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Transcript
CNG280 FORMAL LANGUAGES AND ABSTRACT MACHINES MIDTERM ANSWERS 1. Suppose that the set of all real numbers in the interval [0,1] is countably infinite. That is, there is a way of enumerating all members of this set in the following form: R={r1,r2,r3,…} where each ri in this set is in the form ri = 0.ri1ri2ri3… (rij {0,1} for j N). Now construct a real number r0 = 0.r01r02r03… (r0j {0,1} for j N) as follows: 0 if rjj 1 r0 j 1 if rjj 0 That is, jth bit of r0 is obtained by flipping the corresponding bit of rj. We can see that r0 is also a real number in the interval [0,1] and thus must exist in the set R. But, r 0 cannot be equal to r1 since r01≠r11 according to the construction above (i.e. first bits of r0 and r1 are different); r0 cannot be equal to r2 since r02≠r22; r0 cannot be equal to r3 since r03≠r33; and so on. So, r0 R. This is a contradiction. Therefore, the set of all real numbers in the interval [0,1] is not countable. 2. 0: 0 1: 1 0 1 : e 0 e 1 ( 0 1)* : 0 e e e e e 1 0( 0 1)* : 0 0 e e e e e e 1 0( 0 1)*11: 0 0 e e e e e e e 1 e e 1 e 1 3. First we convert M into the following form where there are no transitions into the initial state nor out of the final state: a a e b e q3 q1 q2 q4 b Calculate R(i,j,k) for 1≤i,j≤4 and k=0,1,2,...: k=0 1 2 3 4 1 a b e 2 b a 3 4 e The automaton corresponding to the transitions for k=0 is the initial automaton, as shown above. k=1 1 2 * 1 a a*b * 2 ba aUba*b 3 a* a*b 4 3 4 e We can eliminate q1 since it has already been considered in the above transitions. Thus, the automaton corresponding to the transitions for k=1 is the following: q3 * ab k=2 2 2 (aUba*b)* * 3 a b(aUba*b)* 4 q2 aUba*b e 3 q4 4 (aUba*b)* * a b(aUba*b)* We can eliminate q2 since it has already been considered in the above transitions. Thus, the automaton corresponding to the transitions for k=2 is the following: a*b(aUba*b)* q3 q4 When the initial automaton is reduced to a two-state automaton, the transition on the arc gives us the desired regular expression. Thus, L(M) = a*b(aUba*b)* 4. Suppose that L={an : n is prime} is regular. Then, according to pumping lemma, there must be an integer n≥1 such that any string w L with |w|≥n can be rewritten as w=xyz such that y≠e, |xy|≤n, and xyiz L for each i≥0. Take x=ap, y=aq and z=ar (p+q+r=n), where p,r≥0 and q>0. By the pumping lemma, xyiz L for each i≥0; that is, p+qi+r is prime for each i≥0. For instance, consider the case i=p+2q+r+2. Then p+qi+r=(q+1)*(p+2q+r). But, this is a product of two natural numbers, each greater than 1; that is, it is not a prime number. This is a contradiction. Therefore, L is not regular. 5. First of all, we can eliminate the rightmost two states, since they are unreachable (i.e. we can never reach them beginning from the initial state): a b a a,b b a,b The equivalence classes for this automaton are: [e]=L, i.e. (ba)*={e,ba,baba,bababa,…} [b]=Lb, i.e. (ba)*b={b,bab,babab,…} [a]=L(aUbb) *, i.e. (ba)*(aUbb)(aUb)*={a,bb,baa,babb,baaa,babba,…} So, the standard automaton is: K={[e], [b], [a]} s=[e] F={[e]} a b [e] [a] [b] [b] [e] [a] [a] [a] [a] a [e] [a] a,b b a [b] b