Pick the correct alternative from among the choices (A), (B), and (C) provided for each question below.
1. Let Aand Bbe disjoint, recursively enumerable languages. Further let A∪Balso be recursively
enumerable. What can you say about Aand B?
(A) It is possible that neither Anor Bis decidable.
(B) At least one among Aand Bis decidable.
(C) Both Aand Bare decidable.
The correct answer is (C). The decision procedure for A(and B) is as follows. Run the TMs for
A,Band A∪Bsimultaneously on the input. One of these is guaranteed to accept. Accept only if
TM for A(B) accepts.
2. Just as we encoded Turing Machines as binary strings, we can also encode DFAs as binary strings.
Let hMidfa be the binary string encoding of DFA M. Consider the following language Ldfa
d=
{hMidfa | hMidfa 6∈ L(M)}. What can we say about Ldfa
d?
(A) Ldfa
dis regular.
(B) Ldfa
dis not regular but it is decidable.
(C) Ldfa
dis not recursively enumerable.
The correct answer is (B). Non-regularity follows for the same reasons that the diagonal language
is not r.e.
3. Just as we encoded Turing Machines as binary strings, we can also encode PDAs as binary strings.
Let hMipda be the binary string encoding of PDA M. Consider the following language Lpda
d=
{hMipda | hMipda 6∈ L(M)}. What can we say about Lpda
d?
(A) Lpda
dis decidable
(B) Lpda
dis not decidable but it is recursively enumerable.
(C) Lpda
dis not recursively enumerable.
The correct answer is (A). Decidability follows from the fact that PDA membership is decidable
(CYK algorithm).
4. Let Lbe decidable. Which of the following is true about L?
(A) If L0⊆Lthen L0is decidable.
(B) If L⊆L0then L0is decidable.
(C) L≤m{0n1n|n≥0}
The correct answer is (C).
5. Let Aand Bbe any languages such that A≤mB. Under what conditions is it the case that A≤mB?
(A) Only when both Aand Bare decidable.
(B) Only when both Aand Bare recursively enumerable.
(C) Always.
The correct answer is (C). Let fbe a reduction from Ato B. Then fis a computable and x∈A
iff f(x)∈Bwhich is the same as x6∈ Aiff f(x)6∈ Bwhich is the same as x∈Aiff f(x)∈B. Thus f
is a reduction from Ato B.
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