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Unformatted text preview: 6.045J/18.400J: Automata, Computability and Complexity Prof. Nancy Lynch Quiz 3 Solutions May 2, 2007 Elena Grigorescu Please write your name in the upper corner of each page. This exam is open book. Problem Points Grade 1 20 2 12 3 8 4 20 5 20 6 20 Total 100 SQ31 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: P = NP. Unknown 2. True, False, or Unknown: 3 2 n is O (2 3 n ) . False 3. True, False, or Unknown: There exists a language that is not decidable in time 2 n but is decidable in time 8 n (where n is the length of the input). True ; by the Time Hierarchy Theorem (see section 9.1 of Sipser). 4. True, False, or Unknown: UHAMPATH, the Hamiltonian path problem for undirected graphs, is in coNP. (Here, UHAMPATH= {( G,s,t ) G is an undirected graph with a Hamiltonian path from s to t } ). (coNP is defined to be the set of languages whose complements are in NP.) Unknown ; UHAMPATH is NPcomplete. If coNP=NP, then UHAMPATH is in coNP. If coNP negationslash = NP, then UHAMPATH is not in coNP. SQ32 5. True, False, or Unknown: If UHAMPATH ∈ P, then P = coNP. True; UHAMPATH is NPcomplete. If UHAMPATH ∈ P then P = NP = coNP 6. True, False, or Unknown: For every language A , P A negationslash = NP A . False. 7. True, False, or Unknown: If B is decidable in time 2 p ( n ) for some polynomial p , and A ≤ p B , then A is decidable in time 2 q ( n ) for some polynomial q . True 8. True, False, or Unknown: Suppose that B is decidable in time 2 p ( n ) by some nondeterministic Tur ing machine, for some polynomial p , but is not decidable in time 2 q ( n ) by any deterministic Turing machine, for any polynomial q . Then P negationslash = NP . True ; Consider the language C = { x # 2 p ( n ) − x   x ∈ B } . Notice that C is decidable in polynomial time by a NTM but it is not decidable in polynomial time by any deterministic TM. SQ33 Problem 2 : (12 points) The proof that SAT is NPcomplete appears in Sipser’s book, p. 276282. Part of the main construction involves constructing a formula φ move , which is expressed as the conjunction of formulas saying that 2 × 3 windows of the tableau are “legal”. For each of the following, state whether it represents a legal window. Assume that the underlying polynomialtime NTM is of the form ( Q, Σ , Γ ,δ,q ,q acc ,q rej ) , where Σ = { a,b,c,d } and Γ = { a,b,c,d, ⊔} ....
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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.
 Spring '07
 NancyLynch

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