CSE Theory of Computing: Homework 3 Regexes and DFA/NFAs

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1 CSE Theory of Computing: Homework 3 Regexes and DFA/NFAs Version 1: Fe. 6, 2018 Instructions Unless otherwise specified, all prolems from the ook are from Version 3. When a prolem in the International Edition is different from Version 3, the prolem will e listed as V3:x.yy/IE:x.zz, where x.zz is the euivalent numer. When Version 2 is different, it will e listed as V3:x.yy/V2:x.zz. If either IE or V2 do not have a matching numer, the prolem text will e duplicated. You can prepare your solutions however you like (handwriting, L A TEX, etc.), ut you must sumit them in legile PDF. You can scan written solutions on the printer in the ack of the classroom, or using a smartphone (with a scanner app like CamScanner). It is up to you to ensure that sumissions are legile. REMEMBER THAT IF WE CAN T READ IT OR SCAN IS CUT OFF, YOU DON T GET A GRADE FOR IT. The prolems marked as TEAM may e solved in a collaorative fashion with up to 2 other students. In such cases, your sumission should have the word TEAM at the start of the prolem, followed y the names of your collaorators (must e other students in this class). When such prolems are graded, the first sumission encountered y the grader for the team will e used for a common grade for all identified team memers. Please give every PDF file a uniue filename. If you re making a complete sumission (all prolems), name your PDF file netid-hw#.pdf, where netid is replaced with your NetID and # is the homework numer. If you re sumitting some prolems now and other prolems later, name your file netid-hw pdf, where is replaced with just the prolems you are sumitting now. This may e useful for team sumissions. If you use the same filename twice, only the most recent version will e graded. The time of sumission is the time the most recent file was uploaded. If you use L A TEX and want to draw something like a state diagram, consider using the tikz package. A reference document is on the wesite under Assignments. You may also find the wesite a useful tool for drawing state diagrams via drop and drag. It will output oth.png image files and latex in the tikz format. Sumit your PDF file in Sakai to the appropriate directory. Don t forget to click the Sumit (or Resumit) utton! Practice Prolems These prolems are from the ook, and most have solutions listed for them. They are listed here for you to practice on as needed and any answers you generate should not e sumitted. You are free to discuss these with others, ut you are not allowed to post solutions to any pulic forum Regex to NFA Regex to strings a DFA to regex regex to NFA 1

2 a,c Pumping Lemma a CLosure Book Exercises These prolems are found in the text ook and are to e answered and sumitted y each student. If they are not marked as TEAM, you are to solve them individually. If they are marked as TEAM you may sumit the same answer as the others in your team. In any case, use of solution manuals from any source or shared solutions is a violation of the ND Honor Code. You are also not allowed to show your solutions to another student not part of your TEAM. 1. (5pt) 1.12 Language to DFA and regular expression We are given the language: D = {w w contains an even numer of a s and an odd numer of s and does not contain the string a}. We are asked to give a DFA with 5 states that recognizes D and a regular expression that generates D. Right away we can note that we will never e ale to have an a followed y a ecause the sustring a will never e accepted. Using the formal definition of a DFA: (Q, Σ, δ, 0, F ) We can make the following DFA: Q = { 1, 2, 3, 4, 5 } Σ = {a, } 0 = 1 F = { 4 } δ = state a Based on this DFA, we can make the regex: ()*(aa)* 2. (TEAM-5pt) Book 1.19a. Regex to NFA 3. (5pt) Book Regex to NFA 4. (TEAM-5pt) V3:1-41;V2:1-41;IE: Show Closure. Hint Rememer that showing a language is regular is euivalent to showing how to construct a DFA/NFA for any language constructed from operator applied to a regular language. Assume M 1 = (Q 1, Σ, δ 1, s 1, F 1 ) accepts A and M 2 = (Q 2, Σ, δ 2, s 2, F 2 ) accepts B. We want to uild M that alternates etween M 1 and M 2. To do this define Q = Q 1 xq 2 x{1, 2} where the third component says are we looking at A or B. So states are now triples. Now δ(( 1, 2, 1), a) = (δ( 1, a), 2, 2) and δ(( 1, 2, 2), a) = ( 1, δ( 2, a), 1) s = (s 1, s 2, 1) F = {( 1, 2, 1)} where 1 is in F 1 and 2 is in F 2 5. (10pt - Part a only is TEAM) V3:1.46a,c; V2:1-46a,c;IE:11.51a,c. Use Pumping Lemma Part a. Try string 0 p 10 p. x must e 0 a for some a etween 0 and p 1, and y is 0 for some from 1 to p. Thus xy i z for i = 0 is 0 ( p )1o p where 1, which is not in L. Also, you could use i = 2 which yields 0 ( p + )10 p = 0 p 0 10 p, which again fails (0 1 is not a valid 1 m Part c. If a language is regular then so is its complement, and vice versa. So let s look at {w w is a palindrone }. Try the string 0 p 10 p. It does not pump. Non-ook Prolems The following prolems are not found in the text ook. If they are not marked as TEAM, you are to solve them individually. If they are marked as TEAM you may sumit the same answer as the 2

3 others in your team. Use of any resource you used other than the text ook or class notes must e cited. You are also not allowed to show your solutions to another student. 7. (TEAM - 5pt) Assume regular languages L 1 and L 2 are accepted y DFAs M 1 and M 2, where M 1 = (S, Σ, δ 1, s 1, F 1 ), S = {s 1,..., s n }, and M 2 = (R, Σ, δ 2, r 1, F 2 )., R = {r 1,..., r m }. Define L 3 = L 1 L 2 as {w w is in L 1 ut not in L 2 }. Show y construction of an NFA that L 3 is regular. Descrie informally how your construction works. We know that L 1 is regular so there exists a machine M 1 = (Q 1, Σ 1, δ 1, 01, F 1 ), and that L 2 is regular so there exists a machine M 2 = (Q 2, Σ 2, δ 2, 02, F 2 ). We define M 3 to e the product of M 1 and M 2 where the final states of M 3 are those states in M 3, ( i, j ) with i Q 1, j Q 2 s.t. ( i, j ) F 3 if i F 1 and j / F 2. What we mean y the product of two machines is that we take Q 1 Q 2 to e the states of the new machine. The start state is the tuple which has oth the start state of M 1 and the start state of M 2. The alphaet is simply the union of the two alphaets of the original machine, and the delta function does element-wise transition, similar to how the NFA to DFA construction works. Informally, this works y simultaneously checking oth machines since we have their product. When we are done, we simply check if we are in a final state of M 1 paired with not a final state of M 2. If oth were accepting, we reject. If neither were accepting, we still reject. And finally, if it was only accepted y M 2 we also reject. We get precisely the strings we desire. Remark Points No proof y construction (5pt) Show using the Pumping Lemma that the language L = {(a) n (cd) n n 0} is not regular. (Ryan Mackey) I will prove y contradiction that language L is not regular. Assume that language L is regular. Thus, L can e represented y DFA M where M = (Q, Σ, s, δ, F ). The amount of states in M is finite, so assume that ased on the pigeonhole principle and Q that the pumping length is p. Now consider a word x L of length greater than p. x can e written in the form x = uvw where uv p and v 1. Then for i Z, uv i L, according to the definition of regular languages. But if x = (a) p (cd) p, u and v will oth lie in the first half of the word (the a region). Thus, with an i = 2, x = (a) (p+v) (cd) p. x / L. This gives a contradiction and proves that language L is not regular. QED 9. (Team - 5 pt) For the following NFA compute the regular expression accepted y converting it to a GNFA and then using the CONVERT algorithm of from Lemma 1.60 on it. Show work. Feel free to use variales like R i to stand for intermediate expressions on an edge (with R i written out elow), ut show entire final answer. You don t need to simplify. a, p a, a, r s Figure 1: State diagram of NFA for uestion 6 Any DFA/NFA can e converted to a GNFA y adding a start state s and a new accepting state a and y making appropriate transitions such that a is the only accepting state. Then, the procedure to convert a k-state GNFA (G) to regular expression is given elow. CONVERT(G): 1. Let k e the numer of states of G 2. If k = 2, then G must consist of only a start state and an accepting state, with a single arrow connecting the states laeled with regular expression R. Return R. 3. If k > 2, we select any state rip different from s and a and let G e the GNFA (Q, Σ, s, δ, a ), where Q = Q { rip }, 3

4 and for any i Q { a } and any j Q { s }, let Step 1: Step 2: Step 3: Step 4: δ ( i, j ) = (R 1 )(R 2 ) (R 3 ) (R 4 ), for R 1 = δ( i, rip ), R 2 = δ( rip, rip ), R 3 = δ( rip, j ), R 4 = δ( i, j ). 4. Compute CONVERT(G ) and return the value. Starting from the given NFA N : (Q, Σ, 0, δ, F ), where Q = {p,, r, s}, Σ = {a, }, 0 = p, δ can e inferred from the state diagram in figure 1, F = {r, s}, we make a six-state GNFA G : (Q, Σ, 0, δ, F ) y adding a start ( s ) and an accepting state ( a ). Therefore, Q = { s, p,, r, s, a }, 0 = s, δ can e inferred from the figure 2, and F = { a }. We also convert the transitions to formal regular expressions. Since the G has more than 2 states, we need to remove a state rip which is different from s and a. Let rip = p. Thus, as per the CONVERT algorithm, we have i = s and j =. R 1 = δ( s, p) =, R 2 = δ(p, p) =, R 3 = δ(p, ) = and R 4 = δ( s, ) =. Thus, δ( s, ) = ()() () ( ) = () (). The state diagram of the reduced GNFA is given in figure 3. Since the G has more than 2 states, we need to remove a state rip which is different from s and a. Let rip =. Thus, as per the CONVERT algorithm, we have i = s and j = r. R 1 = δ( s, ) = () (), R 2 = δ(, ) =, R 3 = δ(, r) = and R 4 = δ( s, r) =. Thus, δ( s, r) = (() ())() () ( ) = (() ())(). The state diagram of the reduced GNFA is given in figure 4. Since the G has more than 2 states, we need to remove a state rip which is different from s and a. Let rip = s. Thus, as per the CONVERT algorithm, we have i = r and j = a. R 1 = δ(r, s) = (), R 2 = δ(s, s) =, R 3 = δ(s, a ) = and R 4 = δ(r, a ) =. Thus, δ(r, a ) = ()() () () = (). The state diagram of the reduced GNFA is given in figure 5. Step 5: Since the G has more than 2 states, we need to remove a state rip which is different from s and a. Let rip = r. Thus, as per the CONVERT algorithm, we have i = s and j = a. R 1 = δ( s, r) = (a ) ()(a ), R 2 = δ(r, r) =, R 3 = δ(r, a ) = (a ) and R 4 = δ( s, a ) =. Thus, δ( s, a ) = () ()()() (() ) ( ) = () ()()(() ). The state diagram of the reduced GNFA is given in figure 6. Step 6: Since the G has 2 states, the CONVERT algorithm terminates and returns δ( s, a ) = (a ) ()()(() ) as the reuired regular expression. Therefore, the reuired regular expression is () ()()(() ), representing the strings over the alphaet {a, } having a in either the second last or the third last position. s p r s a Figure 2: The GNFA G in step 1 s () () r s a - Figure 3: The GNFA G in step 2 4

5 () ()() s r s a Figure 4: The GNFA G in step 3 () ()() () s r a Figure 5: The GNFA G in step 4 s () ()()(() ) a Figure 6: The final GNFA 10. (5pt) Do the same as aove for the NFA in figure 7, i.e. compute the regex. Any DFA/NFA can e converted to a GNFA y adding a start state s and a new accepting state a and y making appropriate transitions such that a is the only accepting state. Then, the procedure to convert a k-state GNFA (G) to regular expression is given elow. CONVERT(G): 1. Let k e the numer of states of G 2. If k = 2, then G must consist of only a start state and an accepting state, with a single arrow connecting the states laeled with regular expression R. Return R. 3. If k > 2, we select any state rip different from s and a and let G e the GNFA (Q, Σ, s, δ, a ), where Q = Q { rip }, and for any i Q { a } and any j Q { s }, let δ ( i, j ) = (R 1 )(R 2 ) (R 3 ) (R 4 ), for R 1 = δ( i, rip ), R 2 = δ( rip, rip ), R 3 = δ( rip, j ), R 4 = δ( i, j ). 4. Compute CONVERT(G ) and return the value. Step 1: Starting from the given NFA N : (Q, Σ, 0, δ, F ), where Q = {p,, r}, Σ = {a, }, 0 = p, δ can e inferred from the state diagram in figure 7, F = {r}, we make a six-state GNFA G : (Q, Σ, 0, δ, F ) y adding a start ( s ) and an accepting state ( a ). Therefore, Q = { s, p,, r, a }, 0 = s, δ can e inferred from the figure 8, and F = { a }. We also convert the transitions to formal regular expressions. Step 2: Step 3: Since the G has more than 2 states, we need to remove a state rip which is different from s and a. Let rip = r. Thus, as per the CONVERT algorithm, we have i = {, p} and j = a. For i =, R 1 = δ(, r) =, R 2 = δ(r, r) =, R 3 = δ(r, a ) = and R 4 = δ(, a ) =. Thus, δ(, a ) = ()( ) () ( ) =. For i = p, R 1 = δ(p, r) = (c ), R 2 = δ(r, r) =, R 3 = δ(r, a ) = and R 4 = δ(p, a ) =. Thus, δ(p, a ) = (c )( ) () ( ) = c. The state diagram of the reduced GNFA is given in figure 9. Since the G has more than 2 states, we need to remove a state rip which is different from s and a. Let rip =. Thus, as per the CONVERT algorithm, we have i = p and j = {p, a }. 5

6 ,,c p r a,c c, Figure 7: NFA for prolem 7 Step 4: For j = p, R 1 = δ(p, ) = ( ), R 2 = δ(, ) = ( c), R 3 = δ(, p) = (a c) and R 4 = δ(p, p) =. Thus, δ(p, p) = ( )( c) (a c) ( ) = ( )( c) (a c). For j = a, R 1 = δ(p, ) = ( ), R 2 = δ(, ) = ( c), R 3 = δ(, a ) =, R 4 = δ(p, a ) = (c ). Thus, δ(p, a ) = ( )( c) () (c ). The state diagram of the reduced GNFA is given in figure 10. Since the G has more than 2 states, we need to remove a state rip which is different from s and a. Let rip = p. Thus, as per the CONVERT algorithm, we have i = s and j = a. Therefore, R 1 = δ( s, p) =, R 2 = δ(p, p) = ( )( c) (a c), R 3 = δ(p, a ) = ( )( c) () (c ), R 4 = δ( s, a ) =. Thus, δ( s, a ) = ()[( )( c) (a c)] ( )( c) () (c ) = [( )( c) (a c)] ( )( c) () (c ). The state diagram of the reduced GNFA is given in figure 11. Therefore, the reuired regular expression is [( )( c) (a c)] ( )( c) () (c ) c s p r a a c c Figure 8: The GNFA G in step 1 c s p a c a c Figure 9: The GNFA G in step 2 Reason Points Correct small rip 1 Two medium large rips 2 2 = 4 Transitions not regex 0.5 Not following CONVERT 5 Not adding the dummy start and end states 0.5 Tale 1: Ruric for prolem 7 - Note: the errors in the GNFA propagate through the steps. They will e penalized accordingly. 6

7 ( )( c) (a c) ( )( c) () (c ) s p a Figure 10: The GNFA G in step 3 s [( )( c) (a c)] ( )( c) () (c ) a Figure 11: The GNFA G in step 4 7