10. The GNFA method is used to show that

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1 CSE 355 Midterm Examination 27 February 27 Last Name Sample ASU ID First Name(s) Ima Exam # Sample Regrading of Midterms If you believe that your grade has not been recorded correctly, return the entire paper to the instructor with a short note indicating what you believe to be the error. Other than for that reason, test grades are almost never changed. If you believe that you did not receive the proper credit, first read these sample solutions carefully to see if you can understand the answer to your concern. If that does not resolve it, write a clear explanation of why you believe the grade is in error and submit that, along with the entire test paper, to the instructor. Please do not discuss in your explanation how your solution is like that of another student, as FERPA legislation makes it impossible for me to discuss one student s work with another. Please take into account that more than 35 papers were graded, and it is quite unfair to change the grade on one paper without giving every other student the same opportunity. If you nevertheless want the paper regraded, be advised that the entire paper will be regraded and the grade may go up, stay the same, or go down. The new grade will be final. It is a violation of the Academic Integrity Policy to request a grade change simply because you need or want a higher grade. If you require a clarification of the sample solutions (not a grade change or review as discussed above), ask in recitations, or in office hours of the TA or instructor. You will be asked whether you have read the sample solution and to indicate what precisely is unclear to you about it, so read these sample solutions carefully first. Note that under no circumstances can anyone change a grade other than the instructor, so do not ask the TAs to do so they are not able to. Grade review requests, whether submitted as described above or not, will not be considered if received after 28 March 27. Instructions. You must write the exam paper that has your name and student number preprinted on it. You must not move to a different seat. 2. Do not open the exam until you are instructed to do so. 3. The exam consists of (a) this cover sheet giving instructions and space for rough work not to be graded; (b) three pieces of paper, printed two-sided, containing five long answer questions (and space to write answers) and space for rough work; (c) three sheets of paper, printed two-sided, containing 4 multiple choice questions; and (d) a Scantron sheet to be completed using the pencil provided giving your answers to the multiple choice questions. You must turn in all sheets including the multiple choice questions. 4. You have minutes to complete the exam. 5. No books, notes, electronic devices, or other aids are permitted. Turn off all wireless devices and place them away from your work space. 6. Write all answers on the examination paper itself. Work on pages indicated rough work only is not graded. 7. BUDGET YOUR TIME WELL! 8. SHOW ALL WORK! Multiple Choice [4 marks in total] Select the most appropriate answer for each, and enter each response (one of a, b, c, d, e) on the Scantron sheet provided using the pencil provided.. If DFA M = (Q, Σ, δ, q, F ) accepts input string w Σ with w = n, a computation of M on w (a) is a sequence of exactly n states (b) is a sequence of exactly n + states (c) may be a sequence of any positive integer number of states

2 Page 2 Sample, Ima () (d) may be a sequence of any integer number of states that is at least n + (e) may be a sequence of any integer number of states that is at least Q 2. If NFA M = (Q, Σ, δ, q, F ) accepts input string w Σ with w = n, a computation of M on w (a) is a sequence of exactly n states (b) is a sequence of exactly n + states (c) may be a sequence of any positive integer number of states (d) may be a sequence of any integer number of states that is at least n + (e) may be a sequence of any integer number of states that is at least Q 3. The powerset method (a) constructs a regular expression from an NFA (b) is not used in polite company. (c) constructs a DFA from a regular expression (d) constructs a DFA for the intersection of two regular languages (e) constructs a DFA from an NFA 4. On input string w Σ, the number of computations of an NFA (Q, Σ, δ, q, F ) is always (a) at least if the NFA accepts w (b) a (finite) nonnegative integer (c) a (finite) positive integer (d) equal to if the NFA accepts w (e) an integer between and Q 5. The pumping lemma for regular languages can be proved by (a) showing that an NFA can be converted to an equivalent DFA. (b) showing that the regular languages are closed under the regular operations. (c) showing that regular languages are context-free and using the pumping lemma for context-free languages. (d) showing that an NFA computation must repeat a state. (e) showing that a DFA computation must repeat a state. 6. To show that a language is not regular, one could (a) show that it has no regular expression. (b) use the pumping lemma for regular languages. (c) use closure properties. (d) find it using Google. (e) (b) or (c) 7. The pumping lemma for regular languages implies that (a) every regular language contains a string that can be pumped (b) all strings in a regular language can be written as uvw so that uv i w is also in the language when i (c) a regular language is infinite if and only if it contains a string that can be pumped (d) regular languages are closed under the regular operations (e) every string can be pumped in at most one way 8. A generalized NFA, or GNFA, (a) can have transitions labelled with or (b) can have more than one start state (c) is a generalized new fangled apparatus (d) can have more than one final state (e) (b) or (d) 9. To rip a state q rip in the GNFA method, when there is a transition from q to q rip labeled A, a transition from q rip to q rip labeled B, a transition from q rip to q labeled C, and a transition from q to q labeled D, we make a transition from q to q labeled (a) (ABC) D (b) ABC D (c) (A(B) C) D (d) D (AC) (e) A B C D. The GNFA method is used to show that

3 Page 3 Sample, Ima () (a) A language is regular if and only if it is described by a regular expression (b) Regular languages are closed under star. (c) Every regular language is described by a regular expression. (d) Every regular expression describes a regular language. (e) NFAs are no more powerful than DFAs.. Which of these does not imply that L is regular? (a) L is closed under the regular operations: i.e, L = LL = L L = L. (b) L is the language of a regular grammar (c) L is finite (d) L is recognized by a DFA (e) the reversal of L is recognized by an NFA 2. The majority A of three languages L, L 2, and L 3 is a language consisting of all strings that are in at least two of the three. Suppose that L, L 2, and L 3 are regular. (a) A is regular because we can run DFAs for L, L 2, and L 3 to recognize A (b) A is regular because we could construct a DFA using the product construction (c) A may not be regular because we need to read the input three times (d) A is regular because A = (L L 2 ) L 3, by closure under union and intersection (e) A may not be regular because it cannot be expressed using unions and intersections. 3. For an NFA M = (Q, Σ, δ, q, F ) and a subset S Q, which of the following is not always true about the ε closure E(S)? (a) E(S) = q S E({q}) (b) there is no ε transition from a state in E(S) to a state not in E(S) (c) there is no ε transition from a state not in E(S) to a state in E(S) (d) S E(S) (e) Given M and S one can calculate E(S) in finite time. 4. {a n b m : n > m } is (a) not regular because a p+ b p cannot be pumped (b) regular because it is a subset of a b (c) not regular because a p b p cannot be pumped (d) regular because it is described by a regular expression (e) not regular because you cannot remember n 5. {(ab) n : n } is (a) not regular because you cannot remember n (b) regular because each string can be pumped (c) regular because it is a b (d) regular because it is (ab) (e) not regular because a p b p cannot be pumped 6. Let L be the regular language () and L = {() n m m : n, m }. Then L L is (a) regular by closure under union (b) not regular because it contains m m (c) regular because you can always pump by setting v to be the leading or (d) not regular because L is not regular, because p p cannot be pumped in L (e) regular because it is described by a regular expression 7. A regular expression describes a language that always (a) is finite when the regular expression contains no (b) is infinite when the regular expression does contain a (c) contains no string with fewer characters than appear in the regular expression (d) may not satisfy the conditions of the pumping lemma (e) is the closure of a regular language under 8. For a DFA (Q, Σ, δ, q, F ), what is the meaning of the statement c = δ(b, a)? (a) It has no meaning, it is just some symbols. (b) When reading an a in state b, the next state is c. (c) b and a are states of the machine. (d) c is a final state when reading an a. (e) When reading a b, and the next character is a, the next state is c.

4 Page 4 Sample, Ima () 9. Which of the following statements is always false about a DFA (Q, Σ, δ, q, F )? (a) q F. (b) Q is a proper subset of F. (c) q loops on some symbol in Σ. (d) Q contains fewer than 355 states. (e) The size of the alphabet is strictly more than the number of states. 2. Given a DFA D = (Q, Σ, δ, q, F ) that accepts every string, what can we infer about D? (a) Every state in D is a final state. (b) There is at least one state in D that is not final. (c) Every reachable state from q in D is a final state. (d) There is only one character in the alphabet. (e) There is at least one final state in D that is not q. 2. Let L be a regular language. How many DFAs have language L? (a) Exactly. (b) Exactly. (c) At most 2. (d) Finitely many. (e) Infinitely many. 22. Which of the following is always true about a given DFA D = (Q, Σ, δ, q, F )? (a) If q / F, D s language contains the empty string. (b) If D does not accept some string w, and the resulting state after the computation is q s q, then if we modify the machine so that q s is the start state and feed w as input, this modified machine accepts w. (c) If there is a transition from a final state to a non-final state, D s language is infinite. (d) If δ(q, ) = q, δ(q, ) = q, and q F, D accepts every string consisting of s with even length. (e) If D has only one final state, D only accepts a finite number of strings. 23. If languages L, L 2 are non-regular languages with alphabet Σ, which of the following is not always true? (a) L, L 2 contain infinitely many strings. (b) L, L 2 Σ. (c) L is not regular. (d) L L is regular. (e) All of the above are always true. 24. Suppose that I have two DFAs and have 5 and 4 states respectively, and 2 and 3 final states respectively. The number of final states in the product DFA for the union of their languages is (a) 2. (b) 7. (c) 6. (d) 5. (e) fewer than Suppose that I have two DFAs and have 5 and 4 states respectively, and 2 and 3 final states respectively. The number of states in the product DFA for the intersection of their languages is (a) at least 2, but can be more. (b) 2. (c) 6. (d) 3. (e) Suppose I have two DFAs D, D 2 and I perform the product construction on them to get a DFA for the union of their languages. What is always true about the resulting DFA D? (a) The number of states in D is equal to the maximum of the number of states in both of D, D 2. (b) If L(D ) = L(D 2 ), then D has every state being final. (c) If L(D ) = L(D 2 ), then D has every state being final. (d) If L(D ) =, then L(D) = L(D 2 ). (e) If L(D ) = Σ, then L(D) = L(D 2 ).

5 27. Given an NFA with Page 5 Sample, Ima () δ(q, ) = {q } δ(q, a) = {q 2 } δ(q, ) = {q } δ(q 2, ) = {q 3 } δ(q 3, a) = {q } δ(q 3, ) = {q 4 } what is E({q, q 2 })? (a) (b) {q, q, q 2 } (c) {q, q 2 } (d) {q 3 } (e) None of the above. 28. A multi-start NFA (MSNFA) is defined like a standard NFA, except that more than one start state is allowed, and we can choose any of them when starting a computation. What can we conclude about the languages that MSNFAs recognize? (a) The languages recognized by MSNFAs are exactly the regular languages. (b) The teaching assistants have active imaginations. (c) Some language that an NFA recognizes is not recognized by an MSNFA because MSNFAs may start at the wrong state. (d) Some language that an MSNFA recognizes is not recognized by an NFA because MSNFAs can choose to start at the right state. (e) (c) and (d) 29. Given two NFAs N = (Q, Σ, δ, q,, F ) and N 2 = (Q 2, Σ, δ 2, q,2, F 2 ), my friend wants to produce an NFA for L(N ) L(N 2 ). He uses a variant of the product construction to produce an NFA N = (Q, Σ, δ, q, F ): Q = Q Q 2, q = (q,, q,2 ), F = F F 2, δ((q, q 2 ), a) = {(q, q 2) q δ (q, a), q 2 δ 2 (q 2, a)} for all q Q, q 2 Q 2, a Σ. δ((q, q 2 ), ) = {(q, q 2) q δ (q, ) {q }, q 2 δ (q 2, ) {q 2 })} for all q Q, q 2 Q 2. Is my friend correct in saying that L(N) = L(N ) L(N 2 )? (a) He is not correct because the transition function is for union, not intersection. (b) He is not correct because the transition function only needs to be defined for at least one character a instead of all of them, and we cannot know that in advance because the machine is inherently nondeterministic. (c) He may be correct for some NFAs, and not for others. (d) He is correct because N satisfies the formal definition of an NFA, and therefore its language is regular. (e) He is correct because one can observe computations in the original two machines and see that if one existed for them, then one exists for the new machine (and vice versa). 3. The set of non-regular languages is closed under which of the following operations? (a) Star (b) Complement (c) Union (d) Intersection (e) Concatenation 3. My friend doesn t like the pumping lemma, but still wants to show that L = {w {, } : w is not of the form n n } is non-regular. To do this, he observes that it is the complement of the non-regular language L = { n n : n }, and uses closure under complementation. Is my friend correct? (a) He is correct because anyone who does not like the pumping lemma must be right. (b) He is not correct because L is not the complement of L. (c) He is not correct because non-regular languages are not closed under complement. (d) He is correct because regular languages are closed under complement.

6 Page 6 Sample, Ima () (e) He is correct because any DFA to recognize L would have to count and check that the number of s and s are different. 32. What is the meaning of a grammar being ambiguous? (a) Some string in the grammar s language has at least two different derivations. (b) All strings in the grammar s language have at least two different parse trees. (c) All strings in the grammar s language have at least two different derivations. (d) Some string in the grammar s language has at least two different parse trees. (e) The grammar is not in CNF. 33. Suppose I have a CFG (not necessarily in CNF) and a string of length 5 in its language. What can I tell about a derivation of this string in the grammar? (a) The derivation involves exactly applications of rules to derive this string. (b) The derivation involves exactly 9 applications of rules to derive this string. (c) The derivation involves at least 5 applications of rules to derive this string, but can be arbitrarily large. (d) The derivation involves some number of applications of rules to derive this string, but is at most 32. (e) None of the above is true. 34. Suppose I have a CFG in CNF and a string of length 5 in its language. What can I tell about a derivation of this string in the grammar? (a) The derivation involves exactly applications of rules to derive this string. (b) The derivation involves exactly 9 applications of rules to derive this string. (c) The derivation involves at least 5 applications of rules to derive this string, but can be arbitrarily large. (d) The derivation involves some number of applications of rules to derive this string, but is at most 32. (e) None of the above is true. 35. DeMorgan s Laws ensure that (a) Closure under intersection and complementation imply closure under union. (b) Union members who meet at an intersection are sent to jail. (c) Closure under intersection and union imply closure under complementation. (d) Closure under union and complementation imply closure under reversal. (e) Closure under any two of union, intersection, and complementation implies closure under all three. 36. The regular operations are (a) star, union, and complementation. (b) star, union, and concatenation. (c) star, union, concatenation, and complementation. (d) union, intersection, and complementation. (e) union, concatenation, and complementation. 37. To show that a language is regular, one could give a DFA for it. One could also (a) give a regular expression. (b) use the pumping lemma for regular languages. (c) use closure properties. (d) (a) or (c) (e) (a), (b), or (c) 38. Which of the following is false? (a) Languages are defined to have a finite number of strings. (b) Regular grammars generate regular languages. (c) Languages are defined over finite alphabets (d) Strings are defined to have finite length (e) NFAs can have any positive integer number of states. 39. My friend thinks that the conversion of CFGs into CNF is done as follows (in order): () Make the start variable not appear on the RHS of any rule, (2) Delete nullable variables, (3) Remove unit rules, (4) Make RHSs consist of either only one terminal or more than one variable, and (5) Break up long RHSs. Which step is not correct?

7 Page 7 Sample, Ima () (a) Step is not correct. (b) Step 2 is not correct. (c) Step 3 is not correct. (d) Step 4 is not correct. (e) Step 5 is not correct. 4. Which of the following statements is correct about the language { i i+j 2 j : i, j }? (a) It is regular, but not context-free. (b) It is regular, and context-free. (c) It is not regular, but is context-free. (d) It is not regular, and not context-free. (e) None of the above. Question. [ marks] In the space provided, give an example of a language L for each of the following if one exists, otherwise check the No column: Requirement Example No L is finite but not empty {a} L = L L is regular and infinite a L is finite but not regular L is infinite but not regular palindromes LL = L and L is regular a LL = L and L is not regular balanced parentheses L is regular but not context-free L is context-free but not regular palindromes L is regular, and is the union of two CFLs {a} {b} Question 2. [ marks] In the space provided, give a BRIEF justification for each statement given. The Statement The Space Provided Regular languages are closed DFA: Swap final and non-final states. under complement. Regular languages are closed Regular grammar: New start S and new rules S S S 2 under union. where S, S 2 old start variables of two regular grammars. DFA: Use product construction. NFA: New start state, ε-transitions to old start stes of two DFA/NFAs. Regex: (R R 2 ) is a regex for union of two regexes. Context-free languages are closed CFG: New start S and new rules S S S 2 under union. where S, S 2 old start variables of two CFGs. Regular languages are closed NFA: ε-transitions from final states of one DFA/NFA to another. under concatenation. Regex: (R R 2 ) is a regex for concatenation of two regexes. Regular grammar: New start S and new rule S S S 2. Regular languages are closed Regular grammar: reverse RHS of each rule (see Question 4.) under reversal. DFA/NFA: reverse all transitions, make old start final, add new start with ε-transitions to all old final states (no longer final) Question 3. [ marks] Let L {a, b} be the language of all strings that contain more substrings equal to abb than to aba. State whether or not L is regular, and show that your answer is correct. All points are for the explanation. This language is NOT {(abb) m (aba) n : m > n}. It is much larger than that. And it definitely is not {abb m aba n : m > n}! We prove that L is not regular. Assume to the contrary that L is regular and let p be its pumping length. Now we choose a string w L with w p and try to pump it. If w starts with a b, it can always be pumped by choosing x = ε and y = b. Otherwise if w starts with aa, it can always be pumped by choosing

8 Page 8 Sample, Ima () x = ε and y = a. Otherwise if w starts with aba, it can always be pumped by choosing x = ε and y = a. So it had better start with abb if this is going to work. Indeed if it has an aba in the first p characters, just pump one of the as and you never leave L. Choose w = (abb) p (aba) p. Writing w = xyz with xy p and y, we find that y must be a nonempty string in the language (ε b bb)(abb) (ε a ab). Almost everyone just did the case when y is a nonempty string in (abb). If y is in the language (ε b bb)(abb) + (ε a ab), then xy z loses at least one abb substrings but cannot lose any aba substrings, so it is not in L, a contradiction in this case. It remains to treat cases when y is not empty and is in (ε b bb)(ε a ab). Now if y starts with a b then again xy z loses at least one abb substrings but cannot lose any aba substrings, so it is not in L, a contradiction. So all that remains is y {a, ab} because it is not empty. Once again, xy z loses at least one abb substrings but cannot lose any aba substrings, so it is not in L, a contradiction. This is all of the cases and every one leads to a contradiction, so w cannot be pumped, and L is not regular. An alternate approach. Suppose to the contrary that L is regular. Then by closure under intersection so is L = L (abb) (aba). Now do pumping for L, and when y (abb) you leave the language L right away, making the cases easier. Question 4. [5 marks] In this question DO NOT USE the equivalences among NFAs, DFAs, regular expressions, and regular grammars, or closure properties, without proving them. My friend has gotten all mixed up about regular grammars. He thinks that their definition says that they can have rules of the form A ε, A a, A B, and A Ba where A and B are variables and a is a terminal. But he has made a mistake! In a regular grammar, the fourth should be A ab instead. So let s call our friend s grammar raluger to keep them straight. So we have Rule Types Allowed Regular Raluger A ε A ε A a A a A B A B A ab A Ba We want to help our friend by showing that a language has a regular grammar if and only if it has a raluger grammar. We are going to do this in four steps. (a) [4 marks] Suppose that L has a regular grammar. Show that L has an NFA. NOTE 4.: The rule types given are not the rules of a specific grammar to be converted! Students who used the set of the four types of regular grammar rules (in the above left column) as a particular four-rule grammar G with L(G) = L were awarded only mark. You were asked to build an NFA for a general regular grammar! Suppose that L has a regular grammar G = (V, Σ, R, S). Then we build an equivalent NFA N as follows. N = (Q, Σ, δ, q, F ), where:. Q = V {D} (where D V ) 2. The alphabet of N is the set of terminals of G. 3. q = S. 4. F = {D} 5. The four regular grammar rule types, depending on how they appear in G, determine the behavior of δ. For all a Σ and A V, δ(a, a) = {B V : A ab R}. For all A V, δ(a, ε) = {B V : A B R}. For all A V and all a Σ such that A a R, δ(a, a) = {D}. Finally, for all A V satisfying A ε R, we have that δ(a, ε) = {D}. (b) [4 marks] Suppose that L has a raluger grammar. Show that L has an NFA. NOTE 4.2: The rule types given are not the rules of a specific grammar to be converted! Students who used the set of the four types of raluger grammar rules (in the above right column) as a particular four-rule

9 Page 9 Sample, Ima () grammar G with L(G) = L were awarded only mark. You were asked to build an NFA for a general raluger grammar! Suppose that L has a raluger grammar G = (V, Σ, R, S). Then we build an equivalent NFA N as follows. N = (Q, Σ, δ, q, F ), where:. Q = V {q } (where q V ) 2. The alphabet of N is the set of terminals of G. 3. The start state q is not in V. 4. F = {S}. 5. The four raluger grammar rule types, depending on how they appear in G, determine the behavior of δ. First, we have that δ(q, ε) = {A V : A ε R}. Next, we have that for all a Σ, δ(q, a) = {A V : A a R}. For all B V, δ(b, ε) = {A V : A B R}. Finally, for all a Σ and B V, we have δ(b, a) = {A V : A Ba R}. (c) [4 marks] Suppose that L has an NFA. Show that L has a regular grammar. Note 4.3: Students who mistook the set of the four types of regular grammar rules (in the above left column) for a particular grammar G with L = L(G) and then defined an NFA for this grammar and then just repeated the rules of G (or some variation thereof) as their solution were awarded only one point. L is just any language that is recognized by some NFA! Proof idea: The grammar we propose is such that its derivation of any terminal string just encodes the computation of the equivalent NFA on that terminal string, from the start state to the final state. Let s suppose the NFA for L is N = (Q, Σ, δ, q, F ). Make a variable V i for each state q i of the NFA. Add the rule V i av j to the grammar if q j δ(q i, a), where a Σ. Add the rule V i V j to the grammar if q j δ(q i, ε). Add the rule V i ε if q i is an accept state of N. Make V the start variable of the grammar, where q is the start state of N. (d) [3 marks] Suppose that L has an NFA. Show that L has a raluger grammar. Note 4.4: Students who mistook the set of the four types of raluger grammar rules (in the above right column) for a particular grammar G with L = L(G) and then defined an NFA for this grammar and then just repeated the rules of G (or some variation thereof) as their solution were awarded only one point. L is just any language that is recognized by some NFA! Proof idea: The grammar we propose in this solution is such that any string generated by it has a derivation that essentially backtracks the NFA computation on that string, from the final state all the way back to the start state. Remember that an NFA computation of a string is just the sequence of states (start-to-final) through which the machine is driven to accept its input. Denote the NFA for L by N = (Q, Σ, δ, Q, F ). For simplicity, let s suppose that the states of Q are capital subscripted letters; i.e., Q = {Q, Q, Q 2,..., Q n } (these states will be variables in our grammar, and by convention we capitalize the variables of a grammar). For N we create an equivalent grammar G = (V, Σ, R, S), where V = Q {S}, the terminals of G are the alphabet characters for N, and R is constructed out of δ as follows. For any Q i, Q j Q, add the rule Q i Q j a to the grammar if Q i δ(q j, a), where a Σ. Add the rule Q i Q j provided Q i δ(q j, ε). For each final state Q f F, add the rule S Q f. Finally, for the start state Q, add the rule Q ε.

10 Page Sample, Ima () Question 5. [5 marks] (a) [7 marks] Using the method from class, produce an NFA for the regular expression (( )). Do not simplify. (Give a transition diagram, and a brief explanation of your method.) q 4 q 6 q q q 2 q 3 q 5 q 7 Explanation: for union, we create a new state that has an -transition to the old start states; for concatenation, we add an -transition from the originally final states of the first NFA to the start state of the second NFA, and for star we add a new final start state, and all final states (other than the new start one) have an -transition to this new start state. Common mistakes include: () not giving an explanation of the method, (2) not following the method, and (3) missing some transitions or states. (b) [8 marks] Using the powerset method from class, produce a DFA for the NFA from the (a) part. Do not simplify the NFA first. (Give a transition diagram, and a brief explanation of your method.), q, q 2,3,4,5 q,,6 q,,7 Explanation: we use the build states as needed construction as described in class, by using -closure not only to find the start state of the converted DFA, but where each transition will go for each of the computed states in this DFA. Common mistakes include: () not giving an explanation of the method, (2) not following the method, and (3) missing some transitions or states. Do not pass Go. Do not collect $2