318 CSc Test 2 Wednesday 25 October 2000 >>>>>>>>>>SUGGESTED ANSWERS<<<<<<<<<<<<<< 1. (15 pts) Prove the following theorem. Theorem. Given DFA's M=(Q, , s, F, ) and N=(Q, , s, F', ). If F F' then L(M) L(N). x L(M) (s,x) F (s,x) F' x L(N) 2. (20 pts) Let A be the regular expression given by A=(a a*)(ab) and let M be the NFA ({s,1,2}, {a,b}, s, {2}, {(s,a,1), (1,a,1), (1,b,2), (2,b,2)}). a. Give the language expression for L(A). >>> ({a} {a}*){a}{b} b. Prove that L(A) L(M) >>>First I note that (clearly) L(A)={a k b: k>0}. For ak b, ({s}, ak b) = ({1}, ak-1 b) =... ({1}, b) = {2} ak b L(M). c. Find some x L(M) - L(A), that is an x in the language of the NFA but not in L(A). >>>> abb L(M) - L(A), 3. (25 pts) Prove that {a, b, c}* is countable. Define the 1-1 (no onto, which is unnecessary) f:{a,b,c}* {0, 1, 2, ...} by f( )=0, f(a)=1, f(b)=2, f(c)=3. In general, f( 1 2.... n)= 1 + 4 2 + 42 3 + ... +4n-1 n Now, to show f that is 1-1, suppose that f( 1 2.... n)= f( 1 2.... m), then 2 n-1 2 n-1 1+ 4 2 + 4 3 + ... +4 n= 1+ 4 2 + 4 3 + ... +4 m . Taking the remainder of both sides after division by 4 gives 1 = 1. Then dividing both sides gives 2+ 4 3 + 42 4 + ... +4n-2 n = 2 + 4 3 + 42 4 + ... +4n-2 m . Repeat the process to get 2= 2 , 3= 3 , 4= 4 , .... and conclude 1 2.... n= 1 2.... m 4. (10 pts) Find a regular expression for {ax: |ax|=3, x {a, b}*}. >>>>>>>>> aaa aab aba abb 5. (15 pts) Find an NFA equivalent to the -NFA ({s,1,2}, {a,b,c}, s, {1}, {(s,a,1), (s, ,1), (1,b,1), (1,b,2), (1, ,2), (2,c,2)}). a b c -------------------------s 1 1 1 1,2 2 2 2 ---------------------------s 2 s 1,2 2 1 2 Now, throw away the -transitions to get: a b c -----------------s 1 1,2 2 1 1,2 2 2 2 (15 pts) Find a DFA equivalent to the NFA ({s,1,2}, {a,b,c}, s, {1}, {(s,a,1), (s,a,2), (s,b,1), (1,b,2), (2,b,s), (2,c,s)}). a b c {s} {1,2} {1} {1} {2} {1,2} {s,2} {s} {2} {s} {s} {s,2} {1,2} {s,1} {s} {s,1} {1,2} {1,2}
6.