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# homework#2 - we can have one by erasing all tapes and reset...

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Problem 3. Suppose SPACE(n) = P. By the Space Hierarchy Theorem, we know that there is a language L SPACE(n 2 ) - SPACE(n); let M be a TM recognizing L in quadratic space. Define language L’ = {(x, 1 2 x )/x L}. Notice that L’ SPACE(n) since we can construct linear-space TM M’ which recognizes L’ by taking input (x, 1 2 x )and simulating M on x in 2 x space (which is linear w.r.t. input (x, 1 2 x ) Since SPACE(n) = P, L’ P. Next, notice that L p L’ trivially since we can simply change an instance x of L to an instance (x, 1 2 x ) of L’. Since L’ P and P is closed under poly-time reductions, L P. But this means L SPACE(n), which contradicts our original definition of L. Problem 5. We construct the configuration graph, each configuration has size S(n), so the entire graph has size 2 )) ( ( n s O . We will compute the “eventually accepting” predicate for each configuration. First we mark the single accept configuration as eventually accepting",
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Unformatted text preview: we can have one by erasing all tapes and reset all tapeheads when we accept. Then we do closure by rounds. In every round, we check for each configuration if it leads to a configuration marked as eventually accepting", and mark it if that is the case. We will need at most 2 )) ( ( n s O rounds to complete the closure, each taking time 2 O(S(n)) . The total time is thus 2 )) ( ( n s O . 2 )) ( ( n s O =2 )) ( ( n s O-------------[1] We know that DTIME(T(n)) ⊆ U L-SIZE(O(T(n) 2 )). So it is enough to show that U L-SIZE(O(T(n) 2 )) ⊆ ASPACE(log T(n)) That is given a machine M that outputs a circuit of size O(T(n) 2 ) in O(log T(n)) space, we must construct a ATM that evaluates the circuit in only O(log(T(n))) space.------[2] Combining [1] and [2] we get, ASPACE(S(n)) = DTIME(2 )) ( ( n s O )...
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