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Lecture 24 Time and Space of NTM

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Time For a NDM M and an input x, Time M (x) = the minimum # of moves leading to accepting x if x ε L(M) = infinity if x not in L(M)

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Time Bound A NTM M is said to have a time bound t(n) if for sufficiently large n and every x ε L(M) With |x|=n, Time M (x) < max {n+1, t(n)}.

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Complexity Classes NTIME(t(n)) = {L(M) | M is a NTM with time bound t(n)} NP = U c > 0 NTIME(n ) c

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Relationship P NP NP ≠ EXP NP EXPOLY

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Theorem Speed Up Theorem still holds. Hierarchy Theorem may not.

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Space For a NTM M and an input x, Space M (x) = the minimum, over all computation paths, of maximum space taken in each work tape on input x if x ε L(M) = infinity otherwise

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Space bound A NTM M is said to have a space bound s(n) if sufficiently large n and every input x with |x|=n, Space M (x) ≤ max{1, s(n)}

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Complexity Classes NSPACE(s(n)) = {L(M) | M is a NTM with space bound s(n)} NSPACE = U c>0 NSPACE(n ) c

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Relationship NP NSPACE PSAPACE = NPSPACE (why?)

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Savich’s Theorem If s(n) ≥ log n, then NSPACE (s(n)) DSPACE(s(n) ) The proof will be given in next lecture! 2

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Theorems Compresion Theorem holds. Hierarchy Theorem may not.

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Translation Lemma Let s 1 (n), s 2 (n) and f(n) be fully space- constructible functions with s 2 (n) > n and f(n) > n. Then NSPACE(s 1 (n)) NSPACE(s 2 (n)) implies NSPACE(s 1 (f(n))) NSPACE(s 2 (f(n)))

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Hierarchy NSPACE(n ) DSPACE(n ) DSPACE(n ) NSPACE(n ) For r > 1 and a > 0, NSPACE(n ) ≠ NSPACE (n ) 48 ≠ 9 9 r r+a

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Proof of Savitch’s Theorem

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