The assignment is not difficult, but quite labour consuming. Do not wait until the very last day.

Transcription

1 CAS 705 CAS 705. Sample solutions to the assignment 1 (many questions have more than one solutions). Total of this assignment is 129 pts. Each assignment is worth 25%. The assignment is not difficult, but quite labour consuming. Do not wait until the very last day. 1.[5] Let M be the following non-deterministic automaton with -moves: Q = {q 0, q 1, q 2 }, = {a,b,c}, q 0 is the initial state, F = {q 1,q 2 }, ä a b c q 0 {q 0 } Ø {q 1 } {q 2 } q 1 Ø Ø {q 1 } Ø q 2 Ø {q 2 } Ø Ø a.[1] Compute CL({q i }) for i=0,1,2. ( CL(X) is -closure of X). b.[1] Give the state diagram (graphical representation) of M. c.[1] Give a state diagram (graphical representation) of a deterministic automaton that is equivalent to M. d.[2] Give a regular expression for L(M). a. CL({q 0 }) = {q 0,q 1 }, CL({q 1 }) = {q 1 }, CL({q 2 }) = {q 2 }. b. 1

2 c. d. a*cc* a*b* 2.[6] Let M be the following non-deterministic automaton with -moves: Q = {q 0, q 1, q 2, q 3 }, = {a,b}, q 0 is the initial state, F = {q 1 }, ä a b q 0 {q 1 } {q 2 } {q 3 } q 1 Ø {q 1,q 2 } Ø q 2 Ø Ø {q 3 } q 3 {q 1,q 3 } {q 2 } Ø a.[1] Compute CL({q i }) for i=0,1,2,3. ( CL(X) is -closure of X). b.[1] Give the state diagram (graphical representation) of M. c.[2] Give a state diagram (graphical representation) of a deterministic automaton that is equivalent to M. d.[2] Give a regular expression for L(M). 2

3 a. CL({q 0 }) = {q 0,q 3 }, CL({q 1 }) = {q 1 }, CL({q 2 }) = {q 2,q 3 }, CL({q 3 }) = {q 3 }. b. c. Note that after standard removal of -moves, the automaton has ä (q 2,a) = ä (q 2,a) = Ø, so we can remove the state q 2 and the language accepted remains the same. Then the standard procedure (restricted to the states that can be reached from the initial state lead to the following solution: d.[4] a(a b)* bb*a(a b)* = (a bb*a)(a b)* 3

4 3.[4] For every string x = a 1 a 2...a n, let x R = a n...a 2 a 1, and R =. a.[1] Provide the formal definition of x R by structural induction. Prove the following: b.[2] For all u,v Ó +, (uv) R =v R u R, c.[1] For all w Ó *, (w R ) R = w. a. R =, for each x Ó *, and a Ó, (xa) R = a x R. b. By induction on the length of v. Clearly (u ) R = u R = ( u) R. From the definition of R operator we also have that for each a Ó, (ua) R = au R = a R u R. Suppose that (uv) R = v R u R. Consider uva. We have (uva) R = a(uv) R = av R u R = (va) R u R, which proves the case. c. By induction on the length of w. Clearly ( R ) R =. Suppose that (w R ) R = w. Consider wa. We have ((wa) R ) R = (aw R ) R = (w R ) R a = wa, which ends the proof. 4.[3] Prove that (L 1 L 2 ) R = L 2R L 1 R for all languages L 1 and L 2. iff means if and only if, we need the results of question 3. x (L 1 L 2 ) R iff x R L 1 L 2 iff ( x R = uv and u L 1, v L 2 ) iff ( x R = uv and u R L 1R,v R L 2R ) iff ( x R = uv and v R u R L 2R L 1R ) iff (by 6b) ( x = (x R ) R = (uv) R and v R u R L 2R L 1R ) iff (by 6a) ( x = v R u R and v R u R L 2R L 1R ) iff x L 2R L 1 R, which ends the proof. 5.[5] Construct a deterministic finite automaton that accept the following language: L = the set of all strings over {1,2,3} the sum of whose elements is divided by 6. You may start with a non-deterministic solution first and then just translate it into the deterministic one. Provide some justification to your solution, could be informal. The states should be numbers {0,1,...,5}, each represent a number modulo 6. In other words, 0 represents {0,6,12,18,...}, 1 represents {1,7,13,19,...}, 2 represents {2,8,14,20,...,}, 3 represents {3,9,15,21,...}, 4 represents {4,10,16,22,...}, 5 represents {5,11,17,23,...}. The transition function represents addition modulo 6, i.e. 4

5 ä(k,l)= (k+l) mod 6, where k=0,...,5, l=1,2,3. All together: M = ({0,1,2,3,4,5}, {1,2,3}, ä, 0, {0}), i.e. initial state is zero, and F={0}, ä [10] Construct a deterministic finite automaton that accept the following language: L = the set of strings over {a,b,c} in which every substring of length four has exactly one b. You may start with a non-deterministic solution first and then just translate it into the deterministic one. Provide some justification to your solution, could be informal. An automaton analyses the string and once a substring of four without b or with more than one b is found, automaton moves to the garbage/sink state. The problem is underspecified since it does not say what to do with the strings of length smaller than 4, so both solutions (they are accepted or not) are o.k. So Ó = {a,b,c}, Q = {s 0,[b],[x],[bx],[xb],[xx],[xxx],[bxx],[xbx],[xxb],[xxxx],[bxxx],[xbxx],[xxbx],[xxxb]}, and s 0 is an initial state, F = { [bxxx], [xbxx], [xxbx], [xxxb]} - if we assume that all strings of length smaller than 4 are rejected, or F = {s 0,[b],[x],[bx],[xb],[xx],[bxx],[xbx],[xxb],[xxx],[bxxx], [xbxx], [xxbx], [xxxb]} - if we assume that all strings of length smaller than 4 are accepted. In the second case only the state [xxxx] is non-final, in both cases it is a sink state. The states are interpreted as follows, for {[b],[x],[bx],[xb],[xx],[xxx],[bxx],[xbx],[xxb]} it means that the all paths from s 0 to s are described by s, for instance ä(s 0,w)=[xxb] means that w=xxb, where x is either a or c, i.e. w {aab,acb,cab,ccb}. For the states from {[bxxx],[xbxx],[xxbx],[xxxb]}, for instance xxbx means that the last four symbols satisfy the pattern xxbx, where x is either a or c. The state [xxxx] means either that the last 4 symbols do not contain b, or that they contain more than one b. 5

6 The function ä is now quite obvious, for instance ä([xxbx], a) = ä([xxbx],c) = [xbxx], ä([xxbx,b) = [xxxx] (as we assume that [xxxx] also represent [xbbx], see above). Formally: ä a b c s 0 [x] [b] [x] [b] [bx] [xxxx] [xb] [x] [xx] [xb] [xx] [bx] [bxx] [xxxx] [bxx] [xx] [xxx] [xxb] [xxx] [xb] [xbx] [xxxx] [xbx] [bxx] [bxxx] [xxxx] [bxxx] [xbx] [xbxx] [xxxx] [xbxx] [xxb] [xxbx] [xxxx] [xxbx] [xxx] [xxxx] [xxxb] [xxxx] [bxxx] [xxxx] [xxxb] [xxxx] [xbxx] [bxxx] [xxxx] [bxxx] [xxbx] [xbxx] [xxxx] [xbxx] [xxxb] [xxbx] [xxxx] [xxbx] [xxxx] [xxxx] [xxxx] [xxxx] 7.[4] Let L be a regular language that does not contain. Show that there exists an nondeterministic finite state automaton without -moves and with a single final state that accept L. Is the assumption that in not in L essential? Is this also true if we replace non-deterministic by deterministic? [3] Let M be any non-deterministic automaton such that L=L(M). Note that any deterministic automaton can be treated as a special case of a non-deterministic automaton. Intuition is given by the figure below: 6

7 Formal proof. Let M=(Q,,ä,q 0,F). Define M =(Q {q F },,ä,q 0, {q F }), where q F Q. Let S be the following set of states: q S iff a. ä(q,a) F Ø. In other words, from S, one may reach a final state in one step. The mapping ä is defined as follows: for all q Q, all a : ä (q,a) = if q S then ä(q,a) else ä(q,a) {q F }, and ä (q F,a)=Ø. Let x, i.e. x=ya. Hence: x=ya L(M) ä(q 0,ya) F Ø ä(q 0,y) S Ø ä (q 0,y) S Ø q F ä (q 0,ya) x=ya L(M ). [1] Yes, the assumption that the language does not contain the empty string is essential. The results is false if L(M). Consider the automaton below: It accepts/generates the language {,a}. The initial state must be a final state since is accepted. If this is the only final state, and a is accepted than there must be an arrow from this state to itself labelled by a. But then every a k is also accepted, a contradiction. 7

8 8.[6] a.[2] How many different deterministic automata with four states s 0,s 1,s 2,s 3 over the alphabet {a,b}, where s 0 is always the start state, do exist? Prove your answer. b.[2] c.[2] How many different non-deterministic automata (without -moves) with four states s 0,s 1,s 2,s 3 over the alphabet {a,b}, where s 0 is always the start state, do exist? Prove your answer. Note that according to the definition, every deterministic automaton is a special case of a non-deterministic automaton. How many different non-deterministic automata with -moves with four states s 0,s 1,s 2,s 3 over the alphabet {a,b}, where s 0 is always the start state, do exist? Prove your answer. Note that according to the definition, every deterministic automaton is a special case of a non-deterministic automaton, and every nodeterministic automaton without -moves is a special case of a non-deterministic automaton with -moves. a Let [Q Ó Q] be the set of all functions from Q Ó to Q, i.e. [Q Ó Q] = { f f:q Ó Q}. If Q =4, Ó =2, then Q Ó = 8 and [Q Ó Q] = Q Q Ó = 4 8 = Hence we have different ä functions. The number of possible sets of final states is the same as the set all subsets of Q, i.e. 2 Q = 2 4 = 16. Finally: = 1,048,456. b. [Q Ó 2 Q ] = 2 Q Q Ó = 16 8, so the number of automata is =16 9. c. [Q (Ó { } 2 Q ] = 2 Q Q (Ó { } = 16 12, so the number of automata is = [7] For integers n,d 1, consider a deterministic finite automaton M n,d = (Q, Ó, ä, s 0, F), where Q = {s 0, s 1, s 2,..., s n-1 }, Ó = {a 0, a 1, a 2,..., a d-1 }, ä(s i, a k ) = s (di+k) mod n, and F = {s 1 }. a.[1] Draw the transition diagram of M n,d with n=7 and d=2. b.[1] Suppose n=7, d=2, a 0 =0, a 1 =1. Find ä(s 3,0101) and ä(s 1,11010). c.[1] Suppose n=7, d=2, a 0 =0, a 1 =1. Find binary string x and y such that ä(s 0, x) = s 5 and ä(s 0, y) = s 6. d.[4] Show that for any state s j Q, there exists a string x Ó * such that ä(s 0, x) = s j. 8

9 a. One just have to transform the below table into a graph: ä b. From either a table or a graph we conclude: ä(3,0101)=4, ä(1,11010)=2. c. The solution is not unique but either form a table or a graph, one may find that the shortest x and y are: x = 101, and y =110. d. Note that in c above we have ä(0,101) = 5, and 101 is 5 in binary notation, while ä(0,110) = 6, and 110 is 6 in binary notation. The simplest proof is merely based on this observation. Each number can be represented using different representations. In general we may say the number m is represented by a sequence a k...a 1 a 0 where a i = 0,1,...,d-1, if m = a k d k a 1 d + a 0, so some k. We shall denote by (m) d its d-representation. For instance (5) 10 = 5, (5) 2 = 101, (5) 3 = 12. We prove that for each i Q, ä(0, (i) d ) = i, so it suffices to take x = (i) d. Let m be a number and let the string w be its d-representation, i.e. (m) d = w and w {0,1,...,d-1}*. Let a {0,1,...,d-1}. What is the value of xa treated as a number in d- representation? Let t be a number such that (t) d = xa. Note that exactly from the definition of d-representation we have: t = dm+a. Now, let i Q, and (i) d = a k...a 1 a 0. Note that k<i, and (remember Horner scheme?) i = a k d k a 1 d + a 0 = (... ((a k d+a k-1 )d+a k-2 )d a 1 )d+a 0 d) = ä(0,a k...a 1 a 0 ), which ends the proof. 9

10 10.[6] Prove that for every language L: a.[2] L * = (L * ) * b.[2] L + = (L + ) + c.[2] Is (L * ) + = (L + ) * for all sets? Prove your answer. a. There are many proofs. Probably the most simple one looks as follows: Let x be a sequence over the alphabet Ó. Suppose that x is a concatenation of x 1,...,x k, some k, i.e. x = x 1 x 2,...,x k. The strings x i, i=1,...,k, are called factors of x. The notion of a factor simplifies the proof a lot. Every string in (L * ) * is made up of factors from L *. Every factor from L * is made up of factors from L. Therefore, every string in (L * ) * is made up of factors from L. Therefore, every string in (L * ) * is also a string in L *. We can write this as (L * ) * L *. On the other hand we have L L *, since in L * we can choose as a string any factor from L. Hence L * (L * ) *, i.e. L * = (L * ) *. b. Almost identically, the only difference is that the length of a factor is in this case at least one, while for the star factor of length zero were allowed. c. Yes, (L + ) * = (L * ) +. (L + ) * (L * ) * = L *. L L +, so L * (L + ) *, i.e. (L + ) * = L *. On the other hand (L * ) + (L * ) * = L *, L * (L * ) +, so (L * ) + = L *, hence (L + ) * = (L * ) + = L *. There are many other proofs as well. 11.[10]Let L be any language. Define even(w) as the string obtained by extracting from w the letters in even-numbered positions; that is if w = a 1 a 2 a 3 a 4..., then even(w) = a 2 a Corresponding to this, we can define a language even(l) = { even(x) x L }. Prove that if L is regular, so is even(l). Hint. Let M be a deterministic automaton such that L(M) = L. Transform M into an automaton (not necessarily deterministic) that accepts even(l). The trick is to use a Deterministic Finite Automaton for L and modify it so that it remembers if it had read an even or an odd number of symbols. This can be done by doubling the number of states and adding O or E to the labels. For example, if part of the DFA is 10

11 its equivalent becomes Now replace all transitions from an E-state to an O-state by -transitions. The new automaton (non-deterministic with -moves) accepts even(l). Let us now do it formally. Let M=(Q,,ä,s 0,F) be a DFA such that L=L(M). Define M 1 =(Q 1,,ä 1,s 01,F 1 ), where Q 1 = { s E s Q } { s O s Q }, s 01 = s 0E, F 1 = { s E s F } { s O s F }, and ä 1 (s E,a) = q O if and only if ä(s,a)=q, and ä 1 (s O,a) = q E if and only if ä(s,a)=q, for each s Q, a Ó. Clearly L(M 1 )=L(M), however to prove it formally one may use induction. Let M 2 =(Q 1,,ä 2,s 01,F 1 ) be the following non-deterministic automaton with -moves: ä 2 (s O,a) = {ä 1 (s O,a)}, for each s Q, a Ó, and ä 2 (s E, ) = {ä 1 (s E,a)}, for each s Q, a Ó. Again L(M 2 )=L(M 1 ), however a formal proof requires using induction. 12.[3] Let M = (Q, Ó, ä, s 0, F) be a deterministic finite automaton. Give a deterministic finite state automaton M such that L(M ) = L(M) - { }. Let M=(Q,,ä,s 0,F) be a DFA such that L=L(M). If L(M), then M =M. If L(M) then s 0 F. Define M = (Q {q 0 },,ä 1,q 0,F), where q 0 Q, and ä 1 is defined as follows: ä 1 (q 0,a) = ä(s 0,a) for each a Ó, ä 1 (s,a) = ä(s,a) for each s Q. Clearly L(M )=L(M)-{ }. 11

12 13.[3] Let L = {a,b} and L 1 * = L *. Prove that L L 1. First we prove that a L 1. Suppose that a L 1. This means that a L 1*, a contradiction since a L L * = L 1*. Identically we prove b L 1. Thus L L 1. An immediate extension of this result (identical proof!) is: for every alphabet Ó and any language L, if L * = Ó *, then Ó L. 14.[10] Give deterministic and non-deterministic finite automata accepting the following languages over the alphabet {a,b}. Give non-deterministic only if it either has less states or is intuitively simpler. a.[5] The set of all strings of length five or more in which the fourth symbol from the right end is different from the leftmost symbol. b.[5] The set of all strings in which the leftmost two symbols and the rightmost two symbols are identical. a. NFA 12

13 DFA: q ini is initial state. F={ q aaaaa, q aaaba, q aabaa, q aabba, q aaaab, q aaabb, q aabab, q aabbb, q bbaaa, q bbaba, q bbbaa, q bbbba, q bbaab, q bbabb, q bbbab, q bbbbb } a b a b a b a b q ini q a q b q aaaa q aaaaa q aaaab q aaaaa q aaaaa q aaaabb q aaaab q aaaba q aaabb q a q aa q ab q aaab q aaaba q aaabb q aaaba q aabaa q aababb q aaabb q aabba q aabbb q b q ba q bb q aaba q aabaa q aabab q aabaa q abaaa q abaab q aabab q ababa q ababb q aa q aaa q aab q aabb q aabba q aabbb q aabba q abbaa q abbab q aabbb q abbba q abbbb q ab q aba q abb q abaa q abaaa q abaab q abaaa q aaaaa q aaaab q abaab q aaaba q aaabb q ba q baa q bab q abab q ababa q ababb q ababa q aabaa q aabab q ababb q aabba q aabbb q bb q bba q bbb q abba q abbaa q abbab q abbaa q abaaa q abaab q abbab q ababa q ababb q aaa q aaaa q aaab q abbb q abbba q abbbb q abbba q abbaa q abbab q abbbb q abbba q abbbb q aab q aaba q aabb q baaa q baaaa q baaab q baaaa q baaaa q baaab q baaab q baaba q baabb q aba q abaa q abab q baab q baaba q baabb q baaba q babaa q babab q baabb q babba q babbb q abb q abba q abbb q baba q babaa q babab q babaa q bbaaa q bbaab q babab q bbaba q bbabb q baa q baaa q baab q babb q babba q babbb q babba q bbbaa q bbbab q babbb q bbbba q bbbbb q bab q baba q babb q bbaa q bbaaa q bbaab q bbaaa q baaaa q baaab q bbaab q baaba q baabb q bba q bbaa q bbab q bbab q bbaba q bbabb q bbaba q babaa q babab q bbabb q babba q babbb q bbb q bbba q bbbb q bbba q bbbaa q bbbab q bbbaa q bbaaa q bbaab q bbbab q bbaba q bbabb q bbbb q bbbba q bbbbb q bbbba q bbbaa q bbbab q bbbbb q bbbba q bbbbb b. DFA q ini is initial state. F={ q aa, q ab, q ba, q bb, q aaa, q bbb, q aaaa, q abab, q baba, q bbbb } a b a b q ini q a q b q aaaa q aaaa q aaab q a q aa q ab q aaab q aaba q aabb q b q ba q bb q aaba q aaaa q aaab q aa q aaa q aab q aabb q aaba q aabb q ab q aba q abb q abaa q abaa q abab q ba q baa q bab q abab q abba q abbb q bb q bba q bbb q abba q abaa q abab q aaa q aaaa q aaab q abbb q abba q abbb q aab q aaba q aabb q baaa q baaa q baab q aba q abaa q abab q baab q baba q babb q abb q abba q abbb q baba q baaa q baab q baa q baaa q baab q babb q baba q babb q bab q baba q babb q bbaa q bbaa q bbab q bba q bbaa q bbab q bbab q bbba q bbbb q bbb q bbba q bbbb q bbba q bbaa q bbab q bbbb q bbba q bbbb 13

14 Note, DFAs for part (a) and (b) above may not be the minimal ones. However, they give clearer intuitive meaning. 15.[10] Consider the automaton below. Construct an equivalent regular expression using the algorithm discussed in class. 14

15 16.[6] Consider the following regular expression: a*b* (a*bb ab*)a*b. a.[2] Using a standard construction construct an equivalent non-deterministic finite state automaton with -moves. b.[2] Transform the automaton from 3a into an equivalent non-deterministic finite state automaton without -moves. c. [2] Transform the automaton from 3b into an equivalent deterministic finite state automaton. Solutions: 15

16 a. Q={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22}, initial_state=1, F={22}, ={a,b}, and is defined below a b 1 {2,8} 2 {3} {4} 3 {2} 4 {5} {22} 5 {4} 6 {7} {9} 7 {6} 8 {6,11} 9 {10} 10 {13} 11 {12} 12 {14} 13 {15} 14 {16} {17} 15 {17} 16 {14} 17 {18} 18 {19} {20} 19 {18} 20 {21} 21 {22} 22 b. Q={1,2,3,4,5,6,7,8,9}, initial_state=1, F={1,2,4,6,9}, and is defined below a b 1 {2,7} {3} 2 {2} {3,4} 3 {5} 4 {4} 5 {5} {6} 6 7 {8} {7,9} 8 {8} {9} 9 16

17 c. Q={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19}, initial_state=1, F={1,2,4,5,6,9,10,12,13,14,15,17,18}, ={a,b}, garbage=19 and is defined below a b [5] Consider the automaton below. Construct an equivalent regular expression without using any formal technique, by just analysing the structure of automaton. Provide some arguments why your reasoning is correct. 17

18 There are many solutions. Here is a possible one. Let s denote the initial state, p denote the second final state, and q denote the only non-final state, Q be a regular expression that corresponds to the automaton derived by erasing the state s and making q initial. Clearly Q = (a ba)*b. Let S be a regular expression describing strings starting from s and ending at s, while P be a the language built from all regular expression describing the language built from all strings starting from s and ending at p. Then the language L = S P, S = (a bqb)*. Let T be a regular expression describing the language built from all the strings that start with s and ending at p but not going thorough s. Clearly T= a*b(a ba)*b and P = S*P. Hence L = (a b(a ba)*bb)* (a b(a ba)*bb)*a*b(a ba)*b. Again, this is not the only solution. 18.[8] Consider the following regular expression: (a b*)(((ba)* bb*b)* a)*b. a.[2] Using a standard construction construct an equivalent non-deterministic finite state automaton with -moves. b.[3] Transform the automaton from 18a into an equivalent non-deterministic finite state automaton without -moves. c.[3] Transform the automaton from 18b into an equivalent deterministic finite state automaton. First note that (a b*)(((ba)* bb*b)* a)*b = (a b*)(ba bb*b a)*b since (r* s)*=(r s)* This simplifies automata a little. 18

19 a. Q={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22}, initial_state=1, F={15}, ={a,b}, and is defined below a b 1 {2,3} 2 {5} 3 {4} {6} 4 {3} 5 {6} 6 {7} 7 {8,9,10,11} 8 {12} 9 {13} 10 {14} 11 {15} 12 {16} 13 {17} 14 {22} {19} 17 {18} {20} 18 {17} 19 {22} 20 {21} 21 {22} 22 b. Q={1,2,3,4,5,6}, initial_state=1, F={3}, and is defined below a b 1 {2,5} {2,3,4,6} 2 {5} {2,3,4,6} 3 4 {5} 5 {5} {3,4,6} 6 {5,6} 19

20 c. Q={1,2,3,4,5,6,7,8}, initial_state=1, F={3,5,6,8}, and is defined below a b [5] Consider the following regular expression (abab)* (aaa* b)*. Find an equivalent deterministic finite state automaton using whatever reasoning you wish. Provide some justification or proof. A possible solution is the following. By observing that aaa* = aa*a, it is rather easy to built deterministic automata defining (abab)* and (aa*a b)*. Then we can construct a non-deterministic automaton with -moves accepting the union of two languages. In principle we only need to add a new initial state and connect it with old two initial states by labelled arrows. Erasing these two labelled arrows is also easy, it does not increase the number of states. Then we have to get rid of non-determinism. A possible DFA looks as follows. Q={1,2,3,4,5,6,7,8,9,10,11}, initial_state=1, F={1,5,7,8,9}, garbage-state=11 and is defined below a b

21 20.[6] Let r and s be regular expressions. a.[3] Prove that : (r* s)* = ((r*s*)* r)* where the equality = above is understood as L((r* s)*) = L(((r*s*)* r)*). b.[3] Show that there are such s and r that : r*(r s)* (r s)*. a. There are many solutions. Below there is a possible one. We will use the following rules: (r s)*=(r*s*)*, (r*)*=r* for every regular expressions r and s. (r* s)* = ((r*)*s*)* = (r*s*)* = (r s)* ((r*s*)* r)*=(r*s* r)*=((r*s*)*s*)*=((r s)*r*)*=(r s r)*=(r s)* b. It cannot be shown since r*(r s)*=(r s)*. Clearly r*(r s)* (r s)*. But (r s)* = (r s)* r*(r s)*, as r*. 21.[5] Give a deterministic finite state automaton that accepts the language: L(ab*a*) L(a*b*a) Method 1. L 1 L 2 = S* - (( * - L 1 ) ( * - L 2 )). So we can construct automata accepting ab*a* and a*b*a, and then those accepting * - L 1 and * - L 2 (this is easy, just replace final with non-final states), then an automaton accepting ( * - L 1 ) ( * - L 2 ), and finally, by reversing final and non-final states again * - (( * - L 1 ) ( * - L 2 )). The solution can then be simplified. Method 2. First we construct DFA accepting ab*a* and a*b*a, and then we construct directly a product automata that accept the intersection. Method 3. Let x L 1 L 2. We have to consider two cases. Case 1. x contains more than one b. Then, since x L 1, there is one a before all b s, and, since x L 2, there is one a after all b s. Hence: x abb*a. Case 2. X contains no b at all. Then x aa*. In short: L 1 L 2 = abb*a aa*. A possible automaton is the following. Q={1,2,3,4,5,6}, initial_state=1, F={2,3,6}, garbage_state=4 and is defined below 21

22 a b [12] Consider the program: z:=0; t:=0; while x>0 do begin x:=x-1; z:=z*y; t:=t+1 end where x and y are integers. Using the same technique we used in class for factorial, prove that the above program is equivalent to begin z:=y x ; t:=x; x:=0; end Note that we must assume that x is non-negative, otherwise the two programs are not equivalent. Let N denote the set of integers, N + ={0,1,2,3,...} and let P, P=, F, G, H be the following functions (interpreted as relations) from N + N N N to N + N N N: P(x,y,z,t) = if x>0 then (x,y,z,t) else undefined P=(x,y,z,t) = if x 0 then (x,y,z,t) else undefined Q(x,y,z,) F 1 (x,y,z,t) = (x,y,1,t) F 2 (x,y,z,t) = (x,y,z,0) G 1 (x,y,z,t) = (x,y,z,t+1) G 2 (x,y,z,t) = (x-1,y,z,t) H(x,y,z,t) = (x,y,z*y,t) Our program may now be modelled by the following regular expression over the domain of relations. F 1 F 2 (P G 1 G 2 H)* P=. Let us calculate the relation (P G 1 G 2 H)* P=. First notice that P G 1 G 2 H (composition as for the relations) can be described as follows: P G 1 G 2 H(x,y,z,t) = if x>0 then (x-1,y,z*y,t+1) else undefined. This means that for every k 0, we have: (P G 1 G 2 H) k (x,y,z,t) = if x>k-1 then (x-k,y,z*y^k, k) else undefined. Every (P G 1 G 2 H) k is a function, but (P G 1 G 2 H) * is not! The relation (P G 1 G 2 H) * is defined as follows: 22

23 (x,y,z,t) (P G 1 G 2 H) * (x=,y=,z=,t=) ( k. x>k-1 x==x-k y==y z==z*y^k t==k) (x=x= y=y= z=z= t==t) Hence we may write: (x,y,z,t) (P G 1 G 2 H) * (x=,y=,z=,t=) P= (x@,y@,z@,t@) ( k. x>k-1 x==x-k y==y z==z*y^k t==k x@=x= x@ 0 y@=y= z@=z= t@=t=) (x=x==x@ y=y==y@ z=z==z@ t=t==t@ x@ 0) {note that x>k-1 x==x-k x@=x= x@ 0 implies x>k=1 x-k 0, so x=k} (x 0 x=x@ y=y@ z=z@ t=t@) (x>0 x@=0 y@=y z@=z*y^x t@=x) This means that (P G 1 G 2 H)* P= is a function and (P G 1 G 2 H)* P=(x,y,z,t) = if x>0 then (0,y,z*y^x,x) else (x,y,z,t) i.e. F 1 F 2 (P G 1 G 2 H)* P=(x,y,z,t) = (0,y,y^x, x) for all integers x,y,z,t such that x 0. 23