings will also be accepted. Therefore, accurate way is draw another state. Declare the new state initial as well as final. Connect the new state with the states originally connected with the old start state with the same transitions as the old start state. Newly drawn diagram will be an NFA representing the language closure of the given FA Q No 4.What is the difference between Union of two FA’s , Concatenation of two FA’s and closure of two FA’s ? Consider two FA’s given below a a b b a b a b Y2+ Y1X2+ X1FA1 FA2 Here FA1 accepts all strings ending in a and FA2 accepts all strings ending b. An FA corresponding to FA1UFA2 will accept all the strings ending in a or ending in b. for example, aba,bbaaab,bbb An FA corresponding to FA1FA2 will accept all the strings whose first substring belongs to FA1 and second substring belongs to FA2. for example, ababab, bbabbb. An FA corresponding to FA1* will accepts all the strings of FA1 including null string. if FA1 represents RE r1 then FA1* will correspond to RE r1*.

FAQ’s about Lectures 21 to 25 Automata Theory FAQ’s about Lectures 21 to 25Q No 1.How Moore and Mealy machine works in Computer Memory what is their importance in Computing ? Mealy & Moore Machines work in computing as incrementing machine & 1’s complement machine etc. These operations as basic computer operations so these machines are very important. Q No 2.What is sequential circuit ? Sequential Circuit: A sequential circuit contains a memory component. The memory component provides a state input. A flip-flop is often used as a memory component. The state variable indicates the states of the sequential machine, i.e. the status or stage or progress of the whole event. The state of a sequential circuit is indicated by the output of a flip-flop. A single flip-flop can be used to indicate two states (q=0 and q=1). When there are more than two states, additional flip-flops are used. Given n flip-flops, a total of 2 n states can be represented. In other words, a sequential machine can be put into a number of different states depending on the particular inputs given. The output is a function of both the Present Inputs and the Present States. In addition to the outputs, the circuit must also generate an update to the memory components so that the state of the machine can also be changed with respect to the new inputs. The update is called the Next State Function and is also a function of the Present Inputs and the Present States. Both the output functions and the Next State Functions are combinational circuits. Z=f(X,St) S=g(X,St) The superscript t indicates the present time period while the superscript (t+1) indicates the next time period. The characteristic of a sequential circuit is completely defined by a state transition diagram that enumerates all possible transitions for every possible input combination. Q No 1.What is the concept of Pumping Lemma I and II and what is the difference between pumping Lemma 1 and pumping Lemma 2 ?In fact PLI & PLII are same (A way to recognize Non Regular language). The only difference is that the conditions in pumping lemma II are more stricter than Pumping Lemma I some language that are difficult to proof Non Regular by Pumping Lemma I are proved Non Regular by pumping Lemma II easily.

Further mare in pumping lemma I we have to generate all words to of a language but in Pumping Lemma II we have to generate a single word to prove a language non regular. Explanation: Some languages like PALINDROME that are proved to be regular by first version due to some of their symmetrical words when we pump these words they remain to be the parts of the language like bbabb By pumping lemma 1 Let y = a Now repeating y three times results in bbaaabb That is also a valid word of PALINDROME so by pumping lemma I PALINDROME can not be proved non regular, so there was the need of pumping lemma version version 2. Now consider for the word bbabb if we take N=2 Then by pumping y (let we take it b ) two times results in bbbbabb That word is not in PALINDROME. But if we take N=3 and y = a Then by pumping y two times results in bbaaabb That word is in PALINDROME. So be careful in taking total no of states of the FA and also the repeating factor (y) to prove an infinite language non regular you need to prove only one word that is not part of the language. Q No 2. What is the significance of Pumping Lemma II ? The significance of 2nd version of ‘pumping lemma’ is that there are some infinite non regular languages like PALINDROME we can built FA that can accept there certain words but if we increase the length of their words that FA don’t accept these words so by pumping lemma version I it is very difficult to prove them non regular but with the

second version we can prove that a language is Non regular even it’s some words may be accepted by some FA’s. See page 195 of the book for further example. Q No 3.Moore and Mealy machine? 1. In order to run a string on a Mealy or Moore machine, you can take directions from transition table. Running string on Mealy or Moore machine is similar to running string on a FA. For example, if want to run abba on the machine, take start from initial state. Check what is the transition for a, what state it goes. After that check what is the path of b from that state and so on. In this way you will be able to run whole of the string. Note that there is no final state in Mealy or Moore machine. So there is no case of acceptance or rejection of string. You just have to determine what the output is. I hope that will clear your mind for further clarification please listens to your lecture carefully. 2. The string is taken for the testing purposes. You can take any sort of string and determine its output using machine.

FAQ’s about Lectures 31 to 35 Q No 1.What is the difference between semiword and word please also give an example regarding this? Word: A word is complete combinations of terminals only e.g. abba or ab or a or null string. Semiword: A semiword is a string of terminals (may be none) concatenated with exactly one nonterminal on the right i.e. a semi word, in general, is of the following form (terminal)(terminal) ————- (terminal)(nonterminal) For example aaaaaaB , aabbaaaA , A. What is the difference between derivation tree and total tree ? A Derivation tree is the one that shows how to derive any specific word of the language described by CFG but Total Language Tree shows all words of the Language described by CFG on it Q No 2.What does mean the LANGUAGE IS CLOSED? When we say that a Language is closed it is always with respect to certain operation. A simple example may be that the set of integers is closed under addition. It means when we take two numbers from set of integers say 3, 7 the result of

their addition would also be in the set of integers. Similarly if the result of an operation on the words of a language results in the word of the same language we say that the language is closed under that operation. Q No 3.What are the Productions? Productions are the grammatical rules and regulations. These rules express the behavior of CFG. Using production in CFG terminals are converted into non-terminals and when all the terminals are converted using productions, a word is acquired. Q No 4.What is the difference between concatenation and intersection of two FA’s also what is the difference among Union of two FA’s and addition of them? In intersection of two FA’s only those strings are accepted which are independently accepted by both FA’s, while in concatenation of two FA’s only those strings will be accepted in which first part of string is accepted by first FA and remaining part of string is accepted by the second FA. While taking union of two FA’s one can represent it using + sign. So (FA1 U FA2) and (FA + FA2) both are same. There is no difference between them.

FAQ’s about Lectures 36 to 40 Automata TheoryFAQ’s about Lectures 36 to 40 Q No 1.What is the Difference between Nullable and Null production? How to make eliminate Nullable and for Null Productions from the CFG ? The production of the form nonterminal ® L is said to be null production. Example: Consider the following CFG S ® aA|bB|L, A ® aa|L, B ® aS Here S ® L and A ® L are null productions. A production is called nullable production there is a derivation that starts at Non Terminal and leads to L i.e.

S ———–> aA | bB | aa A————-> C | bb C————–> L Here A nullable Non Terminal due to Nullable production A—————> C as C leads to null. Example: Consider the following CFG S ® XY, X ® Zb, Y ® bW Z ® AB, W ® Z, A ® aA|bA|L B ®Ba|Bb|L. Here A ® L and B ® L are null productions, while Z ® AB, W ® Z are nullable productions. Method: Delete all the Null productions and add new productions e.g. Consider the following productions of a certain CFG X ® aNbNa, N ® L, delete the production N ® L and using the production X ® aNbNa, add the following new productions X ® aNba, X ® abNa and X ® aba Thus the new CFG will contain the following productions X ® Nba|abNa|aba| aNbNa Note: It is to be noted that X ® aNbNa will still be included in the new CFG. Method: Consider the following CFG S ® XY, X ® Zb, Y ® bW Z ® AB, W ® Z, A ® aA|bA|L B ®Ba|Bb|L. Here A ® L and B ® L are null productions, while Z ® AB, W ® Z are nullable productions. The new CFG after, applying the method, will be S ® XY

X ® Zb|b Y ® bW|b Z ® AB|A|B W®Z A ® aA|a|bA|b B ®Ba|a|Bb|b Note: While adding new productions all Nullable productions should be handled with care. All Nullable productions will be used to add new productions, but only the Null production will be deleted Q No 2. Is it possible to make CFG for infix and postfix expression’s using derivation tree ? Derivation tree is only used to derive words of language that is described by a CFG. Yes, we can create CFG for languages infix expressions, postfix expressions. Q No 3 what is the uses of push down automata in computing ? PDA is just an enhancement in FAs. i.e Memory is attached with machine that recognizes some language. FA is basic structure for most advanced electronic machines such as computer etc. Q No 4 What is difference between PUSH DOWN STACK and PUSH DOWN STORE ? No difference at all. Both are used to describe memory structure attached with FAs to store some characters in it. Q No 5 How we can distinguish between “CFG” and “CNF” in the questions ? Chomsky Normal Form (CNF) If a CFG has only productions of the form nonterminal ———à string of two nonterminals or Nonterminal ———–à one terminal Then the CFG is said to be in Chomsky Normal Form (CNF). Thus if the given CFG is in the form specified above it will be called in CNF. Q No 6.What is meant by the stack consistence and input tape

consistence ? Term Stack consistent means we can pop any character from the top of the stack only. PDA should not be able to pop any character other than that is present on the top of the stack. Term Tape consistent means we can read only the first letter on the tape not any other letter of the tape after the first one. Q No 7 What is the concept of unit production ? The productions of the form one Nonterminal ——–à one Nonterminal Are called unit productions. For example S ——–à A (Unit Produciton) A——–à a | b Here there is no need of Unit Production S® A. we can directly write S —-à a | b Q No 8 Why Context Free Grammars are called “Context Free? Context Free Grammars are called context free because the words of the languages of Context Free Grammars have words like “aaabbb”(PALINDROME). In these words the value of letters (a , b) is the same on whatever position they appear. On the other hand in context sensitive grammars their value depend on the position they appear in the word a simple example may be as follows Suppose we have a decimal number 141 in our language . When compiler reads it, it would be in the form of string. The compiler would calculate its decimal equivalent so that we can perform mathematical functions on it. In calculating its decimal value , weight of first “1” is different than the second “1” it means it is context sensitive (depends on in which position the “1” has appeared). i.e. 1*102 + 4*101 + 1* 100 = 14 (value of one is 100) (value of one is just one) That is not the case with the words of Context Free Languages. (The value of “a” is always same in whatever position “a” appears).

Q No 9. What is Unit Production? The production in which one non-terminal leads to only one non-terminal. Q No 10.What is Left most Derivation in CFG? It is a method of generation of strings from a CFG starting from left most letter of the string.

FAQ’s about Lectures 41 to 45 Q No 1.Give a example of converting a CFG to CNF?Consider the CFG given below S→ ABC A→ aa | b B→ c C→d Its CNF will be S→ DC D → AB A→ EE | b E→a B→c C→d Q No 2.In the lecture 41 ‘s example, we have converted PDA to conversion form and a word ‘aaaabb’ is derived from this conversion form PDA. What are the derivation steps. The PDA converted to conversion form has some specific features that are important to understand first. These features are The states named START, READ, HERE and ACCEPT are called ts of the machine. With the help of the conversion form we have been able to achieve that POP state has only one path out of it and the path taking (multiple paths) decisions take place only on the READ state. The word ‘aaaabb’ is generated as follows from the PDA START-POP4-PUSH $ This step pops $ and then pushes it to ensure that stack contains $ at the beginning. READ1-POP6-PUSH $-PUSH a As first time after reading “a” there is $ at the top of stack so we will follow

path segment READ1-POP6-PUSH $-PUSH a READ1-POP5-PUSH a-PUSH a Now a is on the top of the stack so we will follow READ1-POP5-PUSH a-PUSH a READ1-POP5-PUSH a-PUSH a Again following same segment for a READ1-POP5-PUSH a-PUSH a Again following same segment for a READ1-POP1- HERE-POP2 As we read b on input tape. READ2-POP1-HERE-POP2 As we read b on input tape. READ2-POP3-ACCEPT. As we read ∆ from the input tape Q No 3.How to differentiate between “wanted” and “unwanted branch” ? When we derive a word in Top down parsing beginning with the starting Non Terminal the branches of the tree that do not lead to our required word are left aside these branches are called unwanted branches. For example for CFG S—–>AA A—–>a | b If we want to generate the word “aa” we will leave the branch generated by the production A——>b. Q No 4.What is the difference between intersection and union of a language? Intersection of two languages will consist of all those words which are in both languages while union of two languages will consist of all those words which are present in at least one language. Symbol for intersection is ∩ and for union is U. Q No 5.What is the difference between Context free languages and regular languages?

Regular languages can be represented by FA’s because we do not need any memory to recognize (accept or reject them on FA) them but there is another class of languages that can not be represented by FA’s because these languages require that we have some memory (with the help of memory we can store letters of the string we are checking so that we can compare them with next coming letters in the string). For example language anbn requires that we must store a’s and then compare their count with next coming b’s so that we can check whether a’s are equal to b’s or not. Due to this reason we use Context Free Grammars to represent them because we can5t write RE’s for them. So Context Free Languages represent a broader category this category also include regular languages as subcategory. It means that context free languages include regular languages as well as some other languages. Q No 6.What is the difference between Moore and Mealey machines? In Mealy Machine we read input string letters and generate output while moving along the paths from one state to another while in Moore machine we generate output on reaching the state so the output pattern of Moore machine contains one extra letter because we generated output for state q0 where we read nothing. Q No 7.What does the following mean i. STACK Consistent ii. Y-able Paths iii. Working string iv. Semi Word means Stack consistence means that in the PDA converted in the conversion form, when we follow a path segment (which is formed by combining start, read or here state with next read, here or accept state on the path) along the PDA its pop state should have the path for the same letter that is present on the top of the stack at that stage. If this doesn’t happen our PDA will crash because in conversion form of the PDA the pop state has only one letter path, so if we could not be able to find that letter on the top of the stack our PDA will crash (if will not find path where to go from that state) Working string means the string present on the input tape. Y-able Paths means that when we follow a certain sequence of rows from the row table to generate a path for a word form start state to accept state. The path (sequence of rows) should be stack as well as t consistent it means that rows should end at the same read or here state ( consistency ) and the

rows should be able to pop the letter from the top that is indicated in the pop state of the row. Semi word is the string of terminals it may be null string ending with a Non terminals on the right. For example some semi words are aaS aabbA B

Is Automata Theory is a Programming Subject or theoretical? Automata theory is the study of abstract computing devices, or “machines”. This topic goes back to the days before digital computers and describes what is possible to compute using an abstract machine.These ideas directly apply to creating compilers, programming languages, and deg applications. They also provide a formal framework to analyze new types of computing devices, e.g. biocomputers or quantum computers What are practical Examples of the implications of Automata Theory and the formal Languages? Grammars and languages are closely related to automata theory and are the basis of many important software components like: · Compilers and interpreters · Text editors and processors · Text searching · System verification

What are the Types of Automata? · The Types of Automata Theory are · Finite Automata · Regular Languages · Linear-bounded Automata · Context Sensitive Languages

· Push-Down Automata · Context Free Languages · Turing Machines · Recursively innumerable languages There are others as well like, · Random Access Machines · Parallel Random Access Machines · Arrays of Automata

Question:

How types of Automata Differ? They differ in the following areasComplexity (or Simplicity) Power In the function that can be computed.

Answer:

In the languages that can be accepted.

Question:

What is the difference between the alphabet and an element of a set?

Answer:

Alphabets is a set of letters nothing else but a set of strings (elements) can have more than one letters in one string.

Question:

Difference between Palindrome and Reverse function?

Answer:

The language consisting of Λ and the strings s defined over Σ such that Rev(s)=s.It is to be denoted that the words of PALINDROME are called palindromes. Reverse

=w

Example: Σ={a,b}, PALINDROME={Λ , a, b, aa, bb, aaa, aba, bab, bbb, …} If a is a word in some language L, then reverse (a) is the same string of letters spelled backwards, called the reverse of a. e.g

reverse (xxx) = xxx reverse (623)

= 326

reverse (140)

= 041

Question: Define Strings? Concatenation of finite letters from the alphabet is called a string. e.g If Σ= {a,b} then a language L can be defined asL = {a, abab, aaabb, ababababababababab,……………} Answer:

it’s mean all words with a’s more or equal to b’s

Question:

Define empty or null strings?

Answer:

Concatenation of finite letters from the alphabet is called a string.Sometimes a string with no symbol at all is used, denoted by (Small Greek letter Lambda) λ or (Capital Greek letter Lambda) Λ, is called an empty string or null string.

Question:

Difference between string and word?

Answer:

Any combination of letters of alphabet that follows rules of language is called a word. A string is a finite sequence of symbols from an alphabet.

Question:

There are as many palindromes of length 2n as there are of length 2n1, please explain? If we try to create palindromes then middle elements (2 in even palindromes & 1 in odd palindrome) does not cause any change in no. of palindromesDefining the language PALINDROME, of length 2n and 2n-1 defined over S = {a,b} e.g if we take n= 2 for 2n Length (2n) = 4 and string can be written as {aaaa, abba, baab, bbbb} And if we take n = 2 for 2n-1 Length (2n-1) = 3 and string can be written as

Answer:

{aaa, aba, bab, bbb}

Automata Theory Questions and Answers – Regular Language & Expression – 1 This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “Regular Language & Expression”. 1. A regular language over an alphabet a is one that can be obtained from a) union b) concatenation c) kleene d) All of above View Answer Answer : d Explanation : None.

2. Regular expression {0,1} is equivalent to a) 0 U 1 b) 0 / 1 c) 0 + 1 d) All of above View Answer Answer : d Explanation : All are equivalent to union operation.

3. Precedence of regular expression in decreasing order is a) * , . , + b) . , * , + c) . , + , * d) + , a , * View Answer Answer : a Explanation : None.

4. Regular expression Φ* is equivalent to a) ϵ b) Φ c) 0 d) 1 View Answer Answer : a Explanation : None.

5. a? is equivalent to a) a b) a+Φ c) a+ϵ d) wrong expression View Answer Answer : c Explanation : Zero or one time repetition of previous character .

6. ϵL is equivalent to a) ϵ b) Φ c) L d) Lϵ View Answer Answer : c,d Explanation : None.

7. (a+b)* is equivalent to a) b*a* b) (a*b*)* c) a*b* d) none of above View Answer Answer : b Explanation : None.

8. ΦL is equivalent to a) LΦ b) Φ c) L d) ϵ View Answer Answer : a,b Explanation : None.

9. Which of the following pair of regular expression are not equivalent? a) 1(01)* and (10)*1 b) x(xx)* and (xx)*x c) (ab)* and a*b* d) x+ and x*x+ View Answer

Answer : c Explanation : (ab)*=(a*b*)*.

10. Consider following regular expression i) (a/b)* ii) (a*/b*)* iii) ((ϵ/a)b*)* Which of the following statements is correct a) i,ii are equal and ii,iii are not b) i,ii are equal and i,iii are not c) ii,iii are equal and i,ii are not d) all are equal View Answer Answer : d Explanation : All are equivalent to (a+b)*

Automata Theory Questions and Answers – Regular Language & Expression – 2 This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “Regular Language & Expression”. 1. How many strings of length less than 4 contains the language described by the regular expression (x+y)*y(a+ab)*? a) 7 b) 10 c) 12 d) 11 View Answer Answer : d Explanation : string of length 0 = 1 string of length 1 = 4 string of length 2 = 3 string of length 3 = 3

2. Which of the following is true? a) (01)*0 = 0(10)* b) (0+1)*0(0+1)*1(0+1) = (0+1)*01(0+1)* c) (0+1)*01(0+1)*+1*0* = (0+1)* d) All of the mentioned View Answer Answer : d Explaination : None.

3. A language is regular if and only if a) accepted by DFA b) accepted by PDA c) accepted by LBA d) accepted by Turing machine View Answer Answer : a Explanation : All of above machine can accept regular language but all string accepted by machine is regular only for DFA.

4. Regular grammar is a) context free grammar b) non context free grammar c) english grammar d) none of the mentioned View Answer Answer : a Explanation : Regular grammar is subset of context free grammar.

5. Let the class of language accepted by finite state machine be L1 and the class of languages represented by regular expressions be L2 then a) L1=L2 c) L1 U L2 = .* d) L1=L2 View Answer Answer : d Explanation : Finite state machine and regular expression have same power to express a language.

6. Which of the following is not a regular expression? a) [(a+b)*-(aa+bb)]* b) [(0+1)-(0b+a1)*(a+b)]* c) (01+11+10)* d) (1+2+0)*(1+2)* View Answer Answer : b Explanation : Except b all are regular expression*.

7. Regular expression are a) Type 0 language b) Type 1 language c) Type 2 language

d) Type 3 language View Answer Answer : a Explanation : According to Chomsky hierarchy .

8. Which of the following is true? a) Every subset of a regular set is regular b) Every finite subset of non-regular set is regular c) The union of two non regular set is not regular d) Infinite union of finite set is regular View Answer Answer : b Explanation : None.

9. L and ~L are recursive enumerable then L is a) Regular b) Context free c) Context sensitive d) Recursive View Answer Answer : d Explanation :If L is recursive enumerable and its complement too if and only if L is recursive.

10. Regular expressions are closed under a) Union b) Intersection c) Kleen star d) All of the mentioned View Answer Answer : d Explanation : According to definition of regular expression

Automata Theory Questions and Answers – Regular Expression-Introduction This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “Regular Expression-Introduction”. 1. L is a regular Language if and only If the set of __________ classes of IL is finite. a) Equivalence b) Reflexive

c) Myhill d) Nerode View Answer Answer: a Explanation: According to Myhill Nerode theorem, the corollary proves the given statement correct for equivalence classes.

2. A language can be generated from simple primitive language in a simple way if and only if a) It is recognized by a device of infinite states b) It takes no auxiliary memory c) Both are correct d) Both are wrong View Answer Answer: b Explanation: A language is regular if and only if it can be accepted by a finite automaton. Secondly, It s no concept of auxiliary memory as it loses the data as soon as the device is shut down.

3. Which of the following does not represents the given language? Language: {0,01} a) 0+01 b) {0} U {01} c) {0} U {0}{1} d) {0} ^ {01} View Answer Answer: d Explanation: The given option represents {0, 01} in different forms using set operations and Regular Expressions. The operator like ^, v, etc. are logical operation and they form invalid regular expressions when used.

4. According to the given language, which among the following expressions does it corresponds to? Language L={xϵ{0,1}|x is of length 4 or less} a) (0+1+0+1+0+1+0+1) b) (0+1) c) (01) d) (0+1+ε) View Answer

4

4

4

4

Answer: d Explanation: The extended notation would be (0+1) 4 but however, we may allow some or all the factors to be ε. Thus ε needs to be included in the given regular expression.

5. Which among the following looks similar to the given expression? ((0+1). (0+1)) * a) {xϵ {0,1} *|x is all binary number with even length} b) {xϵ {0,1} |x is all binary number with even length} c) {xϵ {0,1} *|x is all binary number with odd length} d) {xϵ {0,1} |x is all binary number with odd length} View Answer Answer: a Explanation: The given regular expression corresponds to a language of binary strings which is of even length including a length of 0.

6. If R represents a regular language, which of the following represents the Venn-diagram most correctly?

a) An Irregular Set b) R* c) R complement d) R reverse View Answer Answer: b Explanation: The given diagram represents the Kleene operation over the Regular Language R in which the final states become the initial and the initial state becomes final.

7. The given NFA corresponds to which of the following Regular expressions?

a) (0+1) *(00+11) (0+1) * b) (0+1) *(00+11) *(0+1) * c) (0+1) *(00+11) (0+1) d) (0+1) (00+11) (0+1) * View Answer Answer: a Explanation: The transition states shown are the result of breaking down the given regular expression in fragments. For dot operation, we change a state, for union (plus) operation, we diverge into two transitions and for Kleene Operation, we apply a loop.

8. Concatenation Operation refers to which of the following set operations: a) Union b) Dot c) Kleene d) Two of the options are correct View Answer Answer: b Explanation: Two operands are said to be performing Concatenation operation AB = A•B = {xy: x ∈ A & y ∈ B}.

9. Concatenation of R with Ф outputs: a) R b) Ф c) R.Ф d) None of the mentioned View Answer

Answer: b Explanation: By distributive property (Regular expression identities), we can prove the given identity to be Ф.

10. RR* can be expressed in which of the forms: a) R+ b) Rc) R+ U Rd) R View Answer Answer: a Explanation: RR*=R+ as R+ means the occurrence to be at least once.

Automata Theory Questions and Answers – DFA to Regular Expressions This set of Automata Theory Multiple Choice Questions & Answers (MCQs) focuses on “DFA to Regular Expressions”. 1. Which of the following is same as the given DFA?

a) (0+1)*001(0+1)* b) 1*001(0+1)* c) (01)*(0+0+1)(01)* d) None of these View Answer Answer: a Explanation: There needs to be 001 together in the string as an essential substring. Thus, the other components can be anything, 0 or 1 or e.

2. Which of the following statements is not true? a) Every language defined by any of the automata is also defined by a regular expression b) Every language defined by a regular expression can be represented using a DFA c) Every language defined by a regular expression can be represented using NFA with e

moves d) Regular expression is just another representation for any automata definition View Answer Answer: b Explanation: Using NFA with e moves, we can represent all the regular expressions as an automata. As regular expressions include e, we need to use e moves.

3. The total number of states required to automate the given regular expression (00)*(11)* a) 3 b) 4 c) 5 d) 6 View Answer Answer: c Explanation:

4. Which of the given regular expressions correspond to the automata shown?

a) (110+1)*0 b) (11+110)*1 c) (110+11)*0 d) (1+110)*1 View Answer Answer: c Explanation: There is no state change for union operation, but has two different paths while for concatenation or dot operation, we have a state change for every element of the string.

5. Generate a regular expression for the following problem statement: Validation: String should be 8-15 characters long. String must contain a number, an Uppercase letter and a Lower case letter. a) ^(?=.*[a-z])(?=.*[A-Z])(?=.*\d).{8,15}$ b) ^(?=.*[a-z])(?=.*[A-Z])(?=.*\d).{9,16}$ c) ^(?=.[a-z])(?=.[A-Z])(?=.\d).{8,15}$ d) None of the mentioned View Answer Answer: a Explanation: s like abc123, 123XYZ, should not be accepted . If one also wants to include special characters as one of the constraint, one can use the following regular expression: ^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[^\da-za-Z]).{8,15}$

6. Generate a regular expression for the following problem statement: P(x): String of length 6 or less for å={0,1}* a) (1+0+e)6 b) (10)6 c) (1+0)(1+0)(1+0)(1+0)(1+0)(1+0)

d) More than one of the mentioned is correct View Answer Answer: a Explanation: As the input variables are under Kleene Operation, we need to include e,thus option c is not correct,thereby option (a) is the right answer.

7. The minimum number of states required in a DFA (along with a dumping state) to check whether the 3rd bit is 1 or not for |n|>=3 a) 3 b) 4 c) 5 d) 1 View Answer Answer: c Explanation:

8. Which of the regular expressions corresponds to the given problem statement: P(x): Express the identifiers in C Programming language l=letters d=digits a) (l+_)(d+_)* b) (l+d+_)* c) (l+_)(l+d+_)* d) (_+d)(l+d+_)* View Answer Answer: c Explanation: Identifiers in C Programming Language follows the following identifiers rule: a) The name of the identifier should not begin with a digit.

b) It can only begin with a letter or a underscore. c) It can be of length 1 or more.

9. Generate a regular expression for the given language:l L(x): {xÎ{0,1}*| x ends with 1 nd does not contain a substring 01} a) (0+01)* b) (0+01)*1 c) (0+01)*(1+01) d) All of the mentioned View Answer Answer: c Explanation: (a) and (b) are the general cases where we restrict the acceptance of a string witrh substring 00 but we ignore the case where the string needs to end with 1 which therby, does not allows the acceptance of e.

10. The minimum number of transitions to to reach the final state as per the following regular expression is: {a,b}*{baaa} a) 4 b) 5 c) 6 d) 3 View Answer Answer: a Explanation: