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Complexity

Complexity. 13-1. Hierarchy Theorem. Complexity Andrei Bulatov. Complexity. For a language L over an alphabet , we denote the complement of L , the language * - L. Definition The class of languages L such that can be solved by a

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Complexity

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  1. Complexity 13-1 Hierarchy Theorem Complexity Andrei Bulatov

  2. Complexity For a language L over an alphabet , we denote the complement of L, the language * -L Definition The class of languages L such that can be solved by a non-deterministic log-space Turing machine verifier is called coNL 13-2 NL and coNL Theorem (Immerman) NL=coNL

  3. Complexity • if L is NL-complete then is coNL-complete 13-3 Proof The properties of NL and coNL are similar to those of NP and coNP: • if a coNL-complete problem belongs to NL then NL = coNL Reachability is NL-complete. Therefore it is enough to show that No-Reachability is in NL. In order to do this, we have to find a non-deterministic algorithm that proves in log-space that there is a path between two specified vertices in a graph

  4. Complexity Let be the set of all vertices connected to s with a path of length at most i, and Clearly, , and We compute the numbers inductively 13-4 Counting the number of reachable vertices Given a graph G andtwo its vertices s and t; let n be the number of vertices of G First, we count the number c of vertices reachable from s

  5. Complexity Suppose is known. The following algorithm non-deterministically either compute or reject • set • set - check whether or not using non-deterministic walk - if there is the edge (w,v) then set and leave the loop • output 13-5 • for every vertex v from G do • for every vertex w from G non-deterministically do or not do - if not then reject - if yes then set m = m – 1 • if m 0reject

  6. Complexity • set 13-6 Checking Reachability Given G, s, t and c • for every vertex v from G non-deterministically do or not do - check whether or not v is reachable from s using non-deterministic walk - if not then reject - if yes then set m = m – 1 - if v = t then reject • if m 0reject • accept

  7. Complexity 13-7 Complexity Classes We know a number of complexity classes and we how they relate each other L NL  P  NP , coNP  PSPACE However, we do not know if any of them are different Questions P NP and L  NL concern the (possible) difference between determinism and nondeterminism and known to be extremely difficult Complexity classes can be distinguished using another parameter: The amount of time/space available

  8. Complexity Definition A function f:N N, where f(n)  log n, is called space constructable, if the function that maps to the binary representation of f(n) is computable in space O(f(n)). • polynomials • n log n •  13-8 Space Constructable Functions Examples

  9. Complexity 13-9 Hierarchy Theorem Theorem For any space constructable function f:N N, there exists a language L that is decidable in space O(f(n)), but not in space o(f(n)). Corollary If f(n) and g(n) are space constructable functions, and f(n) = o(g(n)), then SPACE[f] SPACE[g] Corollary LPSPACE Corollary NLPSPACE

  10. Complexity 13-10 Proof Idea Diagonalization Method: • apply a Turing Machine to its own description • revert the answer • get a contradiction In our case, a contradiction can be with the claim that something is computable within o(f(n)) Let L = {“M” | M does not accept “M” in f(n) space}

  11. Computability and Complexity • what we can assume about a decider for L is that it works in • O(f(n)); but this means the decider uses fewer than cf(n) cells • for inputs longer than some . What if “M” is shorter than that? 13-11 If M decides L in space f(n) then what can we say about M(“M”)? • if M(“M”) accepts then “M does not accept M in space f(n)” • if M(“M”) rejects then “M accepts M in space f(n)” There are problems • we showed that L cannot be decided in space f(n), while what • need is to show that it is not decidable in space o(f(n))

  12. Complexity L = {“M ” | simulation ofM on a UTM does not accept “M ” in f(n) space} • If x is not of the form M for some M, reject • Simulate M on x while counting the number of steps used in the • simulation. If the count ever exceeds , accept 13-12 Proof In order to kick in asymptotics, change the language The following algorithm decides L in O(f(n)) On input x • Let n be the length of x • Compute f(n) and mark off this much tape. If later stages ever • attempt to use more space, reject • If M accepts, reject. If M rejects, accept

  13. Complexity 13-13 Clearly, this algorithm works in O(f(n)) space The key stage is the simulation of M Our algorithm simulates M with some loss of efficiency, because the alphabet of M can be arbitrary. If M works in g(n) space then our algorithm simulates M using bg(n) space, where b is a constant factor depending on M Thus, bg(n) f(n)

  14. Complexity There is such that for all inputs x with we have Consider Since , the simulation of M either accepts or rejects on this input in space f(n) • If the simulation accepts then does not accept • If the simulation rejects then accepts 13-14 Suppose that there exists a TM M deciding L in space g(n) =o(f(n)) We can simulate M using bg(n) space

  15. Complexity Theorem For any time constructable function f:N N, there exists a language L that is decidable in time O(f(n)), but not in time . Corollary If f(n) and g(n) are space constructable functions, and , then TIME[f] TIME[g] 13-15 Time Hierarchy Theorem

  16. Complexity Definition 16-16 The Class EXPTIME Corollary PEXPTIME

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