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Unformatted text preview: Notes on Complexity Theory Last updated: September, 2011 Lecture 6 Jonathan Katz 1 Space Complexity We define some of the important spacecomplexity classes we will study: Definition 1 PSPACE def = S c space ( n c ) NPSPACE def = S c nspace ( n c ) L def = space (log n ) NL def = nspace (log n ) . We have seen that time ( t ( n )) ntime ( t ( n )) space ( t ( n )). What can we say in the other direction? To study this we look at configurations of a Turing machine, where a configuration consists of all the information necessary to describe the Turing machine at some instant in time. We have the following easy claim. Claim 1 Let M be a (deterministic or nondeterministic) machine using space s ( n ) . The number of configurations C M ( n ) of M on any fixed input of length n is bounded by: C M ( n )  Q M  n s ( n )  M  s ( n ) , (1) where Q M are the states of M and M is the alphabet of M . In particular, when s ( n ) log n we have C M ( n ) = 2 ( s ( n )) . Proof The first term in Eq. (1) comes from the number of states, the second from the possible positions of the input head, the third from the possible positions of the worktape head, and the last from the possible values stored on the work tape. (Note that since the input is fixed and the input tape is readonly, we do not need to consider all possible length n strings that can be written on the input tape.) We can use this to obtain the following relationship between space and time: Theorem 2 Let s ( n ) be space constructible with s ( n ) log n . Then space ( s ( n )) time (2 O ( s ( n )) ) and nspace ( s ( n )) ntime (2 O ( s ( n )) ) . Proof Let L space ( s ( n )), and let M be a machine using space O ( s ( n )) and deciding L . Consider the computation of M ( x ) for some input x of length n . There are at most C M ( n ) = 2 ( s ( n )) configurations of M on x , but if M ( x ) ever repeats a configuration then it would cycle and never halt. Thus, the computation of M ( x ) must terminate in at most C M ( n ) = 2 ( s ( n )) steps. 61 Let L nspace ( s ( n )). Then there is a nondeterministic Turing machine M deciding L and using space O ( s ( n )) on every computation path (i.e., regardless of the nondeterministic choices it makes). Consider a machine M that runs M but only for at most 2 O ( s ( n )) C M ( n ) steps (and rejects if M has not halted by that point); this can be done using a counter of length O ( s ( n )) and so M still uses O ( s ( n )) space. We claim that M still decides L . Clearly if M ( x ) = 0 then M ( x ) = 0....
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This note was uploaded on 01/13/2012 for the course CMSC 652 taught by Professor Staff during the Fall '08 term at Maryland.
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