practice-q3-solutions

# practice-q3-solutions - 6.045J/18.400J: Automata,...

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6.045J/18.400J: Automata, Computability and Complexity Prof. Nancy Lynch Practice Quiz 3 April 25, 2007 Elena Grigorescu Please write your name in the upper corner of each page. Problem Points Grade 1 20 2 10 3 10 4 20 5 20 6 20 Total 100 PQ3-1

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Name: Problem 1 : True, False, or Unknown (20 points). In each case, say whether the given statement is known to be TRUE, known to be FALSE, or currently not known either way. Full credit will be given for correct answers. If you include justification for your answers, you may obtain partial credit for incorrect answers. 1. True, False, or Unknown: 8 n is O (2 n ) . False. 2. True, False, or Unknown: There exists a language that is not decidable in time n 8 but is decidable in time 8 n (where n is the length of the input). True. Invoke the Time-Hierarchy Theorem. 3. True, False, or Unknown: SAT coNP . Unknown. SAT coNP if and only if NP = coNP , which is not known. 4. True, False, or Unknown: The 2COLOR problem is NP-complete. Unknown. 2COLOR P . Therefore, it is NP -complete if and only if P = NP , which is not known.
Name: 5. True, False, or Unknown: The 4-CLIQUE problem, defined as the set of undirected graphs containing a 4-CLIQUE , is in P. True. To decide if a graph has a 4 -clique, check all 4-tuples of nodes to see if any of them form a clique in the graph. If any of them form a clique, accept. Else, reject. 6. True, False, or Unknown: 2SAT (the set of CNF Boolean formulas with at most two literals per clause) p PALINDROMES . True. 2SAT P . Therefore, it trivially reduces to any language L n = Σ . 7. True, False, or Unknown: There is a nontrivial language A such that P A n = NP A . True. Invoke Theorem 9 . 19 of Sipser. 8. Suppose there is a 2 n time-bounded reduction from language L 1 to language L 2 , and L 2 is in TIME ( 2 n ). Then, L 1 is in TIME ( 2 O ( n ) ). False. By Time Hierarchy Theorem, there is a language L 1 in DTIME ( 2 2 n ) but not in DTIME ( 2 1 . 5 n ). Define L 2 def = { xy | | y | = 2 | x | ,x L 1 } . Now, L 2 is in DTIME ( 2 n ). Why ? A decider for L 2 looks at the first log n bits of an n -bit input and decides if the log n -bit string is in L 1 , which can be done in time 2 2 log n = 2 n . L 1 reduces to L 2 in time 2 n via a function f that takes x and outputs xy for an arbitrary y of length 2 | x | . But, we know that L 1 is not in TIME ( 2 O ( n ) ).

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Name: Problem 2 : (10 points) Suppose that A , B , and C are nontrivial languages over Σ = { 0 , 1 } . Assume that:
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## This note was uploaded on 08/12/2009 for the course CS 430 taught by Professor Nancylynch during the Spring '07 term at New Mexico Junior College.

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practice-q3-solutions - 6.045J/18.400J: Automata,...

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