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 LSIZE(O(T(n) 2 )). So it is enough to show that U LSIZE(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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 Spring '10
 hooper
 Vector Space, Space, Computational complexity theory, Closure, Closed set

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