past_exam_03-solutions only

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RMIT School of Computer Science and Information Technology 2006 COSC1105/1107 COMPUTING THEORY EXAM DATE: 13th June TIME ALLOWED: 2 hours TIME: 2.00pm - 4.00pm NUMBER OF PAGES: X TOTAL NUMBER OF MARKS: 100 • Questions may be answered in any order; hence you should start with the questions you consider easiest, and end with the ones you consider hardest. • If you do not understand what is required to answer a particular question, ask for clarification during the reading time. • When asked to “briefly explain” something, or to describe something “informally”, an answer of one or two English sentences is expected. When asked to “explain” or “justify” something, an answer of one or two English paragraphs is expected. • The following may be of use to you. Pumping Lemma Let L be a regular language. The pumping lemma asserts that there is a p such that for all w ∈ L with | w |≥ p then w = xyz , | xy |≤ p , | y | > 0 and xy i z ∈ L for all i ≥ 0.

1. Complexity and Cryptography (3+4 = 8 marks) (a) Explain the complexity classes P , NP and NP − complete . Answer: P is the class of all problems that can be solved within polynomial time. NP is the class of all problems which can be solved in non-deterministic polynomial time. NP − complete problems are NP prob- lems that can be reduced to any other NP problem within polynomial time. (b) Briefly describe the impact developments in quantum computing could have on the tractablity of NP class problems. Answer: A quantum computer with enough qubits could solve an NP problem in polynomial time. 2. Chomsky Hierarchy (6+3+8 = 17 marks) (a) Classify the languages below as either • regular • context-free but not regular • recursively enumerable but not context-free i. { a i b i + j c j | i ≥ j } Answer: recursively enumerable but not context free ii. { a i b i + j c j + k | i, j, k ≥ 0 } Answer: context-free but not regular iii. { w ∈ { a, b } ∗ | w contains an even number of a ’s } Answer: regular iv. { a i b j | i negationslash = j } Answer: context-free but not regular (b) Which of the following statements are true? Briefly explain your answer. i. DFA’s are less powerful than NFA’s. Answer: false, they have equal power ii. Unrestricted gammars are more powerful than non-deterministic PDA’s. Answer: true, unrestricted grammars are as powerful as Turing machines, and so more powerful than any type of PDA iii. Context-sensitive grammars are less powerful than Turing machines. Answer: true, context sensitive grammars are equivalent to PDA’s, which are less powerful than Turing machines. (c) Prove there is no FSA for { a i b j c j | i > j } Answer: let s = xyz = a p +1 b p c p . For | y | > 0 and | xy | < = p to be true, y must consist only of a ’s and z must consist of ab p c p . x may or may not be null. So y = a q , x = a p − q where 1 < = q < = p . The pumping lemma says we can have xy k z for k > = 0 e.g. k = 2 gives xyyz = a p − q a q a q ab p c p = a p + q +1 b p c p which is in L . However k = 0 should also be legal: xz = a p − q ab p c p = a p +1 − q b p c p . Since 1 < = q < = p , we don’t have i > j - so xz is not in L . Since we have failed a condition of the pumping lemma, s is not regular and so there is no FSA. 3. Grammars (3+3+3+9+5 = 23 marks) Consider the grammar G below: S → bSb | aX X → aX | Y Y → Y Z | Z Z → bZc | cZb | a 1

(a) Give a derivation of baaacbbaccbb from G . Answer: S ⇒ bSb ⇒ baXb ⇒ baaXb ⇒ baaaXb ⇒ baaaY b ⇒ baaaZb ⇒ baaacZbb ⇒ baaacbZcbb ⇒ baaacbbZccbb ⇒ baaacbbaccbb (b) Is aabcabccab derivable from G ? If yes, give a derivation, if not explain why. Answer: Yes. S ⇒ aX ⇒ aaX ⇒ aaY ⇒ aaY Z ⇒ aaZZ ⇒ aabZcZ ⇒ aabcZbcZ ⇒ aabcabcZ ⇒ aabcabccZb ⇒ aabcabccab (c) Is bacbab derivable from G ? If yes, give a derivation, if not explain why. Answer: No. We can get to: S ⇒ bSb ⇒ baXb ⇒ baY b ⇒ baY Zb ⇒ baZZb ⇒ bacZbZb but can’t turn cZb into cb . (d) Draw a machine (FSA, PDA, or Turing) to accept G . (e) Consider the grammar H below: S → SaS | SbS | cSa | c Is grammar H ambiguous? Explain why. Answer: Yes it is ambiguous. For example, consider the legal string cacbc , there are two parse trees for this string: S a S S S S c c c b S S S S c c a b S c 2

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