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### lect24

Course: CS 056000, Fall 2009
School: Dallas
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Word Count: 336

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24 Lecture Time and Space of NTM Time For a NDM M and an input x, TimeM(x) = the minimum # of moves leading to accepting x if x L(M) = infinity if x not in L(M) Time Bound A NTM M is said to have a time bound t(n) if for sufficiently large n and every x L(M) With |x|=n, TimeM(x) &lt; max {n+1, t(n)} . Complexity Classes NTIME(t(n)) = {L(M) | M is a NTM with time bound t(n)} NP = U c &gt; 0 NTIME(n )...

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24 Lecture Time and Space of NTM Time For a NDM M and an input x, TimeM(x) = the minimum # of moves leading to accepting x if x L(M) = infinity if x not in L(M) Time Bound A NTM M is said to have a time bound t(n) if for sufficiently large n and every x L(M) With |x|=n, TimeM(x) < max {n+1, t(n)} . Complexity Classes NTIME(t(n)) = {L(M) | M is a NTM with time bound t(n)} NP = U c > 0 NTIME(n ) c Relationship P NP NP EXP NP EXPOLY First, note that EXP DTIME (2 ) DTIME (2 ). For contradiction, suppose NP = EXP. Consider any L DTIME (2 ). Define L' = {x\$ | x L}. Then L' EXP = NP. So, L' NTIME (n c ) for some c > 0. Hence, L NTIME ((n 2 ) c ) NP = EXP. | x|2 -| x| n2 n2 / 2 n2 Theorem Speed Up Theorem still holds. Hierarchy Theorem may not. Space For a NTM M and an input x, SpaceM(x) = the minimum, over all computation paths, of maximum space taken each in work tape on input x if x L(M) = infinity otherwise 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, SpaceM(x) max{1, s(n)} Complexity Classes NSPACE(s(n)) = {L(M) | M is a NTM with space bound s(n)} NSPACE = Uc>0 NSPACE(n c ) Relationship NP NSPACE PSAPACE = NPSPACE (why?) Savich's Theorem If s(n) log n, then NSPACE (s(n)) DSPACE(s(n) ) The proof will be given in next lecture! 2 Theorems Compresion Theorem holds. Hierarchy Theorem may...

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